RBSE Solutions Class 11 Chemistry Chapter 7 Equilibrium

RBSE Class 11 Chemistry Chapter 7 Text Book Questions

RBSE Class 11 Chemistry Chapter 7 Multiple Choice Questions

Question 1. For the reaction \( A + 2B \rightarrow C \), the equilibrium constant will be :
(a) \( \frac{[A][B]^2}{[C]} \)
(b) \( \frac{[A][B]}{[C]} \)
(c) \( \frac{[C]}{[A][B]^2} \)
(d) \( \frac{[C]}{2[A][B]} \)
Answer: (c) \( \frac{[C]}{[A][B]^2} \)
In simple words: The equilibrium constant for a reaction is found by dividing the concentration of products by the concentration of reactants, each raised to the power of their coefficients. For this reaction, 'C' is the product and 'A' and 'B' are reactants.

🎯 Exam Tip: Remember to always place product concentrations in the numerator and reactant concentrations in the denominator, raising each to the power of their stoichiometric coefficients.

Question 3. For the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 + xkJ \), the conditions to make more ammonia are :
(a) high temperature and high pressure
(b) low temperature and high pressure
(c) low temperature and low pressure
(d) high temperature and low pressure
Answer: (b) low temperature and high pressure
In simple words: To get more ammonia from this reaction, you need to use low heat because the reaction makes heat, and high pressure because it brings the gas molecules closer together to form the product.

🎯 Exam Tip: For exothermic reactions like the Haber process, low temperature shifts the equilibrium towards products, and high pressure favors the side with fewer gas moles.

Question 4. The solubility product of two electrolytes AB and \( AB_2 \) is \( 1 \times 10^{-10} \). The molar conductivity of AB is ...... \( AB_2 \).
(a) equal to
(b) more than
(c) less than
(d) no relation with
Answer: (c) less than
In simple words: When you compare how well AB and \( AB_2 \) conduct electricity, AB is less conductive than \( AB_2 \) under these conditions, even if their solubility product is the same. This is because \( AB_2 \) produces more ions in solution.

🎯 Exam Tip: Molar conductivity depends on the number of ions produced per mole of electrolyte. \( AB_2 \) will produce more ions than AB for the same solubility product, leading to higher conductivity.

Question 5. The pH of a solution on addition of 50 ml. water slowly to a solution
(a) 1
(b) 5
(c) 7
(d) 10
Answer: (c) 7
In simple words: Adding plain water to a solution usually dilutes it. If the original solution was neutral or the final dilution makes it neutral, its pH becomes 7.

🎯 Exam Tip: Diluting an acidic solution increases its pH towards 7, while diluting a basic solution decreases its pH towards 7. If the original solution's nature isn't specified, and options include 7, it often points to a neutral outcome after dilution.

RBSE Class 11 Chemistry Chapter 7 Very Short Answer Type Questions

Question 7. What is chemical equilibrium ?
Answer: Chemical equilibrium is a dynamic state where the speed of the forward chemical reaction becomes equal to the speed of the backward reaction. At this point, the concentrations of reactants and products remain constant, though reactions continue to occur. This balance means there is no net change in the system.
In simple words: Chemical equilibrium is a balanced state in a reaction where the forward and backward changes happen at the same speed, making everything look stable.

🎯 Exam Tip: Highlight "dynamic state" and "rates are equal" as key phrases for defining chemical equilibrium.

Question 8. Write the examples of the reaction in which
(a) Product increases on increasing the pressure
(b) Product increases on increasing the temperature.
Answer:
(a) Product increases on increasing the pressure: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3 (g) \). Here, an increase in pressure shifts the equilibrium towards the side with fewer moles of gas (products).
(b) Product increases on increasing the temperature: \( N_2O_4 (g) \rightleftharpoons 2NO_2(g) \). This is an endothermic reaction, so increasing the temperature favors the formation of products.
In simple words: (a) Some reactions make more product when you add pressure, like making ammonia. (b) Other reactions make more product when you add heat, like \( N_2O_4 \) changing to \( NO_2 \).

🎯 Exam Tip: Apply Le Chatelier's principle: pressure changes affect reactions with different numbers of gaseous moles, while temperature changes affect exothermic (heat releasing) and endothermic (heat absorbing) reactions differently.

Question 10. What are the conditions for more yield of ammonia by Haber's process?
Answer: For the Haber's process, \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3 (g) \) (which is an exothermic reaction), the conditions for more ammonia yield are:
1. Low temperature: Since the reaction releases energy (exothermic), a lower temperature helps shift the equilibrium towards the products, increasing ammonia yield.
2. High pressure: The forward reaction reduces the number of gas molecules (4 moles of reactants to 2 moles of product). According to Le Chatelier's principle, high pressure favors the side with fewer gas moles, thus increasing ammonia production. Maintaining optimal conditions helps achieve industrial efficiency.
In simple words: To get more ammonia in the Haber process, you need to use low heat and high squeezing force (pressure).

🎯 Exam Tip: Always remember that the Haber process is an exothermic reaction and involves a decrease in the number of moles of gas; these facts directly dictate the favorable temperature and pressure conditions.

Question 11. Which states of a substance are at equilibrium at its melting point?
Answer: At its melting point, a substance is at equilibrium between its solid and liquid states.
Solid \( \rightleftharpoons \) Liquid. This means that at the melting point, the rate at which the solid melts into liquid is equal to the rate at which the liquid freezes into solid.
In simple words: At the melting point, the solid form and the liquid form of a substance are balanced, changing into each other at the same speed.

🎯 Exam Tip: The melting point is a specific temperature where both solid and liquid phases coexist in equilibrium under a given pressure.

Question 12. Which states of a substance are at equilibrium at its boiling point?
Answer: At its boiling point, a substance is at equilibrium between its liquid and gaseous (vapour) states.
Liquid \( \rightleftharpoons \) Gas (vapour). This equilibrium exists when the vapor pressure of the liquid equals the external atmospheric pressure.
In simple words: At the boiling point, the liquid form and the gas form of a substance are balanced, changing into each other at the same speed.

🎯 Exam Tip: The boiling point is the temperature at which a liquid's vapor pressure equals the surrounding atmospheric pressure, allowing it to rapidly turn into gas.

Question 13. If the degree of dissociation of \( PCl_3 \) and \( Cl_2 \) is x, then how many moles of \( PCl_5 \) is present at equilibrium?
Answer: For the dissociation of \( PCl_5 \):
\( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \)
Initial moles: 1 0 0
Moles reacted: x - -
Moles formed: - x x
Moles at equilibrium: \( (1-x) \) x x
At equilibrium, the number of moles of \( PCl_5 \) present is \( (1-x) \). This value directly shows how much of the original \( PCl_5 \) remains after some has broken apart.
In simple words: If 1 mole of \( PCl_5 \) starts to break down, and 'x' amount breaks, then at the end, \( (1-x) \) moles of \( PCl_5 \) will still be left.

🎯 Exam Tip: When calculating equilibrium moles, always subtract the dissociated amount from the initial amount of the reactant and add the formed amount to the products.

Question 15. Write the conjugate acid of \( NH_2^{-} \).
Answer: The conjugate acid of \( NH_2^{-} \) is \( NH_3 \). A conjugate acid is formed when a base gains a proton (\( H^+ \)).
In simple words: When \( NH_2^{-} \) takes in an \( H^+ \), it becomes \( NH_3 \), which is its conjugate acid.

🎯 Exam Tip: To find the conjugate acid of a base, add one \( H^+ \) and increase the charge by one. To find the conjugate base of an acid, remove one \( H^+ \) and decrease the charge by one.

Question 16. Write the conjugate base of \( HCO_3^{-} \).
Answer: The conjugate base of \( HCO_3^{-} \) is \( CO_3^{2-} \). A conjugate base is formed when an acid donates a proton (\( H^+ \)).
In simple words: When \( HCO_3^{-} \) gives away an \( H^+ \), it becomes \( CO_3^{2-} \), which is its conjugate base.

🎯 Exam Tip: Remember that conjugate acid-base pairs differ by only one proton (\( H^+ \)).

Question 17. Calculate the pH of 0.001 N HCl.
Answer: For a strong acid like HCl, the normality (N) is equal to its molarity (M) and the concentration of \( H^+ \) ions.
Given: Concentration of HCl = 0.001 N = \( 1 \times 10^{-3} \) M
So, \( [H^+] = 1 \times 10^{-3} \) M
The pH is calculated using the formula: \( pH = -\log_{10}[H^+] \)
Substitute the value of \( [H^+] \):
\( pH = -\log_{10}(1 \times 10^{-3}) \)
\( pH = -(-3) \log_{10}(1) \)
\( pH = 3 \times 1 \) (since \( \log_{10}(1) = 0 \), this step is simplified. More accurately, \( -\log_{10}(10^{-3}) = 3 \).)
\( \implies \) \( pH = 3 \). This shows that dilute strong acids have low pH values, indicating high acidity.
In simple words: To find the pH of strong acid like HCl, you just count how many zeros are after the decimal in its concentration, if it's 0.001, the pH is 3.

🎯 Exam Tip: For strong acids, the \( [H^+] \) concentration is directly the molarity of the acid. pH is simply the negative logarithm (base 10) of \( [H^+] \).

Question 18. What will be the effect of presence of HCl on dissociation of \( H_2S \)?
Answer: The dissociation of \( H_2S \) will be suppressed by the presence of HCl due to the common ion effect. When HCl, a strong acid, is added, it increases the concentration of \( H^+ \) ions in the solution. This excess \( H^+ \) shifts the equilibrium of \( H_2S \rightleftharpoons H^+ + HS^{-} \) to the left, reducing the dissociation of \( H_2S \). This principle helps control selective precipitation in qualitative analysis.
In simple words: When HCl is added to \( H_2S \), it puts more \( H^+ \) into the solution. This extra \( H^+ \) makes \( H_2S \) break apart less, because the system tries to use up the extra \( H^+ \).

🎯 Exam Tip: The common ion effect states that the solubility of a sparingly soluble salt or the dissociation of a weak electrolyte decreases when a common ion is added to the solution.

Question 19. What are reversible and irreversible reactions?
Answer:
Reversible reaction: A reaction which can proceed in both forward and backward directions is called a reversible reaction. These reactions typically reach a state of equilibrium where reactants and products coexist. An example is the formation of ammonia from nitrogen and hydrogen.
Irreversible reaction: A reaction in which the entire amount of reactant is converted into products, and the products do not react to form reactants again, is called an irreversible reaction. These reactions usually proceed to completion in one direction. An example is the burning of wood.
In simple words: Reversible reactions can go both ways, turning reactants into products and back again. Irreversible reactions only go one way, turning all reactants into products forever.

🎯 Exam Tip: Reversible reactions are denoted by a double arrow (\( \rightleftharpoons \)), while irreversible reactions are shown with a single arrow (\( \rightarrow \)).

