Maharashtra Board Class 12 Chemistry Chapter 3 Ionic Equilibria Solutions

Class 12 Chemistry Chapter 3 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 3 Exercise Solutions

1. Choose The Most Correct Answer:

Question i. The pH of 10\(^{-8}\) M of HCl is
(a) 8
(b) 7
(c) less than 7
(d) greater than 7
Answer: (c) less than 7
In simple words: Although 10\(^{-8}\) M HCl suggests pH 8, due to the autoionization of water, the actual pH will be slightly less than 7, as HCl is an acid.

🎯 Exam Tip: Remember to consider the autoionization of water for very dilute acid solutions. For acidic solutions, pH is always less than 7, even if calculated value is higher. This demonstrates a deeper understanding of pH calculation in dilute solutions.

Question ii. Which of the following solution will have pH value equal to 1.0?
(a) 50 mL of 0.1M HCl + 50mL of 0.1 M NaOH
(b) 60 mL of 0.1M HCl + 40mL of 0.1 M NaOH
(c) 20 mL of 0.1M HCl + 80mL of 0.1 M NaOH
(d) 75 mL of 0.2M HCl + 25mL of 0.2 M NaOH
Answer: (d) 75 mL of 0.2M HCl + 25mL of 0.2 M NaOH
In simple words: To achieve a pH of 1.0, a solution must have a [H\(^{+}\)] of 0.1 M. The correct option results in an excess of 0.1 M HCl after neutralization, leading to this pH.

🎯 Exam Tip: To solve this, calculate the millimoles of H\(^{+}\) and OH\(^{-}\) in each option. The difference divided by the total volume gives the concentration of the excess ion. For pH=1, [H\(^{+}\)] must be 0.1 M.

Question iii. Which of the following is a buffer solution ?
(a) CH\(_{3}\)COONa + NaCl in water
(b) CH\(_{3}\)COOH + HCl in water
(c) CH\(_{3}\)COOH + CH\(_{3}\)COONa in water
(d) HCl + NH\(_{4}\)Cl in water
Answer: (c) CH\(_{3}\)COOH + CH\(_{3}\)COONa in water
In simple words: A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Acetic acid (weak acid) and sodium acetate (its conjugate base) form a classic acidic buffer.

🎯 Exam Tip: Understand the definition of a buffer: a mixture of a weak acid and its salt with a strong base (acidic buffer), or a weak base and its salt with a strong acid (basic buffer). This is a fundamental concept for buffer chemistry.

Question iv. The solubility product of a sparingly soluble salt AX is 5.2 * 10\(^{-13}\). Its solubility in mol dm\(^{-3}\) is
(a) 7.2 * 10\(^{-7}\)
(b) 1.35 * 10\(^{-4}\)
(c) 7.2 * 10\(^{-8}\)
(d) 13.5 * 10\(^{-8}\)
Answer: (a) 7.2 * 10\(^{-7}\)
In simple words: For a salt of type AX, its solubility (s) is equal to the square root of its solubility product (K\(_{sp}\)). So, s = \(\sqrt{5.2 \times 10^{-13}}\).

🎯 Exam Tip: For a 1:1 sparingly soluble salt like AX, K\(_{sp}\) = s\(^2\). Therefore, solubility s = \(\sqrt{K_{sp}}\). Make sure to correctly perform the square root calculation for scientific notation.

Question v. Blood in human body is highly buffered at pH of
(a) 7.4
(b) 7.0
(c) 6.9
(d) 8.1
Answer: (a) 7.4
In simple words: Human blood maintains a very narrow pH range, primarily around 7.4, due to the presence of various buffer systems like the bicarbonate buffer system.

🎯 Exam Tip: Recognize that the human body's physiological pH is crucial for biochemical processes. The pH of blood (around 7.4) is a commonly tested fact related to buffer systems.

Question vi. The conjugate base of [Zn(H\(_{2}\)O)\(_{4}\)]\(^{2+}\) is
(a) [Zn(H\(_{2}\)O)\(_{4}\)]\(^{2+}\) NH\(_{3}\)
(b) [Zn(H\(_{2}\)O)\(_{3}\)]\(^{2+}\)
(c) [Zn(H\(_{2}\)O)\(_{3}\)OH]\(^{+}\)
(d) [Zn(H\(_{2}\)O)H]\(^{3+}\)
Answer: (c) [Zn(H\(_{2}\)O)\(_{3}\)OH]\(^{+}\)
In simple words: A conjugate base is formed when an acid loses a proton (H\(^{+}\)). Here, the complex ion acts as an acid by donating a proton from one of its coordinated water molecules.

🎯 Exam Tip: For complex ions, a coordinated water molecule can act as an acid, donating an H\(^{+}\) to form a hydroxyl group. Removing H\(^{+}\) from H\(_{2}\)O gives OH\(^{-}\), hence the charge changes from 2+ to 1+.

Question vii. For pH > 7 the hydronium ion concentration would be
(a) 10\(^{-7}\) M
(b) < 10\(^{-7}\) M
(c) > 10\(^{-7}\) M
(d) \(\ge\) 10\(^{-7}\) M
Answer: (b) < 10\(^{-7}\) M
In simple words: A pH greater than 7 indicates a basic solution. In basic solutions, the concentration of hydronium ions ([H\(_{3}\)O\(^{+}\)]) is always less than 10\(^{-7}\) M.

🎯 Exam Tip: Remember the relationship: pH < 7 is acidic ([H\(_{3}\)O\(^{+}\)] > 10\(^{-7}\) M), pH = 7 is neutral ([H\(_{3}\)O\(^{+}\)] = 10\(^{-7}\) M), and pH > 7 is basic ([H\(_{3}\)O\(^{+}\)] < 10\(^{-7}\) M). This is a foundational concept for pH and pOH calculations.

2. Answer The Following In One Sentence:

Question i. Why cations are Lewis acids ?
Answer: Since cations are deficient of electrons they accept a pair of electrons, hence they are Lewis acids.
In simple words: Cations have a positive charge because they have lost electrons, making them electron-deficient. To achieve a stable electron configuration, they tend to accept electron pairs, which is the definition of a Lewis acid.

