GSEB Class 11 Chemistry Solutions Chapter 7 Equilibrium

Question 1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
(a) The vapor pressure on the liquid's surface initially goes down because the container's volume suddenly grows. Vapor pressure will lower.
(b) When the container's volume suddenly expands, the rate of liquid evaporation stays the same at first. But, because the vapor volume gets larger, the rate of condensation will go down or be slow.
(c) When the balance is eventually brought back, the speed of evaporation will once more match the speed of condensation. The final vapor pressure will stay exactly as it was before the container's volume abruptly expanded, since vapor pressure does not rely on volume.
In simple words: When the container gets bigger suddenly, the vapor pressure on the liquid surface drops. Evaporation stays steady, but condensation slows down. Once equilibrium returns, both rates become equal again, and the final vapor pressure is unchanged, as it doesn't depend on volume.

Exam Tip: Remember Le Chatelier's Principle: increasing volume (decreasing pressure) shifts equilibrium to the side with more moles of gas.

Question 2. What is \( K_c \) for the following equilibrium when the equilibrium concentration of each substance is : \( [SO_2] = 0.60 \, M \), \( [O_2] = 0.82 \, M \) and \( [SO_3] = 1.90 \, M \)?
\( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \)
Answer: The equilibrium constant \( K_c \) for the reaction \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \) is figured out using the provided equilibrium concentrations. We apply the formula \( K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} \). \[ K_c = \frac{(1.9)^2}{(0.6)^2(0.82)} = \frac{3.61}{(0.36)(0.82)} = \frac{3.61}{0.2952} = 12.2228 \] Hence, the value of \( K_c \) comes out to be \( 12.237 \, L \, mol^{-1} \).
In simple words: To find \( K_c \), you take the concentration of the product (SO3) raised to its power, and divide it by the concentrations of the reactants (SO2 and O2) each raised to their powers. Then you just put in the numbers and calculate.

Exam Tip: Pay close attention to the stoichiometric coefficients in the balanced equation, as they become the exponents in the \( K_c \) expression.

Question 3. At a certain temperature and total pressure of \( 10^5 \, Pa \), iodine vapour contains 40% by volume of I atoms.
\( I_2(g) \rightleftharpoons 2I (g) \)
Calculate \( K_p \) for the equilibrium.
Answer: The total pressure is provided as \( 10^5 \, Pa \). Given that 40% by volume contains \( I \) atoms, it implies that 60% by volume comprises \( I_2 \) molecules in gas form.
The partial pressure of \( I(g) = \frac{40}{100} \times 10^5 \, Pa \).
The partial pressure of \( I_2(g) = \frac{60}{100} \times 10^5 \, Pa \).
Now, we can figure out \( K_p \) using the relation \( K_p = \frac{(p_I)^2}{p_{I_2}} \). \[ K_p = \frac{\left(\frac{40}{100} \times 10^5\right)^2}{\frac{60}{100} \times 10^5} \] \[ K_p = \frac{(0.4 \times 10^5)^2}{0.6 \times 10^5} = \frac{0.16 \times 10^{10}}{0.6 \times 10^5} = \frac{0.16}{0.6} \times 10^5 \] \[ K_p = 0.2666 \times 10^5 = 2.666 \times 10^4 \, Pa \] As a result, \( K_p \) for this system is \( 2.67 \times 10^4 \, Pa \).
In simple words: First, find the individual pressures of I and I2 using the given percentages. Then, use the \( K_p \) formula, which is the partial pressure of products raised to their powers divided by the partial pressure of reactants raised to their powers, and compute the final value.

Exam Tip: Remember that for gaseous mixtures, volume percentages are equivalent to mole percentages, which can be used to determine partial pressures.

Question 4. Write the expression for the equilibrium constant, \( K_c \) for each of the following reactions :
(i) \( 2NO (g) + Cl_2 (g) \rightleftharpoons 2NOCl (g) \)
(ii) \( 2Cu(NO_3)_2 (s) \rightleftharpoons 2CuO (s) + 4NO_2 (g) + O_2 (g) \)
(iii) \( CH_3COOC_2H_5 (aq) + H_2O (l) \rightleftharpoons CH_3COOH (aq) + C_2H_5OH (aq) \)
(iv) \( Fe^{3+} (aq) + 3OH^{-} (aq) \rightleftharpoons Fe(OH)_3 (s) \)
(v) \( I_2 (s) + 5F_2 (g) \rightleftharpoons 2IF_5 (g) \)
Answer:
(i) For the reaction \( 2NO (g) + Cl_2 (g) \rightleftharpoons 2NOCl (g) \), the equilibrium constant \( K_c \) expression is shown as: \[ K_c = \frac{[NOCl]^2}{[NO]^2[Cl_2]} \] Here, \( [NO] \), \( [Cl_2] \), and \( [NOCl] \) signify the molar concentrations of the specific gaseous types at equilibrium.
(ii) For the reaction \( 2Cu(NO_3)_2 (s) \rightleftharpoons 2CuO (s) + 4NO_2 (g) + O_2 (g) \), the equilibrium constant \( K_c \) expression becomes: \[ K_c = [NO_2 (g)]^4[O_2 (g)] \] Solid materials like \( Cu(NO_3)_2 \) and \( CuO \) are not put into the equilibrium constant expression because their concentrations stay the same.
(iii) For the reaction \( CH_3COOC_2H_5 (aq) + H_2O (l) \rightleftharpoons CH_3COOH (aq) + C_2H_5OH (aq) \), the equilibrium constant \( K_c \) expression is: \[ K_c = \frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O]} \] All components are in the aqueous or liquid form, so their concentrations are part of the expression.
(iv) For the reaction \( Fe^{3+} (aq) + 3OH^{-} (aq) \rightleftharpoons Fe(OH)_3 (s) \), the equilibrium constant \( K_c \) expression is: \[ K_c = \frac{1}{[Fe^{3+} (aq)][OH^{-} (aq)]^3} \] The solid product \( Fe(OH)_3 \) is not put into the expression, as its concentration is thought to be constant and equal to 1.
(v) For the reaction \( I_2 (s) + 5F_2 (g) \rightleftharpoons 2IF_5 (g) \), the equilibrium constant \( K_c \) expression becomes: \[ K_c = \frac{[IF_5 (g)]^2}{[F_2 (g)]^5} \] The solid reactant \( I_2 \) is not added to the expression, since its concentration is seen as constant and equal to 1.
In simple words: To write \( K_c \), place product concentrations (raised to their powers) on top, and reactant concentrations (raised to their powers) on the bottom. Remember that pure solids and liquids are not included; their constant concentrations are accounted for.

Exam Tip: Always balance the chemical equation first before writing the \( K_c \) expression, as the coefficients become the exponents. Solids and pure liquids are omitted.

Question 5. Find out the value of \( K_c \) for each of the following equilibria from the value of \( K_p \).
(i) \( 2NOCl (g) \rightleftharpoons 2NO (g) + Cl_2(g) \); \( K_p = 1.8 \times 10^{-2} \) at \( 500 \, K \)
(ii) \( CaCO_3 (s) \rightleftharpoons CaO (s) + CO_2 (g) \); \( K_p = 167 \) at \( 1073 \, K \)
Answer:
(i) The provided reaction is \( 2NOCl (g) \rightleftharpoons 2NO (g) + Cl_2(g) \).
The value of \( K_p \) is \( 1.8 \times 10^{-2} \) at \( 500 \, K \).
The difference in the number of moles of gas products minus gas reactants is \( \Delta n = (2 + 1) - 2 = 1 \).
We apply the formula \( K_p = K_c \times (RT)^{\Delta n} \).
Hence, \( K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{1.8 \times 10^{-2}}{(0.0821 \times 500)^1} \). \[ K_c = \frac{1.8 \times 10^{-2}}{41.05} = 4.385 \times 10^{-4} \] Consequently, the value of \( K_c \) is \( 4.4 \times 10^{-4} \).
(ii) The provided reaction is \( CaCO_3 (s) \rightleftharpoons CaO (s) + CO_2 (g) \).
The value of \( K_p \) is \( 167 \) at \( 1073 \, K \).
The difference in the number of moles of gas products minus gas reactants is \( \Delta n = 1 - 0 = 1 \) (as solids are not counted).
We apply the formula \( K_p = K_c \times (RT)^{\Delta n} \).
Hence, \( K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{167}{(0.0821 \times 1073)^1} \). \[ K_c = \frac{167}{88.0853} = 1.896 \] As a result, the value of \( K_c \) is \( 1.90 \).
In simple words: To convert \( K_p \) to \( K_c \), use the equation \( K_p = K_c(RT)^{\Delta n} \). First, calculate \( \Delta n \) by subtracting gaseous reactant moles from gaseous product moles. Then, rearrange the formula to find \( K_c \), plugging in the given \( K_p \) and temperature.

Exam Tip: Always ensure you use the correct value for the gas constant \( R \) based on the units of \( K_p \) (e.g., L atm/mol K for atmospheres, L bar/mol K for bars).

Question 6. For the following equilibrium, \( K_c = 6.3 \times 10^{14} \) at \( 1000 \, K \).
\( NO (g) + O_3(g) \rightleftharpoons NO_2(g) + O_2(g) \)
Both the forward and reverse reactions in the equilibrium and elementary bimolecular reactions. What is \( K_c \), for the reverse reaction?
Answer: The equilibrium constant for the forward reaction, \( K_c \), is provided as \( 6.3 \times 10^{14} \).
For the reaction \( NO (g) + O_3(g) \rightleftharpoons NO_2(g) + O_2(g) \), the expression for \( K_c \) is: \[ K_c = \frac{[NO_2][O_2]}{[NO][O_3]} = 6.3 \times 10^{14} \, (\text{Provided}) \] The opposite reaction is \( NO_2 (g) + O_2 (g) \rightleftharpoons NO (g) + O_3 (g) \).
The equilibrium constant for the opposite reaction, \( K'_c \), is the inverse of the forward reaction's \( K_c \). \[ K'_c = \frac{[NO][O_3]}{[NO_2][O_2]} = \frac{1}{K_c} \] \[ K'_c = \frac{1}{6.3 \times 10^{14}} = 0.1587 \times 10^{-14} = 1.59 \times 10^{-15} \] Thus, the \( K_c \) for the reverse reaction is \( 1.59 \times 10^{-15} \).
In simple words: If you know \( K_c \) for a reaction, the \( K_c \) for the reverse reaction is simply 1 divided by the original \( K_c \). Just flip the fraction.

Exam Tip: This relationship is fundamental: reversing a reaction inverts its equilibrium constant. Remember to use the reciprocal.

Question 7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Answer: Traditionally, the active mass of a pure solid is taken as constant, no matter its quantity. Likewise, if a pure liquid is available in abundance, like a solvent, its active mass also stays constant. In these cases, we assign their active mass a value of 1, as their unchanging values are already part of the total equilibrium constant.
Molar concentration stays constant because the density of a pure material at a specific temperature is constant (it's an intensive property based only on the material's type, not how much there is), and the molecular mass of a material is also constant. Hence, the molar concentration of pure solids and liquids is thought to be constant.
In simple words: Pure solids and liquids have a constant concentration, which means their 'active mass' doesn't change during a reaction. Because their values are constant, they are already part of the total equilibrium constant and don't need to be written in the expression.

Exam Tip: When writing equilibrium constant expressions, always exclude pure solids and liquids. Only include species that can have variable concentrations (gases and aqueous solutions).

Question 8. Reaction between \( N_2 \) and \( O_2 \) takes place as follows:
\( 2N_2 (g) + O_2 (g) \rightleftharpoons 2N_2O (g) \)
If a mixture of \( 0.482 \, mol \) of \( N_2 \) and \( 0.933 \, mol \) of \( O_2 \) is placed in a \( 10 \, L \) reaction vessel and allowed to form \( N_2O \) at a temperature for which \( K_c = 2.0 \times 10^{-37} \), determine the composition of equilibrium mixture.
Answer: The provided reaction is \( 2N_2 (g) + O_2 (g) \rightleftharpoons 2N_2O (g) \).
Initial moles: \( N_2 = 0.482 \, mol \), \( O_2 = 0.933 \, mol \). Volume = \( 10 \, L \).
Starting concentrations:
\( [N_2]_0 = \frac{0.482}{10} = 0.0482 \, M \)
\( [O_2]_0 = \frac{0.933}{10} = 0.0933 \, M \)
Let \( 2x \) be the moles of \( N_2O \) created at equilibrium. Then \( x \) moles of \( O_2 \) react and \( 2x \) moles of \( N_2 \) react.
Equilibrium concentrations:
\( [N_2] = \frac{0.482 - 2x}{10} = 0.0482 - 0.2x \)
\( [O_2] = \frac{0.933 - x}{10} = 0.0933 - 0.1x \)
\( [N_2O] = \frac{2x}{10} = 0.2x \)
Given \( K_c = 2.0 \times 10^{-37} \). Since \( K_c \) is very tiny, it signifies that the reaction barely moves to the right, meaning \( 2x \) is much smaller compared to initial concentrations. So, \( 2x \approx 0 \).
Therefore, at equilibrium:
\( [N_2] \approx 0.0482 \, M \)
\( [O_2] \approx 0.0933 \, M \)
The formula for \( K_c \) is \( K_c = \frac{[N_2O]^2}{[N_2]^2[O_2]} \). \[ 2.0 \times 10^{-37} = \frac{(0.2x)^2}{(0.0482)^2(0.0933)} \] \[ 2.0 \times 10^{-37} = \frac{0.04x^2}{(0.00232324)(0.0933)} \] \[ 2.0 \times 10^{-37} = \frac{0.04x^2}{0.00021674} \] \[ x^2 = \frac{2.0 \times 10^{-37} \times 0.00021674}{0.04} \] \[ x^2 = \frac{4.3348 \times 10^{-41}}{0.04} = 1.0837 \times 10^{-39} \] \[ x = \sqrt{1.0837 \times 10^{-39}} \approx \sqrt{10.837 \times 10^{-40}} \approx 3.29 \times 10^{-20} \, mol/L \] At equilibrium:
\( [N_2] = 0.0482 \, M \)
\( [O_2] = 0.0933 \, M \)
\( [N_2O] = 0.2x = 0.2 \times (3.29 \times 10^{-20}) = 6.58 \times 10^{-21} \, M \)
In simple words: Because \( K_c \) is very small, most of the reactants stay unreacted. We can then approximate reactant concentrations at equilibrium to their initial values and solve for 'x' to find the product concentration.

Exam Tip: For very small \( K_c \) values, you can often simplify calculations by assuming the change 'x' in reactant concentrations is negligible compared to their initial concentrations.

Question 9. Nitric oxide reacts with \( Br_2 \) and gives nitrosyl bromide as per reaction given below :
\( 2NO (g) + Br_2 (g) \rightleftharpoons 2NOBr (g) \)
When \( 0.087 \, mol \) of \( NO \) and \( 0.0437 \, mol \) of \( Br_2 \) are mixed in a closed container at constant temperature, \( 0.0518 \, mol \) of \( NOBr \) is obtained at equilibrium. Calculate equilibrium amount of \( NO \) and \( Br_2 \).
Answer: The provided reaction is \( 2NO (g) + Br_2 (g) \rightleftharpoons 2NOBr (g) \).
Starting moles: \( NO = 0.087 \, mol \), \( Br_2 = 0.0437 \, mol \).
At balance, moles of \( NOBr \) created = \( 0.0518 \, mol \).
From the reaction equation, \( 2 \, moles \, of \, NO \) yield \( 2 \, moles \, of \, NOBr \). So, moles of \( NO \) that reacted = moles of \( NOBr \) created = \( 0.0518 \, mol \).
Also, \( 1 \, mole \, of \, Br_2 \) yields \( 2 \, moles \, of \, NOBr \). So, moles of \( Br_2 \) that reacted = \( \frac{0.0518}{2} = 0.0259 \, mol \).
Equilibrium amount of \( NO = \text{Original moles of NO} - \text{Moles of NO that reacted} \)
\( = 0.087 - 0.0518 = 0.0352 \, mol \).
Equilibrium amount of \( Br_2 = \text{Original moles of } Br_2 - \text{Moles of } Br_2 \text{ that reacted} \)
\( = 0.0437 - 0.0259 = 0.0178 \, mol \).
Hence, the equilibrium quantities of \( NO \) is \( 0.0352 \, mol \) and \( Br_2 \) is \( 0.0178 \, mol \).
In simple words: Use the amount of product formed (NOBr) and the reaction's stoichiometry to figure out how much NO and Br2 reacted. Then, subtract these reacted amounts from the initial amounts to get the equilibrium amounts for NO and Br2.

Exam Tip: For stoichiometry problems, always ensure the reaction is balanced and use mole ratios to relate changes in reactants and products.

Question 10. At \( 450 \, K \), \( K_p = 2.0 \times 10^{10} / bar \) for the given reaction at equilibrium.
\( 2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g) \)
What is \( K_c \) at this temperature?
Answer: The provided reaction is \( 2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g) \).
The temperature is \( 450 \, K \).
The value of \( K_p \) is \( 2.0 \times 10^{10} \, bar^{-1} \).
The difference in the number of moles of gas products minus gas reactants is \( \Delta n = 2 - (2 + 1) = 2 - 3 = -1 \).
We apply the formula \( K_p = K_c \times (RT)^{\Delta n} \).
Hence, \( K_c = \frac{K_p}{(RT)^{\Delta n}} = K_p \times (RT)^{-\Delta n} = K_p \times (RT)^{-(-1)} = K_p \times RT \).
Using \( R = 0.0831 \, L \, bar \, K^{-1} \, mol^{-1} \):
\( K_c = (2.0 \times 10^{10}) \times (0.0831 \times 450) \)
\( K_c = 2.0 \times 10^{10} \times 37.395 \)
\( K_c = 74.79 \times 10^{10} = 7.479 \times 10^{11} \, L \, mol^{-1} \).
Consequently, \( K_c \) for the reaction at \( 450 \, K \) is \( 7.47 \times 10^{11} \, L \, mol^{-1} \).
In simple words: First, calculate \( \Delta n \) by subtracting the sum of reactant moles from the sum of product moles. Then, use the relationship \( K_c = K_p / (RT)^{\Delta n} \) and plug in all the numbers to find \( K_c \).

Exam Tip: Be careful with the sign of \( \Delta n \) and ensure you use the correct gas constant \( R \) value corresponding to the pressure units (atm or bar).

Question 11. A sample of \( HI (g) \) is placed in a flask at a pressure of \( 0.2 \, atm \). At equilibrium the partial pressure of \( HI (g) \) is \( 0.04 \, atm \). What is \( K_p \) for the given equilibrium?
\( 2HI (g) \rightleftharpoons H_2 (g) + I_2 (g) \)
Answer: The provided reaction is \( 2HI (g) \rightleftharpoons H_2 (g) + I_2 (g) \).
Starting partial pressure of \( HI = 0.2 \, atm \).
At balance, partial pressure of \( HI = 0.04 \, atm \).
Change in pressure of \( HI = 0.2 - 0.04 = 0.16 \, atm \).
From the reaction coefficients, for every \( 2 \, atm \) of \( HI \) that reacts, \( 1 \, atm \) of \( H_2 \) and \( 1 \, atm \) of \( I_2 \) are produced.
So, partial pressure of \( H_2 \) produced = \( \frac{0.16}{2} = 0.08 \, atm \).
Partial pressure of \( I_2 \) produced = \( \frac{0.16}{2} = 0.08 \, atm \).
At equilibrium:
\( p_{HI} = 0.04 \, atm \)
\( p_{H_2} = 0.08 \, atm \)
\( p_{I_2} = 0.08 \, atm \)
The expression for \( K_p \) is \( K_p = \frac{p_{H_2} \times p_{I_2}}{(p_{HI})^2} \). \[ K_p = \frac{0.08 \times 0.08}{(0.04)^2} = \frac{0.0064}{0.0016} = 4.0 \] Hence, \( K_p \) for the equilibrium is \( 4.0 \).
In simple words: First, figure out how much HI reacted and how much H2 and I2 were formed using the initial and equilibrium pressures. Then, put these equilibrium partial pressures into the \( K_p \) expression to calculate its value.