Question. Both the forward and backward reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the backward reaction?
Answer:
Given forward reaction: \( NO(g) + O_3 (g) \rightarrow NO_2 (g) + O_2 (g) \)
For this forward reaction, \( K_c = 6.3 \times 10^{14} \)
For the backward reaction, the equilibrium constant \( K_c' \) is the reciprocal of the forward equilibrium constant \( K_c \). This means that if a reaction goes in reverse, its equilibrium constant is inverted.
\( K_c' = \frac{1}{K_c} \)
Substitute the value of \( K_c \):
\( \implies \) \( K_c' = \frac{1}{6.3 \times 10^{14}} \)
\( \implies \) \( K_c' = 0.1587 \times 10^{-14} \)
\( \implies \) \( K_c' = 1.587 \times 10^{-15} \). This very small value indicates that the backward reaction is not favored.
In simple words: If you know the balance number (Kc) for a reaction going forward, the balance number for the reaction going backward is simply one divided by the forward balance number.

🎯 Exam Tip: For any reverse reaction, its equilibrium constant is the reciprocal of the equilibrium constant for the forward reaction: \( K_{reverse} = \frac{1}{K_{forward}} \).

RBSE Class 11 Chemistry Chapter 7 Short Answer Type Questions

Question 21. 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) \( COCl_2 \rightleftharpoons CO(g) + Cl_2(g) \)
(ii) \( CH_4 (g) + 2S_2 (g) \rightleftharpoons CS_2 (g) + 2H_2S(g) \)
(iii) \( CO_2(g) + C(s) \rightleftharpoons 2CO(g) \)
(iv) \( 2H_2 (g) + CO(g) \rightleftharpoons CH_3OH(g) \)
(v) \( CaCO_3 (s) \rightleftharpoons CaO(s) + CO_2 (g) \)
(vi) \( 4NH_3 (g) + 5O_2 (g) \rightleftharpoons 4NO(g) + 6H_2O (g) \)
Answer: According to Le Chatelier's principle, an increase in pressure favors the side with fewer moles of gas. Reactions with an unequal number of gaseous moles on reactant and product sides are affected by pressure changes.
(i) \( COCl_2 (g) \rightleftharpoons CO(g) + Cl_2(g) \)
(Moles of gas: Reactants = 1, Products = 2). If pressure increases, the reaction will go into the backward direction (towards fewer moles of gas).
(ii) \( CH_4 (g) + 2S_2 (g) \rightleftharpoons CS_2 (g) + 2H_2S(g) \)
(Moles of gas: Reactants = 3, Products = 3). There is no change in the number of gaseous moles, so pressure change will not affect this reaction.
(iii) \( CO_2(g) + C(s) \rightleftharpoons 2CO(g) \)
(Moles of gas: Reactants = 1, Products = 2; solid carbon does not count for pressure). If pressure increases, the reaction will go into the backward direction.
(iv) \( 2H_2 (g) + CO(g) \rightleftharpoons CH_3OH(g) \)
(Moles of gas: Reactants = 3, Products = 1). If pressure increases, the reaction will go into the forward direction (towards fewer moles of gas).
(v) \( CaCO_3 (s) \rightleftharpoons CaO(s) + CO_2 (g) \)
(Moles of gas: Reactants = 0, Products = 1; solids do not count for pressure). If pressure increases, the reaction will go into the backward direction.
(vi) \( 4NH_3 (g) + 5O_2 (g) \rightleftharpoons 4NO(g) + 6H_2O (g) \)
(Moles of gas: Reactants = 9, Products = 10). If pressure increases, the reaction will go into the backward direction.
In simple words: Pressure changes affect reactions only when the number of gas particles on one side is different from the other. If pressure goes up, the reaction moves to the side with fewer gas particles.

🎯 Exam Tip: When analyzing the effect of pressure, only consider the moles of gaseous reactants and products. Solids and liquids do not significantly contribute to the total pressure in the system.

Question 22. Describe the factors which affect chemical equilibrium?
Answer: Chemical equilibrium can be influenced by several factors, as explained by Le Chatelier's principle:
1. Concentration: If the concentration of reactants is increased at equilibrium, the rate of the forward reaction rises, producing more product. Conversely, increasing product concentration boosts the backward reaction, yielding more reactant. The system tries to reduce the added component.
2. Pressure: Pressure affects equilibrium when gaseous reactants and products have different numbers of molecules. For example, in the Haber process: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \). Here, 4 moles of gas react to form 2 moles of gas. Increasing pressure shifts the equilibrium to the right, favoring the side with fewer moles of gas (ammonia formation), which helps reduce the overall pressure. If the number of gas molecules is equal on both sides, pressure changes have no effect.
3. Temperature: Changes in temperature affect the equilibrium constant. For exothermic reactions (which release heat), a decrease in temperature shifts the equilibrium towards the product side. For endothermic reactions (which absorb heat), an increase in temperature shifts the equilibrium towards the product side. The system attempts to counteract the temperature change.
4. Catalyst: A catalyst speeds up both the forward and backward reactions equally by providing a new, lower-energy pathway. It helps the system reach equilibrium faster but does not change the equilibrium position or the final amounts of reactants and products. Catalysts are essential in industrial processes to optimize reaction rates.
In simple words: How much stuff you have (concentration), how much you squeeze the gases (pressure), how hot it is (temperature), and using a helper (catalyst) all change where the reaction balances out. A catalyst just makes it balance faster.

🎯 Exam Tip: Understand Le Chatelier's principle thoroughly, as it's the core concept for explaining how external changes affect equilibrium. Always state the direction of shift and the reason (e.g., favoring fewer moles, using up excess heat).

The Conditions For More Synthesis Of SO3 Are Given As Below:

(i) Increase in concentration of \( SO_2 \) or \( O_2 \)
If the concentration of \( SO_2 \) or \( O_2 \) increases, according to Le Chatelier's principle, the equilibrium will shift in the forward direction, producing more \( SO_3 \). This is because the system tries to consume the added reactants.
(ii) Decrease in temperature
A decrease in temperature favors the exothermic reaction. For the synthesis of \( SO_3 \), which is exothermic, a lower temperature shifts the equilibrium towards the forward direction, leading to a greater amount of \( SO_3 \) being formed. This helps to remove excess heat.
(iii) Increase in Pressure :
An increase in pressure favors reactions that reduce the number of gaseous molecules. If the forward reaction for \( SO_3 \) synthesis decreases the total number of gaseous moles, then increasing the pressure will shift the equilibrium to the right, and the yield of \( SO_3 \) will increase. The system attempts to relieve the added pressure.
(iv) Removal of \( SO_3 \) from reaction vessel
If \( SO_3 \) is continuously removed from the reaction vessel, the concentration of product decreases. According to Le Chatelier's principle, the equilibrium will shift in the forward direction to replenish the \( SO_3 \), thus increasing the yield.
In simple words: To make more \( SO_3 \), you can add more starting materials, make it cooler, squeeze it more, or keep taking out the \( SO_3 \) you've made.

🎯 Exam Tip: Remember to consider all four factors: concentration, temperature, pressure, and removal of products. Each factor can be manipulated to maximize the yield of a desired product in industrial processes.

Question 24. The equilibrium for the synthesis of ammonia is
\( N_2+3H_2 \rightleftharpoons NH_3 + x kJ \),
What is the effect of pressure, temperature and concentration on this equilibrium ?
Answer: For the equilibrium \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3 (g) + x kJ \):
Effect of pressure:
High pressure favors the formation of ammonia. An increase in pressure will push the reaction towards the side with fewer gas molecules. Since 4 moles of gas on the reactant side form 2 moles of gas on the product side, increasing the pressure shifts the equilibrium to the right, thereby increasing the yield of ammonia. This helps reduce the system's overall pressure.
Effect of temperature:
Low temperature favors the formation of ammonia. Since the reaction is exothermic (releases heat), a decrease in temperature will shift the equilibrium to the forward direction. This helps to generate more heat and reach a new balance, increasing the yield of ammonia.
Effect of concentration:
High concentration of reactants (\( N_2 \) and \( H_2 \)) favors the formation of ammonia. According to Le Chatelier's principle, if the concentration of \( N_2 \) or \( H_2 \) is increased, the equilibrium will shift to the forward direction. This happens because the system attempts to consume the added reactants, leading to an increased yield of ammonia.
In simple words: To get more ammonia, you need lots of starting gases, low heat, and high squeezing pressure. These conditions help the reaction make more product.

🎯 Exam Tip: Always analyze the reaction's \( \Delta n_g \) (change in moles of gas) for pressure effects and its enthalpy change for temperature effects. For concentration, adding reactants or removing products always drives the reaction forward.

Question 26. Write the conjugate acid of the following:
\( S^{2-}, NH_3, H_2PO_4^{-}, CH_3NH_2 \)
Answer: To find the conjugate acid, add one \( H^+ \) to each species:

🎯 Exam Tip: Remember to adjust the charge accordingly when adding an \( H^+ \). For example, \( S^{2-} \) (charge -2) becomes \( HS^{-} \) (charge -1) after gaining an \( H^+ \).

Question 27. Write any two factors which affect ionization.
Answer: Two factors that affect ionization are:

• Concentration: In a solution, increasing the concentration of an electrolyte generally leads to a decrease in its degree of ionization. This means the extent of ionization is inversely proportional to the solution's concentration, as described by Ostwald's dilution law.
• Temperature: The degree of ionization usually increases with a rise in temperature. Higher temperatures provide more thermal energy, which can overcome the forces holding ions together in the undissociated molecule.

🎯 Exam Tip: For weak electrolytes, dilution (decreasing concentration) increases the degree of ionization, while for both weak and strong electrolytes, increasing temperature generally enhances ionization due to increased kinetic energy.

Question 28. In precipitation of saturated solution of common salt, HCl gas is added and not HCl acid. Explain why?
Answer: In the precipitation of a saturated common salt (NaCl) solution, HCl gas is added instead of HCl acid. This is because HCl gas generates a high concentration of \( Cl^- \) ions in the solution, without significantly increasing the volume or diluting the solution. Adding \( Cl^- \) ions shifts the equilibrium of \( NaCl(s) \rightleftharpoons Na^+(aq) + Cl^-(aq) \) to the left, causing NaCl to precipitate due to the common ion effect. If HCl acid (aqueous solution) were added, it would also introduce \( H^+ \) ions and significantly dilute the saturated NaCl solution, which might hinder precipitation or require a larger volume of acid. HCl gas is a more efficient way to increase the chloride ion concentration.
In simple words: To make salt come out of a salty water solution, we use HCl gas. The gas adds many chloride ions which forces the salt to settle out. If we used liquid HCl, it would also add water, which would just make the solution less concentrated.

🎯 Exam Tip: HCl gas is preferred because it provides a high concentration of the common ion (\( Cl^- \)) without diluting the solution, which is crucial for maximizing precipitation in a saturated solution.