🎯 Exam Tip: The key to defining a Lewis acid is its ability to accept an electron pair. Cations, having lost electrons, are prime examples of species that readily accept electron pairs.

Question ii. Why is KCl solution neutral to litmus?
Answer: 1. Since KCl is a salt of strong base KOH and strong acid HCl, it does not undergo hydrolysis in its aqueous solution.
2. Due to strong acid and strong base, concentrations [H\(_{3}\)O\(^{+}\)] = [OH\(^{-}\)] and the solution is neutral.
In simple words: KCl is formed from a strong acid (HCl) and a strong base (KOH). When dissolved in water, neither the K\(^{+}\) nor the Cl\(^{-}\) ions react significantly with water (hydrolysis), so the solution remains neutral, with equal concentrations of H\(_{3}\)O\(^{+}\) and OH\(^{-}\) ions.

🎯 Exam Tip: Salts formed from strong acids and strong bases do not hydrolyze and thus produce neutral solutions. This is a common application of hydrolysis concepts.

Question iii. How are basic buffer solutions prepared?
Answer: 1. Basic buffer solution is prepared by mixing aqueous solutions of a weak base like NH\(_{4}\)OH and its salt of a strong acid like NH\(_{4}\)Cl.
2. A weak base is selected according to the required pH or pOH of the solution and dissociation constant of the weak base.
In simple words: A basic buffer is made by combining a weak base, like ammonia, with a salt of that weak base and a strong acid, such as ammonium chloride. The choice of weak base depends on the desired pH or pOH of the buffer.

🎯 Exam Tip: Clearly state the components: a weak base and its salt with a strong acid. Mentioning specific examples like NH\(_{4}\)OH and NH\(_{4}\)Cl is helpful for illustrating the concept.

Question iv. Dissociation constant of acetic acid is 1.8 * 10\(^{-5}\). Calculate percent dissociation of acetic acid in 0.01 M solution.
Answer: Given : K\(_{a}\) = 1.8 * 10\(^{-5}\); C = 0.01 M
Percent dissociation = ?
K\(_{a}\) = \(\frac{Ca^2}{1-\alpha}\) \(\approx\) \(Ca^2\) (since \(\alpha\) is very small for weak acids)
\(\therefore\) \(\alpha\) = \(\sqrt{\frac{K_a}{C}}\)
= \(\left(\frac{1.8 * 10^{-5}}{0.01}\right)^{\frac{1}{2}}\)
= \(\left(1.8 * 10^{-3}\right)^{\frac{1}{2}}\)
= \(\sqrt{0.0018}\)
= 0.04242
Percent dissociation = \(\alpha\) * 100
= 0.04242 * 100
= 4.242%
In simple words: To find the percent dissociation, we first calculate the degree of dissociation (\(\alpha\)) using the formula \(\alpha\) = \(\sqrt{K_a/C}\) for weak acids. Then, multiply \(\alpha\) by 100 to get the percentage.

🎯 Exam Tip: For weak acids, the approximation K\(_{a}\) \(\approx\) C\(\alpha\)\(^{2}\) is often valid if \(\alpha\) is small (less than 5%). Clearly show the calculation steps for \(\alpha\) and then for percent dissociation. Ensure correct use of exponents.

Question v. Write one property of a buffer solution.
Answer: Properties (or advantages) of a buffer solution :
• The pH of a buffer solution is maintained appreciably constant.
• By addition of a small amount of an acid or a base pH does not change.
• On dilution with water, pH of the solution doesn't change.
In simple words: A key property of a buffer solution is its ability to resist significant changes in pH when small amounts of acid or base are added, or upon dilution.

🎯 Exam Tip: The most important property to mention is the resistance to pH change upon addition of small amounts of acid/base or dilution. Any one of the listed points is sufficient for a one-sentence answer.

Question vi. The pH of a solution is 6.06. Calculate its H\(^{+}\) ion concentration.
Answer: Given: pH = 6.06, [H\(^{+}\)] = ?
pH = -log\(_{10}\) [H\(^{+}\)]
6.06 = -log\(_{10}\) [H\(^{+}\)]
log\(_{10}\) [H\(^{+}\)] = -6.06
[H\(^{+}\)] = Antilog(-6.06)
[H\(^{+}\)] = Antilog(\(\overline{7}\).94)
[H\(^{+}\)] = 8.71 * 10\(^{-7}\) M
In simple words: To find the H\(^{+}\) ion concentration from a given pH, you need to calculate the antilog of the negative pH value, i.e., [H\(^{+}\)] = 10\(^{-\text{pH}}\).

🎯 Exam Tip: Remember the inverse relationship: if pH = -log[H\(^{+}\)], then [H\(^{+}\)] = 10\(^{-\text{pH}}\). Practice using a calculator for antilogarithms (usually 10\(^{x}\) function) to avoid errors.

Question vii. Calculate the pH of 0.01 M sulphuric acid.
Answer: Given : C = 0.01 M H\(_{2}\)SO\(_{4}\), pH = ?
H\(_{2}\)SO\(_{4(aq)}\) \(\implies\) 2H\(_{(aq)}^{+}\) + SO\(_{4(aq)}^{2-}\)
\(\therefore\) [H\(_{3}\)O\(^{+}\)] = 2 * 0.01 = 0.02 M
PH = -log\(_{10}\) [H\(_{3}\)O\(^{+}\)]
= -log\(_{10}\) 0.02
= -(log\(_{10}\) 2 + log\(_{10}\) 10\(^{-2}\))
= -(0.3010 - 2)
= -(-1.6990)
= 1.6990
pH = 1.6990.
In simple words: Sulphuric acid is a strong diprotic acid, meaning it releases two H\(^{+}\) ions per molecule. So, for a 0.01 M solution, the H\(^{+}\) concentration is 2 * 0.01 = 0.02 M. Then, calculate pH using pH = -log[H\(^{+}\)].