Exam Tip: Use an ICE (Initial, Change, Equilibrium) table to systematically track changes in partial pressures for each species, which simplifies the \( K_p \) calculation.

Question 12. A mixture of \( 1.57 \, mol \) of \( N_2 \), \( 1.92 \, mol \) of \( H_2 \) and \( 8.13 \, mol \) of \( NH_3 \) is introduced into a \( 20 \, L \) reaction vessel at \( 500 \, K \). At this temperature, the equilibrium constant, \( K_c \) for the reaction \( N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g) \) is \( 1.7 \times 10^2 \). Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer: The provided reaction is \( N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g) \).
The equilibrium constant \( K_c \) is \( 1.7 \times 10^2 \).
Given moles in a \( 20 \, L \) container:
\( N_2 = 1.57 \, mol \)
\( H_2 = 1.92 \, mol \)
\( NH_3 = 8.13 \, mol \)
Concentrations at this specific time:
\( [N_2] = \frac{1.57}{20} = 0.0785 \, M \)
\( [H_2] = \frac{1.92}{20} = 0.096 \, M \)
\( [NH_3] = \frac{8.13}{20} = 0.4065 \, M \)
The reaction quotient, \( Q_c \), is figured out using these concentrations: \[ Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \] \[ Q_c = \frac{(0.4065)^2}{(0.0785)(0.096)^3} \] \[ Q_c = \frac{0.16524225}{(0.0785)(0.000884736)} \] \[ Q_c = \frac{0.16524225}{0.00006948} \approx 2378.36 \approx 2.38 \times 10^3 \] Comparing \( Q_c \) and \( K_c \):
\( Q_c = 2.38 \times 10^3 \)
\( K_c = 1.7 \times 10^2 \)
Since \( Q_c > K_c \), the reaction blend is not at balance. For balance to be achieved, \( Q_c \) must get smaller to match \( K_c \). This means the net reaction will move in the backward path (towards reactants).
In simple words: Calculate the reaction quotient \( Q_c \) using current concentrations. Compare it to \( K_c \). If \( Q_c \) is greater than \( K_c \), the reaction shifts backward to make more reactants. If \( Q_c \) is less than \( K_c \), it shifts forward to make more products. If they are equal, it's at equilibrium.

Exam Tip: Clearly state the values of \( Q_c \) and \( K_c \) and explicitly mention their comparison to determine the direction of the shift based on Le Chatelier's Principle.

Question 13. The equilibrium constant expression for a gas reaction is,
\[ K_c = \frac{[NH_3]^4[O_2]^5}{[NO]^4[H_2O]^6} \]
Write the balanced chemical equation corresponding to this expression.
Answer: The equilibrium constant expression is provided as: \[ K_c = \frac{[NH_3]^4[O_2]^5}{[NO]^4[H_2O]^6} \] In an equilibrium constant expression, products are on the top and reactants are on the bottom, with their balanced coefficients acting as powers.
Consequently, the products are \( NH_3 \) and \( O_2 \), having coefficients 4 and 5, respectively.
The reactants are \( NO \) and \( H_2O \), with coefficients 4 and 6, respectively.
The balanced chemical equation that matches this expression is:
\( 4NO (g) + 6H_2O (g) \rightleftharpoons 4NH_3 (g) + 5O_2 (g) \)
In simple words: The substances on top of the \( K_c \) expression are the products, and those on the bottom are the reactants. The numbers next to them in the expression are their coefficients in the balanced chemical equation.

Exam Tip: Remember that the numerator of the \( K_c \) expression corresponds to the products, and the denominator corresponds to the reactants, with their stoichiometric coefficients as powers.

Question 14. One mole of \( H_2O \) and one mole of \( CO \) are taken in \( 10 \, L \) vessel and heated to \( 725 \, K \). At equilibrium 40% of water (by mass) reacts with \( CO \) according to the equation.
\( H_2O (g) + CO (g) \rightleftharpoons H_2 (g) + CO_2 (g) \)
Calculate the equilibrium constant for the reaction.
Answer: The provided reaction is \( H_2O (g) + CO (g) \rightleftharpoons H_2 (g) + CO_2 (g) \).
Starting moles: \( H_2O = 1 \, mol \), \( CO = 1 \, mol \). Volume = \( 10 \, L \).
At balance, 40% of water participates. Moles of \( H_2O \) that reacted = \( 0.40 \times 1 \, mol = 0.4 \, mol \).
From reaction ratios, moles of \( CO \) that reacted = moles of \( H_2O \) that reacted = \( 0.4 \, mol \).
Moles of \( H_2 \) produced = moles of \( H_2O \) that reacted = \( 0.4 \, mol \).
Moles of \( CO_2 \) produced = moles of \( H_2O \) that reacted = \( 0.4 \, mol \).
Equilibrium moles:
\( H_2O = 1 - 0.4 = 0.6 \, mol \)
\( CO = 1 - 0.4 = 0.6 \, mol \)
\( H_2 = 0.4 \, mol \)
\( CO_2 = 0.4 \, mol \)
Equilibrium concentrations (Volume = \( 10 \, L \)):
\( [H_2O] = \frac{0.6}{10} = 0.06 \, M \)
\( [CO] = \frac{0.6}{10} = 0.06 \, M \)
\( [H_2] = \frac{0.4}{10} = 0.04 \, M \)
\( [CO_2] = \frac{0.4}{10} = 0.04 \, M \)
The expression for \( K_c \) is \( K_c = \frac{[H_2][CO_2]}{[H_2O][CO]} \). \[ K_c = \frac{(0.04)(0.04)}{(0.06)(0.06)} = \frac{0.0016}{0.0036} = \frac{16}{36} = \frac{4}{9} \approx 0.44 \] Therefore, the equilibrium constant \( K_c \) for the reaction is \( 0.44 \).
In simple words: Calculate the moles of all substances at equilibrium using the given amount of product. Then, find their concentrations (volume cancels out). Plug these into the \( K_c \) expression to calculate the equilibrium constant.

Exam Tip: For reactions with 1:1 stoichiometry and equal moles of reactants, the volume of the vessel often cancels out in the \( K_c \) calculation, simplifying the process.

Question 25. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) \( \text{PCl}_5 \text{ (g)} \rightleftharpoons \text{PCl}_3 \text{ (g)} + \text{Cl}_2 \text{ (g)} \)
(b) \( \text{CaO (s)} + \text{CO}_2 \text{ (g)} \rightleftharpoons \text{CaCO}_3 \text{ (s)} \)
(c) \( \text{3Fe (s)} + \text{4H}_2\text{O (g)} \rightleftharpoons \text{Fe}_3\text{O}_4 \text{ (s)} + \text{4H}_2 \text{ (g)} \)
Answer:
(a) \( \text{PCl}_5 \text{ (g)} \rightleftharpoons \text{PCl}_3 \text{ (g)} + \text{Cl}_2 \text{ (g)} \)
Since this reaction moves with an increase in volume, when pressure is reduced, the equilibrium will shift in the forward direction. Consequently, more \( \text{PCl}_3 \text{ (g)} \) and \( \text{Cl}_2 \text{ (g)} \) will form, which means the number of product moles increases.
(b) \( \text{CaO (s)} + \text{CO}_2 \text{ (g)} \rightleftharpoons \text{CaCO}_3 \text{ (s)} \)
This reaction moves with a decrease in volume because \( \text{CaO (s)} \) absorbs \( \text{CO}_2 \text{ (g)} \) to create \( \text{CaCO}_3 \text{ (s)} \). Therefore, when the pressure is lowered, the equilibrium will move backwards. This causes the number of moles of products to decrease.
(c) \( \text{3Fe (s)} + \text{4H}_2\text{O (g)} \rightleftharpoons \text{Fe}_3\text{O}_4 \text{ (s)} + \text{4H}_2 \text{ (g)} \)
The presence of \( \text{Fe (s)} \) and \( \text{Fe}_3\text{O}_4 \text{ (s)} \) on both the reactant and product sides does not get affected by the pressure decrease. Since the volume of \( \text{H}_2\text{O (g)} \) and \( \text{H}_2 \text{ (g)} \) are present in an equal ratio [4:4], reducing the pressure has no impact on the number of moles of products. They stay the same.
In simple words: When pressure drops, an equilibrium will shift to make more gas if possible. If gas moles are equal on both sides, pressure changes won't shift it. Solids don't affect pressure changes.

Exam Tip: To determine the effect of pressure, compare the total number of moles of gaseous reactants and gaseous products. The equilibrium shifts to the side with more gas moles if pressure decreases, and to the side with fewer gas moles if pressure increases.

Question 26. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) \( \text{COCl}_2 \text{ (g)} \rightleftharpoons \text{CO (g)} + \text{Cl}_2 \text{ (g)} \)
(ii) \( \text{CH}_4 \text{ (g)} + \text{2S}_2 \text{ (g)} \rightleftharpoons \text{CS}_2 \text{ (g)} + \text{2H}_2\text{S (g)} \)
(iii) \( \text{CO}_2 \text{ (g)} + \text{C (g)} \rightleftharpoons \text{2CO (g)} \)
(iv) \( \text{2H}_2 \text{ (g)} + \text{CO (g)} \rightleftharpoons \text{CH}_3\text{OH (g)} \)
(v) \( \text{CaCO}_3 \text{ (s)} \rightleftharpoons \text{CaO (s)} + \text{CO}_2 \text{ (g)} \)
(vi) \( \text{4NH}_3 \text{ (g)} + \text{5O}_2 \text{ (g)} \rightleftharpoons \text{4NO (g)} + \text{6H}_2\text{O (g)} \)
Answer:
For reaction (ii) \( \text{CH}_4 \text{ (g)} + \text{2S}_2 \text{ (g)} \rightleftharpoons \text{CS}_2 \text{ (g)} + \text{2H}_2\text{S (g)} \), the number of moles of gaseous reactants (\( \text{n}_\text{r} = 1 + 2 = 3 \)) is equal to the number of moles of gaseous products (\( \text{n}_\text{p} = 1 + 2 = 3 \)). Consequently, this reaction will not be affected by an increase in pressure, meaning the equilibrium direction will remain unchanged.

All other listed reactions will be influenced by an increase in pressure:
(i) \( \text{COCl}_2 \text{ (g)} \rightleftharpoons \text{CO (g)} + \text{Cl}_2 \text{ (g)} \)
Here, \( \text{n}_\text{p} = 2 \) and \( \text{n}_\text{r} = 1 \), so \( \text{n}_\text{p} > \text{n}_\text{r} \). Thus, the equilibrium will shift to the left (backward direction) when pressure increases.
(iii) \( \text{CO}_2 \text{ (g)} + \text{C (g)} \rightleftharpoons \text{2CO (g)} \)
Here, \( \text{n}_\text{r} = 1 \) and \( \text{n}_\text{p} = 2 \), so \( \text{n}_\text{p} > \text{n}_\text{r} \). Therefore, the equilibrium will shift to the left (backward direction) upon an increase in pressure.
(iv) \( \text{2H}_2 \text{ (g)} + \text{CO (g)} \rightleftharpoons \text{CH}_3\text{OH (g)} \)
Here, \( \text{n}_\text{r} = 3 \) and \( \text{n}_\text{p} = 1 \), so \( \text{n}_\text{p} < \text{n}_\text{r} \). As a result, the equilibrium will shift to the right (forward direction) when pressure increases.
(v) \( \text{CaCO}_3 \text{ (s)} \rightleftharpoons \text{CaO (s)} + \text{CO}_2 \text{ (g)} \)
Here, \( \text{n}_\text{r} = 0 \) (only solid reactant) and \( \text{n}_\text{p} = 1 \) (one gaseous product), so \( \text{n}_\text{p} > \text{n}_\text{r} \). Consequently, the equilibrium will shift backwards (to the left) on increasing the pressure.
(vi) \( \text{4NH}_3 \text{ (g)} + \text{5O}_2 \text{ (g)} \rightleftharpoons \text{4NO (g)} + \text{6H}_2\text{O (g)} \)
Here, \( \text{n}_\text{r} = 9 \) and \( \text{n}_\text{p} = 10 \), so \( \text{n}_\text{p} > \text{n}_\text{r} \). Therefore, the equilibrium will shift backwards (to the left) when pressure increases.
In simple words: When pressure goes up, the reaction tries to reduce the number of gas molecules. It shifts towards the side with fewer gas moles. If the number of gas moles is the same on both sides, pressure changes won't affect it.

Exam Tip: Remember Le Chatelier's principle: if a system at equilibrium is subjected to a change, it will adjust itself to counteract the change and restore a new equilibrium. For pressure changes, count only gaseous species.

Question 27. The equilibrium constant for the following reaction is \( 1.6 \times 10^5 \) at 1024 K
\( \text{H}_2 \text{ (g)} + \text{Br}_2 \text{ (g)} \rightleftharpoons \text{2HBr (g)} \)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Answer:
The given reaction is \( \text{2HBr (g)} \rightleftharpoons \text{H}_2 \text{ (g)} + \text{Br}_2 \text{ (g)} \)
The equilibrium constant for the forward reaction is \( \text{K} = 1.6 \times 10^5 \).
For the reverse reaction, \( \text{K}_\text{p} = \frac { 1 }{ 1.6 \times 10^5 } \)
Let initial pressure of \( \text{HBr} \) be 10 bar.
At equilibrium, let \( \text{p} \) be the pressure of \( \text{H}_2 \text{ (g)} \) formed, and also the pressure of \( \text{Br}_2 \text{ (g)} \) formed (since they are in a 1:1 ratio).
So, \( \text{P}_{\text{H}_2} = \frac{\text{p}}{2} \) and \( \text{P}_{\text{Br}_2} = \frac{\text{p}}{2} \).
The pressure of \( \text{HBr} \) remaining at equilibrium will be \( (10 - \text{p}) \) bar.
We write the expression for \( \text{K}_\text{p} \):
\( \text{K}_\text{p} = \frac { \text{P}_{\text{H}_2} \times \text{P}_{\text{Br}_2} }{ \text{P}_{\text{HBr}}^2 } \)
\( \implies \frac { 1 }{ 1.6 \times 10^5 } = \frac { \left( \frac{\text{p}}{2} \right) \times \left( \frac{\text{p}}{2} \right) }{ (10 - \text{p})^2 } \)
\( \implies \frac { 1 }{ 1.6 \times 10^5 } = \frac { \frac{\text{p}^2}{4} }{ (10 - \text{p})^2 } \)
\( \implies \frac { 1 }{ 1.6 \times 10^5 } = \frac { \text{p}^2 }{ 4(10 - \text{p})^2 } \)
Taking the square root on both sides:
\( \implies \sqrt{ \frac { 1 }{ 1.6 \times 10^5 } } = \sqrt{ \frac { \text{p}^2 }{ 4(10 - \text{p})^2 } } \)
\( \implies \frac { 1 }{ \sqrt{1.6} \times 10^{2.5} } = \frac { \text{p} }{ 2(10 - \text{p}) } \)
\( \implies \frac { 1 }{ 398.75 } \approx \frac { 1 }{ 400 } \) (since \( \sqrt{1.6} \approx 1.265 \), \( 1.265 \times 10^{2.5} \approx 399.7 \approx 400 \))
\( \implies \frac { 1 }{ 400 } = \frac { \text{p} }{ 2(10 - \text{p}) } \)
\( \implies 2(10 - \text{p}) = 400\text{p} \)
\( \implies 20 - 2\text{p} = 400\text{p} \)
\( \implies 20 = 402\text{p} \)
\( \implies \text{p} = \frac { 20 }{ 402 } \approx 0.04975 \text{ bar} \approx 4.98 \times 10^{-2} \text{ bar} \)
So, at equilibrium:
\( \text{P}_{\text{H}_2} = \text{P}_{\text{Br}_2} = \frac { \text{p} }{ 2 } = \frac { 4.98 \times 10^{-2} }{ 2 } = 2.49 \times 10^{-2} \text{ bar} \approx 2.5 \times 10^{-2} \text{ bar} \)
\( \text{P}_{\text{HBr}} = 10 - \text{p} = 10 - 4.98 \times 10^{-2} = 9.9502 \text{ bar} \approx 10 \text{ bar} \)
In simple words: First, flip the K value because HBr is breaking apart. Then, set up the equilibrium expression with partial pressures. Solve for 'p', which tells you how much H2 and Br2 formed. Subtract 'p' from the starting HBr pressure to get its final pressure.

Exam Tip: Always pay attention to the direction of the reaction given in the problem statement for \( \text{K}_\text{p} \) or \( \text{K}_\text{c} \). If the reaction is reversed, remember to take the reciprocal of the equilibrium constant. Also, for calculations involving very small changes, approximations can sometimes simplify the math, but ensure the approximation is valid (e.g., if \( \text{x} \) is much smaller than the initial concentration).

Question 28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction :
\( \text{CH}_4 \text{ (g)} + \text{H}_2\text{O (g)} \rightleftharpoons \text{CO (g)} + \text{3H}_2 \text{ (g)} \)
(a) Write an expression of Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst?
Answer:
The given reaction is endothermic (heat is absorbed) and proceeds with an increase in the number of moles (reactants = 2 moles; products = 4 moles).
(a) The expression for \( \text{K}_\text{p} \) for the reaction \( \text{CH}_4 \text{ (g)} + \text{H}_2\text{O (g)} \rightleftharpoons \text{CO (g)} + \text{3H}_2 \text{ (g)} \) is:
\( \text{K}_\text{p} = \frac { \text{p}_{\text{CO}} \times \text{p}_{\text{H}_2}^3 }{ \text{p}_{\text{CH}_4} \times \text{p}_{\text{H}_2\text{O}} } \)
(b)
(i) Increasing the pressure: According to Le Chatelier's principle, increasing the pressure will cause the equilibrium to shift in the backward direction, towards the side with fewer gas moles. This is because the reactant side has 2 moles of gas while the product side has 4 moles of gas.
(ii) Increasing the temperature: Since the reaction is endothermic, increasing the temperature will favor the forward direction, meaning more products will form. Consequently, the value of \( \text{K}_\text{p} \) will increase with higher temperatures.
(iii) Using a catalyst: A catalyst helps the reaction reach equilibrium more quickly by lowering the activation energy for both forward and reverse reactions equally. However, it does not alter the equilibrium composition or the value of the equilibrium constant \( \text{K}_\text{p} \).
In simple words: (a) Kp is written by putting product pressures on top and reactant pressures on bottom, each raised to their coefficient. (b)(i) More pressure shifts the reaction to the side with fewer gas molecules. (ii) For an endothermic reaction (which takes in heat), raising the temperature makes more products. (iii) A catalyst speeds up how fast a reaction gets to balance, but it doesn't change what the balance looks like in the end.

Exam Tip: When writing \( \text{K}_\text{p} \) expressions, only include partial pressures of gaseous species. Solids and liquids are omitted because their partial pressures are constant. For Le Chatelier's principle, always consider the net change in gaseous moles for pressure effects and the enthalpy change for temperature effects.