Question 31. Experimentally how it is proved that chemical equilibrium is dynamic in nature?
Answer: Chemical equilibrium is proved to be dynamic in nature through experiments involving radioactive isotopes. Consider the reaction:
\( H_2(g) + I_2(g) \rightleftharpoons 2HI (g) \)
If we start with a mixture of non-radioactive hydrogen (\( H_2 \)) and non-radioactive iodine (\( I_2 \)), and then at equilibrium, we introduce a small amount of radioactive iodine isotope (\( ^{131}I_2 \)), we observe some key changes. While the overall concentrations of \( H_2 \), \( I_2 \), and HI remain constant (because it's equilibrium), after some time, radioactive iodine can be detected in the hydrogen iodide (HI) molecules. This shows that even though the concentrations seem unchanging, the forward reaction (forming HI) and the backward reaction (breaking HI) are still actively occurring at equal rates. The radioactive atoms are continuously exchanged between the reactants and products, confirming that the equilibrium is not static but a state of continuous microscopic activity.
In simple words: Even though a reaction looks still at equilibrium, we can prove things are still moving by using special, traceable atoms (radioactive isotopes). If we put in a radioactive atom, it will soon show up in all the products and reactants, even if the total amounts don't change. This means both forward and backward reactions are always happening.

🎯 Exam Tip: The use of radioactive tracers is a classic method to demonstrate the dynamic nature of equilibrium, as it allows observation of ongoing molecular processes without disturbing the macroscopic balance.

Question 32. Establish the relation between concentration equilibrium constant \( K_c \) and pressure equilibrium constant \( K_p \) for the following homogeneous reaction-
\( 4NH_3 (g) + 5O_2 (g) \rightleftharpoons 4NO (g) + 6H_2O (g) \)
Answer: The relationship between \( K_p \) and \( K_c \) is given by the equation: \( K_p = K_c (RT)^{\Delta n} \)
Where:
- \( K_p \) is the pressure equilibrium constant
- \( K_c \) is the concentration equilibrium constant
- R is the ideal gas constant
- T is the absolute temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

For the given reaction: \( 4NH_3 (g) + 5O_2 (g) \rightleftharpoons 4NO (g) + 6H_2O (g) \)
Calculate \( \Delta n \):
Moles of gaseous products = moles of \( NO(g) \) + moles of \( H_2O(g) \) = \( 4 + 6 = 10 \)
Moles of gaseous reactants = moles of \( NH_3(g) \) + moles of \( O_2(g) \) = \( 4 + 5 = 9 \)
\( \implies \) \( \Delta n = \) (moles of gaseous products) - (moles of gaseous reactants)
\( \implies \) \( \Delta n = 10 - 9 = 1 \)

Now substitute \( \Delta n = 1 \) into the \( K_p \) and \( K_c \) relationship:
\( \implies \) \( K_p = K_c (RT)^1 \)
\( \implies \) \( K_p = K_c (RT) \). This equation shows how the equilibrium constants relate for this specific gaseous reaction.
In simple words: \( K_p \) and \( K_c \) are related by a formula that uses R (a gas number), T (temperature), and \( \Delta n \). For this reaction, \( \Delta n \) is 1, so \( K_p \) is simply \( K_c \) multiplied by RT.

🎯 Exam Tip: Accurately calculate \( \Delta n \) by only including gaseous species, as solids and liquids do not affect the pressure term in the \( K_p \) expression.

Question 33. Explain law of mass Action by taking the following example,
\( CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 +H_2O \)
Answer:
Law of Mass Action:
The Law of Mass Action states that the rate of a chemical reaction at a constant temperature is directly proportional to the product of the molar concentrations (or active masses) of the reactants, with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation. The term "active mass" refers to the molar concentration of a substance in moles per litre of solution.

Example: For the reaction:
\( CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O \)
According to the Law of Mass Action:
The rate of the forward reaction is proportional to \( [CH_3COOH] [C_2H_5OH] \)
\( \implies \) Rate of forward reaction \( = k_f [CH_3COOH] [C_2H_5OH] \)
Where \( k_f \) is the rate constant for the forward reaction.

Similarly, for the backward reaction:
The rate of the backward reaction is proportional to \( [CH_3COOC_2H_5] [H_2O] \)
\( \implies \) Rate of backward reaction \( = k_b [CH_3COOC_2H_5] [H_2O] \)
Where \( k_b \) is the rate constant for the backward reaction.

At equilibrium, the rate of the forward reaction equals the rate of the backward reaction:
\( k_f [CH_3COOH] [C_2H_5OH] = k_b [CH_3COOC_2H_5] [H_2O] \)
Rearranging this gives the equilibrium constant \( K_c \):
\( K_c = \frac{k_f}{k_b} = \frac{[CH_3COOC_2H_5] [H_2O]}{[CH_3COOH] [C_2H_5OH]} \). This constant indicates the extent of the reaction at equilibrium.
In simple words: The Law of Mass Action says that how fast a reaction happens depends on how much of each starting material is present. If you have more of something, the reaction goes faster. At balance, the speeds of the forward and backward reactions are equal.

🎯 Exam Tip: Clearly define "active mass" as molar concentration. Show how to derive the rate expressions for both forward and backward reactions, and then combine them to get the equilibrium constant expression.

Question 34. For the system, \( N_2 + O_2 \rightleftharpoons 2NO - 44 kcal \), at equilibrium, what will be the effect of the following
(i) Increasing the temperature
(ii) Decreasing the pressure
(iii) Increasing the concentration of NO
(iv) Presence of catalyst
Answer: The given reaction is \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) - 44 kcal \). The "- 44 kcal" indicates that the forward reaction is endothermic (absorbs heat).
(i) Increasing the temperature: Since the forward reaction is endothermic, increasing the temperature will favor the forward reaction, shifting the equilibrium to the right. This will increase the yield of NO, as the system tries to absorb the added heat.
(ii) Decreasing the pressure: In this reaction, the number of moles of gaseous reactants (1 mole \( N_2 \) + 1 mole \( O_2 \) = 2 moles) is equal to the number of moles of gaseous products (2 moles NO). Therefore, decreasing the pressure will not affect the equilibrium position. The system experiences no net change in gas volume.
(iii) Increasing the concentration of NO: If the concentration of product NO is increased, the equilibrium will shift to the backward direction (left). This will reduce the yield of NO and increase the concentration of \( N_2 \) and \( O_2 \), as the system tries to consume the excess product.
(iv) Presence of catalyst: A catalyst increases the rate of both the forward and backward reactions equally. It helps the system reach equilibrium faster but does not change the equilibrium position or the final amounts of reactants and products. Catalysts only affect the speed of the reaction, not its ultimate balance.
In simple words: (i) Making it hotter makes more NO because the reaction uses heat. (ii) Changing pressure does nothing because there are equal gas particles on both sides. (iii) Adding more NO makes the reaction go backward. (iv) A catalyst just makes everything happen faster, but doesn't change how much NO you get in the end.

🎯 Exam Tip: Always identify if a reaction is exothermic or endothermic to determine temperature effects. For pressure effects, count the moles of gaseous species on both sides. Catalysts speed up equilibrium but don't shift its position.

Question 35. What will be the effect of temperature and pressure on the following equilibrium?
\( N_2 + 3H_2 \rightleftharpoons 2NH_3 \)
Answer: The given equilibrium is for the synthesis of ammonia, \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3 (g) \). This reaction is exothermic, meaning it releases heat.
Effect of temperature:
A decrease in temperature favors the forward reaction because the reaction is exothermic. The system will shift to the right to produce more heat, thereby increasing the yield of ammonia. High temperatures would favor the decomposition of ammonia.
Effect of pressure:
The number of moles of gaseous reactants (1 mole \( N_2 \) + 3 moles \( H_2 \) = 4 moles) is greater than the number of moles of gaseous product (2 moles \( NH_3 \)). Therefore, increasing the pressure will favor the forward reaction, shifting the equilibrium to the right. This results in a higher yield of ammonia, as the system seeks to reduce the total number of gas molecules and relieve the pressure.
In simple words: To make more ammonia, you should use low heat because the reaction itself gives off heat, and high squeezing force (pressure) because the product takes up less space as a gas.

🎯 Exam Tip: The Haber process is a classic example for illustrating Le Chatelier's principle. Remember that an exothermic reaction is favored by low temperature, and a reaction that reduces the number of gas moles is favored by high pressure.

RBSE Class 11 Chemistry Chapter 7 Long Answer Type Questions

Question 36. (i) Explain equilibrium in physical process and chemical process with examples.
(ii) Prove that degree of dissociation of \( \text{PCl}_5 \) is inversely proportional to square root of its pressure.
Answer:
(i) **Equilibrium in Physical Processes:**
A physical process involves only a change in the physical state of a substance, not its chemical identity. Physical equilibrium happens when the rate of the forward physical change equals the rate of the backward physical change. For example, solid-liquid equilibrium occurs when a solid melts into a liquid at the same rate the liquid freezes back into a solid. This is often seen at the melting point, like with ice and water at 273K and atmospheric pressure. At this point, the mass of ice and water stays constant, but molecules are constantly moving between the solid and liquid states, making it a dynamic equilibrium. For instance, `\( \text{H}_2\text{O}(\text{s}) \rightleftharpoons \text{H}_2\text{O}(\text{l}) \)`. The freezing point is the specific temperature where a pure substance's solid and liquid phases are in equilibrium under atmospheric pressure.

**Equilibrium in Chemical Processes:**
Chemical equilibrium is a state where the rate of the forward chemical reaction equals the rate of the backward (reverse) reaction. At this point, the concentrations of reactants and products remain constant over time, even though the reactions are still occurring. This is also a dynamic equilibrium. For example, `\( \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}) \)`. If a radioactive isotope of iodine is added to this equilibrium, it quickly becomes part of the HI molecules, showing that the reactions are still actively happening in both directions.
In simple words: Physical equilibrium is when a substance changes form (like solid to liquid) at equal rates in both directions. Chemical equilibrium is when a chemical reaction goes forward and backward at the same speed. Both are always moving, but look stable.

🎯 Exam Tip: When explaining dynamic equilibrium, always mention that the processes are still occurring, but their rates are equal, leading to constant macroscopic properties.

Answer:
(ii) **Proof that degree of dissociation of \( \text{PCl}_5 \) is inversely proportional to square root of its pressure:**
Let's consider the dissociation of \( \text{PCl}_5 \) into \( \text{PCl}_3 \) and \( \text{Cl}_2 \):
`\( \text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \)`
Suppose we start with 1 mole of \( \text{PCl}_5 \). Let \( x \) be the degree of dissociation at equilibrium.

Total number of moles at equilibrium \( = (1-x) + x + x = 1+x \).
We know that partial pressure of a component is its mole fraction multiplied by the total pressure \( (P) \).
Mole fraction is the number of moles of that component divided by the total number of moles in the mixture.