🎯 Exam Tip: For strong acids, assume complete dissociation. For polyprotic acids like H\(_{2}\)SO\(_{4}\), remember to multiply the molarity by the number of dissociable protons to find the total [H\(^{+}\)] before calculating pH.

Question viii. The dissociation of H\(_{2}\)S is suppressed in the presence of HCl. Name the phenomenon.
Answer: The weak dibasic acid H\(_{2}\)S is dissociated as follows :
H\(_{2}\)S\(_{(aq)}\) + 2H\(_{2}\)O\(_{(l)}\) \(\leftrightharpoons\) 2H\(_{3}\)O\(_{(aq)}^{+}\) + S\(_{(aq)}^{2-}\)
K\(_{a}\) = \(\frac{[H_3O^+]^2 \times [S^{2-}]}{[H_2S]}\)
When HCl is added, it increases the concentration of common ion H\(_{3}\)O\(^{+}\).
HCl\(_{(aq)}\) + H\(_{2}\)O\(_{(l)}\) \(\implies\) H\(_{3}\)O\(_{(aq)}^{+}\) + Cl\(_{(aq)}^{-}\)
Hence by Le Chatelier's principle, the equilibrium is shifted from right to left, suppressing the dissociation of weak electrolyte H\(_{2}\)S.
In simple words: This phenomenon is called the common ion effect. Adding HCl, a strong acid, increases the concentration of H\(_{3}\)O\(^{+}\) ions, which is a product of H\(_{2}\)S dissociation. According to Le Chatelier's principle, this excess H\(_{3}\)O\(^{+}\) shifts the H\(_{2}\)S equilibrium backward, reducing its dissociation.

🎯 Exam Tip: The common ion effect is a crucial concept. Clearly state the name of the phenomenon and explain how the added common ion (H\(_{3}\)O\(^{+}\) in this case) shifts the equilibrium of the weak electrolyte (H\(_{2}\)S) to reduce its dissociation, as per Le Chatelier's principle.

Question ix. Why is it necessary to add H\(_{2}\)SO\(_{4}\) while preparing the solution of CuSO\(_{4}\)?
Answer: CuSO\(_{4}\) is a salt of strong acid H\(_{2}\)SO\(_{4}\) and weak base Cu(OH)\(_{2}\). CuSO\(_{4}\) in aqueous solution undergoes hydrolysis and forms a precipitate of Cu(OH)\(_{2}\) and solution becomes turbid.
CuSO\(_{4}\) + 2H\(_{2}\)O \(\leftrightharpoons\) Cu(OH)\(_{2}\)\(\downarrow\) + H\(_{2}\)SO\(_{4}\)
OR
CuSO\(_{4}\) + 4H\(_{2}\)O \(\leftrightharpoons\) Cu(OH)\(_{2}\) + 2H\(_{3}\)O\(^{+}\) + SO\(_{4}^{2-}\)
When H\(_{2}\)SO\(_{4}\) is added, the hydrolysis equilibrium is shifted to left hand side and Cu(OH)\(_{2}\) dissolves giving clear solution.
In simple words: Copper sulfate undergoes hydrolysis in water to form insoluble copper hydroxide, making the solution turbid. Adding H\(_{2}\)SO\(_{4}\) (a product of hydrolysis) shifts the equilibrium to the left, preventing the formation of Cu(OH)\(_{2}\) precipitate and keeping the solution clear.

🎯 Exam Tip: Explain that CuSO\(_{4}\) hydrolyzes to form insoluble Cu(OH)\(_{2}\). The addition of H\(_{2}\)SO\(_{4}\) introduces a common ion (H\(_{3}\)O\(^{+}\)) or simply reverses the hydrolysis by providing the strong acid, thus preventing precipitation and ensuring a clear solution. Writing the equilibrium reaction is essential.

Question x. Classify the following buffers into different types :
a. CH\(_{3}\)COOH + CH\(_{3}\)COONa
b. NH\(_{4}\)OH + NH\(_{4}\)Cl
c. Sodium benzoate + benzoic acid
d. Cu(OH)\(_{2}\) + CuCl\(_{2}\)
Answer: (a) Acidic buffer (CH\(_{3}\)COOH + CH\(_{3}\)COONa)
(b) Basic buffer (NH\(_{4}\)OH + NH\(_{4}\)Cl)
(c) Acidic buffer (Sodium benzoate + benzoic acid)
(d) Basic buffer (Cu(OH)\(_{2}\) + CuCl\(_{2}\))
[Note: Cu(OH)\(_{2}\) being insoluble is not used to prepare a buffer solution.]
In simple words: Buffers are classified as acidic if they contain a weak acid and its salt, and basic if they contain a weak base and its salt. CH\(_{3}\)COOH/CH\(_{3}\)COONa and Sodium benzoate/benzoic acid are acidic. NH\(_{4}\)OH/NH\(_{4}\)Cl and Cu(OH)\(_{2}\)/CuCl\(_{2}\) are basic, though Cu(OH)\(_{2}\) is often excluded due to insolubility.

🎯 Exam Tip: To classify buffers, identify the components: weak acid/conjugate base for acidic buffers, and weak base/conjugate acid for basic buffers. Remember that the components must be soluble for effective buffer action.

3. Answer The Following In Brief:

Question i. What are acids and bases according to Arrhenius theory ?
Answer: According to Arrhenius theory :
Acid : It is a substance which contains hydrogen and on dissolving in water produces hydrogen ions (H\(^{+}\)) E.g. HCl
HCl\(_{(aq)}\) \(\leftrightharpoons\) H\(_{(aq)}^{+}\) + Cl\(_{(aq)}^{-}\)
Base : It is a substance which contains OH group and on dissolving in water produces hydroxyl ions (OH\(^{-}\)). E.g. NaOH
NaOH\(_{(aq)}\) \(\leftrightharpoons\) Na\(_{(aq)}^{+}\) + OH\(_{(aq)}^{-}\)
In simple words: The Arrhenius theory defines acids as substances that release H\(^{+}\) ions in water, and bases as substances that release OH\(^{-}\) ions in water.