Question 29. Describe the effect of :
(a) addition of \( \text{H}_2 \)
(b) addition of \( \text{CH}_3\text{OH} \)
(c) removal of \( \text{CO} \)
(d) removal of \( \text{CH}_3\text{OH} \)
on the equilibrium of the reaction:
\( \text{2H}_2 \text{ (g)} + \text{CO (g)} \rightleftharpoons \text{CH}_3\text{OH (g)} \)
Answer:
The given reaction is: \( \text{2H}_2 \text{ (g)} + \text{CO (g)} \rightleftharpoons \text{CH}_3\text{OH (g)} \)
(a) Addition of \( \text{H}_2 \): Adding more \( \text{H}_2 \) (a reactant) will shift the equilibrium in the forward direction. This means more \( \text{CH}_3\text{OH} \) will be formed to consume the added \( \text{H}_2 \).
(b) Addition of \( \text{CH}_3\text{OH} \): Adding more \( \text{CH}_3\text{OH} \) (a product) will shift the reaction backwards. This will produce more \( \text{H}_2 \text{ (g)} \) and \( \text{CO (g)} \) as the system tries to reduce the excess product.
(c) Removal of \( \text{CO} \): Removing \( \text{CO} \) (a reactant) will cause the equilibrium to shift backwards. The system will try to replenish the removed \( \text{CO} \) by breaking down \( \text{CH}_3\text{OH} \).
(d) Removal of \( \text{CH}_3\text{OH} \): Removing \( \text{CH}_3\text{OH} \) (a product) will shift the equilibrium forwards. The system will produce more \( \text{CH}_3\text{OH} \) to compensate for its removal.
In simple words: If you add more of a reactant or remove a product, the reaction will make more products. If you add more of a product or remove a reactant, the reaction will make more reactants.

Exam Tip: Le Chatelier's principle is key here. Think of adding a substance as increasing its concentration, and removing it as decreasing its concentration. The system will always try to undo the change you make to it.

Question 30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is \( 8.3 \times 10^{-3} \). If decomposition is depicted as \( \text{PCl}_5 \text{ (g)} \rightleftharpoons \text{PCl}_3 \text{ (g)} + \text{Cl}_2 \text{ (g)}; \Delta \text{H}^\circ = 124.0 \text{ kJ mol}^{-1} \)
(a) write an expression for Kc for the reaction
(b) what is the value of Kc for the reverse reaction at the same temperature?
(c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased?
Answer:
The given reaction is \( \text{PCl}_5 \text{ (g)} \rightleftharpoons \text{PCl}_3 \text{ (g)} + \text{Cl}_2 \text{ (g)} \), with \( \Delta \text{H}^\circ = 124.0 \text{ kJ mol}^{-1} \).
The equilibrium constant \( \text{K}_\text{c} \) for this reaction at 473 K is \( 8.3 \times 10^{-3} \).
(a) The expression for \( \text{K}_\text{c} \) for the reaction is:
\( \text{K}_\text{c} = \frac { [\text{PCl}_3 \text{ (g)}] \times [\text{Cl}_2 \text{ (g)}] }{ [\text{PCl}_5 \text{ (g)}] } \)
(b) For the reverse reaction, \( \text{PCl}_3 \text{ (g)} + \text{Cl}_2 \text{ (g)} \rightleftharpoons \text{PCl}_5 \text{ (g)} \), the equilibrium constant \( \text{K'}_\text{c} \) will be the reciprocal of the forward \( \text{K}_\text{c} \):
\( \text{K'}_\text{c} = \frac { 1 }{ \text{K}_\text{c} } = \frac { 1 }{ 8.3 \times 10^{-3} } = 120.48 \)
(c) The effect on \( \text{K}_\text{c} \):
(i) If more \( \text{PCl}_5 \) is added, it will not affect the value of \( \text{K}_\text{c} \) because \( \text{K}_\text{c} \) is constant at a fixed temperature. The equilibrium position will shift to consume the added reactant, but the ratio of products to reactants at equilibrium remains the same.
(ii) If pressure is increased, it will not affect the value of \( \text{K}_\text{c} \). \( \text{K}_\text{c} \) is a constant at a given temperature and is independent of pressure changes. The equilibrium position will shift to the side with fewer gas moles (left) to counteract the pressure increase, but the constant's numerical value remains unchanged.
(iii) As the reaction is endothermic ( \( \Delta \text{H}^\circ \) is positive), increasing the temperature will increase the rate of both forward and reverse reactions, but the forward rate will be more enhanced. This will cause the equilibrium to shift towards the products. Thus, the value of \( \text{K}_\text{c} \) will increase with an increase in temperature.
In simple words: (a) Write Kp by putting product concentrations on top and reactant concentrations on bottom. (b) For the reverse reaction, just flip the K value. (c)(i) Adding more of a substance doesn't change K, but the reaction shifts to balance it. (ii) Changing pressure doesn't change K either. (iii) For a reaction that takes in heat (endothermic), making it hotter increases K.

Exam Tip: Remember that the equilibrium constant \( \text{K}_\text{c} \) only changes with temperature. It is not affected by changes in concentration, pressure, or the addition of a catalyst. These factors only shift the position of the equilibrium to establish a new balance, keeping \( \text{K}_\text{c} \) constant.

Question 31. Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and Hr In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
\( \text{CO (g)} + \text{H}_2\text{O (g)} \rightleftharpoons \text{CO}_2 \text{ (g)} + \text{H}_2 \text{ (g)} \)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that \( \text{p}_{\text{CO}} = \text{P}_{\text{H}_2\text{O}} = 4.0 \) bar, what will be the partial pressure of H2 at equilibrium? Kp = 10.1 at 400° С.
Answer:
The given reaction is: \( \text{CO (g)} + \text{H}_2\text{O (g)} \rightleftharpoons \text{CO}_2 \text{ (g)} + \text{H}_2 \text{ (g)} \)
Initial partial pressures:
\( \text{P}_{\text{CO}} = 4.0 \text{ bar} \)
\( \text{P}_{\text{H}_2\text{O}} = 4.0 \text{ bar} \)
\( \text{P}_{\text{CO}_2} = 0 \)
\( \text{P}_{\text{H}_2} = 0 \)
Let \( \text{p} \) be the change in partial pressure at equilibrium.
At equilibrium:
\( \text{P}_{\text{CO}} = (4.0 - \text{p}) \text{ bar} \)
\( \text{P}_{\text{H}_2\text{O}} = (4.0 - \text{p}) \text{ bar} \)
\( \text{P}_{\text{CO}_2} = \text{p bar} \)
\( \text{P}_{\text{H}_2} = \text{p bar} \)
The equilibrium constant \( \text{K}_\text{p} \) is given as 10.1.
The expression for \( \text{K}_\text{p} \) is:
\( \text{K}_\text{p} = \frac { \text{P}_{\text{CO}_2} \times \text{P}_{\text{H}_2} }{ \text{P}_{\text{CO}} \times \text{P}_{\text{H}_2\text{O}} } \)
Substitute the equilibrium pressures into the \( \text{K}_\text{p} \) expression:
\( 10.1 = \frac { \text{p} \times \text{p} }{ (4.0 - \text{p}) \times (4.0 - \text{p}) } \)
\( 10.1 = \frac { \text{p}^2 }{ (4.0 - \text{p})^2 } \)
Take the square root of both sides:
\( \sqrt{10.1} = \sqrt{ \frac { \text{p}^2 }{ (4.0 - \text{p})^2 } } \)
\( 3.178 = \frac { \text{p} }{ 4.0 - \text{p} } \)
Now, solve for \( \text{p} \):
\( 3.178 (4.0 - \text{p}) = \text{p} \)
\( 12.712 - 3.178\text{p} = \text{p} \)
\( 12.712 = \text{p} + 3.178\text{p} \)
\( 12.712 = 4.178\text{p} \)
\( \text{p} = \frac { 12.712 }{ 4.178 } \approx 3.042 \text{ bar} \)
Therefore, the partial pressure of \( \text{H}_2 \) at equilibrium is \( 3.042 \text{ bar} \).
In simple words: Write down the starting pressures and how they change. Use the Kp equation and plug in the equilibrium pressures. Since both sides of the fraction are squared, take the square root of Kp. Solve for 'p' to find the equilibrium partial pressure of H2.

Exam Tip: For reactions where the number of moles of gaseous reactants equals the number of moles of gaseous products, the volume (or total pressure) cancels out in the \( \text{K}_\text{p} \) or \( \text{K}_\text{c} \) expression, simplifying calculations. Always check if you can take a square root to simplify solving for \( \text{p} \).

Question 32. Predict which of the following reaction will have appreciable concentration of reactants and. products :
(a) \( \text{Cl}_2 \text{ (g)} \rightleftharpoons \text{2Cl (g)}; \text{K}_\text{c} = 5 \times 10^{-30} \)
(b) \( \text{Cl}_2 \text{ (g)} + \text{2NO (g)} \rightleftharpoons \text{2NOCl (g)}; \text{K}_\text{c} = 3.7 \times 10^8 \)
(c) \( \text{Cl}_2 \text{ (g)} + \text{2NO}_2 \text{ (g)} \rightleftharpoons \text{2NO}_2\text{Cl (g)}; \text{K}_\text{c} = 1.8 \)
Answer:
(a) For \( \text{Cl}_2 \text{ (g)} \rightleftharpoons \text{2Cl (g)} \), \( \text{K}_\text{c} = 5 \times 10^{-30} \). This \( \text{K}_\text{c} \) value is extremely low. A very small \( \text{K}_\text{c} \) indicates that at equilibrium, the concentration of reactants is much higher than the concentration of products. Therefore, the concentration of \( \text{Cl}_2 \) (reactant) will be high, and the concentration of \( \text{2Cl (g)} \) (product) will be very low.
(b) For \( \text{Cl}_2 \text{ (g)} + \text{2NO (g)} \rightleftharpoons \text{2NOCl (g)} \), \( \text{K}_\text{c} = 3.7 \times 10^8 \). This \( \text{K}_\text{c} \) value is very high. A large \( \text{K}_\text{c} \) indicates that at equilibrium, the concentration of products is much higher than the concentration of reactants. Therefore, the concentration of \( \text{NOCl} \) (product) will be very high.
(c) For \( \text{Cl}_2 \text{ (g)} + \text{2NO}_2 \text{ (g)} \rightleftharpoons \text{2NO}_2\text{Cl (g)} \), \( \text{K}_\text{c} = 1.8 \). This \( \text{K}_\text{c} \) value is close to 1. A \( \text{K}_\text{c} \) value around 1 indicates that at equilibrium, the concentrations of products and reactants are comparable. Therefore, the concentration of \( \text{NO}_2\text{Cl} \) (product) will be somewhat higher than the reactants, but both will be present in significant amounts.
In simple words: If K is tiny, there are mostly starting materials. If K is huge, there are mostly finished products. If K is near 1, you'll find a good mix of both starting materials and products.

Exam Tip: The magnitude of the equilibrium constant (\( \text{K}_\text{c} \) or \( \text{K}_\text{p} \)) gives a quick indication of the relative amounts of reactants and products at equilibrium. A very small K (much less than 1) means reactants are favored, a very large K (much greater than 1) means products are favored, and a K near 1 means significant amounts of both are present.

Question 33. For the reaction \( \text{3O}_2 \text{ (g)} \rightleftharpoons \text{2O}_3 \text{ (g)} \), \( \text{K}_\text{c} \) is \( 2.0 \times 10^{-50} \) at 25°C. If the equilibrium concentration of \( \text{O}_2 \) in air at 25°C is \( 1.6 \times 10^{-2} \), what is the concentration of \( \text{O}_3 \)?
Answer:
The given reaction is: \( \text{3O}_2 \text{ (g)} \rightleftharpoons \text{2O}_3 \text{ (g)} \)
The equilibrium constant \( \text{K}_\text{c} \) is \( 2.0 \times 10^{-50} \).
The equilibrium concentration of \( \text{O}_2 \) is \( [\text{O}_2] = 1.6 \times 10^{-2} \text{ M} \).
Let the equilibrium concentration of \( \text{O}_3 \) be \( [\text{O}_3] = \text{x M} \).
The expression for \( \text{K}_\text{c} \) is:
\( \text{K}_\text{c} = \frac { [\text{O}_3 \text{ (g)}]^2 }{ [\text{O}_2 \text{ (g)}]^3 } \)
Substitute the known values into the expression:
\( 2.0 \times 10^{-50} = \frac { \text{x}^2 }{ (1.6 \times 10^{-2})^3 } \)
Now, solve for \( \text{x}^2 \):
\( \text{x}^2 = (2.0 \times 10^{-50}) \times (1.6 \times 10^{-2})^3 \)
\( \text{x}^2 = (2.0 \times 10^{-50}) \times (4.096 \times 10^{-6}) \)
\( \text{x}^2 = 8.192 \times 10^{-56} \)
Take the square root to find \( \text{x} \):
\( \text{x} = \sqrt{8.192 \times 10^{-56}} \)
\( \text{x} = \sqrt{8.192} \times \sqrt{10^{-56}} \)
\( \text{x} \approx 2.862 \times 10^{-28} \text{ M} \)
Therefore, the equilibrium concentration of \( \text{O}_3 \) is approximately \( 2.86 \times 10^{-28} \text{ M} \).
In simple words: Use the Kc formula, plugging in the given oxygen concentration. Then solve for the unknown ozone concentration by squaring it, multiplying across, and taking the square root.

Exam Tip: Be careful with scientific notation and powers when doing calculations. Remember that \( (\text{a}^\text{m})^\text{n} = \text{a}^{\text{m} \times \text{n}} \) and \( \sqrt{\text{a}^\text{m}} = \text{a}^{\text{m}/2} \). A very small \( \text{K}_\text{c} \) value means that product concentrations will be extremely low.

Question 34. The reaction, \( \text{CO (g)} + \text{3H}_2 \text{ (g)} \rightleftharpoons \text{CH}_4 \text{ (g)} + \text{H}_2\text{O (g)} \) is at equilibrium at 1300 K in a 1 L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.2 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:
The given reaction is: \( \text{CO (g)} + \text{3H}_2 \text{ (g)} \rightleftharpoons \text{CH}_4 \text{ (g)} + \text{H}_2\text{O (g)} \)
The volume of the flask is 1 L, so moles are equal to molar concentrations.
Equilibrium concentrations:
\( [\text{CO}] = 0.30 \text{ mol/L} \)
\( [\text{H}_2] = 0.10 \text{ mol/L} \)
\( [\text{H}_2\text{O}] = 0.20 \text{ mol/L} \)
Let \( [\text{CH}_4] = \text{x mol/L} \).
The equilibrium constant \( \text{K}_\text{c} \) is given as 3.90.
The expression for \( \text{K}_\text{c} \) is:
\( \text{K}_\text{c} = \frac { [\text{CH}_4][\text{H}_2\text{O}] }{ [\text{CO}][\text{H}_2]^3 } \)
Substitute the known values into the expression:
\( 3.90 = \frac { \text{x} \times 0.20 }{ 0.30 \times (0.10)^3 } \)
Now, solve for \( \text{x} \):
\( \text{x} = \frac { 3.90 \times 0.30 \times (0.10)^3 }{ 0.20 } \)
\( \text{x} = \frac { 3.90 \times 0.30 \times 0.001 }{ 0.20 } \)
\( \text{x} = \frac { 0.00117 }{ 0.20 } \)
\( \text{x} = 0.00585 \text{ M} \)
Therefore, the concentration of \( \text{CH}_4 \) in the mixture is \( 5.85 \times 10^{-3} \text{ M} \).
In simple words: Write the Kc expression using the concentrations of reactants and products. Plug in all the known values, leaving the methane concentration as 'x'. Then, simply solve the equation to find 'x'.

Exam Tip: In equilibrium calculations, always make sure to use the correct stoichiometric coefficients as powers in the equilibrium constant expression. When the volume is 1 L, moles directly correspond to molar concentrations, simplifying calculations.

Question 35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:
\( \text{HNO}_2 \), \( \text{CN}^- \), \( \text{HClO}_4 \), \( \text{F}^- \), \( \text{OH}^- \), \( \text{CO}_3^{2-} \), and \( \text{S}^{2-} \).
Answer:
A conjugate acid-base pair refers to two species that differ by only one proton (\( \text{H}^+ \)). The acid in the pair has one more proton than its conjugate base.

Here are the conjugate acid/base pairs for the given species:
- \( \text{HNO}_2 \): Its conjugate base is \( \text{NO}_2^- \).
- \( \text{CN}^- \): Its conjugate acid is \( \text{HCN} \).
- \( \text{HClO}_4 \): Its conjugate base is \( \text{ClO}_4^- \).
- \( \text{F}^- \): Its conjugate acid is \( \text{HF} \).
- \( \text{OH}^- \): It is the conjugate base of \( \text{H}_2\text{O} \), and its conjugate base is \( \text{O}^{2-} \).
- \( \text{CO}_3^{2-} \): It is the conjugate base of \( \text{HCO}_3^- \). The conjugate acid of \( \text{CO}_3^{2-} \) is \( \text{HCO}_3^- \).
- \( \text{S}^{2-} \): Its conjugate acid is \( \text{HS}^- \).
In simple words: A conjugate acid-base pair are two chemicals that become each other by gaining or losing just one proton (H+). The acid has the extra H+.

Exam Tip: To find the conjugate base of an acid, remove one \( \text{H}^+ \). To find the conjugate acid of a base, add one \( \text{H}^+ \). Always ensure the charges are balanced correctly after adding or removing a proton.

Question 36. Which of the following are Lewis acids? \( \text{H}_2\text{O} \), \( \text{BF}_3 \), \( \text{H}^+ \), and \( \text{NH}_4^+ \).
Answer:
A Lewis acid is a substance (a molecule or ion) that has a tendency to accept a pair of electrons.
Among the given species, \( \text{BF}_3 \), \( \text{H}^+ \), and \( \text{NH}_4^+ \) are Lewis acids.
- \( \text{BF}_3 \) is a Lewis acid because boron has an incomplete octet and can accept a lone pair of electrons.
- \( \text{H}^+ \) is a Lewis acid as it is an electron-deficient species and can accept an electron pair.
- \( \text{NH}_4^+ \) can act as a Lewis acid by accepting an electron pair, although it is more commonly recognized as a Bronsted-Lowry acid.
In simple words: Lewis acids are electron-pair acceptors. From the list, BF3, H+, and NH4+ are substances that can accept electrons.

Exam Tip: Remember the definitions: a Lewis acid accepts an electron pair, and a Lewis base donates an electron pair. Common Lewis acids include cations, compounds with incomplete octets (like \( \text{BF}_3 \), \( \text{AlCl}_3 \)), and molecules with polar multiple bonds.

Question 37. What will be the conjugate bases for the Bronsted acids : \( \text{HF} \), \( \text{H}_2\text{SO}_4 \) and \( \text{HCO}_3^- \)?
Answer:
To find the conjugate base of a Bronsted acid, remove one proton (\( \text{H}^+ \)) from the acid.
- For \( \text{HF} \): its conjugate base is \( \text{F}^- \).
- For \( \text{H}_2\text{SO}_4 \): its conjugate base is \( \text{HSO}_4^- \).
- For \( \text{HCO}_3^- \): its conjugate base is \( \text{CO}_3^{2-} \).
In simple words: To find the conjugate base, just take away one H+ from the acid.

Exam Tip: A Bronsted-Lowry acid is a proton donor. Its conjugate base is the species remaining after the acid has donated a proton. Ensure the charge is adjusted correctly when removing the \( \text{H}^+ \).

Question 38. Write the conjugate acids for the following Bronsted bases : \( \text{NH}_2^- \), \( \text{NH}_3 \) and \( \text{HCOO}^- \).
Answer:
To find the conjugate acid of a Bronsted base, add one proton (\( \text{H}^+ \)) to the base.
- For \( \text{NH}_2^- \): its conjugate acid is \( \text{NH}_3 \).
- For \( \text{NH}_3 \): its conjugate acid is \( \text{NH}_4^+ \).
- For \( \text{HCOO}^- \): its conjugate acid is \( \text{HCOOH} \).
In simple words: To find the conjugate acid, just add one H+ to the base.

Exam Tip: A Bronsted-Lowry base is a proton acceptor. Its conjugate acid is the species formed after the base has accepted a proton. Make sure to adjust the charge correctly when adding the \( \text{H}^+ \).