So, the partial pressures are:
`\( P_{\text{PCl}_5} = \left( \frac{1-x}{1+x} \right) P \)`
`\( P_{\text{PCl}_3} = \left( \frac{x}{1+x} \right) P \)`
`\( P_{\text{Cl}_2} = \left( \frac{x}{1+x} \right) P \)`

The equilibrium constant \( K_p \) for the reaction is:
`\( K_p = \frac{P_{\text{PCl}_3} \times P_{\text{Cl}_2}}{P_{\text{PCl}_5}} \)`
Substitute the partial pressures into the \( K_p \) expression:
`\( K_p = \frac{\left( \frac{x}{1+x} P \right) \times \left( \frac{x}{1+x} P \right)}{\left( \frac{1-x}{1+x} \right) P} \)`
Simplify the expression:
`\( K_p = \frac{\frac{x^2}{(1+x)^2} P^2}{\frac{1-x}{1+x} P} \)`
`\( K_p = \frac{x^2 P^2}{(1+x)^2} \times \frac{1+x}{(1-x)P} \)`
`\( K_p = \frac{x^2 P}{(1+x)(1-x)} \)`
`\( K_p = \frac{x^2 P}{1-x^2} \)`

If the degree of dissociation \( x \) is very small (i.e., \( x \ll 1 \)), then \( x^2 \) can be ignored compared to 1 in the denominator.
So, `\( 1-x^2 \approx 1 \)`
This simplifies the expression for \( K_p \) to:
`\( K_p \approx x^2 P \)`
Now, rearrange to solve for \( x \):
`\( x^2 = \frac{K_p}{P} \)`
`\( x = \sqrt{\frac{K_p}{P}} \)`
This shows that \( x \) is proportional to `\( \frac{1}{\sqrt{P}} \)`. Therefore, the degree of dissociation \( x \) is inversely proportional to the square root of the total pressure \( P \). Increasing the pressure will shift the equilibrium to the side with fewer moles of gas, which is the reactant side in this case, thus decreasing the dissociation.
In simple words: When \( \text{PCl}_5 \) breaks apart, how much it breaks depends on the pressure. If the pressure goes up, less of it breaks apart. This is because the reaction tries to make fewer gas molecules to reduce the pressure. So, the breaking apart (dissociation) is linked to the inverse of the square root of the pressure.

🎯 Exam Tip: Remember to clearly define the degree of dissociation (\( x \)) and show all steps when deriving the relationship between \( K_p \) and pressure. State the approximation \( x \ll 1 \) and its implication.

Question 37. What is buffer solution? Write any two properties of buffer solution. Derive the formula for calculating the pH for acidic buffer. Write any two examples of simple buffer.
Answer:
**Buffer Solution:** A buffer solution is a mixture that resists changes in its \( \text{pH} \) when small amounts of an acid or a base are added to it, or when it is diluted. These solutions are crucial in many biological and chemical systems where a stable \( \text{pH} \) is needed. The buffer capacity measures how well a solution can resist these \( \text{pH} \) changes.

**Properties of Buffer Solution:**
* Buffer solutions resist changes in \( \text{pH} \) even upon addition of small amounts of strong acid or strong base.
* They are used to prepare solutions with a specific, stable \( \text{pH} \).

**Derivation of \( \text{pH} \) for Acidic Buffer (Henderson-Hasselbalch Equation):**
An acidic buffer is typically formed by a weak acid and its salt (conjugate base). Let's consider a weak acid, `\( \text{HA} \)`, dissociating in water:
`\( \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^- \)`
The acid dissociation constant `\( K_a \)` is given by:
`\( K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \)`
Rearranging this equation to solve for `\( [\text{H}_3\text{O}^+] \)`:
`\( [\text{H}_3\text{O}^+] = K_a \frac{[\text{HA}]}{[\text{A}^-]} \)`
Now, taking the negative logarithm of both sides:
`\( -\log[\text{H}_3\text{O}^+] = -\log K_a - \log \frac{[\text{HA}]}{[\text{A}^-]} \)`
We know that `\( \text{pH} = -\log[\text{H}_3\text{O}^+] \)` and `\( \text{p}K_a = -\log K_a \)`:
`\( \text{pH} = \text{p}K_a - \log \frac{[\text{HA}]}{[\text{A}^-]} \)`
Using the logarithm property `\( -\log(x/y) = \log(y/x) \)`:
`\( \text{pH} = \text{p}K_a + \log \frac{[\text{A}^-]}{[\text{HA}]} \)`
In an acidic buffer, the concentration of the weak acid `\( [\text{HA}] \)` is approximately the concentration of the undissociated acid, and `\( [\text{A}^-] \)` is approximately the concentration of the salt (which fully dissociates).
So, the Henderson-Hasselbalch equation for an acidic buffer is:
`\( \text{pH} = \text{p}K_a + \log \frac{[\text{Salt}]}{[\text{Acid}]} \)`

**Examples of Simple Buffers:**
* `\( \text{CH}_3\text{COONH}_4 \)` (Ammonium acetate)
* `\( (\text{NH}_4)_2\text{CO}_3 \)` (Ammonium carbonate)
In simple words: A buffer solution keeps its pH almost the same even if you add a little acid or base. This is because it contains a weak acid and its partner salt. The formula helps calculate its pH. Common examples are special types of ammonium salts.

🎯 Exam Tip: Clearly state the components of an acidic buffer (weak acid and its salt/conjugate base) and remember the Henderson-Hasselbalch equation. Practice rearranging log terms correctly.

Question 38. What is solubility product? Establish the relation between solubility and solubility product for CdS type compounds. In third group analysis, \( \text{NH}_4\text{Cl} \) is added before \( \text{NH}_4\text{OH} \), why?
Answer:
**Solubility Product (`\( K_{sp} \)`):**
The solubility product is a special equilibrium constant that describes the extent to which an ionic solid dissolves in water. For a sparingly soluble ionic compound, `\( K_{sp} \)` is defined as the product of the molar concentrations of its ions, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation, at a given temperature in a saturated solution. For example, its symbol is `\( K_{sp} \)`. This helps us understand how much of a solid can dissolve.

**Relation between Solubility and Solubility Product for \( \text{CdS} \) Type Compounds:**
Let \( x \) be the molar solubility of `\( \text{CdS} \)`. When `\( \text{CdS} \)` dissolves, it dissociates as:
`\( \text{CdS}(\text{s}) \rightleftharpoons \text{Cd}^{2+}(\text{aq}) + \text{S}^{2-}(\text{aq}) \)`
If \( x \) moles of `\( \text{CdS} \)` dissolve per liter, then at equilibrium, the concentrations of the ions will be:
`\( [\text{Cd}^{2+}] = x \text{ mol/L} \)`
`\( [\text{S}^{2-}] = x \text{ mol/L} \)`
The solubility product `\( K_{sp} \)` is then given by:
`\( K_{sp} = [\text{Cd}^{2+}][\text{S}^{2-}] \)`
Substitute the concentrations:
`\( K_{sp} = (x)(x) \)`
`\( K_{sp} = x^2 \)`
So, for `\( \text{CdS} \)`, the solubility is `\( x = \sqrt{K_{sp}} \)`.

**Reason for Adding \( \text{NH}_4\text{Cl} \) Before \( \text{NH}_4\text{OH} \) in Third Group Analysis:**
In third group qualitative analysis, `\( \text{NH}_4\text{Cl} \)` is added before `\( \text{NH}_4\text{OH} \)` to control the concentration of hydroxide ions (`\( \text{OH}^- \)`). `\( \text{NH}_4\text{OH} \)` is a weak base that dissociates to produce `\( \text{OH}^- \)` ions. `\( \text{NH}_4\text{Cl} \)` is a strong electrolyte that provides a high concentration of `\( \text{NH}_4^+ \)` ions. According to Le-Chatelier's principle, the excess `\( \text{NH}_4^+ \)` ions from `\( \text{NH}_4\text{Cl} \)` suppress the ionization of `\( \text{NH}_4\text{OH} \)` (common ion effect). This significantly reduces the `\( \text{OH}^- \)` ion concentration. A low `\( \text{OH}^- \)` concentration is important because it allows only the hydroxides of Group III metals (like `\( \text{Fe}^{3+} \)`, `\( \text{Al}^{3+} \)`, `\( \text{Cr}^{3+} \)`), which have very low solubility products, to precipitate. Other metal hydroxides with higher solubility products, which would precipitate at high `\( \text{OH}^- \)` concentrations, remain in solution. This ensures clean separation of the metal ions.
In simple words: Solubility product tells us how much of a solid can dissolve in water. For a substance like `\( \text{CdS} \)`, its solubility is the square root of its solubility product. In chemical tests, we add `\( \text{NH}_4\text{Cl} \)` first to make sure only certain metal hydroxides fall out of solution by keeping the `\( \text{OH}^- \)` level low.

🎯 Exam Tip: When defining solubility product, specify "molar concentrations" and "saturated solution". For the `\( \text{NH}_4\text{Cl} \)` and `\( \text{NH}_4\text{OH} \)` question, emphasize the common ion effect and selective precipitation.

Question 39. Explain the law of mass action using a suitable chemical reaction. Derive the relationship between \( \text{K}_\text{p} \) and \( \text{K}_\text{c} \) for a homogeneous reaction.
Answer:
**Law of Mass Action:**
The law of mass action states that the rate of a chemical reaction is directly proportional to the product of the molar concentrations (or active masses) of the reactants, with each concentration term raised to the power of its stoichiometric coefficient in the balanced chemical equation, at a constant temperature. Molar concentration is also known as active mass, which means the number of moles of a substance dissolved in one liter of solution. This law helps predict how fast a reaction will proceed.

Consider a general reversible homogeneous reaction:
`\( \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} \)`

According to the law of mass action:
Rate of forward reaction `\( \propto [\text{A}][\text{B}] \)`
`\( \text{Rate}_{\text{f}} = k_1 [\text{A}][\text{B}] \)`
Here, `\( k_1 \)` is the rate constant for the forward reaction.

Rate of backward (reverse) reaction `\( \propto [\text{C}][\text{D}] \)`
`\( \text{Rate}_{\text{b}} = k_2 [\text{C}][\text{D}] \)`
Here, `\( k_2 \)` is the rate constant for the backward reaction.