🎯 Exam Tip: Clearly state the definition for both acids and bases according to Arrhenius, focusing on the release of H\(^{+}\) and OH\(^{-}\) ions in an aqueous solution. Provide a simple example for each.

Question ii. What is meant by conjugate acid-base pair?
Answer: Conjugate acid-base pair : A pair of an acid and a base differing by a proton is called a conjugate acid-base pair.
HCl\(_{(aq)}\) + H\(_{2}\)O\(_{(l)}\) \(\leftrightharpoons\) H\(_{3}\)O\(_{(aq)}^{+}\) + Cl\(_{(aq)}^{-}\)
Acid 1 Base 2 Acid 2 Base 1
\( \implies \)
conjugate acid-base pair 2
\( \implies \)
conjugate acid-base pair 1
In simple words: A conjugate acid-base pair consists of two species that are related by the gain or loss of a single proton (H\(^{+}\)). The acid has one more proton than its conjugate base.

🎯 Exam Tip: The definition should clearly state that the pair differs by *one proton*. Providing a reaction example (like HCl/Cl\(^{-}\) and H\(_{2}\)O/H\(_{3}\)O\(^{+}\)) and labeling the pairs is crucial for demonstrating understanding.

Question iii. Label the conjugate acid-base pair in the following reactions
a. HCl + H\(_{2}\)O \(\leftrightharpoons\) H\(_{3}\)O\(^{+}\) + Cl\(^{-}\)
b. CO\(_{3}^{2-}\) + H\(_{2}\)O \(\leftrightharpoons\) OH\(^{-}\) + HCO\(_{3}^{-}\)
Answer: (a) HCl + H\(_{2}\)O \(\leftrightharpoons\) H\(_{3}\)O\(^{+}\) + Cl\(^{-}\)
Acid-1 Base-2 Acid-2 Base-1
(b) CO\(_{3}^{2-}\) + H\(_{2}\)O \(\leftrightharpoons\) OH\(^{-}\) + HCO\(_{3}^{-}\)
Base-1 Acid-2 Base-2 Acid-1
In simple words: In a reaction, an acid donates a proton to form its conjugate base, and a base accepts a proton to form its conjugate acid. We identify these pairs based on the transfer of a single H\(^{+}\).

🎯 Exam Tip: For each reaction, correctly identify which species is donating a proton (acid) and which is accepting (base). The resulting species are their respective conjugates. Labeling them systematically helps avoid confusion.

Question iv. Write a reaction in which water acts as a base.
Answer: H\(_{2}\)O\(_{(l)}\) + HCl \(\leftrightharpoons\) H\(_{3}\)O\(^{+}\) + Cl\(^{-}\)
Base\(_{1}\) Acid\(_{2}\) Acid\(_{1}\) Base\(_{2}\)
Since water accepts a proton, it acts as a base.
In simple words: Water can act as a base by accepting a proton from an acid like HCl, forming a hydronium ion (H\(_{3}\)O\(^{+}\)).

🎯 Exam Tip: To show water acting as a base, it must accept a proton from another species (an acid). The example with HCl is a classic illustration. Make sure to correctly balance the equation and charges.

Question v. Ammonia serves as a Lewis base whereas AlCl\(_{3}\) is Lewis acid. Explain.
Answer: • Since ammonia molecule, NH\(_{3}\) has a lone pair of electrons to donate it acts as a Lewis base.
• AlCl\(_{3}\) is a molecule with incomplete octet hence it is electron deficient and acts as a Lewis acid.
In simple words: Ammonia has a lone pair of electrons it can donate, making it a Lewis base. AlCl\(_{3}\) has an incomplete octet and can accept an electron pair, making it a Lewis acid.

🎯 Exam Tip: For Lewis acids and bases, focus on electron pair donation or acceptance. A Lewis base donates an electron pair (like NH\(_{3}\) with its lone pair), while a Lewis acid accepts an electron pair (like AlCl\(_{3}\) with its empty orbitals).

Question vi. Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid.
Answer: Given : C = 0.1 M; Dissociation = 5%, K\(_{a}\)=?
\(\alpha\) = \(\frac{\text{Percent dissociation}}{100}\)
= \(\frac{5}{100}\) = 0.05
K\(_{a}\) = \(\frac{Ca^2}{1-\alpha}\)
= \(\frac{0.1 \times (0.05)^2}{1-0.05}\)
= \(\frac{0.1 \times 0.0025}{0.95}\)
= \(\frac{0.00025}{0.95}\)
= 0.000263
= 2.63 * 10\(^{-4}\)
Dissociation constant = K\(_{a}\) = 2.63 * 10\(^{-4}\)
In simple words: First, convert the percent ionization into the degree of dissociation (\(\alpha\)). Then, use the formula for the dissociation constant K\(_{a}\) = \(\frac{C\alpha^2}{1-\alpha}\), plugging in the given concentration (C) and calculated \(\alpha\).

🎯 Exam Tip: Be careful with unit conversions (percentage to decimal for \(\alpha\)) and algebraic manipulation of the K\(_{a}\) expression. For accurate results, avoid the approximation (1-\(\alpha\)) \(\approx\) 1 unless specifically allowed or if \(\alpha\) is extremely small.