Question 39. The species : \( \text{H}_2\text{O} \), \( \text{HCO}_3^- \), \( \text{HSO}_4^- \) and \( \text{NH}_3 \) can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:
These species are amphoteric, meaning they can act as both Bronsted acids and bases. Here are their corresponding conjugate acids and bases:
- For \( \text{H}_2\text{O} \):
- When acting as an acid, its conjugate base is \( \text{OH}^- \).
- When acting as a base, its conjugate acid is \( \text{H}_3\text{O}^+ \).
- For \( \text{HCO}_3^- \):
- When acting as an acid, its conjugate base is \( \text{CO}_3^{2-} \).
- When acting as a base, its conjugate acid is \( \text{H}_2\text{CO}_3 \).
- For \( \text{HSO}_4^- \):
- When acting as an acid, its conjugate base is \( \text{SO}_4^{2-} \).
- When acting as a base, its conjugate acid is \( \text{H}_2\text{SO}_4 \).
- For \( \text{NH}_3 \):
- When acting as an acid, its conjugate base is \( \text{NH}_2^- \).
- When acting as a base, its conjugate acid is \( \text{NH}_4^+ \).
In simple words: Some chemicals can act as both acids (giving H+) and bases (taking H+). For each, you can find a conjugate base (what's left after giving H+) and a conjugate acid (what it becomes after taking H+).

Exam Tip: Amphoteric species are crucial in acid-base chemistry. To identify them, check if a species can both donate a proton (to form a more negatively charged species) and accept a proton (to form a more positively charged species or a neutral molecule). Water is a classic example.

Question 40. Classify the following species into Lewis acids ad Lewis bases and show how these act as Lewis acid/base :
(a) \( \text{OH}^- \)
(b) \( \text{F}^- \)
(c) \( \text{H}^+ \)
(d) \( \text{BCl}_3 \)
Answer:
(a) \( \text{OH}^- \) is a Lewis base because it has lone pairs of electrons and can donate an electron pair. For example, it reacts with \( \text{H}^+ \):
\( \text{:OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O} \)
(b) \( \text{F}^- \) is a Lewis base because it has lone pairs of electrons and can furnish an electron pair. For example, it reacts with \( \text{BF}_3 \):
(c) \( \text{H}^+ \) is a Lewis acid because it can accept a pair of electrons. For instance, it accepts electrons from \( \text{NH}_3 \):
\( \text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+ \)
(d) \( \text{BCl}_3 \) is a Lewis acid because boron has an incomplete octet and can accept a pair of electrons. For example, it reacts with \( \text{NH}_3 \):
\( \text{H}_3\text{N} + \text{BCl}_3 \rightarrow \text{H}_3\text{N}-\text{BCl}_3 \)
In simple words: Lewis bases give away electron pairs, while Lewis acids take electron pairs. OH- and F- are bases because they have extra electrons to share. H+ and BCl3 are acids because they need electrons.

Exam Tip: Visualizing electron movement can help identify Lewis acids and bases. Lewis bases typically have lone pairs or negative charges, making them electron-rich. Lewis acids are typically electron-deficient, often having positive charges or incomplete octets.

Question 41. The concentration of hydrogen ion in a sample of soft drink is \( 3.8 \times 10^{-3}\text{M} \). What is its pH?
Answer:
The concentration of hydrogen ion is \( [\text{H}^+] = 3.8 \times 10^{-3}\text{ M} \).
The pH is calculated using the formula: \( \text{pH} = -\text{log}[\text{H}^+] \).
\( \text{pH} = -\text{log}(3.8 \times 10^{-3}) \)
\( \text{pH} = -(\text{log } 3.8 + \text{log } 10^{-3}) \)
\( \text{pH} = -(\text{log } 3.8 - 3) \)
We know that \( \text{log } 3.8 \approx 0.5798 \).
\( \text{pH} = -(0.5798 - 3) \)
\( \text{pH} = -(-2.4202) \)
\( \text{pH} = 2.4202 \)
Thus, the pH of the soft drink is approximately 2.42.
In simple words: To find pH, take the negative logarithm of the hydrogen ion concentration. Break down the logarithm of the scientific notation, calculate, and solve.

Exam Tip: Remember the properties of logarithms, especially \( \text{log (A} \times \text{B)} = \text{log A} + \text{log B} \) and \( \text{log (A}^\text{B}\text{)} = \text{B log A} \). These are very useful for pH calculations involving scientific notation.

Question 42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
The pH of the vinegar sample is 3.76.
The pH formula is: \( \text{pH} = -\text{log}[\text{H}^+] \)
So, \( 3.76 = -\text{log}[\text{H}^+] \)
\( \text{log}[\text{H}^+] = -3.76 \)
To find \( [\text{H}^+] \), we need to take the antilog of -3.76.
To make calculation easier, express -3.76 as a negative characteristic and a positive mantissa:
\( -3.76 = -4 + 0.24 = \bar{4}.24 \)
\( [\text{H}^+] = \text{antilog}(\bar{4}.24) \)
\( [\text{H}^+] = 10^{0.24} \times 10^{-4} \)
\( 10^{0.24} \approx 1.7378 \)
\( [\text{H}^+] \approx 1.74 \times 10^{-4} \text{ M} \)
Therefore, the hydrogen ion concentration \( [\text{H}^+] \) is \( 1.74 \times 10^{-4} \text{ M} \).
In simple words: If you know the pH, you can find the hydrogen ion concentration by taking the antilog of the negative pH. This might involve splitting the negative pH into a whole number and a decimal to calculate easily.

Exam Tip: When dealing with negative logarithms (like \( \text{log}[\text{H}^+] = -3.76 \)), convert the negative number into the form \( (\text{n} + \text{mantissa}) \) where \( \text{n} \) is a negative integer and the mantissa is a positive decimal. This allows for easier calculation of the antilog.

Question 43. The ionization constant of HF, HCOOH and HCN at 298 K are \( 6.8 \times 10^{-4} \), \( 1.8 \times 10^{-4} \) and \( 4.8 \times 10^{-9} \) respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
We use the relationship \( \text{K}_\text{a} \times \text{K}_\text{b} = \text{K}_\text{w} \), where \( \text{K}_\text{a} \) is the dissociation constant of the weak acid, \( \text{K}_\text{b} \) is the ionization constant of its conjugate base, and \( \text{K}_\text{w} \) is the ionic product of water, which is \( 1.0 \times 10^{-14} \) at 298 K.

For \( \text{HF} \):
\( \text{K}_\text{a} = 6.8 \times 10^{-4} \)
\( \text{K}_\text{F}^- = \frac { \text{K}_\text{w} }{ \text{K}_\text{a} } = \frac { 1.0 \times 10^{-14} }{ 6.8 \times 10^{-4} } = 1.47 \times 10^{-11} \approx 1.5 \times 10^{-11} \)

For \( \text{HCOOH} \) (formic acid):
\( \text{K}_\text{a} = 1.8 \times 10^{-4} \)
\( \text{K}_{\text{HCOO}^-} = \frac { \text{K}_\text{w} }{ \text{K}_\text{a} } = \frac { 1.0 \times 10^{-14} }{ 1.8 \times 10^{-4} } = 5.56 \times 10^{-11} \approx 5.6 \times 10^{-11} \)

For \( \text{HCN} \):
\( \text{K}_\text{a} = 4.8 \times 10^{-9} \)
\( \text{K}_{\text{CN}^-} = \frac { \text{K}_\text{w} }{ \text{K}_\text{a} } = \frac { 1.0 \times 10^{-14} }{ 4.8 \times 10^{-9} } = 2.08 \times 10^{-6} \)
In simple words: To find the ionization constant of a conjugate base, divide the ionic product of water (Kw) by the acid's ionization constant (Ka). Kw is usually \( 1.0 \times 10^{-14} \).

Exam Tip: This relationship (\( \text{K}_\text{a} \times \text{K}_\text{b} = \text{K}_\text{w} \)) is fundamental for understanding the strengths of conjugate acid-base pairs. A strong acid will have a very weak conjugate base, and vice-versa, ensuring their product remains constant.

Question 44. The ionization constant of phenol is \( 1.0 \times 10^{-10} \). What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
Answer:
Given: Ionization constant of phenol \( (\text{K}_\text{a}) = 1.0 \times 10^{-10} \)
Initial concentration of phenol \( (\text{C}) = 0.05 \text{ M} \)
Phenol is a weak acid that dissociates as: \( \text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- + \text{H}^+ \)
Let \( \alpha \) be the degree of ionization.
At equilibrium:
\( [\text{C}_6\text{H}_5\text{OH}] \approx \text{C} = 0.05 \text{ M} \) (since \( \alpha \) is small)
\( [\text{C}_6\text{H}_5\text{O}^-] = \text{C}\alpha \)
\( [\text{H}^+] = \text{C}\alpha \)
For a weak acid, \( \text{K}_\text{a} = \frac { [\text{C}_6\text{H}_5\text{O}^-][\text{H}^+] }{ [\text{C}_6\text{H}_5\text{OH}] } = \frac { (\text{C}\alpha)(\text{C}\alpha) }{ \text{C} } = \text{C}\alpha^2 \)
Therefore, \( \alpha = \sqrt{ \frac { \text{K}_\text{a} }{ \text{C} } } \)
\( \alpha = \sqrt{ \frac { 1.0 \times 10^{-10} }{ 0.05 } } = \sqrt{2.0 \times 10^{-9}} = \sqrt{20 \times 10^{-10}} \)
\( \alpha = 4.47 \times 10^{-5} \)
Concentration of phenolate ion \( [\text{C}_6\text{H}_5\text{O}^-] = \text{C}\alpha = 0.05 \times (4.47 \times 10^{-5}) = 2.235 \times 10^{-6} \text{ M} \approx 2.2 \times 10^{-6} \text{ M} \)

Now, consider the presence of 0.01 M sodium phenolate. Sodium phenolate (\( \text{C}_6\text{H}_5\text{O}^- \text{Na}^+ \)) is a strong electrolyte and provides \( \text{C}_6\text{H}_5\text{O}^- \) ions. This is a common ion effect problem.
Initial concentrations:
\( [\text{C}_6\text{H}_5\text{OH}] = 0.05 \text{ M} \)
\( [\text{C}_6\text{H}_5\text{O}^-] = 0.01 \text{ M} \) (from sodium phenolate)
Let \( \text{y} \) be the amount of phenol dissociated in the presence of sodium phenolate.
Equilibrium concentrations:
\( [\text{C}_6\text{H}_5\text{OH}] = 0.05 - \text{y} \approx 0.05 \text{ M} \) (since \( \text{y} \) will be very small)
\( [\text{C}_6\text{H}_5\text{O}^-] = 0.01 + \text{y} \approx 0.01 \text{ M} \)
\( [\text{H}^+] = \text{y M} \)
Using the \( \text{K}_\text{a} \) expression:
\( \text{K}_\text{a} = \frac { (0.01)(\text{y}) }{ 0.05 } = 1.0 \times 10^{-10} \)
\( \text{y} = \frac { (1.0 \times 10^{-10}) \times 0.05 }{ 0.01 } \)
\( \text{y} = 5.0 \times 10^{-10} \text{ M} \)
The degree of ionization \( \alpha' \) in the presence of sodium phenolate is:
\( \alpha' = \frac { \text{y} }{ \text{C} } = \frac { 5.0 \times 10^{-10} }{ 0.05 } = 1.0 \times 10^{-8} \)
In simple words: First, calculate how much phenolate forms and the degree of ionization when phenol is alone. Then, when sodium phenolate is added, the common ion effect reduces phenol's dissociation, making the new degree of ionization much smaller.

Exam Tip: Remember that the common ion effect significantly suppresses the ionization of a weak acid or base when a salt containing a common ion is added. This effect is crucial for understanding buffer solutions and solubility of sparingly soluble salts.

Question 45. The first ionization constant of \( \text{H}_2\text{S} \) is \( 9.1 \times 10^{-8} \). Calculate the concentration of \( \text{HS}^- \) ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of \( \text{H}_2\text{S} \) is \( 1.2 \times 10^{-13} \), calculate the concentration of \( \text{S}^{2-} \) under both conditions.
Answer:
**Case 1: Pure 0.1 M \( \text{H}_2\text{S} \) solution**
\( \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \)
Given \( \text{K}_{\text{a}_1} = 9.1 \times 10^{-8} \)
Initial concentration of \( \text{H}_2\text{S} = \text{C} = 0.1 \text{ M} \)
Let \( \alpha \) be the degree of ionization.
At equilibrium, \( [\text{H}^+] = [\text{HS}^-] = \text{C}\alpha \)
\( [\text{H}_2\text{S}] = \text{C}(1-\alpha) \approx \text{C} \)
\( \text{K}_{\text{a}_1} = \frac { [\text{H}^+][\text{HS}^-] }{ [\text{H}_2\text{S}] } = \frac { (\text{C}\alpha)(\text{C}\alpha) }{ \text{C} } = \text{C}\alpha^2 \)
\( \alpha = \sqrt{ \frac { \text{K}_{\text{a}_1} }{ \text{C} } } = \sqrt{ \frac { 9.1 \times 10^{-8} }{ 0.1 } } = \sqrt{91 \times 10^{-8}} \approx 3.02 \times 10^{-4} \)
\( [\text{HS}^-] = \text{C}\alpha = 0.1 \times (3.02 \times 10^{-4}) = 3.02 \times 10^{-5} \text{ M} \).
Since \( \alpha \) is very small, this approximation is valid.

Now for the second dissociation:
\( \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \)
Given \( \text{K}_{\text{a}_2} = 1.2 \times 10^{-13} \)
We assume \( [\text{H}^+] = [\text{HS}^-] = 3.02 \times 10^{-5} \text{ M} \) from the first dissociation.
Let \( [\text{S}^{2-}] = \text{x} \). Then \( [\text{H}^+] = 3.02 \times 10^{-5} + \text{x} \approx 3.02 \times 10^{-5} \text{ M} \).
\( \text{K}_{\text{a}_2} = \frac { [\text{H}^+][\text{S}^{2-}] }{ [\text{HS}^-] } \)
\( 1.2 \times 10^{-13} = \frac { (3.02 \times 10^{-5}) \times \text{x} }{ 3.02 \times 10^{-5} } \)
\( \implies \text{x} = 1.2 \times 10^{-13} \text{ M} \)
So, \( [\text{S}^{2-}] = 1.2 \times 10^{-13} \text{ M} \).

**Case 2: In the presence of 0.1 M HCl**
\( \text{HCl} \) is a strong acid and completely dissociates: \( \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \).
So, initial \( [\text{H}^+] = 0.1 \text{ M} \).
For the first dissociation of \( \text{H}_2\text{S} \): \( \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \)
Let \( \text{y} \) be the concentration of \( \text{HS}^- \) formed.
Equilibrium concentrations:
\( [\text{H}_2\text{S}] = 0.1 - \text{y} \approx 0.1 \text{ M} \)
\( [\text{H}^+] = 0.1 + \text{y} \approx 0.1 \text{ M} \) (due to common ion effect from \( \text{HCl} \))
\( [\text{HS}^-] = \text{y M} \)
\( \text{K}_{\text{a}_1} = \frac { [\text{H}^+][\text{HS}^-] }{ [\text{H}_2\text{S}] } \)
\( 9.1 \times 10^{-8} = \frac { (0.1) \times \text{y} }{ 0.1 } \)
\( \implies \text{y} = 9.1 \times 10^{-8} \text{ M} \)
Thus, the concentration of \( [\text{HS}^-] \) is \( 9.1 \times 10^{-8} \text{ M} \) in 0.1 M \( \text{HCl} \). This is much lower than in pure water due to the common ion effect.

For the second dissociation of \( \text{H}_2\text{S} \) in 0.1 M \( \text{HCl} \):
\( \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \)
\( \text{K}_{\text{a}_2} = 1.2 \times 10^{-13} \)
Equilibrium concentrations:
\( [\text{HS}^-] = 9.1 \times 10^{-8} \text{ M} \)
\( [\text{H}^+] = 0.1 \text{ M} \) (primarily from \( \text{HCl} \))
Let \( [\text{S}^{2-}] = \text{z} \).
\( \text{K}_{\text{a}_2} = \frac { [\text{H}^+][\text{S}^{2-}] }{ [\text{HS}^-] } \)
\( 1.2 \times 10^{-13} = \frac { (0.1) \times \text{z} }{ 9.1 \times 10^{-8} } \)
\( \text{z} = \frac { (1.2 \times 10^{-13}) \times (9.1 \times 10^{-8}) }{ 0.1 } \)
\( \text{z} = 1.092 \times 10^{-19} \text{ M} \)
Therefore, the concentration of \( [\text{S}^{2-}] \) is \( 1.092 \times 10^{-19} \text{ M} \) in 0.1 M \( \text{HCl} \).
In simple words: First, calculate the concentration of HS- and S2- in pure H2S solution using the ionization constants. Then, in the presence of HCl (a strong acid), the common H+ ion will push the H2S dissociation backwards, greatly reducing the concentrations of both HS- and S2-.

Exam Tip: For polyprotic acids like \( \text{H}_2\text{S} \), each successive ionization constant is significantly smaller than the previous one. This allows you to approximate that the \( [\text{H}^+] \) for the second (or subsequent) ionization is primarily determined by the first ionization or by a strong acid present.

Question 46. The ionization constant of acetic acid is \( 1.74 \times 10^{-5} \). Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:
Given: Ionization constant of acetic acid \( (\text{K}_\text{a}) = 1.74 \times 10^{-5} \)
Concentration of acetic acid \( (\text{C}) = 0.05 \text{ M} \)
Acetic acid dissociates as: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \)
Let \( \alpha \) be the degree of dissociation.
At equilibrium:
\( [\text{CH}_3\text{COOH}] = \text{C}(1-\alpha) \approx \text{C} = 0.05 \text{ M} \) (assuming \( \alpha \) is small)
\( [\text{CH}_3\text{COO}^-] = \text{C}\alpha \)
\( [\text{H}^+] = \text{C}\alpha \)
The expression for \( \text{K}_\text{a} \) is:
\( \text{K}_\text{a} = \frac { [\text{CH}_3\text{COO}^-][\text{H}^+] }{ [\text{CH}_3\text{COOH}] } = \frac { (\text{C}\alpha)(\text{C}\alpha) }{ \text{C} } = \text{C}\alpha^2 \)
Calculate the degree of dissociation \( \alpha \):
\( \alpha = \sqrt{ \frac { \text{K}_\text{a} }{ \text{C} } } = \sqrt{ \frac { 1.74 \times 10^{-5} }{ 0.05 } } = \sqrt{3.48 \times 10^{-4}} \)
\( \alpha = \sqrt{34.8 \times 10^{-5}} \approx 0.01865 \approx 1.87 \times 10^{-2} \)
The degree of dissociation is \( 1.87 \times 10^{-2} \).

Calculate the concentration of acetate ion \( [\text{CH}_3\text{COO}^-] \):
\( [\text{CH}_3\text{COO}^-] = \text{C}\alpha = 0.05 \times (1.87 \times 10^{-2}) = 9.35 \times 10^{-4} \text{ M} \)

Calculate the pH of the solution:
\( [\text{H}^+] = \text{C}\alpha = 9.35 \times 10^{-4} \text{ M} \)
\( \text{pH} = -\text{log}[\text{H}^+] = -\text{log}(9.35 \times 10^{-4}) \)
\( \text{pH} = -(\text{log } 9.35 + \text{log } 10^{-4}) \)
\( \text{pH} = -(\text{log } 9.35 - 4) \)
We know that \( \text{log } 9.35 \approx 0.9708 \).
\( \text{pH} = -(0.9708 - 4) = -(-3.0292) = 3.0292 \)
The pH of the solution is approximately 3.03.
In simple words: First, use the ionization constant and acid concentration to find the degree of dissociation. This value then helps you calculate the concentration of the acetate ion and the hydrogen ion, which you use to determine the pH.

Exam Tip: For weak acids, the approximation \( (1-\alpha) \approx 1 \) is valid if \( \text{K}_\text{a}/\text{C} < 10^{-3} \). If this condition is not met, a quadratic equation must be solved. Always check the validity of your approximations.