At equilibrium, the rate of the forward reaction equals the rate of the backward reaction:
`\( \text{Rate}_{\text{f}} = \text{Rate}_{\text{b}} \)`
`\( k_1 [\text{A}][\text{B}] = k_2 [\text{C}][\text{D}] \)`
Rearranging this gives:
`\( \frac{k_1}{k_2} = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} \)`
The ratio of the rate constants `\( \frac{k_1}{k_2} \)` is defined as the equilibrium constant in terms of concentrations, `\( K_c \)`:
`\( K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} \)`

**Relationship between \( \text{K}_\text{p} \) and \( \text{K}_\text{c} \) for a Homogeneous Reaction:**
Consider the general homogeneous gaseous reaction:
`\( a\text{A}(\text{g}) + b\text{B}(\text{g}) \rightleftharpoons c\text{C}(\text{g}) + d\text{D}(\text{g}) \)`
The equilibrium constant `\( K_c \)` is expressed as:
`\( K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \)`
The equilibrium constant `\( K_p \)` is expressed in terms of partial pressures:
`\( K_p = \frac{P_\text{C}^c P_\text{D}^d}{P_\text{A}^a P_\text{B}^b} \)`

For ideal gases, the partial pressure `\( P_i \)` of a component `\( i \)` is related to its molar concentration `\( [C_i] \)` by the ideal gas equation: `\( P_i V = n_i R T \implies P_i = \frac{n_i}{V} R T \implies P_i = C_i R T \)`
Where `\( C_i = \frac{n_i}{V} \)` is the molar concentration of component `\( i \)`.

Substitute `\( P_i = C_i R T \)` into the `\( K_p \)` expression:
`\( K_p = \frac{(C_\text{C}RT)^c (C_\text{D}RT)^d}{(C_\text{A}RT)^a (C_\text{B}RT)^b} \)`
`\( K_p = \frac{C_\text{C}^c C_\text{D}^d}{C_\text{A}^a C_\text{B}^b} \times \frac{(RT)^c (RT)^d}{(RT)^a (RT)^b} \)`
`\( K_p = \left( \frac{C_\text{C}^c C_\text{D}^d}{C_\text{A}^a C_\text{B}^b} \right) (RT)^{(c+d)-(a+b)} \)`
We know that the term in the parenthesis is `\( K_c \)`:
`\( K_p = K_c (RT)^{\Delta n_g} \)`
Where `\( \Delta n_g = (c+d) - (a+b) \)` is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants. This relationship helps us convert between `\( K_p \)` and `\( K_c \)` when dealing with gases.
In simple words: The law of mass action says that a reaction happens faster if there are more starting materials. At equilibrium, the speed of going forward is the same as going backward. We can also link two types of equilibrium constants, \( \text{K}_\text{p} \) (for pressures) and \( \text{K}_\text{c} \) (for concentrations), using a simple formula that involves temperature and the change in gas moles.

🎯 Exam Tip: Clearly state the Law of Mass Action definition. For the Kp-Kc derivation, ensure all steps from the ideal gas law to the final formula `\( K_p = K_c (RT)^{\Delta n_g} \)` are shown and `\( \Delta n_g \)` is defined.

Question 40. What is an indicator? Explain Ostwald's theory of indicators. Explain the titration between \( \text{HNO}_3 \) and \( \text{KOH} \) using a suitable indicator with the help of a curve.
Answer:
**Indicator:**
An indicator is a chemical substance, usually a weak acid or weak base, that changes color depending on the \( \text{pH} \) of the solution. These color changes help us determine the end point of a titration, which is the point where the reaction is complete. Indicators have different colors in acidic and basic environments.

**Ostwald's Theory of Indicators:**
Ostwald's theory explains how indicators change color. It proposes that indicators are weak organic acids or bases that exist in two different tautomeric forms (one un-ionized and one ionized) which have different colors. The color change depends on the `\( \text{pH} \)` of the solution because it shifts the equilibrium between these two forms.

Let's consider a weak acidic indicator, `\( \text{HIn} \)` (e.g., phenolphthalein), which dissociates as:
`\( \text{HIn}(\text{color 1}) \rightleftharpoons \text{H}^+ + \text{In}^-(\text{color 2}) \)`
(Colorless) (Pink)
The equilibrium constant `\( K_{In} \)` is:
`\( K_{In} = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]} \)`

* **In an acidic solution (high `\( [\text{H}^+] \)`):** According to Le-Chatelier's principle, the equilibrium shifts to the left, favoring the un-ionized `\( \text{HIn} \)` form. So, the solution shows Color 1 (e.g., colorless for phenolphthalein).
* **In a basic solution (low `\( [\text{H}^+] \)`):** The `\( \text{OH}^- \)` ions from the base react with `\( \text{H}^+ \)` ions, reducing their concentration. This shifts the equilibrium to the right, favoring the ionized `\( \text{In}^- \)` form. So, the solution shows Color 2 (e.g., pink for phenolphthalein).

**Titration between \( \text{HNO}_3 \) (Strong Acid) and \( \text{KOH} \) (Strong Base):**
This is a strong acid-strong base titration. In this titration, a strong base like `\( \text{KOH} \)` is slowly added to a strong acid like `\( \text{HNO}_3 \)`. The `\( \text{pH} \)` of the solution is measured as the volume of the base added increases.

Initially, the solution is highly acidic with a very low `\( \text{pH} \)`. As `\( \text{KOH} \)` is added, the `\( \text{pH} \)` changes slowly at first. However, near the equivalence point (where the moles of acid equal the moles of base), the `\( \text{pH} \)` changes very rapidly, jumping from about `\( \text{pH } 3 \)` to `\( \text{pH } 10 \)`. The equivalence point for a strong acid-strong base titration is at `\( \text{pH } 7 \)` (neutral).

**Titration Curve and Indicator Choice:**
A titration curve plots `\( \text{pH} \)` against the volume of base added. For a strong acid-strong base titration, the curve shows a sharp vertical rise around the equivalence point. Due to this steep change, many indicators can be used, including phenolphthalein (colorless to pink, `\( \text{pH} \) 8.2-10) and methyl orange (red to yellow, `\( \text{pH} \) 3.1-4.4). Both change color within this steep `\( \text{pH} \)` range, allowing for an accurate determination of the end point. For example, if phenolphthalein is used, the titration is stopped when the solution turns faintly pink.

In simple words: An indicator is a special chemical that changes color when the solution's acid or base level changes. Ostwald's theory says this happens because the indicator changes its form. When you mix a strong acid and a strong base, the pH changes very fast at the exact point they become equal, and an indicator like phenolphthalein shows this by changing color.

🎯 Exam Tip: When drawing a titration curve, clearly label the axes, show the steep change around the equivalence point, and mark where common indicators would change color.

RBSE Class 11 Chemistry Chapter 7 Numerical Problems

Question 41. The first ionization constant of \( \text{H}_2\text{S} \) is \( 9.1 \times 10^{-8} \). Calculate the concentration of \( \text{HS}^- \) ions in its \( 0.1 \text{ M} \) solution. How will this concentration be affected if the solution is \( 0.1 \text{ M} \) in \( \text{HCl} \) 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:
**To calculate the concentration of \( \text{HS}^- \) ion:**

**Case I: In the absence of \( \text{HCl} \)**
The first dissociation of \( \text{H}_2\text{S} \) is:
`\( \text{H}_2\text{S}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{HS}^-(\text{aq}) \)`
Initial concentration: `\( 0.1 \text{ M} \)` \( 0 \)\( 0 \)
At equilibrium: `\( 0.1-x \)` \( x \)\( x \)

The first ionization constant `\( K_{a1} \)` is:
`\( K_{a1} = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]} \)`
`\( 9.1 \times 10^{-8} = \frac{(x)(x)}{(0.1-x)} \)`
Since \( K_{a1} \) is very small, we can approximate `\( 0.1-x \approx 0.1 \)`:
`\( 9.1 \times 10^{-8} = \frac{x^2}{0.1} \)`
`\( x^2 = 9.1 \times 10^{-8} \times 0.1 \)`
`\( x^2 = 9.1 \times 10^{-9} \)`
`\( x = \sqrt{9.1 \times 10^{-9}} = \sqrt{91 \times 10^{-10}} = 9.54 \times 10^{-5} \text{ M} \)`
Therefore, `\( [\text{HS}^-] = 9.54 \times 10^{-5} \text{ M} \)` in the absence of \( \text{HCl} \).

**Case II: In the presence of \( 0.1 \text{ M HCl} \)**
When `\( 0.1 \text{ M HCl} \)` is present, it provides `\( 0.1 \text{ M H}^+ \)` ions (since `\( \text{HCl} \)` is a strong acid and dissociates completely). This is a common ion effect.
The dissociation of `\( \text{H}_2\text{S} \)` is:
`\( \text{H}_2\text{S}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{HS}^-(\text{aq}) \)`
Initial concentrations: `\( 0.1 \text{ M} \)` `\( 0.1 \text{ M} \)` \( 0 \)
At equilibrium: `\( 0.1-y \)` `\( 0.1+y \)` `\( y \)`
The `\( K_{a1} \)` expression is:
`\( K_{a1} = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]} \)`
`\( 9.1 \times 10^{-8} = \frac{(0.1+y)(y)}{(0.1-y)} \)`
Since `\( K_{a1} \)` is very small, `\( y \)` will be much smaller than `\( 0.1 \)`. So, we can approximate `\( 0.1+y \approx 0.1 \)` and `\( 0.1-y \approx 0.1 \)`:
`\( 9.1 \times 10^{-8} = \frac{(0.1)(y)}{0.1} \)`
`\( 9.1 \times 10^{-8} = y \)`
Therefore, `\( [\text{HS}^-] = 9.1 \times 10^{-8} \text{ M} \)` in the presence of `\( 0.1 \text{ M HCl} \)`. The concentration of `\( \text{HS}^- \)` decreases significantly due to the common ion effect.

**To calculate the concentration of \( \text{S}^{2-} \) under both conditions:**
The second dissociation of `\( \text{H}_2\text{S} \)` is:
`\( \text{HS}^-(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{S}^{2-}(\text{aq}) \)`
The second ionization constant `\( K_{a2} \)` is `\( 1.2 \times 10^{-13} \)`:
`\( K_{a2} = \frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]} \)`
`\( 1.2 \times 10^{-13} = \frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]} \)`

**Case I: In the absence of \( \text{HCl} \)**
From the first dissociation, we found:
`\( [\text{H}^+] = 9.54 \times 10^{-5} \text{ M} \)`
`\( [\text{HS}^-] = 9.54 \times 10^{-5} \text{ M} \)`
Let `\( [\text{S}^{2-}] = x' \)`
`\( 1.2 \times 10^{-13} = \frac{(9.54 \times 10^{-5})(x')}{9.54 \times 10^{-5}} \)`
`\( 1.2 \times 10^{-13} = x' \)`
Therefore, `\( [\text{S}^{2-}] = 1.2 \times 10^{-13} \text{ M} \)` in the absence of `\( \text{HCl} \)`.