Question vii. Derive the relation pH + pOH = 14.
Answer: The ionic product of water, K\(_{w}\) is given by,
K\(_{w}\) = [H\(_{3}\)O\(^{+}\)] * [OH\(^{-}\)]
At 298 K, K\(_{w}\) = 1 * 10\(^{-14}\)
\(\therefore\) pK\(_{w}\) = -log\(_{10}\)K\(_{w}\) = -log\(_{10}\) (1 * 10\(^{-14}\)) = 14
\(\therefore\) [H\(_{3}\)O\(^{+}\)] * [OH\(^{-}\)] = 1 * 10\(^{-14}\)
Taking logarithm to base 10 of both sides,
log\(_{10}\) [H\(_{3}\)O\(^{+}\)] + log\(_{10}\) [OH\(^{-}\)] = log\(_{10}\) (1 * 10\(^{-14}\))
Multiplying both the sides by -1,
-log\(_{10}\) [H\(_{3}\)O\(^{+}\)] – log\(_{10}\) [OH\(^{-}\)] = -log\(_{10}\) (1 * 10\(^{-14}\))
\(\therefore\) pH = -log\(_{10}\) [H\(_{3}\)O\(^{+}\)]; pOH = -log\(_{10}\) [OH\(^{-}\)];
pK\(_{w}\) = -log\(_{10}\) K\(_{w}\)
\(\therefore\) pH + pOH = pK\(_{w}\)
OR pH + pOH = 14
In simple words: The derivation starts from the ionic product of water, K\(_{w}\) = [H\(^{+}\)][OH\(^{-}\)], which is 10\(^{-14}\) at 25°C. Taking the negative logarithm of both sides and using the definitions of pH and pOH, we arrive at the relationship pH + pOH = 14.

🎯 Exam Tip: This is a standard derivation. Ensure you clearly define K\(_{w}\), pH, and pOH, and show all logarithmic steps correctly, especially multiplying by -1. State the temperature at which K\(_{w}\) = 10\(^{-14}\).

Question viii. Aqueous solution of sodium carbonate is alkaline whereas aqueous solution of ammonium chloride is acidic. Explain.
Answer: (A) (i) Sodium carbonate is a salt of weak acid (H\(_{2}\)CO\(_{3}\)) and strong base (NaOH).
(ii) In aqueous solution it undergoes hydrolysis.
Na\(_{2}\)CO\(_{3(aq)}\) + 2H\(_{2}\)O\(_{(l)}\) \(\leftrightharpoons\) 2NaOH\(_{(aq)}\) + H\(_{2}\)CO\(_{3(aq)}\)
(Salt of strong base and weak acid)
2Na\(_{(aq)}^{+}\) + CO\(_{3(aq)}^{2-}\) + 2H\(_{2}\)O\(_{(l)}\) \(\leftrightharpoons\) 2Na\(_{(aq)}^{+}\) + 2OH\(_{(aq)}^{-}\) + H\(_{2}\)CO\(_{3(aq)}\)
(iii) Strong base dissociates completely while weak acid dissociates partially since [OH\(^{-}\)] > [H\(_{3}\)O\(^{+}\)], the solution is basic.
(B) (i) Ammonium chloride is a salt of strong acid (HCl) and weak base (NH\(_{4}\)OH).
(ii) In aqueous solution it undergoes hydrolysis
NH\(_{4}\)Cl\(_{(aq)}\) + H\(_{2}\)O\(_{(l)}\) \(\leftrightharpoons\) NH\(_{4}\)OH\(_{(aq)}\) + HCl\(_{(aq)}\)
(Salt of strong acid and weak base)
NH\(_{4(aq)}^{+}\) + Cl\(_{(aq)}^{-}\) + H\(_{2}\)O\(_{(l)}\) \(\leftrightharpoons\) NH\(_{4}\)OH\(_{(aq)}\) + H\(_{(aq)}^{+}\) + Cl\(_{(aq)}^{-}\)
(iii) Since [H\(^{+}\)] or [H\(_{3}\)O\(^{+}\)] > [OH\(^{-}\)] the solution is acidic.
In simple words: Sodium carbonate is basic because its carbonate ion hydrolyzes water to produce OH\(^{-}\) ions. Ammonium chloride is acidic because its ammonium ion hydrolyzes water to produce H\(^{+}\) ions.

🎯 Exam Tip: For salts, determine the strength of the parent acid and base. Hydrolysis of the ion from the weak parent produces either H\(^{+}\) (acidic) or OH\(^{-}\) (basic). Write the hydrolysis reactions and compare [H\(^{+}\)] and [OH\(^{-}\)] concentrations for a complete explanation.

Question ix. pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.
Answer: Given : pH = 3.2; C = 0.02 M; K\(_{a}\) = ?
pH = -log\(_{10}\) [H\(^{+}\)]
\(\therefore\) [H\(^{+}\)] = Antilog – pH
= Antilog – 3.2
= Antilog(\(\overline{4}\).8)
= 6.31 * 10\(^{-4}\)M
For a weak monobasic acid HA,
HA\(_{(aq)}\) \(\leftrightharpoons\) H\(^{+}\) + A\(^{-}\)
Initial conc. c(1-\(\alpha\))
Conc. at equilibrium c c\(\alpha\) c\(\alpha\)
\(\therefore\) [H\(^{+}\)] = c\(\alpha\)
\(\alpha\) = \(\frac{[H^+]}{C}\) = \(\frac{6.31 * 10^{-4}}{0.02}\)
= 0.03155
K\(_{a}\) = \(\frac{C\alpha^2}{1-\alpha}\)
= \(\frac{0.02 \times (0.03155)^2}{1-0.03155}\)
= \(\frac{0.02 \times 0.000995}{0.96845}\)
= \(\frac{1.99 \times 10^{-5}}{0.96845}\)
= 2.05 * 10\(^{-5}\)
Dissociation constant = K\(_{a}\) = 2.05 * 10\(^{-5}\)
In simple words: First, convert the given pH to [H\(^{+}\)]. Then, use the relationship [H\(^{+}\)] = C\(\alpha\) to find the degree of dissociation (\(\alpha\)). Finally, use K\(_{a}\) = \(\frac{C\alpha^2}{1-\alpha}\) to calculate the dissociation constant.

🎯 Exam Tip: This problem involves multiple steps. Accurately converting pH to [H\(^{+}\)] using antilog is critical. Ensure correct calculation of \(\alpha\) and then substitute all values into the K\(_{a}\) expression, paying attention to significant figures and scientific notation.