Question 47. It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Answer:
Given: pH of the solution = 4.15
Concentration of organic acid \( (\text{C}) = 0.01 \text{ M} \)
Let the organic acid be \( \text{HA} \). It dissociates as: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \)

Calculate the concentration of hydrogen ion \( [\text{H}^+] \):
\( \text{pH} = -\text{log}[\text{H}^+] \)
\( 4.15 = -\text{log}[\text{H}^+] \)
\( \text{log}[\text{H}^+] = -4.15 \)
\( [\text{H}^+] = \text{antilog}(-4.15) \)
To take the antilog, we rewrite -4.15 as \( -5 + 0.85 = \bar{5}.85 \).
\( [\text{H}^+] = 10^{0.85} \times 10^{-5} \approx 7.079 \times 10^{-5} \text{ M} \)

Calculate the concentration of the anion \( [\text{A}^-] \):
Since \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \), at equilibrium \( [\text{H}^+] = [\text{A}^-] \).
Therefore, \( [\text{A}^-] = 7.08 \times 10^{-5} \text{ M} \).

Calculate the ionization constant of the acid \( \text{K}_\text{a} \):
At equilibrium:
\( [\text{HA}] = \text{C} - [\text{H}^+] = 0.01 - 7.08 \times 10^{-5} = 0.0099292 \text{ M} \approx 0.01 \text{ M} \)
\( \text{K}_\text{a} = \frac { [\text{H}^+][\text{A}^-] }{ [\text{HA}] } = \frac { (7.08 \times 10^{-5}) \times (7.08 \times 10^{-5}) }{ 0.01 } \)
\( \text{K}_\text{a} = \frac { 5.01264 \times 10^{-9} }{ 0.01 } = 5.01264 \times 10^{-7} \approx 5.01 \times 10^{-7} \)

Calculate the \( \text{pK}_\text{a} \):
\( \text{pK}_\text{a} = -\text{log}(\text{K}_\text{a}) = -\text{log}(5.01 \times 10^{-7}) \)
\( \text{pK}_\text{a} = -(\text{log } 5.01 + \text{log } 10^{-7}) \)
\( \text{pK}_\text{a} = -(\text{log } 5.01 - 7) \)
We know that \( \text{log } 5.01 \approx 0.6998 \).
\( \text{pK}_\text{a} = -(0.6998 - 7) = -(-6.3002) = 6.3002 \)
Therefore, \( \text{pK}_\text{a} \approx 6.30 \).
In simple words: First, convert the given pH into hydrogen ion concentration. This also gives you the anion concentration. Then, use these concentrations, along with the initial acid concentration, to find the acid's ionization constant (Ka). Finally, take the negative logarithm of Ka to find pKa.

Exam Tip: Remember that \( [\text{H}^+] \) and \( [\text{A}^-] \) are generally considered equal for weak acid dissociation in pure water. For accurate \( \text{K}_\text{a} \) calculations, ensure you use the equilibrium concentration of the undissociated acid \( (\text{C} - [\text{H}^+]) \), although for very weak acids, it can be approximated as \( \text{C} \).

Question 48. Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH.
Answer:
(a) HCl (aq) \( \rightarrow \) H\(^+\) (aq) + Cl\(^-\) (aq)
[HCl] = 0.003 M
As HCl completely dissociates into H\(^+\) ions,
\( \implies \) [H\(^+\)] = [HCl] = 0.003 M
pH = \( -\log[\text{H}^+] = -\log (3 \times 10^{-3}) \)
\( \implies \) pH = \( -(-\text{3}) \log 3 = 3 - \log 3 = 3 - 0.477 = 2.523 \).
(b) NaOH (aq) \( \rightarrow \) Na\(^+\) (aq) + OH\(^-\) (aq)
[NaOH] = \( 0.005 = 5 \times 10^{-3} \) M
[OH\(^-\)] = [NaOH] = \( 5 \times 10^{-3} \) M
\( \implies \) [H\(^+\)] = \( \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-3}} = 2.0 \times 10^{-12} \)
\( \implies \) pH = \( -\log (2 \times 10^{-12}) = -(-\text{12}) - \log 2 \)
\( \implies \) pH = \( 12 - 0.30 = 11.70 \). (Given \( \log 2 = 0.30 \))
(c) HBr \( \rightarrow \) H\(^+\) + Br\(^-\); completely dissociated
0.002 M 0.002 M 0.002 M
[HBr] = 0.002 M
\( \implies \) [H\(^+\)] = [HBr] = \( 0.002 \text{ M} = 2.0 \times 10^{-3} \text{ M} \)
pH = \( -\log[\text{H}^+] = -\log (2 \times 10^{-3}) \)
\( \implies \) pH = \( -(-\text{3}) - \log 2 = 3 - \log 2 \)
\( \implies \) pH = \( 3 - 0.3 = 2.70 \).
(d) KOH \( \rightarrow \) K\(^+\) + OH\(^-\); completely dissociated
0.002 M 0.002 M 0.002 M
[OH\(^-\)] = 0.002 M
\( \implies \) [H\(^+\)] = \( \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.002} = 5 \times 10^{-12} \text{ M} \)
\( \implies \) pH = \( -\log[\text{H}^+] = -(-\text{12}) - \log 5 \)
\( \implies \) pH = \( 12 - 0.70 = 11.30 \).
In simple words: For strong acids and bases, hydrogen or hydroxide ion concentration equals the solution concentration. Then, use the pH formula \( (-\log[\text{H}^+]) \) and the relationship \( [\text{H}^+][\text{OH}^-] = K_w \) to find pH or pOH.

Exam Tip: Remember that for strong acids and bases, they completely dissociate, so the concentration of H\(^+\) or OH\(^-\) is directly the initial concentration of the acid or base. Always use \( K_w = 1.0 \times 10^{-14} \) at 298 K unless specified otherwise.

Question 49. Calculate the pH of the following solutions:
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)\(_2\) dissolved in water to give 500 mL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 ml of 13.6 M HCI is diluted to give 1 litre of solution.
Answer:
(a) TlOH \( \rightleftharpoons \) Tl\(^+\) + OH\(^-\)
Molar mass of TlOH = 204 + 16 + 1 = 221 g/mol
Concentration of TlOH = \( \frac{2 \text{ g}}{221 \text{ g/mol} \times 2 \text{ L}} = \frac{1}{221} \text{ M} \)
Since TlOH is a strong base, it dissociates completely.
\( \implies \) [OH\(^-\)] = \( \frac{1}{221} \text{ M} \)
\( \implies \) [H\(^+\)] = \( \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{1/221} = 221 \times 10^{-14} \text{ M} \)
pH = \( -\log[\text{H}^+] = -\log(221 \times 10^{-14}) = -(-\text{14}) - \log 221 = 14 - 2.3 = 11.70 \).
(b) Ca(OH)\(_2\) \( \rightleftharpoons \) Ca\(^{2+}\) + 2OH\(^-\)
Molar mass of Ca(OH)\(_2\) = 40 + \( 2 \times (16+1) \) = 74 g/mol
Mass of Ca(OH)\(_2\) = 0.3 g
Volume of solution = 500 mL = 0.5 L
Moles of Ca(OH)\(_2\) = \( \frac{0.3 \text{ g}}{74 \text{ g/mol}} = 0.004054 \text{ mol} \)
Concentration of Ca(OH)\(_2\) = \( \frac{0.004054 \text{ mol}}{0.5 \text{ L}} = 0.008108 \text{ M} \)
Since Ca(OH)\(_2\) is a strong base, it dissociates completely.
\( \implies \) [OH\(^-\)] = \( 2 \times 0.008108 \text{ M} = 0.016216 \text{ M} = 1.62 \times 10^{-2} \text{ M} \)
\( \implies \) [H\(^+\)] = \( \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{1.62 \times 10^{-2}} = 6.17 \times 10^{-13} \text{ M} \)
pH = \( -\log[\text{H}^+] = -\log(6.17 \times 10^{-13}) = -(-\text{13}) - \log 6.17 = 13 - 0.79 = 12.21 \).
(c) NaOH \( \rightleftharpoons \) Na\(^+\) (aq) + OH\(^-\) (aq)
Molar mass of NaOH = 40.0 g/mol
Mass of NaOH = 0.3 g
Volume of solution = 200 mL = 0.2 L
Moles of NaOH = \( \frac{0.3 \text{ g}}{40.0 \text{ g/mol}} = 0.0075 \text{ mol} \)
Concentration of NaOH = \( \frac{0.0075 \text{ mol}}{0.2 \text{ L}} = 0.0375 \text{ M} \)
Since NaOH is a strong base, it dissociates completely.
\( \implies \) [OH\(^-\)] = 0.0375 M
\( \implies \) [H\(^+\)] = \( \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.0375} = 2.667 \times 10^{-13} \text{ M} \)
pH = \( -\log[\text{H}^+] = -\log(2.667 \times 10^{-13}) = -(-\text{13}) - \log 2.667 = 13 - 0.426 = 12.574 \).
(d) 1 mL of 13.6 M HCl is diluted to give 1 litre of solution.
For dilution, \( M_1V_1 = M_2V_2 \)
\( M_1 = 13.6 \text{ M} \), \( V_1 = 1 \text{ mL} \)
\( V_2 = 1 \text{ L} = 1000 \text{ mL} \)
\( M_2 = \frac{13.6 \text{ M} \times 1 \text{ mL}}{1000 \text{ mL}} = 0.0136 \text{ M} \)
Since HCl is a strong acid, it dissociates completely.
\( \implies \) [H\(^+\)] = 0.0136 M
pH = \( -\log[\text{H}^+] = -\log(0.0136) = -\log(1.36 \times 10^{-2}) \)
\( \implies \) pH = \( -(-\text{2}) - \log 1.36 = 2 - 0.13 = 1.87 \).
In simple words: To calculate pH, first find the molar concentration of the acid or base. For strong electrolytes, the ion concentration is simple. Then use the pH formula or \( K_w \) to determine the pH of the solution.

Exam Tip: Be careful with units (mL vs L) and the stoichiometry for bases that produce more than one OH\(^-\) ion, like Ca(OH)\(_2\). Always double-check your logarithmic calculations.

Question 50. The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Answer:
Given: Degree of ionization \( (\alpha) = 0.132 \)
Molar concentration \( (c) = 0.1 \text{ M} \)
Bromoacetic acid is a weak acid: BrCH\(_2\)COOH \( \rightleftharpoons \) BrCH\(_2\)COO\(^-\) + H\(^+\)
Initial conc.: \( c \) 0 0
At equilibrium: \( c(1-\alpha) \) \( c\alpha \) \( c\alpha \)
[H\(^+\)] = \( c\alpha = 0.1 \times 0.132 = 0.0132 \)
pH = \( -\log[\text{H}^+] = -\log(0.0132) = -\log(1.32 \times 10^{-2}) \)
\( \implies \) pH = \( -(-\text{2}) - \log 1.32 = 2 - 0.12 = 1.88 \).
For a weak acid, the ionization constant \( K_a \) is given by:
\( K_a = \frac{[\text{BrCH}_2\text{COO}^-][\text{H}^+]}{[\text{BrCH}_2\text{COOH}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha} \)
\( K_a = \frac{0.1 \times (0.132)^2}{1 - 0.132} = \frac{0.1 \times 0.017424}{0.868} = 0.002007 \)
\( \implies K_a \approx 2.01 \times 10^{-3} \).
pKa = \( -\log K_a = -\log(2.01 \times 10^{-3}) = -(-\text{3}) - \log 2.01 = 3 - 0.303 = 2.697 \).
In simple words: First, use the given concentration and degree of ionization to find the hydrogen ion concentration. This lets you compute the pH. Then, use the ionization formula for weak acids to find \( K_a \), and finally, calculate pKa using \( -\log K_a \).

Exam Tip: For weak acids, remember to use the \( \alpha \) (degree of ionization) in your calculations for equilibrium concentrations. When \( \alpha \) is small, you can sometimes simplify the denominator to 1, but always check if this approximation is valid. In this case, \( 1-\alpha \) is substantial.

Question 51. The pH of 0.005 M codeine [C\(_{18}\)H\(_{21}\)NO\(_{3}\)] solution is 9.95. Calculate its ionization constant and pKb.
Answer:
Given: Molar concentration of codeine \( (c) = 0.005 \text{ M} \)
pH = 9.95
Since pH + pOH = 14,
pOH = \( 14 - 9.95 = 4.05 \).
pOH = \( -\log[\text{OH}^+] \)
\( \implies -\log[\text{OH}^+] = 4.05 \)
\( \implies \log[\text{OH}^+] = -4.05 = \bar{5}.95 \)
\( \implies [\text{OH}^+] = \text{antilog}(\bar{5}.95) = 8.91 \times 10^{-5} \text{ M} \).
Codeine is a weak base, so let its ionization be represented as:
B + H\(_{2}\)O \( \rightleftharpoons \) BH\(^+\) + OH\(^-\)
Initial conc: \( c \) 0 0
At equilibrium: \( c - [\text{OH}^-] \) \( [\text{OH}^-] \) \( [\text{OH}^-] \)
Since \( [\text{OH}^+] = 8.91 \times 10^{-5} \text{ M} \),
\( [\text{B}] = 0.005 - 8.91 \times 10^{-5} = 0.0049109 \text{ M} \).
\( K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} = \frac{(8.91 \times 10^{-5})^2}{0.0049109} \)
\( K_b = \frac{7.938 \times 10^{-9}}{0.0049109} = 1.616 \times 10^{-6} \).
\( \implies K_b \approx 1.62 \times 10^{-6} \).
pKb = \( -\log K_b = -\log(1.616 \times 10^{-6}) = -(-\text{6}) - \log 1.616 = 6 - 0.208 = 5.792 \).
In simple words: First, calculate pOH from the given pH. Use pOH to find the concentration of hydroxide ions. Then, using the initial concentration of codeine and the hydroxide ion concentration, determine the base ionization constant \( K_b \). Finally, calculate pKb by taking the negative logarithm of \( K_b \).

Exam Tip: For weak bases, always start by converting pH to pOH. Then calculate the OH\(^-\) concentration, which helps determine the extent of ionization. Remember that the equilibrium concentration of the base should be \( c - [\text{OH}^-] \).

Question 52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline is \( 4.27 \times 10^{-10} \). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
Given: Concentration of aniline solution \( (c) = 0.001 \text{ M} \)
\( K_b \) (aniline) = \( 4.27 \times 10^{-10} \)
Aniline is a weak base (B). The ionization reaction is:
B + H\(_{2}\)O \( \rightleftharpoons \) BH\(^+\) + OH\(^-\)
Let \( [\text{OH}^-] = x \). Then \( [\text{BH}^+] = x \), and \( [\text{B}] = c - x \).
\( K_b = \frac{x^2}{c-x} \). Since \( K_b \) is small, we can approximate \( c-x \approx c \).
\( K_b = \frac{x^2}{c} \implies x^2 = K_b \times c \)
\( x = \sqrt{K_b \times c} = \sqrt{4.27 \times 10^{-10} \times 0.001} \)
\( x = \sqrt{4.27 \times 10^{-13}} = \sqrt{42.7 \times 10^{-14}} = 6.534 \times 10^{-7} \text{ M} \).
Therefore, \( [\text{OH}^-] = 6.534 \times 10^{-7} \text{ M} \).
pOH = \( -\log[\text{OH}^-] = -\log(6.534 \times 10^{-7}) = 7 - 0.8152 = 6.1848 \).
pH = \( 14 - \text{pOH} = 14 - 6.1848 = 7.8152 \).
Degree of ionization \( (\alpha) = \frac{[\text{OH}^-]}{c} = \frac{6.534 \times 10^{-7}}{0.001} = 6.534 \times 10^{-4} \).
For a conjugate acid-base pair, \( K_a \times K_b = K_w \).
\( K_a \) (conjugate acid of aniline) = \( \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{4.27 \times 10^{-10}} = 2.3419 \times 10^{-5} \).
\( \implies K_a \approx 2.34 \times 10^{-5} \).
In simple words: For a weak base, first find the hydroxide ion concentration using \( K_b \) and the base concentration. Then calculate pOH and convert it to pH. The degree of ionization is the hydroxide ion concentration divided by the total base concentration. Finally, use the relationship \( K_a \times K_b = K_w \) to find the ionization constant of the conjugate acid.

Exam Tip: When \( K_b \) is very small (less than \( 10^{-4} \)), the approximation \( c-x \approx c \) is generally valid, simplifying the calculation for \( x \). Remember the relationship between \( K_a \), \( K_b \), and \( K_w \) for conjugate pairs.

Question 53. Calculate the degree of ionization of 0.05 M acetic acid if its pK value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M HCl (b) 0.1 M in HCl?
Answer:
Given: pKa for CH\(_3\)COOH = 4.74
Concentration of acetic acid \( (c) = 0.05 \text{ M} \)
From pKa, we find \( K_a \):
\( -\log K_a = 4.74 \implies K_a = 10^{-4.74} = 10^{0.26} \times 10^{-5} = 1.82 \times 10^{-5} \).
For a weak acid, \( K_a = c\alpha^2 \) (assuming \( 1-\alpha \approx 1 \) for small \( \alpha \)).
\( \alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.82 \times 10^{-5}}{0.05}} = \sqrt{3.64 \times 10^{-4}} = 1.91 \times 10^{-2} \).
Degree of ionization = \( 1.91 \times 10^{-2} \).
(a) In presence of 0.01 M HCl:
HCl is a strong acid, so it completely dissociates, giving \( [\text{H}^+] = 0.01 \text{ M} \).
Let \( \alpha' \) be the new degree of ionization for acetic acid.
CH\(_3\)COOH \( \rightleftharpoons \) CH\(_3\)COO\(^-\) + H\(^+\)
Initial conc: 0.05 0 0.01 (from HCl)
At equilibrium: \( 0.05 - 0.05\alpha' \) \( 0.05\alpha' \) \( 0.01 + 0.05\alpha' \)
\( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{(0.05\alpha')(0.01 + 0.05\alpha')}{0.05(1-\alpha')} \)
Since \( \alpha' \) is expected to be even smaller, we can approximate \( 1-\alpha' \approx 1 \) and \( 0.01 + 0.05\alpha' \approx 0.01 \).
\( 1.82 \times 10^{-5} = \frac{0.05\alpha' \times 0.01}{0.05} = 0.01\alpha' \)
\( \alpha' = \frac{1.82 \times 10^{-5}}{0.01} = 1.82 \times 10^{-3} \).
The degree of ionization decreases from \( 1.91 \times 10^{-2} \) to \( 1.82 \times 10^{-3} \).
(b) In presence of 0.1 M HCl:
Similarly, \( [\text{H}^+] = 0.1 \text{ M} \). Using the same approximation:
\( 1.82 \times 10^{-5} = 0.1\alpha'' \)
\( \alpha'' = \frac{1.82 \times 10^{-5}}{0.1} = 1.82 \times 10^{-4} \).
The degree of ionization further decreases to \( 1.82 \times 10^{-4} \).
In simple words: First, calculate the original degree of ionization of acetic acid from its pKa and concentration. When a strong acid like HCl is added, it increases the hydrogen ion concentration, which shifts the acetic acid equilibrium backwards. This effect, known as the common ion effect, causes the degree of ionization of the weak acid to decrease. The higher the concentration of HCl, the more the ionization is suppressed.

Exam Tip: This question demonstrates the "common ion effect." The presence of a strong acid providing H\(^+\) ions suppresses the dissociation of a weak acid (acetic acid), reducing its degree of ionization. Always consider the contribution of H\(^+\) from the strong acid when setting up the ICE table.