**Case II: In the presence of \( 0.1 \text{ M HCl} \)**
In this case, the `\( \text{H}^+ \)` concentration is primarily from `\( \text{HCl} \)`:
`\( [\text{H}^+] = 0.1 \text{ M} \)`
And `\( [\text{HS}^-] = 9.1 \times 10^{-8} \text{ M} \)` (from the calculation above).
Let `\( [\text{S}^{2-}] = x'' \)`
`\( 1.2 \times 10^{-13} = \frac{(0.1)(x'')}{9.1 \times 10^{-8}} \)`
`\( x'' = \frac{(1.2 \times 10^{-13}) \times (9.1 \times 10^{-8})}{0.1} \)`
`\( x'' = \frac{10.92 \times 10^{-21}}{0.1} \)`
`\( x'' = 1.092 \times 10^{-19} \text{ M} \)`
Therefore, `\( [\text{S}^{2-}] = 1.092 \times 10^{-19} \text{ M} \)` in the presence of `\( 0.1 \text{ M HCl} \)`.
In simple words: First, we find how much `\( \text{HS}^- \)` is formed from `\( \text{H}_2\text{S} \)`. When `\( \text{HCl} \)` is added, it reduces the `\( \text{HS}^- \)` amount because of a shared ion effect. Then, we use these values to calculate `\( \text{S}^{2-} \)` under both conditions. The `\( \text{S}^{2-} \)` concentration also drops a lot when `\( \text{HCl} \)` is present.

🎯 Exam Tip: Remember to apply the common ion effect when calculating ion concentrations in the presence of a strong acid or base. Always clearly state any approximations made (e.g., \( x \ll 0.1 \)).

Question 42. The ionization constant of propanoic acid is \( 1.32 \times 10^{-5} \). Calculate the degree of ionization of the acid in its \( 0.05 \text{ M} \) solution and also its \( \text{pH} \). What will be its degree of ionization if the solution is \( 0.01 \text{ M HCl} \) also?
Answer:
**Part 1: In \( 0.05 \text{ M} \) propanoic acid solution (without \( \text{HCl} \))**
Let propanoic acid be `\( \text{HA} \)` and its ionization constant `\( K_a = 1.32 \times 10^{-5} \)`.
The dissociation is:
`\( \text{C}_2\text{H}_5\text{COOH}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq}) \)`
Initial concentration: `\( C \)` \( 0 \)\( 0 \)
At equilibrium: `\( C(1-\alpha) \)` `\( C\alpha \)` `\( C\alpha \)`
Where `\( C = 0.05 \text{ M} \)` and `\( \alpha \)` is the degree of ionization.

`\( K_a = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{C}_2\text{H}_5\text{COOH}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \)`
Since `\( K_a \)` is small, we can assume `\( \alpha \ll 1 \)`, so `\( 1-\alpha \approx 1 \)`:
`\( K_a \approx C\alpha^2 \)`
`\( \alpha^2 = \frac{K_a}{C} \)`
`\( \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} \)`
`\( \alpha = 1.62 \times 10^{-2} \)`
**Degree of ionization `\( \alpha = 0.0162 \)`**

Now, calculate `\( \text{pH} \)`:
`\( [\text{H}_3\text{O}^+] = C\alpha = (0.05)(1.62 \times 10^{-2}) \)`
`\( [\text{H}_3\text{O}^+] = 0.081 \times 10^{-2} = 8.1 \times 10^{-4} \text{ M} \)`
`\( \text{pH} = -\log[\text{H}_3\text{O}^+] = -\log(8.1 \times 10^{-4}) \)`
`\( \text{pH} = 4 - \log(8.1) = 4 - 0.908 = 3.092 \)`
**pH of the solution `\( \approx 3.09 \)`**

**Part 2: If the solution is \( 0.01 \text{ M HCl} \) also**
`\( \text{HCl} \)` is a strong acid, so it provides `\( 0.01 \text{ M H}^+ \)` ions. This is a common ion effect, which will suppress the dissociation of propanoic acid.
`\( \text{C}_2\text{H}_5\text{COOH}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{C}_2\text{H}_5\text{COO}^-(\text{aq}) \)`
Initial concentration: `\( 0.05 \text{ M} \)` `\( 0.01 \text{ M} \)` \( 0 \)
At equilibrium: `\( 0.05 - 0.05\alpha' \)` `\( 0.01 + 0.05\alpha' \)` `\( 0.05\alpha' \)`
Where `\( \alpha' \)` is the new degree of ionization.

`\( K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_5\text{COO}^-]}{[\text{C}_2\text{H}_5\text{COOH}]} \)`
`\( 1.32 \times 10^{-5} = \frac{(0.01 + 0.05\alpha')(0.05\alpha')}{(0.05 - 0.05\alpha')} \)`
Since `\( \alpha' \)` will be very small due to the common ion effect, we can approximate:
`\( 0.01 + 0.05\alpha' \approx 0.01 \)`
`\( 0.05 - 0.05\alpha' \approx 0.05 \)`
So, the equation simplifies to:
`\( 1.32 \times 10^{-5} = \frac{(0.01)(0.05\alpha')}{0.05} \)`
`\( 1.32 \times 10^{-5} = (0.01)\alpha' \)`
`\( \alpha' = \frac{1.32 \times 10^{-5}}{0.01} \)`
`\( \alpha' = 1.32 \times 10^{-3} \)`
**Degree of ionization `\( \alpha' = 0.00132 \)`** if `\( 0.01 \text{ M HCl} \)` is present. As expected, the degree of ionization decreases significantly.
In simple words: First, we find how much propanoic acid breaks apart and what its acidity level (pH) is. Then, if we add another acid like `\( \text{HCl} \)`, it makes the propanoic acid break apart much less because of a shared ingredient, which is the hydrogen ion.

🎯 Exam Tip: Remember to use the approximation \( 1-\alpha \approx 1 \) for weak acids when \( \alpha \) is small. For common ion effect problems, ensure to consider the initial concentration of the common ion from the strong electrolyte.

Question 43. Calculate the \( \text{pH} \) of the following mixtures:
Answer:
(i) **\( 25 \text{ mL} \) of \( 0.1 \text{ M NaOH} + 25 \text{ mL} \) of \( 0.1 \text{ M HCl} \)**
Moles of `\( \text{H}_3\text{O}^+ \)` from `\( \text{HCl} \)` `= (0.1 \text{ M}) \times (25 \text{ mL}) = 0.1 \times \frac{25}{1000} = 0.0025 \text{ mol} \)`
Moles of `\( \text{OH}^- \)` from `\( \text{NaOH} \)` `= (0.1 \text{ M}) \times (25 \text{ mL}) = 0.1 \times \frac{25}{1000} = 0.0025 \text{ mol} \)`
Since moles of `\( \text{H}_3\text{O}^+ \)` and `\( \text{OH}^- \)` are equal, the solution is neutral after reaction.
Total volume `\( = 25 \text{ mL} + 25 \text{ mL} = 50 \text{ mL} \)`
At neutralization, `\( \text{pH} = 7 \)`.

(ii) **\( 10 \text{ mL} \) of \( 0.01 \text{ M H}_2\text{SO}_4 + 10 \text{ mL} \) of \( 0.01 \text{ M Ca(OH)}_2 \)**
`\( \text{H}_2\text{SO}_4 \)` is a strong acid, providing 2 `\( \text{H}^+ \)` ions per molecule.
Moles of `\( \text{H}_3\text{O}^+ \)` from `\( \text{H}_2\text{SO}_4 \)` `= 2 \times (0.01 \text{ M}) \times (10 \text{ mL}) = 2 \times 0.01 \times \frac{10}{1000} = 0.0002 \text{ mol} \)`
`\( \text{Ca(OH)}_2 \)` is a strong base, providing 2 `\( \text{OH}^- \)` ions per molecule.
Moles of `\( \text{OH}^- \)` from `\( \text{Ca(OH)}_2 \)` `= 2 \times (0.01 \text{ M}) \times (10 \text{ mL}) = 2 \times 0.01 \times \frac{10}{1000} = 0.0002 \text{ mol} \)`
Since moles of `\( \text{H}_3\text{O}^+ \)` and `\( \text{OH}^- \)` are equal, the solution is neutral after reaction.
Total volume `\( = 10 \text{ mL} + 10 \text{ mL} = 20 \text{ mL} \)`
At neutralization, `\( \text{pH} = 7 \)`.

(iii) **\( 10 \text{ mL} \) of \( 0.1 \text{ M H}_2\text{SO}_4 + 10 \text{ mL} \) of \( 0.1 \text{ M KOH} \)**
Moles of `\( \text{H}_3\text{O}^+ \)` from `\( \text{H}_2\text{SO}_4 \)` `= 2 \times (0.1 \text{ M}) \times (10 \text{ mL}) = 2 \times 0.1 \times \frac{10}{1000} = 0.002 \text{ mol} \)`
Moles of `\( \text{OH}^- \)` from `\( \text{KOH} \)` `= (0.1 \text{ M}) \times (10 \text{ mL}) = 0.1 \times \frac{10}{1000} = 0.001 \text{ mol} \)`
Excess moles of `\( \text{H}_3\text{O}^+ \)` `= 0.002 - 0.001 = 0.001 \text{ mol} \)`
Total volume `\( = 10 \text{ mL} + 10 \text{ mL} = 20 \text{ mL} \)`
Concentration of excess `\( [\text{H}_3\text{O}^+] = \frac{0.001 \text{ mol}}{20 \text{ mL}} = \frac{0.001}{0.020 \text{ L}} = 0.05 \text{ M} \)`
`\( \text{pH} = -\log[\text{H}_3\text{O}^+] = -\log(0.05) = -\log(5 \times 10^{-2}) \)`
`\( \text{pH} = -(\log 5 + \log 10^{-2}) = -(\log 5 - 2) = - (0.699 - 2) = -(-1.301) = 1.301 \)`
**pH of the mixture `\( \approx 1.3 \)`**
In simple words: For each mix, we first find out how many active acid or base particles there are. If they cancel each other out, the pH is 7. If one is left over, we calculate its amount in the total liquid to find the final pH.

🎯 Exam Tip: For pH calculations involving mixtures, always calculate the moles of H+ and OH- first to determine if there is an excess of acid or base, and then use the total volume to find the final concentration.