Question x. In NaOH solution [OH\(^{-}\)] is 2.87 * 10\(^{-4}\). Calculate the pH of solution.
Answer: Given : [OH\(^{-}\)] = 2.87 * 10\(^{-4}\) M, pH = ?
pOH = -log\(_{10}\) [OH\(^{-}\)]
= -log\(_{10}\) (2.87 * 10\(^{-4}\))
= -(log\(_{10}\) 2.87 + log\(_{10}\) 10\(^{-4}\))
= -(0.4579 - 4)
= -(-3.5421)
= 3.5421
\(\therefore\) pH + pOH = 14
\(\therefore\) pH = 14 – pOH
= 14 – 3.5421
= 10.4579
pH = 10.4579.
In simple words: First, calculate the pOH from the given [OH\(^{-}\)] using pOH = -log[OH\(^{-}\)]. Then, use the relationship pH + pOH = 14 to find the pH of the solution.

🎯 Exam Tip: This problem requires calculating pOH first, then using the pOH-pH relationship. Ensure proper use of logarithm rules and correct calculation of antilog if needed, though here direct log calculation suffices.

4. Answer The Following:

Question i. Define degree of dissociation. Derive Ostwald's dilution law for the CH\(_{3}\)COOH.
Answer: (A) Degree of dissociation :
It is defined as a fraction of total number of moles of an electrolyte that dissociate into its ions at equilibrium.
It is denoted by \(\alpha\) and represented by,
\(\alpha\) = \(\frac{\text{number of moles dissociated}}{\text{total number of moles of an electrolyte}}\)
Or \(\alpha\) = \(\frac{\text{Per cent dissociation}}{100}\)
\(\therefore\) Per cent dissociation = \(\alpha\) * 100
(B) Consider V dm\(^{-3}\) of a solution containing one mole of CH\(_{3}\)COOH. Then the concentration of acid is, C = \(\frac{1}{V}\) mol dm\(^{-3}\). Let \(\alpha\) be the degree of dissociation
CH\(_{3}\)COOH\(_{(aq)}\) \(\leftrightharpoons\) CH\(_{3}\)COO\(_{(aq)}^{-}\) + H\(_{(aq)}^{+}\)

Initial moles :	1	0	0
Moles at equilibrium :	(1-\(\alpha\))	\(\alpha\)	\(\alpha\)
Concentration at equilibrium (mol dm\(^{-3}\)):	\(\frac{(1-\alpha)}{V}\)	\(\frac{\alpha}{V}\)	\(\frac{\alpha}{V}\)

If K\(_{a}\) is dissociation constant, then
K\(_{a}\) = \(\frac{[CH_3COO^-] \times [H^+]}{[CH_3COOH]}\)
K\(_{a}\) = \(\frac{\frac{\alpha}{V} \times \frac{\alpha}{V}}{\frac{(1-\alpha)}{V}}\) = \(\frac{\alpha^2}{(1-\alpha)V}\)
\(\therefore\) C = \(\frac{1}{V}\) mol dm\(^{-3}\)
K\(_{a}\) = \(\frac{C\alpha^2}{1-\alpha}\)
As the electrolyte is weak, \(\alpha\) is very small as compared to unity, \(\therefore\) (1 – \(\alpha\)) \(\approx\) 1.
\(\therefore\) K = \(\frac{\alpha^2}{V}\)
\(\implies\) \(\alpha\) = \(\sqrt{KV}\)
\(\therefore\) \(\alpha\) \(\propto\) \(\sqrt{V}\)
C = \(\frac{1}{V}\) where C = concentration in mol dm\(^{-3}\)
\(\therefore\) K = \(\alpha^2C\)
\(\therefore\) \(\alpha\) = \(\sqrt{\frac{K}{C}}\)
\(\therefore\) \(\alpha\) = \(\sqrt{K \times V}\)
\(\therefore\) C = \(\frac{1}{V}\) or V = \(\frac{1}{C}\)
\(\alpha\) = \(\sqrt{\frac{K}{C}}\)
This is the expression of Ostwald's dilution law. Thus, the degree of dissociation of a weak electrolyte is directly proportional to the square root of the volume of the solution containing 1 mole of an electrolyte.
In simple words: Degree of dissociation (\(\alpha\)) is the fraction of electrolyte molecules that break into ions. Ostwald's dilution law states that for a weak electrolyte, \(\alpha\) is inversely proportional to the square root of its concentration or directly proportional to the square root of the solution's volume.

🎯 Exam Tip: Define degree of dissociation clearly. For the derivation of Ostwald's dilution law, start with the equilibrium of a weak acid, set up initial and equilibrium concentrations, write the K\(_{a}\) expression, and then apply the approximation for weak electrolytes. Show the final relationship between \(\alpha\), K\(_{a}\), and C (or V).

Question ii. Define pH and pOH. Derive relationship between pH and pOH.
Answer: (1) pH : The negative logarithm, to the base 10, of the molar concentration of hydrogen ions, H\(^{+}\) is known as the pH of a solution.
pH = -log\(_{10}\) [H\(^{+}\)]
(2) pOH : The negative logarithm, to the base 10, of the molar concentration of hydroxyl ions, OH\(^{-}\) is known as the pOH of a solution.
pOH = -log\(_{10}\) [OH\(^{-}\)]
Relationship between pH and pOH:
The ionic product of water, K\(_{w}\) is given by,
K\(_{w}\) = [H\(_{3}\)O\(^{+}\)] * [OH\(^{-}\)]
At 298 K, K\(_{w}\) = 1 * 10\(^{-14}\)
\(\therefore\) pK\(_{w}\) = -log\(_{10}\)K\(_{w}\) = -log\(_{10}\) (1 * 10\(^{-14}\)) = 14
\(\therefore\) [H\(_{3}\)O\(^{+}\)] * [OH\(^{-}\)] = 1 * 10\(^{-14}\)
Taking logarithm to base 10 of both sides,
log\(_{10}\) [H\(_{3}\)O\(^{+}\)] + log\(_{10}\) [OH\(^{-}\)] = log\(_{10}\) (1 * 10\(^{-14}\))
Multiplying both the sides by -1,
-log\(_{10}\) [H\(_{3}\)O\(^{+}\)] – log\(_{10}\) [OH\(^{-}\)] = -log\(_{10}\) (1 * 10\(^{-14}\))
\(\therefore\) pH = -log\(_{10}\) [H\(_{3}\)O\(^{+}\)]; pOH = -log\(_{10}\) [OH\(^{-}\)];
pK\(_{w}\) = -log\(_{10}\) K\(_{w}\)
\(\therefore\) pH + pOH = pK\(_{w}\)
OR pH + pOH = 14
In simple words: pH measures the acidity ([H\(^{+}\)]) and pOH measures the basicity ([OH\(^{-}\)]). Their sum is 14 at 25°C, derived from the ionic product of water, K\(_{w}\) = [H\(^{+}\)][OH\(^{-}\)], which is 10\(^{-14}\).