Question 54. The ionization constant of dimethylamine is \( 5.4 \times 10^{-4} \). Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?
Answer:
Given: \( K_b \) of dimethylamine = \( 5.4 \times 10^{-4} \)
Concentration of dimethylamine \( (c) = 0.02 \text{ M} \).
Dimethylamine (CH\(_{3}\))\(_2\)NH is a weak base. Let \( \alpha \) be its degree of ionization.
(CH\(_{3}\))\(_2\)NH + H\(_{2}\)O \( \rightleftharpoons \) (CH\(_{3}\))\(_2\)NH\(_{2}\)\(^+\) + OH\(^-\)
\( K_b = c\alpha^2 \) (assuming \( 1-\alpha \approx 1 \) for approximation, but let's check).
\( \alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{5.4 \times 10^{-4}}{0.02}} = \sqrt{2.7 \times 10^{-2}} = \sqrt{0.027} \approx 0.1643 \).
Since \( \alpha \) is not very small (more than 5% of 1), the approximation \( 1-\alpha \approx 1 \) is not ideal. Let's use the full equation:
\( K_b = \frac{c\alpha^2}{1-\alpha} \)
\( 5.4 \times 10^{-4} = \frac{0.02\alpha^2}{1-\alpha} \)
\( 5.4 \times 10^{-4} (1-\alpha) = 0.02\alpha^2 \)
\( 0.02\alpha^2 + 5.4 \times 10^{-4}\alpha - 5.4 \times 10^{-4} = 0 \)
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( \alpha = \frac{-5.4 \times 10^{-4} + \sqrt{(5.4 \times 10^{-4})^2 - 4(0.02)(-5.4 \times 10^{-4})}}{2(0.02)} \)
\( \alpha = \frac{-5.4 \times 10^{-4} + \sqrt{2.916 \times 10^{-7} + 4.32 \times 10^{-5}}}{0.04} \)
\( \alpha = \frac{-5.4 \times 10^{-4} + \sqrt{4.34916 \times 10^{-5}}}{0.04} \)
\( \alpha = \frac{-5.4 \times 10^{-4} + 0.006594}{0.04} = \frac{0.006054}{0.04} = 0.15135 \).
So, the degree of ionization \( \alpha \approx 0.151 \).

If the solution is also 0.1 M in NaOH:
NaOH is a strong base, so it completely dissociates, giving \( [\text{OH}^-] = 0.1 \text{ M} \).
Let \( x \) be the amount of dimethylamine dissociated in the presence of NaOH.
(CH\(_{3}\))\(_2\)NH + H\(_{2}\)O \( \rightleftharpoons \) (CH\(_{3}\))\(_2\)NH\(_{2}\)\(^+\) + OH\(^-\)
Initial conc: 0.02 0 0.1 (from NaOH)
At equilibrium: \( 0.02 - x \) \( x \) \( 0.1 + x \)
\( K_b = \frac{x(0.1+x)}{0.02-x} \)
Since the common ion OH\(^-\) will suppress dissociation, \( x \) will be very small compared to 0.02 and 0.1. So, \( 0.1+x \approx 0.1 \) and \( 0.02-x \approx 0.02 \).
\( 5.4 \times 10^{-4} = \frac{x \times 0.1}{0.02} \)
\( x = \frac{5.4 \times 10^{-4} \times 0.02}{0.1} = 1.08 \times 10^{-4} \text{ M} \).
Percentage of dimethylamine ionized = \( \frac{x}{c} \times 100\% = \frac{1.08 \times 10^{-4}}{0.02} \times 100\% = 0.0054 \times 100\% = 0.54\% \).
In simple words: First, calculate the degree of ionization for dimethylamine alone using its \( K_b \) and concentration, solving a quadratic equation if needed. When NaOH is added, it introduces a common ion (OH\(^-\)), which suppresses the ionization of the weak base. This reduces the amount of dimethylamine that ionizes, resulting in a much lower percentage of ionization.

Exam Tip: Always assess if the "small \( \alpha \)" approximation is valid (\( \alpha < 0.05 \)). If not, use the full quadratic equation. When a common ion is present, the degree of ionization of the weak electrolyte significantly decreases, making the approximations \( c-x \approx c \) and \( [\text{common ion}] + x \approx [\text{common ion}] \) more valid.

Question 55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83
(b) Human stomach fluid, 1.2
(c) Human blood, 7.38
(d) Human saliva, 6.4.
Answer:
The general formula to find hydrogen ion concentration from pH is: \( [\text{H}^+] = 10^{-\text{pH}} \).
(a) Human muscle-fluid: pH = 6.83
\( [\text{H}^+] = 10^{-6.83} = 10^{0.17} \times 10^{-7} = 1.479 \times 10^{-7} \text{ M} \).
(b) Human stomach fluid: pH = 1.2
\( [\text{H}^+] = 10^{-1.2} = 10^{0.8} \times 10^{-2} = 6.309 \times 10^{-2} \text{ M} \).
(c) Human blood: pH = 7.38
\( [\text{H}^+] = 10^{-7.38} = 10^{0.62} \times 10^{-8} = 4.168 \times 10^{-8} \text{ M} \).
(d) Human saliva: pH = 6.4
\( [\text{H}^+] = 10^{-6.4} = 10^{0.6} \times 10^{-7} = 3.981 \times 10^{-7} \text{ M} \).
In simple words: To find the hydrogen ion concentration from pH, you just need to calculate 10 to the power of negative pH. This inverse operation gives you the concentration directly.

Exam Tip: Remember the relationship \( \text{pH} = -\log[\text{H}^+] \) and its inverse \( [\text{H}^+] = 10^{-\text{pH}} \). For non-integer pH values, use the mantissa and characteristic properly with antilogs or a calculator.

Question 56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Answer:
The general formula to find hydrogen ion concentration from pH is: \( [\text{H}^+] = 10^{-\text{pH}} \).
(a) Milk: pH = 6.8
\( [\text{H}^+] = 10^{-6.8} = 10^{0.2} \times 10^{-7} = 1.58 \times 10^{-7} \text{ M} \).
(b) Black coffee: pH = 5.0
\( [\text{H}^+] = 10^{-5.0} = 1 \times 10^{-5} \text{ M} \).
(c) Tomato juice: pH = 4.2
\( [\text{H}^+] = 10^{-4.2} = 10^{0.8} \times 10^{-5} = 6.31 \times 10^{-5} \text{ M} \).
(d) Lemon juice: pH = 2.2
\( [\text{H}^+] = 10^{-2.2} = 10^{0.8} \times 10^{-3} = 6.31 \times 10^{-3} \text{ M} \).
(e) Egg-white: pH = 7.8
\( [\text{H}^+] = 10^{-7.8} = 10^{0.2} \times 10^{-8} = 1.58 \times 10^{-8} \text{ M} \).
In simple words: To find the hydrogen ion concentration for each item, simply take 10 raised to the power of the negative pH value. This mathematical operation converts the pH scale back into concentration units for hydrogen ions.

Exam Tip: Remember that pH is a logarithmic scale. A lower pH indicates a higher hydrogen ion concentration (more acidic), and a higher pH indicates a lower hydrogen ion concentration (more basic).

Question 57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer:
Given: Mass of KOH = 0.561 g
Volume of solution = 200 mL = 0.2 L
Temperature = 298 K (at this temperature, \( K_w = 1.0 \times 10^{-14} \))
Molar mass of KOH = 39 + 16 + 1 = 56.1 g/mol
Moles of KOH = \( \frac{0.561 \text{ g}}{56.1 \text{ g/mol}} = 0.01 \text{ mol} \)
Molar concentration of KOH = \( \frac{0.01 \text{ mol}}{0.2 \text{ L}} = 0.05 \text{ M} \).
KOH is a strong electrolyte, so it completely ionizes in aqueous solution:
KOH (aq) \( \rightarrow \) K\(^+\) (aq) + OH\(^-\) (aq)
\( \implies \) Concentration of K\(^+\) ions = \( [\text{K}^+] = 0.05 \text{ M} \)
\( \implies \) Concentration of hydroxyl ions = \( [\text{OH}^-] = 0.05 \text{ M} \).
Now, calculate the hydrogen ion concentration using \( K_w \):
\( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.05} = 2.0 \times 10^{-13} \text{ M} \).
Finally, calculate the pH:
pH = \( -\log[\text{H}^+] = -\log(2.0 \times 10^{-13}) = -(-\text{13}) - \log 2 = 13 - 0.30 = 12.70 \).
In simple words: First, find the molarity of KOH. Since it's a strong base, the concentrations of K\(^+\) and OH\(^-\) ions are equal to the KOH molarity. Then, use the ion product of water to calculate the hydrogen ion concentration. Finally, compute the pH from the hydrogen ion concentration.

Exam Tip: Always convert given mass and volume to molar concentration first. For strong electrolytes, assume 100% dissociation. Remember that \( K_w \) relates \( [\text{H}^+] \) and \( [\text{OH}^-] \) concentrations at a given temperature.

Question 58. The solubility of Sr[OH]\(_2\) at 298 K is 19.23 g/L of solution. Calculate corresponding hydrogen ion concentration in each.
Answer:
Given: Solubility of Sr(OH)\(_2\) = 19.23 g/L at 298 K
Molar mass of Sr(OH)\(_2\) = 87.6 + \( 2 \times (16+1) \) = 87.6 + 34 = 121.6 g/mol
Molar solubility (s) of Sr(OH)\(_2\) = \( \frac{19.23 \text{ g/L}}{121.6 \text{ g/mol}} = 0.1581 \text{ mol/L} \).
Strontium hydroxide dissociates as:
Sr(OH)\(_2\) (s) \( \rightleftharpoons \) Sr\(^{2+}\) (aq) + 2OH\(^-\) (aq)
So, \( [\text{Sr}^{2+}] = s = 0.1581 \text{ M} \)
\( [\text{OH}^-] = 2s = 2 \times 0.1581 = 0.3162 \text{ M} \).
Now, calculate the hydrogen ion concentration using \( K_w = 1.0 \times 10^{-14} \) at 298 K:
\( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.3162} = 3.1625 \times 10^{-14} \text{ M} \).
The pH can also be calculated:
pH = \( -\log[\text{H}^+] = -\log(3.1625 \times 10^{-14}) = -(-\text{14}) - \log 3.1625 = 14 - 0.50 = 13.50 \).
In simple words: First, convert the given solubility in grams per liter to molar solubility by dividing by the molar mass. Then, use the stoichiometry of the dissociation reaction to find the hydroxide ion concentration. Finally, use the ion product of water to determine the hydrogen ion concentration.

Exam Tip: Pay close attention to the stoichiometry of the dissociation reaction (e.g., Sr(OH)\(_2\) produces 2 OH\(^-\) ions for every mole dissolved). This is crucial for correctly determining ion concentrations from molar solubility.

Question 59. The ionization constant of propanoic acid is \( 1.32 \times 10^{-5} \). Calculate the degree of ionization of its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?
Answer:
Given: \( K_a \) of propanoic acid = \( 1.32 \times 10^{-5} \)
Concentration of propanoic acid \( (c) = 0.05 \text{ M} \).
Propanoic acid (CH\(_3\)CH\(_2\)COOH) is a weak acid. Let \( \alpha \) be its degree of ionization.
CH\(_3\)CH\(_2\)COOH \( \rightleftharpoons \) CH\(_3\)CH\(_2\)COO\(^-\) + H\(^+\)
\( K_a = \frac{c\alpha^2}{1-\alpha} \). Since \( K_a \) is small, we can approximate \( 1-\alpha \approx 1 \).
\( K_a = c\alpha^2 \implies \alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} = 0.01625 \).
Degree of ionization \( \alpha \approx 1.62 \times 10^{-2} \).
Now, calculate pH:
\( [\text{H}^+] = c\alpha = 0.05 \times 0.01625 = 8.125 \times 10^{-4} \text{ M} \).
pH = \( -\log[\text{H}^+] = -\log(8.125 \times 10^{-4}) = -(-\text{4}) - \log 8.125 = 4 - 0.909 = 3.091 \).

If the solution is also 0.01 M in HCl:
HCl is a strong acid, so it completely dissociates, giving \( [\text{H}^+] = 0.01 \text{ M} \).
Let \( \alpha' \) be the new degree of ionization for propanoic acid.
CH\(_3\)CH\(_2\)COOH \( \rightleftharpoons \) CH\(_3\)CH\(_2\)COO\(^-\) + H\(^+\)
Initial conc: 0.05 0 0.01 (from HCl)
At equilibrium: \( 0.05 - 0.05\alpha' \) \( 0.05\alpha' \) \( 0.01 + 0.05\alpha' \)
\( K_a = \frac{(0.05\alpha')(0.01 + 0.05\alpha')}{0.05(1-\alpha')} \)
Since \( \alpha' \) is expected to be even smaller due to the common ion effect, we can approximate \( 1-\alpha' \approx 1 \) and \( 0.01 + 0.05\alpha' \approx 0.01 \).
\( 1.32 \times 10^{-5} = \frac{0.05\alpha' \times 0.01}{0.05} = 0.01\alpha' \)
\( \alpha' = \frac{1.32 \times 10^{-5}}{0.01} = 1.32 \times 10^{-3} \).
The degree of ionization decreases from \( 1.62 \times 10^{-2} \) to \( 1.32 \times 10^{-3} \).
In simple words: First, calculate the degree of ionization for propanoic acid in pure water and then its pH. When HCl is added, it introduces extra hydrogen ions, which pushes the propanoic acid dissociation equilibrium to the left, reducing its degree of ionization. This is because of the common ion effect.

Exam Tip: For calculations involving weak acids in the presence of a strong acid, remember to account for the initial concentration of H\(^+\) from the strong acid. The common ion effect will always decrease the degree of ionization of the weak electrolyte.

Question 60. The pH of 0.1 M solution of cyanic acid [HCNO] is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer:
Given: Concentration of cyanic acid \( (c) = 0.1 \text{ M} \)
pH = 2.34
From pH, calculate hydrogen ion concentration:
\( [\text{H}^+] = 10^{-\text{pH}} = 10^{-2.34} = 10^{0.66} \times 10^{-3} = 4.57 \times 10^{-3} \text{ M} \).
Cyanic acid (HCNO) is a weak acid. Let \( \alpha \) be its degree of ionization.
HCNO \( \rightleftharpoons \) H\(^+\) + CNO\(^-\)
At equilibrium: \( [\text{H}^+] = c\alpha \)
\( \implies \alpha = \frac{[\text{H}^+]}{c} = \frac{4.57 \times 10^{-3}}{0.1} = 0.0457 \).
Degree of ionization \( = 0.0457 \).
Now, calculate the ionization constant \( K_a \):
\( K_a = \frac{[\text{H}^+][\text{CNO}^-]}{[\text{HCNO}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha} \)
\( K_a = \frac{0.1 \times (0.0457)^2}{1 - 0.0457} = \frac{0.1 \times 0.002088}{0.9543} = 2.188 \times 10^{-4} \).
\( \implies K_a \approx 2.19 \times 10^{-4} \).
In simple words: First, use the given pH and acid concentration to determine the hydrogen ion concentration. This allows you to find the degree of ionization. Then, use the degree of ionization and the acid concentration to calculate the ionization constant \( K_a \) for cyanic acid.

Exam Tip: Always calculate \( [\text{H}^+] \) from the given pH first, as it is the key to finding both the degree of ionization and \( K_a \). Remember that \( [\text{H}^+] = c\alpha \) for a weak acid, and \( K_a = \frac{c\alpha^2}{1-\alpha} \).

Question 61. The ionization constant of nitrous acid is \( 4.5 \times 10^{-4} \). Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Given: \( K_a \) of nitrous acid (HNO\(_{2}\)) = \( 4.5 \times 10^{-4} \)
Concentration of sodium nitrite (NaNO\(_{2}\)) solution \( (c) = 0.04 \text{ M} \).
Sodium nitrite is the salt of a strong base (NaOH) and a weak acid (HNO\(_{2}\)). The nitrite ion (NO\(_{2}\)\(^-\)) will undergo hydrolysis:
NO\(_{2}\)\(^-\) + H\(_{2}\)O \( \rightleftharpoons \) HNO\(_{2}\) + OH\(^-\)
The hydrolysis constant \( K_h \) for the conjugate base of a weak acid is given by:
\( K_h = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11} \).
Let \( h \) be the degree of hydrolysis.
\( K_h = \frac{c h^2}{1-h} \). Since \( K_h \) is very small, we can approximate \( 1-h \approx 1 \).
\( K_h = ch^2 \implies h = \sqrt{\frac{K_h}{c}} = \sqrt{\frac{2.22 \times 10^{-11}}{0.04}} = \sqrt{5.55 \times 10^{-10}} = 2.355 \times 10^{-5} \).
Degree of hydrolysis \( (h) \approx 2.36 \times 10^{-5} \).
Now, calculate \( [\text{OH}^-] \):
\( [\text{OH}^-] = ch = 0.04 \times 2.355 \times 10^{-5} = 9.42 \times 10^{-7} \text{ M} \).
pOH = \( -\log[\text{OH}^-] = -\log(9.42 \times 10^{-7}) = 7 - 0.974 = 6.026 \).
pH = \( 14 - \text{pOH} = 14 - 6.026 = 7.974 \).
In simple words: First, find the hydrolysis constant \( K_h \) for the nitrite ion using \( K_w \) and the \( K_a \) of nitrous acid. Then, calculate the degree of hydrolysis using \( K_h \) and the solution concentration. With the degree of hydrolysis, determine the hydroxide ion concentration, then compute pOH, and finally the pH of the sodium nitrite solution.

Exam Tip: For salts of a strong base and a weak acid, the solution will be basic due to the hydrolysis of the anion. Remember the relationship \( K_h = K_w / K_a \) and that \( [\text{OH}^-] = ch \) where \( h \) is the degree of hydrolysis.

Question 62. A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Given: Concentration of pyridinium hydrochloride \( (c) = 0.02 \text{ M} \)
pH = 3.44
Pyridinium hydrochloride (C\(_{5}\)H\(_{5}\)NHCl) is a salt of a weak base (pyridine, C\(_{5}\)H\(_{5}\)N) and a strong acid (HCl). The pyridinium ion (C\(_{5}\)H\(_{5}\)NH\(^+\)) will undergo hydrolysis:
C\(_{5}\)H\(_{5}\)NH\(^+\) + H\(_{2}\)O \( \rightleftharpoons \) C\(_{5}\)H\(_{5}\)N + H\(_{3}\)O\(^+\)
From pH, calculate hydrogen ion concentration:
\( [\text{H}^+] = 10^{-\text{pH}} = 10^{-3.44} = 10^{0.56} \times 10^{-4} = 3.63 \times 10^{-4} \text{ M} \).
For hydrolysis of a cation from a weak base and strong acid, the hydrolysis constant \( K_h \) is equal to \( K_a \) of the conjugate acid. Also, \( K_h = \frac{K_w}{K_b} \).
Let \( [\text{H}^+] = x \). Then \( [\text{C}_5\text{H}_5\text{N}] = x \), and \( [\text{C}_5\text{H}_5\text{NH}^+] = c - x \).
\( K_h = \frac{[\text{C}_5\text{H}_5\text{N}][\text{H}_3\text{O}^+]}{[\text{C}_5\text{H}_5\text{NH}^+]} = \frac{x^2}{c-x} \)
\( K_h = \frac{(3.63 \times 10^{-4})^2}{0.02 - 3.63 \times 10^{-4}} = \frac{1.31769 \times 10^{-7}}{0.019637} = 6.71 \times 10^{-6} \).
Now, calculate the ionization constant \( K_b \) of pyridine:
\( K_b = \frac{K_w}{K_h} = \frac{1.0 \times 10^{-14}}{6.71 \times 10^{-6}} = 1.49 \times 10^{-9} \).
\( \implies K_b \approx 1.49 \times 10^{-9} \).
In simple words: First, use the pH to find the hydrogen ion concentration. Then, set up the equilibrium expression for the hydrolysis of the pyridinium ion and calculate the hydrolysis constant \( K_h \). Finally, use the relationship \( K_h = K_w / K_b \) to determine the ionization constant of pyridine, the weak base.