Question 44. The solubility product of \( \text{Ag}_2\text{CrO}_4 \) and \( \text{AgBr} \) are \( 1.1 \times 10^{-12} \) and \( 5.0 \times 10^{-13} \) respectively. Calculate the ratio of their saturated solutions.
Answer:
**For \( \text{Ag}_2\text{CrO}_4 \):**
The dissociation is: `\( \text{Ag}_2\text{CrO}_4(\text{s}) \rightleftharpoons 2\text{Ag}^+(\text{aq}) + \text{CrO}_4^{2-}(\text{aq}) \)`
Let `\( S \)` be the molar solubility of `\( \text{Ag}_2\text{CrO}_4 \)`.
`\( [\text{Ag}^+] = 2S \)`
`\( [\text{CrO}_4^{2-}] = S \)`
The solubility product `\( K_{sp} \)` is:
`\( K_{sp}(\text{Ag}_2\text{CrO}_4) = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2S)^2(S) = 4S^3 \)`
Given `\( K_{sp}(\text{Ag}_2\text{CrO}_4) = 1.1 \times 10^{-12} \)`
`\( 1.1 \times 10^{-12} = 4S^3 \)`
`\( S^3 = \frac{1.1 \times 10^{-12}}{4} = 0.275 \times 10^{-12} \)`
`\( S = \sqrt[3]{0.275 \times 10^{-12}} = (0.275)^{1/3} \times 10^{-4} \)`
`\( S \approx 0.650 \times 10^{-4} = 6.5 \times 10^{-5} \text{ M} \)`

**For \( \text{AgBr} \):**
The dissociation is: `\( \text{AgBr}(\text{s}) \rightleftharpoons \text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \)`
Let `\( S' \)` be the molar solubility of `\( \text{AgBr} \)`.
`\( [\text{Ag}^+] = S' \)`
`\( [\text{Br}^-] = S' \)`
The solubility product `\( K_{sp} \)` is:
`\( K_{sp}(\text{AgBr}) = [\text{Ag}^+][\text{Br}^-] = (S')(S') = (S')^2 \)`
Given `\( K_{sp}(\text{AgBr}) = 5.0 \times 10^{-13} \)`
`\( (S')^2 = 5.0 \times 10^{-13} = 50 \times 10^{-14} \)`
`\( S' = \sqrt{50 \times 10^{-14}} = 7.07 \times 10^{-7} \text{ M} \)`

**Ratio of their saturated solutions (solubilities):**
`\( \text{Ratio} = \frac{S(\text{Ag}_2\text{CrO}_4)}{S'(\text{AgBr})} = \frac{6.5 \times 10^{-5}}{7.07 \times 10^{-7}} \)`
`\( \text{Ratio} = \frac{6.5 \times 10^{-5}}{0.0707 \times 10^{-5}} \approx 91.93 \)`
The ratio of the molar solubilities of `\( \text{Ag}_2\text{CrO}_4 \)` to `\( \text{AgBr} \)` is approximately `\( 91.9 \)`.
In simple words: We calculate how much of each silver compound dissolves in water by using their solubility product values. Then, we divide these amounts to find how much more soluble one is compared to the other.

🎯 Exam Tip: Be careful with the stoichiometry when setting up the \( K_{sp} \) expression (e.g., \( (2S)^2(S) \) for `\( \text{Ag}_2\text{CrO}_4 \)`). Double-check the square root or cube root calculations.

Question 45. The ionization constant of benzoic acid is \( 6.46 \times 10^{-5} \) and \( \text{K}_\text{sp} \) for silver benzoate is \( 2.5 \times 10^{-13} \). How many times is silver benzoate more soluble in a buffer of \( \text{pH } 3.9 \) compared to its solubility in pure water?
Answer:
**1. Solubility of Silver Benzoate in Pure Water:**
Let the solubility of silver benzoate (`\( \text{AgC}_6\text{H}_5\text{COO} \)` or `\( \text{AgBz} \)` for simplicity) be `\( x' \)`.
`\( \text{AgBz}(\text{s}) \rightleftharpoons \text{Ag}^+(\text{aq}) + \text{Bz}^-(\text{aq}) \)`
`\( K_{sp} = [\text{Ag}^+][\text{Bz}^-] = (x')(x') = (x')^2 \)`
Given `\( K_{sp} = 2.5 \times 10^{-13} \)`
`\( (x')^2 = 2.5 \times 10^{-13} = 25 \times 10^{-14} \)`
`\( x' = \sqrt{25 \times 10^{-14}} = 5 \times 10^{-7} \text{ mol/L} \)`
So, solubility in pure water is `\( 5 \times 10^{-7} \text{ mol/L} \)`.

**2. Solubility of Silver Benzoate in a Buffer of \( \text{pH } 3.9 \):**
In an acidic buffer, the benzoate ion (`\( \text{Bz}^- \)` or `\( \text{C}_6\text{H}_5\text{COO}^- \)` ) reacts with `\( \text{H}^+ \)` ions to form undissociated benzoic acid (`\( \text{HBz} \)` or `\( \text{C}_6\text{H}_5\text{COOH} \)`):
`\( \text{Bz}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightleftharpoons \text{HBz}(\text{aq}) \)`
The total solubility (`\( x \)` ) of silver benzoate in the buffer will be `\( x = [\text{Ag}^+] \)` and also `\( x = [\text{Bz}^-] + [\text{HBz}] \)`.
From `\( \text{pH} = 3.9 \)`, we can find `\( [\text{H}^+] \)`:
`\( [\text{H}^+] = 10^{-\text{pH}} = 10^{-3.9} \approx 1.259 \times 10^{-4} \text{ M} \)`
The ionization of benzoic acid is: `\( \text{HBz} \rightleftharpoons \text{H}^+ + \text{Bz}^- \)`
`\( K_a = \frac{[\text{H}^+][\text{Bz}^-]}{[\text{HBz}]} \)`
Given `\( K_a = 6.46 \times 10^{-5} \)`.
Rearranging for the ratio `\( \frac{[\text{HBz}]}{[\text{Bz}^-]} \)`:
`\( \frac{[\text{HBz}]}{[\text{Bz}^-]} = \frac{[\text{H}^+]}{K_a} = \frac{1.259 \times 10^{-4}}{6.46 \times 10^{-5}} \approx 1.949 \)`
So, `\( [\text{HBz}] = 1.949 [\text{Bz}^-] \)`

Now, the total concentration of benzoate species is `\( x = [\text{Bz}^-] + [\text{HBz}] \)`:
`\( x = [\text{Bz}^-] + 1.949 [\text{Bz}^-] = 2.949 [\text{Bz}^-] \)`
`\( [\text{Bz}^-] = \frac{x}{2.949} \)`

From the `\( K_{sp} \)` expression, `\( K_{sp} = [\text{Ag}^+][\text{Bz}^-] \)`.
Since `\( [\text{Ag}^+] = x \)` (the total solubility), we have:
`\( K_{sp} = (x)\left(\frac{x}{2.949}\right) \)`
`\( K_{sp} = \frac{x^2}{2.949} \)`
`\( x^2 = K_{sp} \times 2.949 \)`
`\( x^2 = (2.5 \times 10^{-13}) \times 2.949 = 7.3725 \times 10^{-13} \)`
`\( x = \sqrt{7.3725 \times 10^{-13}} = \sqrt{73.725 \times 10^{-14}} \)`
`\( x \approx 8.586 \times 10^{-7} \text{ mol/L} \)`
So, solubility in buffer is `\( 8.586 \times 10^{-7} \text{ mol/L} \)`.

**3. Ratio of Solubilities:**
`\( \text{Ratio} = \frac{\text{Solubility in buffer}}{\text{Solubility in pure water}} = \frac{8.586 \times 10^{-7}}{5 \times 10^{-7}} \approx 1.717 \)`
Therefore, silver benzoate is approximately `\( 1.72 \)` times more soluble in the buffer of `\( \text{pH } 3.9 \)` compared to its solubility in pure water. This is an unexpected result as solubility usually increases in buffers if the common ion is removed. Let's recheck the calculation. `\( \text{pH } 3.9 \)` means `\( [\text{H}^+] \)` is high, which shifts `\( \text{Bz}^- + \text{H}^+ \rightleftharpoons \text{HBz} \)` to the right, decreasing `\( [\text{Bz}^-] \)` and thus increasing `\( \text{AgBz} \)` solubility. The calculation seems consistent with this logic.
In simple words: We find how much silver benzoate dissolves in plain water first. Then, we figure out how much dissolves in a special "buffer" liquid with a certain acidity (pH). The buffer makes the silver benzoate dissolve a bit more, because the acid in the buffer uses up one of the parts of the silver benzoate, letting more dissolve.

🎯 Exam Tip: For solubility in buffers, remember that the solubility increases if one of the ions reacts with the buffer components (e.g., a basic anion reacting with H+ in an acidic buffer). Use the Ka and pH to determine the distribution of the conjugate acid/base forms.

Question 46. In a reaction, \( \text{A} + 2\text{B} \rightarrow 2\text{C} + \text{D} \), \( \text{A} \) and \( \text{B} \) are heated in a flask at \( 25^\circ\text{C} \). The initial concentration of \( \text{B} \) is \( 1.5 \) times the concentration of \( \text{A} \). At equilibrium, the concentration of \( \text{A} \) and \( \text{D} \) is the same. Calculate the equilibrium constant at this temperature.
Answer:
Let the initial concentration of `\( \text{A} \)` be `\( [\text{A}]_{\text{initial}} = a \)`.
Then the initial concentration of `\( \text{B} \)` is `\( [\text{B}]_{\text{initial}} = 1.5a \)`.
The reaction is: `\( \text{A} + 2\text{B} \rightleftharpoons 2\text{C} + \text{D} \)`

Let `\( x \)` be the change in concentration of `\( \text{A} \)` at equilibrium.

At equilibrium, `\( [\text{A}] = [\text{D}] \)`
So, `\( a-x = x \)`
`\( a = 2x \)`
This implies `\( x = \frac{a}{2} = 0.5a \)`.

Now substitute `\( x = 0.5a \)` into the equilibrium concentrations:
`\( [\text{A}] = a - 0.5a = 0.5a \)`
`\( [\text{B}] = 1.5a - 2(0.5a) = 1.5a - a = 0.5a \)`
`\( [\text{C}] = 2(0.5a) = a \)`
`\( [\text{D}] = 0.5a \)`

The equilibrium constant `\( K_c \)` for the reaction is:
`\( K_c = \frac{[\text{C}]^2[\text{D}]}{[\text{A}][\text{B}]^2} \)`
Substitute the equilibrium concentrations:
`\( K_c = \frac{(a)^2(0.5a)}{(0.5a)(0.5a)^2} \)`
`\( K_c = \frac{a^2 \times 0.5a}{0.5a \times 0.25a^2} \)`
`\( K_c = \frac{0.5a^3}{0.125a^3} \)`
`\( K_c = \frac{0.5}{0.125} = 4 \)`
The equilibrium constant `\( K_c \)` at `\( 25^\circ\text{C} \)` is `\( 4 \)`.
In simple words: We set up the reaction with starting amounts of A and B, knowing B is 1.5 times A. At the end, the amount of A left is the same as the amount of D created. Using these facts, we can calculate how much of each substance is present at equilibrium and then find the equilibrium constant, which tells us how much product is formed at balance.

🎯 Exam Tip: Always set up an ICE (Initial, Change, Equilibrium) table to clearly track concentrations. Pay close attention to the stoichiometric coefficients when calculating the change and when writing the `\( K_c \)` expression.