🎯 Exam Tip: Define pH and pOH clearly with their formulas. The derivation of pH + pOH = 14 is standard and involves starting with K\(_{w}\), taking the negative logarithm of both sides, and substituting the definitions of pH and pOH. Be precise with mathematical steps.

Question iii. What is meant by hydrolysis ? A solution of CH\(_{3}\)COONH\(_{4}\) is neutral. why ?
Answer: Hydrolysis : A reaction in which the cations or anions or both the ions of a salt react with water to produce acidity or basicity or sometimes neutrality is called hydrolysis.
A salt of weak acid and weak base for which K\(_{a}\) = K\(_{b}\):
Consider hydrolysis of CH\(_{3}\)COONH\(_{4}\).
CH\(_{3}\)COO\(_{(aq)}^{-}\) + NH\(_{4(aq)}^{+}\) \(\leftrightharpoons\) CH\(_{3}\)COOH\(_{(aq)}\) +
weak acid
NH\(_{4}\)OH\(_{(aq)}\)
weak base
Since K\(_{a}\) = K\(_{b}\), the weak acid CH\(_{3}\)COOH and weak base NH\(_{4}\)OH dissociate to the same extent, hence, [H\(_{3}\)O\(^{+}\)] = [OH\(^{-}\)] and the solution reacts neutral after hydrolysis.
In simple words: Hydrolysis is when a salt's ions react with water to form an acidic, basic, or neutral solution. CH\(_{3}\)COONH\(_{4}\) is neutral because both its acetate (from weak acid) and ammonium (from weak base) ions hydrolyze, but their dissociation constants (K\(_{a}\) and K\(_{b}\)) are roughly equal, balancing the H\(^{+}\) and OH\(^{-}\) production.

🎯 Exam Tip: Define hydrolysis precisely. For CH\(_{3}\)COONH\(_{4}\), explain that both cation and anion hydrolyze. The key reason for neutrality is that the K\(_{a}\) of the weak acid formed (CH\(_{3}\)COOH) is approximately equal to the K\(_{b}\) of the weak base formed (NH\(_{4}\)OH), resulting in equal [H\(^{+}\)] and [OH\(^{-}\)] in the solution.

Question iv. Dissociation of HCN is suppressed by the addition of HCl. Explain.
Answer: The weak acid HCN is dissociated as follows :
HCN\(_{(aq)}\) + H\(_{2}\)O \(_{(l)}\) \(\leftrightharpoons\) H\(_{3}\)O\(_{(aq)}^{+}\) + CN\(_{(aq)}^{-}\)
The dissociation constant K\(_{a}\) is represented as,
K\(_{a}\) = \(\frac{[H_3O^+] \times [CN^-]}{[HCN]}\)
When HCl is added, it increases the concentration of H\(_{3}\)O\(^{+}\), hence in order to keep the ratio constant, then by Le Chatelier's principle, the equilibrium is shifted from right to left, suppressing the dissociation of HCN.
In simple words: This is an example of the common ion effect. Adding HCl increases the concentration of H\(_{3}\)O\(^{+}\) ions (a common product with HCN dissociation). According to Le Chatelier's principle, the equilibrium for HCN dissociation shifts to the left, reducing the extent of HCN dissociation.

🎯 Exam Tip: Name the common ion effect. Write the equilibrium for HCN dissociation. Explain that HCl is a strong acid that adds H\(_{3}\)O\(^{+}\) (a common ion). Then, clearly state how Le Chatelier's principle dictates the shift in equilibrium and the suppression of dissociation.

Question v. Derive the relationship between degree of dissociation and dissociation constant in weak electrolytes.
Answer: Expression of Ostwald's dilution law in the case of a weak electrolyte : Consider the dissociation of a weak electrolyte BA. Let V dm\(^{-3}\) of a solution contain one mole of the electrolyte. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm\(^{-3}\). Let \(\alpha\) be the degree of dissociation of the electrolyte.
BA \(\leftrightharpoons\) B\(^{+}\) + A\(^{-}\)

Initial moles :	1	0	0
Moles at equilibrium :	(1-\(\alpha\))	\(\alpha\)	\(\alpha\)
Concentration at equilibrium (mol dm\(^{-3}\)):	\(\frac{(1-\alpha)}{V}\)	\(\frac{\alpha}{V}\)	\(\frac{\alpha}{V}\)

Applying the law of mass action to this dissociation equilibrium, we have,
K = \(\frac{[B^+][A^-]}{[BA]}\)
\(\therefore\) K = \(\frac{\frac{\alpha}{V} \times \frac{\alpha}{V}}{\frac{(1-\alpha)}{V}}\) = \(\frac{\alpha^2}{(1-\alpha)V}\)
As the electrolyte is weak, \(\alpha\) is very small as compared to unity, \(\therefore\) (1 – \(\alpha\)) \(\approx\) 1.
\(\therefore\) K = \(\frac{\alpha^2}{V}\)
\(\implies\) \(\alpha\) = \(\sqrt{KV}\)
\(\therefore\) \(\alpha\) \(\propto\) \(\sqrt{V}\)
Since C = \(\frac{1}{V}\), then V = \(\frac{1}{C}\)
\(\therefore\) K = \(\alpha^2C\)
\(\therefore\) \(\alpha\) = \(\sqrt{\frac{K}{C}}\)
This is the expression of Ostwald's dilution law. Thus, the degree of dissociation of a weak electrolyte is directly proportional to the square root of the volume of the solution containing 1 mole of an electrolyte.
In simple words: Ostwald's dilution law relates the degree of dissociation (\(\alpha\)) of a weak electrolyte to its dissociation constant (K) and concentration (C). For weak electrolytes, \(\alpha\) is inversely proportional to the square root of its concentration (C) and directly proportional to the square root of its volume (V).