Exam Tip: For salts of a weak base and a strong acid, the solution will be acidic due to cation hydrolysis. Remember that \( K_h = K_w / K_b \) for the hydrolysis of the conjugate acid of a weak base.

Question 63. Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH\(_{4}\)NO\(_{3}\), NaNO\(_{2}\) and KF.
Answer:
A salt's solution pH depends on the hydrolysis of its constituent ions. A strong acid/strong base salt does not hydrolyze and is neutral. A strong acid/weak base salt is acidic. A weak acid/strong base salt is basic. A weak acid/weak base salt's pH depends on the relative strengths of the acid and base.

• **NaCl**: Formed from a strong acid (HCl) and a strong base (NaOH). Its solution is **neutral**.


• **KBr**: Formed from a strong acid (HBr) and a strong base (KOH). Its solution is **neutral**.


• **NaCN**: Formed from a weak acid (HCN) and a strong base (NaOH). The cyanide ion (CN\(^-\)) hydrolyzes to produce OH\(^-\). Its solution is **basic**.


• **NH\(_{4}\)NO\(_{3}\)**: Formed from a strong acid (HNO\(_{3}\)) and a weak base (NH\(_{4}\)OH). The ammonium ion (NH\(_{4}\)\(^+\)) hydrolyzes to produce H\(^+\). Its solution is **acidic**.


• **NaNO\(_{2}\)**: Formed from a weak acid (HNO\(_{2}\)) and a strong base (NaOH). The nitrite ion (NO\(_{2}\)\(^-\)) hydrolyzes to produce OH\(^-\). Its solution is **basic**.


• **KF**: Formed from a weak acid (HF) and a strong base (KOH). The fluoride ion (F\(^-\)) hydrolyzes to produce OH\(^-\). Its solution is **basic**.

Exam Tip: To predict the pH of a salt solution, identify the parent acid and base. A strong component (acid or base) will dominate the solution's pH if the other component is weak. If both are strong, the solution is neutral; if both are weak, compare their \( K_a \) and \( K_b \) values.

Question 64. The ionization constant of chloroacetic acid is, \( 1.35 \times 10^{-3} \). What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?
Answer:
Given: \( K_a \) of chloroacetic acid (CH\(_{2}\)ClCOOH) = \( 1.35 \times 10^{-3} \)

**pH of 0.1 M chloroacetic acid solution:**
Let \( c = 0.1 \text{ M} \). Chloroacetic acid is a weak acid.
CH\(_{2}\)ClCOOH \( \rightleftharpoons \) CH\(_{2}\)ClCOO\(^-\) + H\(^+\)
Let \( x = [\text{H}^+] \).
\( K_a = \frac{x^2}{c-x} \)
\( 1.35 \times 10^{-3} = \frac{x^2}{0.1-x} \)
Since \( K_a \) is not extremely small, we should solve the quadratic equation:
\( x^2 + (1.35 \times 10^{-3})x - (1.35 \times 10^{-3} \times 0.1) = 0 \)
\( x^2 + (1.35 \times 10^{-3})x - (1.35 \times 10^{-4}) = 0 \)
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-1.35 \times 10^{-3} + \sqrt{(1.35 \times 10^{-3})^2 - 4(1)(-1.35 \times 10^{-4})}}{2} \)
\( x = \frac{-1.35 \times 10^{-3} + \sqrt{1.8225 \times 10^{-6} + 5.4 \times 10^{-4}}}{2} \)
\( x = \frac{-1.35 \times 10^{-3} + \sqrt{5.418225 \times 10^{-4}}}{2} \)
\( x = \frac{-1.35 \times 10^{-3} + 0.023277}{2} = \frac{0.021927}{2} = 0.01096 \text{ M} \).
\( \implies [\text{H}^+] = 0.01096 \text{ M} \).
pH = \( -\log(0.01096) = -\log(1.096 \times 10^{-2}) = -(-\text{2}) - \log 1.096 = 2 - 0.0398 = 1.96 \).

**pH of 0.1 M sodium chloroacetate solution:**
Sodium chloroacetate is the salt of a strong base (NaOH) and a weak acid (chloroacetic acid). The chloroacetate ion (CH\(_{2}\)ClCOO\(^-\)) will hydrolyze.
CH\(_{2}\)ClCOO\(^-\) + H\(_{2}\)O \( \rightleftharpoons \) CH\(_{2}\)ClCOOH + OH\(^-\)
The hydrolysis constant \( K_h \) is:
\( K_h = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.35 \times 10^{-3}} = 7.407 \times 10^{-12} \).
Let \( x = [\text{OH}^-] \). We can approximate \( K_h = \frac{x^2}{c} \) as \( h \) is very small.
\( x = \sqrt{K_h \times c} = \sqrt{7.407 \times 10^{-12} \times 0.1} = \sqrt{7.407 \times 10^{-13}} = \sqrt{74.07 \times 10^{-14}} \)
\( x = 8.607 \times 10^{-7} \text{ M} \).
\( \implies [\text{OH}^-] = 8.607 \times 10^{-7} \text{ M} \).
pOH = \( -\log(8.607 \times 10^{-7}) = 7 - 0.9348 = 6.0652 \).
pH = \( 14 - \text{pOH} = 14 - 6.0652 = 7.9348 \).
\( \implies \text{pH} \approx 7.93 \).
In simple words: For the chloroacetic acid solution, first calculate the hydrogen ion concentration using the acid dissociation constant \( K_a \) and concentration, solving a quadratic equation. This gives the pH. For the sodium salt solution, calculate the hydrolysis constant \( K_h \) from \( K_w \) and \( K_a \). Use \( K_h \) to find the hydroxide ion concentration, then convert it to pOH and finally to pH.

Exam Tip: Always decide whether to use approximations for \( x \) (or \( \alpha \)) based on the value of \( K_a \) or \( K_b \). For \( K_a \) values like \( 10^{-3} \) or higher, the quadratic equation is generally necessary for accurate pH calculation. For salts, remember the hydrolysis mechanism to determine if the solution will be acidic or basic.

Question 65. Ionic product of water at 310 K is \( 2.7 \times 10^{-14} \). What is the pH of neutral water at this temperature?
Answer:
Given: Ionic product of water \( (K_w) = 2.7 \times 10^{-14} \) at 310 K.
For neutral water, the concentration of hydrogen ions \( [\text{H}^+] \) is equal to the concentration of hydroxide ions \( [\text{OH}^-] \).
Let \( [\text{H}^+] = [\text{OH}^-] = x \).
We know that \( K_w = [\text{H}^+][\text{OH}^-] \).
\( \implies K_w = x \times x = x^2 \).
\( \implies x^2 = 2.7 \times 10^{-14} \)
\( \implies x = \sqrt{2.7 \times 10^{-14}} = 1.643 \times 10^{-7} \text{ M} \).
So, \( [\text{H}^+] = 1.643 \times 10^{-7} \text{ M} \).
Now, calculate the pH:
pH = \( -\log[\text{H}^+] = -\log(1.643 \times 10^{-7}) = -(-\text{7}) - \log 1.643 = 7 - 0.215 = 6.785 \).
In simple words: For neutral water, the amounts of hydrogen ions and hydroxide ions are equal. Since the ionic product of water is given, you can find the hydrogen ion concentration by taking the square root of \( K_w \). Once you have the hydrogen ion concentration, calculate the pH using the negative logarithm formula.

Exam Tip: The pH of neutral water is 7 only at 25°C. For other temperatures, \( K_w \) changes, and consequently, the pH of neutral water also changes. Always use the given \( K_w \) value for the specified temperature.

Question 66. Calculate the pH of the resultant mixtures:
(a) 10 mL of 0.2 M Ca[OH]\(_2\) + 25 mL of 0.1 M HCl
(b) 10 mL of 0.01 M H\(_{2}\)SO\(_{4}\) + 10 mL of 0.01 M Ca[OH]\(_2\)
(c) 10 ml of 0.1 M H\(_{2}\)SO\(_{4}\) + 10 mL of 0.1 M KOH.
Answer:
(a) 10 mL of 0.2 M Ca(OH)\(_2\) + 25 mL of 0.1 M HCl
Moles of OH\(^-\) from Ca(OH)\(_2\) = \( 2 \times \text{Volume (L)} \times \text{Molarity (M)} \)
\( = 2 \times 0.010 \text{ L} \times 0.2 \text{ M} = 0.004 \text{ mol} \).
Moles of H\(^+\) from HCl = \( \text{Volume (L)} \times \text{Molarity (M)} \)
\( = 0.025 \text{ L} \times 0.1 \text{ M} = 0.0025 \text{ mol} \).
Since moles of OH\(^-\) (\( 0.004 \text{ mol} \)) > moles of H\(^+\) (\( 0.0025 \text{ mol} \)), the solution will be basic.
Excess moles of OH\(^-\) = \( 0.004 - 0.0025 = 0.0015 \text{ mol} \).
Total volume of solution = \( 10 \text{ mL} + 25 \text{ mL} = 35 \text{ mL} = 0.035 \text{ L} \).
\( [\text{OH}^-] = \frac{0.0015 \text{ mol}}{0.035 \text{ L}} = 0.042857 \text{ M} \).
pOH = \( -\log(0.042857) = -\log(4.2857 \times 10^{-2}) = -(-\text{2}) - \log 4.2857 = 2 - 0.632 = 1.368 \).
pH = \( 14 - \text{pOH} = 14 - 1.368 = 12.632 \).

(b) 10 mL of 0.01 M H\(_{2}\)SO\(_{4}\) + 10 mL of 0.01 M Ca[OH]\(_2\)
Moles of H\(^+\) from H\(_{2}\)SO\(_{4}\) = \( 2 \times \text{Volume (L)} \times \text{Molarity (M)} \)
\( = 2 \times 0.010 \text{ L} \times 0.01 \text{ M} = 0.0002 \text{ mol} \).
Moles of OH\(^-\) from Ca(OH)\(_2\) = \( 2 \times \text{Volume (L)} \times \text{Molarity (M)} \)
\( = 2 \times 0.010 \text{ L} \times 0.01 \text{ M} = 0.0002 \text{ mol} \).
Since moles of H\(^+\) = moles of OH\(^-\), the solution will be neutral.
pH = 7.

(c) 10 mL of 0.1 M H\(_{2}\)SO\(_{4}\) + 10 mL of 0.1 M KOH
Moles of H\(^+\) from H\(_{2}\)SO\(_{4}\) = \( 2 \times \text{Volume (L)} \times \text{Molarity (M)} \)
\( = 2 \times 0.010 \text{ L} \times 0.1 \text{ M} = 0.002 \text{ mol} \).
Moles of OH\(^-\) from KOH = \( \text{Volume (L)} \times \text{Molarity (M)} \)
\( = 0.010 \text{ L} \times 0.1 \text{ M} = 0.001 \text{ mol} \).
Since moles of H\(^+\) (\( 0.002 \text{ mol} \)) > moles of OH\(^-\) (\( 0.001 \text{ mol} \)), the solution will be acidic.
Excess moles of H\(^+\) = \( 0.002 - 0.001 = 0.001 \text{ mol} \).
Total volume of solution = \( 10 \text{ mL} + 10 \text{ mL} = 20 \text{ mL} = 0.020 \text{ L} \).
\( [\text{H}^+] = \frac{0.001 \text{ mol}}{0.020 \text{ L}} = 0.05 \text{ M} \).
pH = \( -\log(0.05) = -\log(5 \times 10^{-2}) = -(-\text{2}) - \log 5 = 2 - 0.70 = 1.30 \).
In simple words: For each mixture, calculate the total moles of hydrogen ions from the acid and hydroxide ions from the base. Compare these moles to determine if the solution is acidic or basic, or neutral. Then, find the concentration of the excess ion in the total volume and calculate the final pH.

Exam Tip: Remember to account for the stoichiometry of polyprotic acids (like H\(_{2}\)SO\(_{4}\)) and bases (like Ca(OH)\(_2\)) that produce more than one H\(^+\) or OH\(^-\) ion per molecule. Always calculate moles, not just molarity, before mixing to determine the net excess of ions.

Question 67. Determine the solubilities of silver chromate, barium* chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given in Table 7.9. Determine also the molarities of individual ions.
Answer:
We'll calculate the solubility (s) and ion molarities for each compound using the given \( K_{sp} \) values (Table 7.9 would provide these, assuming standard values here).

**1. Silver Chromate (Ag\(_{2}\)CrO\(_{4}\))**
Ag\(_{2}\)CrO\(_{4}\) (s) \( \rightleftharpoons \) 2Ag\(^+\) (aq) + CrO\(_{4}\)\(^{2-}\) (aq)
If s is the molar solubility, then \( [\text{Ag}^+] = 2s \) and \( [\text{CrO}_4^{2-}] = s \).
\( K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2s)^2(s) = 4s^3 \).
Assuming \( K_{sp} = 1.1 \times 10^{-12} \) (a common value for Ag\(_{2}\)CrO\(_{4}\)):
\( 4s^3 = 1.1 \times 10^{-12} \)
\( s^3 = \frac{1.1 \times 10^{-12}}{4} = 0.275 \times 10^{-12} \)
\( s = (0.275 \times 10^{-12})^{1/3} = (275 \times 10^{-15})^{1/3} = 6.5 \times 10^{-5} \text{ M} \).
Solubility = \( 6.5 \times 10^{-5} \text{ M} \).
Ion Molarities: \( [\text{Ag}^+] = 2s = 2 \times 6.5 \times 10^{-5} = 1.3 \times 10^{-4} \text{ M} \)
\( [\text{CrO}_4^{2-}] = s = 6.5 \times 10^{-5} \text{ M} \).

**2. Barium Chromate (BaCrO\(_{4}\))**
BaCrO\(_{4}\) (s) \( \rightleftharpoons \) Ba\(^{2+}\) (aq) + CrO\(_{4}\)\(^{2-}\) (aq)
If s is the molar solubility, then \( [\text{Ba}^{2+}] = s \) and \( [\text{CrO}_4^{2-}] = s \).
\( K_{sp} = [\text{Ba}^{2+}][\text{CrO}_4^{2-}] = s \times s = s^2 \).
Assuming \( K_{sp} = 1.2 \times 10^{-10} \) (a common value for BaCrO\(_{4}\)):
\( s^2 = 1.2 \times 10^{-10} \)
\( s = \sqrt{1.2 \times 10^{-10}} = 1.095 \times 10^{-5} \text{ M} \).
Solubility = \( 1.095 \times 10^{-5} \text{ M} \).
Ion Molarities: \( [\text{Ba}^{2+}] = s = 1.095 \times 10^{-5} \text{ M} \)
\( [\text{CrO}_4^{2-}] = s = 1.095 \times 10^{-5} \text{ M} \).

**3. Ferric Hydroxide (Fe(OH)\(_3\))**
Fe(OH)\(_3\) (s) \( \rightleftharpoons \) Fe\(^{3+}\) (aq) + 3OH\(^-\) (aq)
If s is the molar solubility, then \( [\text{Fe}^{3+}] = s \) and \( [\text{OH}^-] = 3s \).
\( K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 = (s)(3s)^3 = 27s^4 \).
Assuming \( K_{sp} = 1.0 \times 10^{-38} \) (a common value for Fe(OH)\(_3\)):
\( 27s^4 = 1.0 \times 10^{-38} \)
\( s^4 = \frac{1.0 \times 10^{-38}}{27} = 0.037 \times 10^{-38} = 3.7 \times 10^{-40} \)
\( s = (3.7 \times 10^{-40})^{1/4} = 1.39 \times 10^{-10} \text{ M} \).
Solubility = \( 1.39 \times 10^{-10} \text{ M} \).
Ion Molarities: \( [\text{Fe}^{3+}] = s = 1.39 \times 10^{-10} \text{ M} \)
\( [\text{OH}^-] = 3s = 3 \times 1.39 \times 10^{-10} = 4.17 \times 10^{-10} \text{ M} \).

**4. Lead Chloride (PbCl\(_{2}\))**
PbCl\(_{2}\) (s) \( \rightleftharpoons \) Pb\(^{2+}\) (aq) + 2Cl\(^-\) (aq)
If s is the molar solubility, then \( [\text{Pb}^{2+}] = s \) and \( [\text{Cl}^-] = 2s \).
\( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (s)(2s)^2 = 4s^3 \).
Assuming \( K_{sp} = 1.6 \times 10^{-5} \) (a common value for PbCl\(_{2}\)):
\( 4s^3 = 1.6 \times 10^{-5} \)
\( s^3 = \frac{1.6 \times 10^{-5}}{4} = 0.4 \times 10^{-5} = 4 \times 10^{-6} \)
\( s = (4 \times 10^{-6})^{1/3} = 1.587 \times 10^{-2} \text{ M} \).
Solubility = \( 1.587 \times 10^{-2} \text{ M} \).
Ion Molarities: \( [\text{Pb}^{2+}] = s = 1.587 \times 10^{-2} \text{ M} \)
\( [\text{Cl}^-] = 2s = 2 \times 1.587 \times 10^{-2} = 3.174 \times 10^{-2} \text{ M} \).

**5. Mercurous Iodide (Hg\(_{2}\)I\(_{2}\))**
Hg\(_{2}\)I\(_{2}\) (s) \( \rightleftharpoons \) Hg\(_{2}\)\(^{2+}\) (aq) + 2I\(^-\) (aq)
If s is the molar solubility, then \( [\text{Hg}_2^{2+}] = s \) and \( [\text{I}^-] = 2s \).
\( K_{sp} = [\text{Hg}_2^{2+}][\text{I}^-]^2 = (s)(2s)^2 = 4s^3 \).
Assuming \( K_{sp} = 4.5 \times 10^{-29} \) (a common value for Hg\(_{2}\)I\(_{2}\)):
\( 4s^3 = 4.5 \times 10^{-29} \)
\( s^3 = \frac{4.5 \times 10^{-29}}{4} = 1.125 \times 10^{-29} = 11.25 \times 10^{-30} \)
\( s = (11.25 \times 10^{-30})^{1/3} = 2.24 \times 10^{-10} \text{ M} \).
Solubility = \( 2.24 \times 10^{-10} \text{ M} \).
Ion Molarities: \( [\text{Hg}_2^{2+}] = s = 2.24 \times 10^{-10} \text{ M} \)
\( [\text{I}^-] = 2s = 2 \times 2.24 \times 10^{-10} = 4.48 \times 10^{-10} \text{ M} \).
In simple words: For each compound, write out its dissociation equation and the expression for \( K_{sp} \) in terms of molar solubility 's'. Use the given \( K_{sp} \) value to solve for 's'. Once 's' is known, multiply it by the stoichiometric coefficients to find the molarity of each individual ion.

Exam Tip: Correctly writing the solubility product expression (K\(_{sp}\)) in terms of molar solubility (s) is crucial. Pay careful attention to the stoichiometric coefficients, especially when squaring or cubing terms, as errors here will lead to incorrect calculations of 's' and ion concentrations.

Question 68. The solubility product constant of Ag\(_{2}\)CrO\(_{4}\) and AgBr are \( 1.1 \times 10^{-12} \) and \( 5.0 \times 10^{-13} \) respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
**1. For Silver Bromide (AgBr):**
AgBr (s) \( \rightleftharpoons \) Ag\(^+\) (aq) + Br\(^-\) (aq)
Let \( s_1 \) be the molar solubility of AgBr.
\( K_{sp} = [\text{Ag}^+][\text{Br}^-] = s_1 \times s_1 = s_1^2 \).
Given \( K_{sp} \) for AgBr = \( 5.0 \times 10^{-13} \).
\( s_1^2 = 5.0 \times 10^{-13} \)
\( s_1 = \sqrt{5.0 \times 10^{-13}} = \sqrt{50 \times 10^{-14}} = 7.071 \times 10^{-7} \text{ M} \).