Question 47. At 25°C and one atmospheric pressure, 20% of \( \text{N}_2\text{O}_4 \) is dissociated into \( \text{NO}_2 \). Calculate equilibrium constant \( \text{K}_\text{p} \) for this equilibrium.
Answer: The reaction for the dissociation of \( \text{N}_2\text{O}_4 \) into \( \text{NO}_2 \) is:
\( \text{N}_2\text{O}_4 (\text{g}) \rightleftharpoons 2\text{NO}_2 (\text{g}) \)
If we start with 1 mole of \( \text{N}_2\text{O}_4 \) and 0 moles of \( \text{NO}_2 \):
Initial moles: \( \text{N}_2\text{O}_4 = 1 \), \( \text{NO}_2 = 0 \)
Since 20% of \( \text{N}_2\text{O}_4 \) dissociates, \( \text{x} = 0.2 \).
Moles at equilibrium:
\( \text{N}_2\text{O}_4 = 1 - \text{x} = 1 - 0.2 = 0.8 \) moles
\( \text{NO}_2 = 2\text{x} = 2 \times 0.2 = 0.4 \) moles
Total moles at equilibrium \( = 0.8 + 0.4 = 1.2 \) moles.
The total pressure is 1 atmosphere.
Partial pressure of \( \text{N}_2\text{O}_4 \) (\( \text{P}_{\text{N}_2\text{O}_4} \)) \( = \frac{ \text{Moles of N}_2\text{O}_4 }{ \text{Total moles} } \times \text{Total pressure} = \frac{ 0.8 }{ 1.2 } \times 1 = \frac{ 2 }{ 3 } \)
Partial pressure of \( \text{NO}_2 \) (\( \text{P}_{\text{NO}_2} \)) \( = \frac{ \text{Moles of NO}_2 }{ \text{Total moles} } \times \text{Total pressure} = \frac{ 0.4 }{ 1.2 } \times 1 = \frac{ 1 }{ 3 } \)
The equilibrium constant \( \text{K}_\text{p} \) is given by:
\( \text{K}_\text{p} = \frac{ (\text{P}_{\text{NO}_2})^2 }{ \text{P}_{\text{N}_2\text{O}_4} } \)
Substitute the partial pressures:
\( \text{K}_\text{p} = \frac{ (1/3)^2 }{ 2/3 } \)
\( \text{K}_\text{p} = \frac{ 1/9 }{ 2/3 } \)
\( \text{K}_\text{p} = \frac{ 1 }{ 9 } \times \frac{ 3 }{ 2 } \)
\( \text{K}_\text{p} = \frac{ 3 }{ 18 } = \frac{ 1 }{ 6 } \)
\( \text{K}_\text{p} \approx 0.167 \)
In simple words: We calculated how many moles of each gas were present after some \( \text{N}_2\text{O}_4 \) broke apart. Then, we used these amounts to find each gas's share of the total pressure. Finally, we put these partial pressures into the \( \text{K}_\text{p} \) formula to get the equilibrium constant, which tells us how far the reaction goes towards products.

🎯 Exam Tip: Remember to calculate the total moles at equilibrium before finding the partial pressures when dealing with gas-phase equilibrium constants.

Question 48. At normal temperature and pressure, 5.29 mL hydrogen reacts with 0.040 gm iodine at 444°C and gives 6.35 mL HI. Calculate equilibrium constant for the synthesis of HI at this temperature.
Answer: The reaction for the synthesis of HI is:
\( \text{H}_2 (\text{g}) + \text{I}_2 (\text{g}) \rightleftharpoons 2\text{HI} (\text{g}) \)
We are given the initial and equilibrium volumes/masses:
Initial volume of \( \text{H}_2 = 5.29 \) mL
Initial mass of \( \text{I}_2 = 0.040 \) gm
Since iodine is a gas at 444°C, we need to convert its mass to volume or moles to compare with hydrogen. However, the problem gives an equilibrium volume for HI, so it is easier to work with volumes if we assume ideal gas behavior and that volumes are proportional to moles.
Let's assume initial volume of \( \text{H}_2 = 5.29 \) and initial volume of \( \text{I}_2 \) needs to be found.
The problem states that \( \text{I}_2 \) is 0.040 gm. To find its volume, we need the density of iodine vapor at 444°C and normal pressure, which is not provided directly in a usable way for volume calculation here. However, the provided solution uses values 5.29 and 3.53 for initial volumes, and 6.35 for HI at equilibrium. This implies the 0.040 gm of iodine corresponds to 3.53 mL at the given conditions, which is a common simplification in such problems where some initial data might be slightly off or implied.
Let's use the volumes as moles (or concentrations) for \( \text{K}_\text{c} \) calculation directly, as commonly done when units cancel out.
Let \( \text{V}_{\text{H}_2} = 5.29 \) and \( \text{V}_{\text{I}_2} = 3.53 \).
At equilibrium, volume of HI formed \( = 6.35 \).
Since 2 moles of HI are formed from 1 mole of \( \text{H}_2 \) and 1 mole of \( \text{I}_2 \), the change in moles (or volume) for HI is \( 2\text{x} \).
So, \( 2\text{x} = 6.35 \implies \text{x} = 3.175 \).
At equilibrium:
Volume of \( \text{H}_2 = 5.29 - \text{x} = 5.29 - 3.175 = 2.115 \)
Volume of \( \text{I}_2 = 3.53 - \text{x} = 3.53 - 3.175 = 0.355 \)
Volume of \( \text{HI} = 2\text{x} = 6.35 \)
The equilibrium constant \( \text{K}_\text{c} \) for the reaction is:
\( \text{K}_\text{c} = \frac{ [\text{HI}]^2 }{ [\text{H}_2][\text{I}_2] } \)
Using the equilibrium volumes (which can be treated as proportional to concentrations in this case, as volume terms cancel out):
\( \text{K}_\text{c} = \frac{ (6.35)^2 }{ (2.115) \times (0.355) } \)
\( \text{K}_\text{c} = \frac{ 40.3225 }{ 0.750825 } \)
\( \text{K}_\text{c} \approx 53.76 \)
In simple words: We used the given volumes of hydrogen, iodine, and hydrogen iodide at equilibrium to find the constant that describes how much product is formed when the reaction settles down. The calculation shows that at this temperature, a lot more product (HI) is favored.

🎯 Exam Tip: For reactions where the number of moles of gaseous reactants equals the number of moles of gaseous products, \( \text{K}_\text{c} \) (concentration constant) and \( \text{K}_\text{p} \) (pressure constant) are numerically equal.

Question 49. The molecular mass of \( \text{PCl}_5 \) is 208.3. At 200°C, the mass of partial dissociated vapours is 62 times that of mass of hydrogen. Calculate degree of dissociation of \( \text{PCl}_5 \).
Answer: The reaction for the dissociation of \( \text{PCl}_5 \) is:
\( \text{PCl}_5 (\text{g}) \rightleftharpoons \text{PCl}_3 (\text{g}) + \text{Cl}_2 (\text{g}) \)
In this reaction, one mole of reactant forms two moles of products. So, \( \text{n} = 2 \).
The degree of dissociation \( \text{x} \) can be calculated using the formula related to vapor densities:
\( \text{x} = \frac{ \text{D} - \text{d} }{ \text{d} (\text{n} - 1) } \)
Where:
\( \text{D} \) is the theoretical vapor density of the undissociated substance.
\( \text{d} \) is the observed vapor density of the dissociated mixture at equilibrium.
\( \text{n} \) is the number of moles of gaseous products formed from one mole of reactant.
Given:
Molecular mass of \( \text{PCl}_5 = 208.3 \)
Theoretical vapor density \( \text{D} = \frac{ \text{Molecular mass} }{ 2 } = \frac{ 208.3 }{ 2 } = 104.15 \)
The problem states that the mass of partial dissociated vapors is 62 times that of the mass of hydrogen. Since the mass of hydrogen is 1 (in atomic mass units or relative to other gases for vapor density comparison), this means the observed vapor density \( \text{d} = 62 \).
Now, substitute the values into the formula:
\( \text{x} = \frac{ 104.15 - 62 }{ 62 (2 - 1) } \)
\( \text{x} = \frac{ 42.15 }{ 62 \times 1 } \)
\( \text{x} = \frac{ 42.15 }{ 62 } \)
\( \text{x} \approx 0.6798 \)
To express as a percentage:
Percentage degree of dissociation \( = 0.6798 \times 100 \approx 67.98\% \approx 68\% \)
In simple words: We found how much \( \text{PCl}_5 \) breaks down into smaller parts by comparing its normal weight in vapor form to its weight when it's partly broken. We calculated that about 68% of the \( \text{PCl}_5 \) has dissociated at that temperature.

🎯 Exam Tip: Always remember the definition of theoretical vapor density (molecular mass divided by 2) and observed vapor density (from experimental data) for dissociation problems.

Question 50. For the system \( \text{H}_2\text{O(g)} + \text{CO(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{CO}_2\text{(g)} \), if at equilibrium, \( \text{H}_2\text{O} = 0.6 \) (since 40% reacts), \( \text{CO} = 0.6 \), \( \text{H}_2 = 0.4 \) and \( \text{CO}_2 = 0.4 \), calculate the equilibrium constant \( \text{K}_\text{c} \).
Answer: The chemical reaction at equilibrium is:
\( \text{H}_2\text{O}(\text{g}) + \text{CO}(\text{g}) \rightleftharpoons \text{H}_2(\text{g}) + \text{CO}_2(\text{g}) \)
The equilibrium concentrations (or values proportional to concentrations, like moles if volume is constant) are given:
\( [\text{H}_2\text{O}] = 0.6 \)
\( [\text{CO}] = 0.6 \)
\( [\text{H}_2] = 0.4 \)
\( [\text{CO}_2] = 0.4 \)
The expression for the equilibrium constant \( \text{K}_\text{c} \) is:
\( \text{K}_\text{c} = \frac{ [\text{H}_2][\text{CO}_2] }{ [\text{H}_2\text{O}][\text{CO}] } \)
Now, substitute the given equilibrium values into the expression:
\( \text{K}_\text{c} = \frac{ (0.4) \times (0.4) }{ (0.6) \times (0.6) } \)
\( \text{K}_\text{c} = \frac{ 0.16 }{ 0.36 } \)
To simplify the fraction, we can divide both numerator and denominator by 0.04:
\( \text{K}_\text{c} = \frac{ 0.16 \div 0.04 }{ 0.36 \div 0.04 } = \frac{ 4 }{ 9 } \)
\( \text{K}_\text{c} \approx 0.44 \)
In simple words: We took the amounts of all the substances when the reaction stopped changing and put them into a special formula. This gave us a number, \( \text{K}_\text{c} \), which shows how much of the original substances turned into new ones.

🎯 Exam Tip: For reactions where the total number of moles of gaseous reactants equals the total number of moles of gaseous products, the volume term (if using moles/volume) cancels out in the \( \text{K}_\text{c} \) expression.