**2. For Silver Chromate (Ag\(_{2}\)CrO\(_{4}\)):**
Ag\(_{2}\)CrO\(_{4}\) (s) \( \rightleftharpoons \) 2Ag\(^+\) (aq) + CrO\(_{4}\)\(^{2-}\) (aq)
Let \( s_2 \) be the molar solubility of Ag\(_{2}\)CrO\(_{4}\).
\( K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2s_2)^2(s_2) = 4s_2^3 \).
Given \( K_{sp} \) for Ag\(_{2}\)CrO\(_{4}\) = \( 1.1 \times 10^{-12} \).
\( 4s_2^3 = 1.1 \times 10^{-12} \)
\( s_2^3 = \frac{1.1 \times 10^{-12}}{4} = 0.275 \times 10^{-12} \)
\( s_2 = (0.275 \times 10^{-12})^{1/3} = 6.5 \times 10^{-5} \text{ M} \).

**Ratio of molarities of their saturated solutions:**
Ratio = \( \frac{s_2}{s_1} = \frac{6.5 \times 10^{-5}}{7.071 \times 10^{-7}} = \frac{650 \times 10^{-7}}{7.071 \times 10^{-7}} = 91.92 \).
The ratio of molarities (solubilities) of Ag\(_{2}\)CrO\(_{4}\) to AgBr is approximately 91.9.
In simple words: First, calculate the molar solubility for silver bromide using its \( K_{sp} \) and the dissociation expression. Then, do the same for silver chromate. Finally, divide the molar solubility of silver chromate by that of silver bromide to get the required ratio.

Exam Tip: When comparing solubilities based on K\(_{sp}\) values, remember that directly comparing K\(_{sp}\) values is only valid for compounds with the same stoichiometry (e.g., AB and CD). For compounds with different stoichiometries (like AB and A\(_{2}\)B), you must calculate their individual molar solubilities 's' before comparing.

Question 68. The solubility product constant of Ag2CrO4 and AgBr are \( 1.1 \times 10^{-12} \) and \( 5.0 \times 10^{-13} \) respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer: For AgBr:
Suppose the solubility of AgBr is \( s_1 \) M.
\( \text{AgBr (s)} \rightleftharpoons \text{Ag}^+ \text{(aq)} + \text{Br}^- \text{(aq)} \)
\( s_1 \quad s_1 \)
\( K_{sp} = [Ag^+][Br^-] = s_1 \times s_1 = s_1^2 \)
\( 5.0 \times 10^{-13} = s_1^2 \)
\( s_1 = \sqrt{5.0 \times 10^{-13}} = \sqrt{50 \times 10^{-14}} = 7.07 \times 10^{-7} \text{ M} \)
For Ag2CrO4:
Suppose the solubility of Ag2CrO4 is \( s_2 \) M.
\( \text{Ag}_2\text{CrO}_4 \text{(s)} \rightleftharpoons 2\text{Ag}^+ \text{(aq)} + \text{CrO}_4^{2-} \text{(aq)} \)
\( 2s_2 \quad s_2 \)
\( K_{sp} = [Ag^+]^2[CrO_4^{2-}] = (2s_2)^2(s_2) = 4s_2^3 \)
\( 1.1 \times 10^{-12} = 4s_2^3 \)
\( s_2^3 = \frac{1.1 \times 10^{-12}}{4} = 0.275 \times 10^{-12} \)
\( s_2 = (0.275 \times 10^{-12})^{1/3} = (275 \times 10^{-15})^{1/3} = 6.49 \times 10^{-5} \text{ M} \)
Ratio of molarities: \( \frac{s_2}{s_1} = \frac{6.49 \times 10^{-5}}{7.07 \times 10^{-7}} = \frac{6.49 \times 100}{7.07} = \frac{649}{7.07} \approx 91.8 \)
The ratio of their molarities in their saturated solutions is approximately 91.8, with Ag2CrO4 being more soluble.
In simple words: First, calculate how much of each substance dissolves in water using its solubility product. Then, divide the solubility of Ag2CrO4 by the solubility of AgBr to get the ratio.

Exam Tip: Remember the stoichiometry when setting up the solubility product expression for compounds like Ag2CrO4, as the coefficients affect the powers in the Ksp formula.

Question 69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate \( K_{sp} = 7.4 \times 10^{-8} \)).
Answer: The reaction is:
\( 2\text{NaIO}_3 + \text{Cu}(\text{ClO}_3)_2 \rightarrow \text{Cu}(\text{IO}_3)_2 + 2\text{NaClO}_3 \)
When equal volumes of two solutions are mixed, the concentration of each reactant is halved.
Initial concentration of \( \text{NaIO}_3 = 0.002 \text{ M} \)
Initial concentration of \( \text{Cu}(\text{ClO}_3)_2 = 0.002 \text{ M} \)
After mixing equal volumes:
\( [\text{IO}_3^-] = \frac{0.002 \text{ M}}{2} = 0.001 \text{ M} \)
\( [\text{Cu}^{2+}] = \frac{0.002 \text{ M}}{2} = 0.001 \text{ M} \)
For the formation of copper iodate \( \text{Cu}(\text{IO}_3)_2 \):
\( \text{Cu}(\text{IO}_3)_2 \text{(s)} \rightleftharpoons \text{Cu}^{2+} \text{(aq)} + 2\text{IO}_3^- \text{(aq)} \)
The ionic product (Qsp) is calculated as:
\( Q_{sp} = [\text{Cu}^{2+}][\text{IO}_3^-]^2 \)
\( Q_{sp} = (0.001)(0.001)^2 \)
\( Q_{sp} = (1 \times 10^{-3})(1 \times 10^{-3})^2 \)
\( Q_{sp} = 1 \times 10^{-3} \times 1 \times 10^{-6} = 1 \times 10^{-9} \)
Given \( K_{sp} \) for cupric iodate is \( 7.4 \times 10^{-8} \).
Comparing Qsp and Ksp:
\( Q_{sp} = 1 \times 10^{-9} \)
\( K_{sp} = 7.4 \times 10^{-8} \)
Since \( Q_{sp} < K_{sp} \) ( \( 1 \times 10^{-9} \) is smaller than \( 7.4 \times 10^{-8} \)), no precipitation will occur.
In simple words: When the two solutions mix, their concentrations are cut in half. We then calculate a value called the 'ionic product' (Qsp) using these new concentrations. If this calculated Qsp is smaller than the given 'solubility product' (Ksp), then nothing will settle out of the solution. Here, Qsp is smaller, so no solid copper iodate will form.

Exam Tip: Remember to halve the concentrations of ions when equal volumes of two solutions are mixed. Compare Qsp with Ksp to determine if precipitation occurs; if Qsp < Ksp, no precipitation.

Question 70. The ionization constant of benzoic acid is \( 6.46 \times 10^{-5} \) and \( K_{sp} \) for silver benzoate is \( 2.5 \times 10^{-13} \). How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Answer: Let the solubility of silver benzoate in pure water be \( x \) mol \( \text{L}^{-1} \).
\( \text{C}_6\text{H}_5\text{COOAg (s)} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- \text{(aq)} + \text{Ag}^+ \text{(aq)} \)
At equilibrium in pure water: \( [\text{C}_6\text{H}_5\text{COO}^-] = x \text{ mol L}^{-1} \) and \( [\text{Ag}^+] = x \text{ mol L}^{-1} \)
\( K_{sp} = [\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-] = x \times x = x^2 \)
\( 2.5 \times 10^{-13} = x^2 \)
\( x = \sqrt{2.5 \times 10^{-13}} = \sqrt{25 \times 10^{-14}} = 5.0 \times 10^{-7} \text{ mol L}^{-1} \)
Now, in a buffer solution of pH 3.19:
pH = \( -\log[\text{H}^+] = 3.19 \)
\( \log[\text{H}^+] = -3.19 = \overline{4}.81 \)
\( [\text{H}^+] = \text{antilog}(\overline{4}.81) = 6.457 \times 10^{-4} \text{ M} \)
The benzoic acid equilibrium is:
\( \text{C}_6\text{H}_5\text{COOH (aq)} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- \text{(aq)} + \text{H}^+ \text{(aq)} \)
The ionization constant \( K_a \) for benzoic acid is \( 6.46 \times 10^{-5} \).
From the \( K_a \) expression: \( K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \)
Rearranging to find the ratio of ionized to unionized acid:
\( \frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]} = \frac{K_a}{[\text{H}^+]} = \frac{6.46 \times 10^{-5}}{6.457 \times 10^{-4}} \approx 0.100 \)
This means \( [\text{C}_6\text{H}_5\text{COOH}] = \frac{[\text{C}_6\text{H}_5\text{COO}^-]}{0.100} = 10 \times [\text{C}_6\text{H}_5\text{COO}^-] \)
Let the solubility of silver benzoate in the buffer solution be \( y \) mol \( \text{L}^{-1} \).
So, \( [\text{Ag}^+] = y \)
In the buffer, the total concentration of benzoate species is \( [\text{C}_6\text{H}_5\text{COO}^-]_{\text{total}} = [\text{C}_6\text{H}_5\text{COO}^-] + [\text{C}_6\text{H}_5\text{COOH}] \)
Since the silver ions come from the dissolution of silver benzoate, the total benzoate species must equal the silver ion concentration:
\( y = [\text{C}_6\text{H}_5\text{COO}^-] + [\text{C}_6\text{H}_5\text{COOH}] \)
Substitute \( [\text{C}_6\text{H}_5\text{COOH}] = 10 \times [\text{C}_6\text{H}_5\text{COO}^-] \):
\( y = [\text{C}_6\text{H}_5\text{COO}^-] + 10 \times [\text{C}_6\text{H}_5\text{COO}^-] = 11 \times [\text{C}_6\text{H}_5\text{COO}^-] \)
So, \( [\text{C}_6\text{H}_5\text{COO}^-] = \frac{y}{11} \)
Now, use the \( K_{sp} \) expression for silver benzoate:
\( K_{sp} = [\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-] \)
\( 2.5 \times 10^{-13} = y \times \frac{y}{11} = \frac{y^2}{11} \)
\( y^2 = 11 \times 2.5 \times 10^{-13} = 27.5 \times 10^{-13} = 2.75 \times 10^{-12} \)
\( y = \sqrt{2.75 \times 10^{-12}} = 1.66 \times 10^{-6} \text{ mol L}^{-1} \)
The ratio of solubility in buffer to solubility in pure water is:
\( \frac{y}{x} = \frac{1.66 \times 10^{-6} \text{ mol L}^{-1}}{5.0 \times 10^{-7} \text{ mol L}^{-1}} = \frac{1.66 \times 10}{5.0} = \frac{16.6}{5.0} = 3.32 \)
Thus, silver benzoate is 3.32 times more soluble in the buffer solution.
In simple words: First, calculate how much silver benzoate dissolves in plain water. Then, calculate how much dissolves in the special "buffer" water, which has a set pH. The buffer's pH changes how much of the benzoate ion turns into benzoic acid, which makes more silver benzoate dissolve. Finally, divide the solubility in the buffer by the solubility in plain water to find out how many times more it dissolves.

Exam Tip: For problems involving common ion effect or pH influence on solubility, always consider the acid-base equilibrium of the anion. An acidic buffer will protonate the anion of a sparingly soluble salt of a weak acid, thereby increasing its solubility.

Question 71. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, \( K_{sp} = 6.3 \times 10^{-18} \)).
Answer: Let the initial concentration of both ferrous sulphate \( (\text{FeSO}_4) \) and sodium sulphide \( (\text{Na}_2\text{S}) \) be \( x \) mol \( \text{L}^{-1} \).
When equal volumes of these solutions are mixed, their concentrations are halved.
So, after mixing:
\( [\text{FeSO}_4] = \frac{x}{2} \text{ M} \)
\( [\text{Na}_2\text{S}] = \frac{x}{2} \text{ M} \)
Ferrous sulphate dissociates as: \( \text{FeSO}_4 \rightarrow \text{Fe}^{2+} + \text{SO}_4^{2-} \)
Sodium sulphide dissociates as: \( \text{Na}_2\text{S} \rightarrow 2\text{Na}^+ + \text{S}^{2-} \)
So, in the mixed solution, the concentrations of the relevant ions are:
\( [\text{Fe}^{2+}] = \frac{x}{2} \text{ M} \)
\( [\text{S}^{2-}] = \frac{x}{2} \text{ M} \)
For no precipitation of iron sulphide \( (\text{FeS}) \), the ionic product (Qsp) must be less than or equal to the solubility product \( (K_{sp}) \).
\( \text{FeS (s)} \rightleftharpoons \text{Fe}^{2+} \text{(aq)} + \text{S}^{2-} \text{(aq)} \)
\( Q_{sp} = [\text{Fe}^{2+}][\text{S}^{2-}] \)
To avoid precipitation, \( Q_{sp} \le K_{sp} \):
\( (\frac{x}{2})(\frac{x}{2}) \le 6.3 \times 10^{-18} \)
\( \frac{x^2}{4} \le 6.3 \times 10^{-18} \)
\( x^2 \le 4 \times 6.3 \times 10^{-18} \)
\( x^2 \le 25.2 \times 10^{-18} \)
\( x \le \sqrt{25.2 \times 10^{-18}} \)
\( x \le 5.02 \times 10^{-9} \text{ M} \)
Therefore, the maximum initial concentration of the equimolar solutions for no precipitation is \( 5.02 \times 10^{-9} \text{ M} \).
In simple words: We want to find the highest concentration of two equal solutions we can mix without forming a solid. When you mix equal amounts, the concentrations of the chemicals get cut in half. We set up an equation where the product of these halved concentrations is less than or equal to the given solubility product, then we solve for the original concentration.

Exam Tip: Remember to adjust concentrations for dilution when mixing solutions. For precipitation to occur, the ionic product must exceed Ksp; for no precipitation, it must be less than or equal to Ksp.

Question 72. What is the maximum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, \( K_{sp} \) is \( 9.1 \times 10^{-6} \)).
Answer: Calcium sulphate \( (\text{CaSO}_4) \) dissociates as:
\( \text{CaSO}_4 \text{(s)} \rightleftharpoons \text{Ca}^{2+} \text{(aq)} + \text{SO}_4^{2-} \text{(aq)} \)
Let \( s \) be the molar solubility of \( \text{CaSO}_4 \) in mol \( \text{L}^{-1} \).
Then, \( [\text{Ca}^{2+}] = s \) and \( [\text{SO}_4^{2-}] = s \)
\( K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2 \)
Given \( K_{sp} = 9.1 \times 10^{-6} \).
\( s^2 = 9.1 \times 10^{-6} \)
\( s = \sqrt{9.1 \times 10^{-6}} = 3.017 \times 10^{-3} \text{ mol L}^{-1} \)
This means that 3.017 \( \times 10^{-3} \) moles of \( \text{CaSO}_4 \) can dissolve in 1 liter of water.
Now, we need to find the molar mass of \( \text{CaSO}_4 \):
Molar mass of Ca = 40.08 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of \( \text{CaSO}_4 = 40.08 + 32.07 + (4 \times 16.00) = 40.08 + 32.07 + 64.00 = 136.15 \text{ g/mol} \)
Now, calculate the mass of \( \text{CaSO}_4 \) that can dissolve in 1 liter:
Mass = Molar solubility \( \times \) Molar mass
Mass = \( (3.017 \times 10^{-3} \text{ mol L}^{-1}) \times (136.15 \text{ g/mol}) = 0.4108 \text{ g L}^{-1} \)
So, 0.4108 g of \( \text{CaSO}_4 \) dissolves in 1 L of water.
We need to dissolve 1 g of \( \text{CaSO}_4 \).
Volume of water required = \( \frac{\text{Mass to be dissolved}}{\text{Solubility in g/L}} \)
Volume = \( \frac{1 \text{ g}}{0.4108 \text{ g/L}} \approx 2.43 \text{ L} \)
Therefore, the maximum volume of water required to dissolve 1 g of calcium sulphate is approximately 2.43 L.
In simple words: First, we use the \( K_{sp} \) to find out how many moles of calcium sulphate can dissolve in one liter of water. Then, we change that amount from moles to grams per liter using its molecular weight. Finally, we figure out how many liters of water are needed to dissolve exactly 1 gram of the substance.

Exam Tip: Carefully convert between molar solubility and solubility in g/L using the molar mass. Ensure you understand the units at each step of the calculation.

Question 73. The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is \( 1.0 \times 10^{-19} \text{ M} \). If 10 mL of this is added to 5 mL of 0.04 solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2 in which of these solutions precipitation will take place?
Answer: Given concentration of sulphide ion \( [\text{S}^{2-}] = 1.0 \times 10^{-19} \text{ M} \).
This sulphide solution (10 mL) is mixed with 5 mL of metal salt solutions (0.04 M).
First, calculate the new concentration of \( [\text{S}^{2-}] \) after mixing:
Total volume = 10 mL (sulphide solution) + 5 mL (metal solution) = 15 mL
Using dilution formula \( C_1V_1 = C_2V_2 \):
\( (1.0 \times 10^{-19} \text{ M})(10 \text{ mL}) = C_{2\text{,S}^{2-}}(15 \text{ mL}) \)
\( C_{2\text{,S}^{2-}} = \frac{1.0 \times 10^{-19} \times 10}{15} = \frac{10}{15} \times 10^{-19} = \frac{2}{3} \times 10^{-19} = 0.667 \times 10^{-19} = 6.67 \times 10^{-20} \text{ M} \)
Now, calculate the new concentration of metal ions after mixing:
Initial metal ion concentration = 0.04 M
\( (0.04 \text{ M})(5 \text{ mL}) = C_{2\text{,metal}}(15 \text{ mL}) \)
\( C_{2\text{,metal}} = \frac{0.04 \times 5}{15} = \frac{0.20}{15} = 0.0133 \text{ M} = 1.33 \times 10^{-2} \text{ M} \)
The ionic product \( (Q_{sp}) \) for metal sulphides is given by \( [\text{M}^{2+}][\text{S}^{2-}] \).
\( Q_{sp} = (1.33 \times 10^{-2} \text{ M}) \times (6.67 \times 10^{-20} \text{ M}) \)
\( Q_{sp} = 8.87 \times 10^{-22} \)
Now, compare this calculated \( Q_{sp} \) with the given \( K_{sp} \) values for various metal sulphides (from Table 7.9, which is not provided but implied in the problem context):
Given \( K_{sp} \) values (typically, these would be provided in the question or a reference table):
FeS: \( K_{sp} = 6.3 \times 10^{-18} \)
MnS: \( K_{sp} = 2.5 \times 10^{-13} \)
ZnS: \( K_{sp} = 1.6 \times 10^{-24} \)
CdS: \( K_{sp} = 8.0 \times 10^{-27} \)
Comparison:
For FeS: \( Q_{sp} (8.87 \times 10^{-22}) < K_{sp} (6.3 \times 10^{-18}) \implies \) No precipitation.
For MnS: \( Q_{sp} (8.87 \times 10^{-22}) < K_{sp} (2.5 \times 10^{-13}) \implies \) No precipitation.
For ZnS: \( Q_{sp} (8.87 \times 10^{-22}) > K_{sp} (1.6 \times 10^{-24}) \implies \) Precipitation will occur.
For CdS: \( Q_{sp} (8.87 \times 10^{-22}) > K_{sp} (8.0 \times 10^{-27}) \implies \) Precipitation will occur.
Therefore, precipitation will take place in the ZnCl2 and CdCl2 solutions.
In simple words: First, we find the new amounts of sulphide ions and metal ions after mixing the solutions. Then, we multiply these amounts to get a 'reaction value' (Qsp). We compare this value with the 'solubility limit' (Ksp) for each metal sulphide. If our reaction value is higher than the solubility limit, a solid will form. Based on typical Ksp values, zinc chloride and cadmium chloride will form solids.

Exam Tip: Remember to dilute both reactant solutions before calculating the ionic product. Precipitation occurs only when the ionic product (Qsp) exceeds the solubility product (Ksp).