🎯 Exam Tip: Clearly define the terms and set up the equilibrium expression for a generic weak electrolyte BA. Show the concentrations at equilibrium in terms of \(\alpha\) and V (or C). Apply the law of mass action, simplify the expression using the weak electrolyte approximation (1-\(\alpha\)) \(\approx\) 1, and derive the final relationship, highlighting its proportionality to \(\sqrt{V}\) or \(\frac{1}{\sqrt{C}}\).

Question vi. Sulfides of cation of group II are precipitated in acidic solution (H\(_{2}\)S + HCl) whereas sulfides of cations of group IIIB are precipitated in ammoniacal solution of H\(_{2}\)S. Comment on the relative values of solubility product of sulfides of these.
Answer: (1) In qualitative analysis, the cations of group II are precipitated as sulfides, namely HgS, CuS, PbS, etc., while cations of group IIIB are precipitated as sulfides, namely, CoS, NiS, ZnS.
(2) The sulfides of group II have extremely low solubility product (K\(_{sp}\)) about 10\(^{-29}\) to 10\(^{-53}\) while the sulfides of group IIIB have slightly higher K\(_{sp}\) values about 10\(^{-20}\) to 10\(^{-23}\).
(3) In group II, sulfides are precipitated by adding H\(_{2}\)S in acidic solution while in IIIB group they are precipitated in a basic solution like ammonical solution.
(4) In acidic medium due to common ion H\(^{+}\), H\(_{2}\)S is dissociated to very less extent but gives sufficient S\(^{2-}\) ion to exceed solubility product of group II sulfides of cations and precipitate them.
HCl\(_{(aq)}\) \(\implies\) H\(_{(aq)}^{+}\) + Cl\(_{(aq)}^{-}\); H\(_{2}\)S\(_{(aq)}\) \(\leftrightharpoons\) 2H\(_{(aq)}^{+}\) + S\(_{(aq)}^{2-}\)
In simple words: Group II sulfides (e.g., CuS) have very low solubility products (K\(_{sp}\)), allowing them to precipitate even with low S\(^{2-}\) concentrations achieved in acidic H\(_{2}\)S solutions due to the common ion effect. Group IIIB sulfides (e.g., ZnS) have higher K\(_{sp}\) values, requiring higher S\(^{2-}\) concentrations, which are achieved in basic (ammoniacal) H\(_{2}\)S solutions where H\(_{2}\)S dissociation is enhanced.

🎯 Exam Tip: The key is to correlate K\(_{sp}\) values with the conditions required for precipitation. Lower K\(_{sp}\) values mean less S\(^{2-}\) is needed for precipitation, which is provided by H\(_{2}\)S in acidic medium (common ion effect). Higher K\(_{sp}\) values require more S\(^{2-}\), which is provided by H\(_{2}\)S in basic medium (where H\(^{+}\) is removed, shifting H\(_{2}\)S dissociation to the right). Mention specific K\(_{sp}\) ranges for each group.

Question viii. The pH of rain water collected in a certain region of Maharashtra on particular day was 5.1. Calculate the H3O+ ion concentration of the rain water and its percent dissociation.

Answer:
Given : pH = 5.1, [H3O+] = ?
PH = \( -\log_{10} [\text{H}^+] \)
\( \implies \) \( \log_{10} [\text{H}^+] = -\text{pH} \)
\( \implies \) \( [\text{H}^+] = \text{Antilog} - \text{pH} \)
\( \implies \) \( [\text{H}^+] = \text{Antilog} - 5.1 \)
\( \implies \) \( [\text{H}^+] = \text{Antilog } \overline{6}.9 \)
\( \implies \) \( [\text{H}^+] = 7.943 \times 10^{-6} \text{ M} \)
(H3O+ in rainwater is due to dissolved gases, CO2, SO2, etc. forming acids which dissociate giving H3O+ and acidity to rainwater.)
In simple words: Given the pH of rainwater, we use the pH formula to find the hydronium ion concentration. The negative logarithm of [H+] gives pH, so we reverse this calculation using antilogarithm to find [H+].

🎯 Exam Tip: Remember the relationship between pH and [H+] and how to use logarithms for calculation. Pay attention to significant figures in your final answer.

Question ix. Explain the relation between ionic product and solubility product to predict whether a precipitate will form when two solutions are mixed?

Answer:
If ionic product and solubility product are indicated by IP and Ksp respectively then,
(I) When IP = Ksp, the solution is saturated.
(II) When IP > Ksp, the solution is supersaturated and hence precipitation will occur, when two solutions are mixed.
(III) When IP < Ksp, the solution is unsaturated and precipitation will not occur, when two solutions are mixed.
In simple words: The ionic product (IP) compares the current concentration of ions in a solution to the solubility product (Ksp), which is the maximum concentration at saturation. If IP is greater than Ksp, a precipitate will form; if equal, the solution is saturated; if less, no precipitate forms.

🎯 Exam Tip: Clearly distinguish between Ionic Product (IP) and Solubility Product (Ksp) and their implications for precipitation. Providing clear conditions for each outcome (saturated, supersaturated, unsaturated) is crucial for scoring.

12th Chemistry Digest Chapter 3 Ionic Equilibria Intext Questions And Answers

Use your brain power (Textbook Page No. 47)

Question 1. Which of the following is a strong electrolyte ? HF, AgCl, CuSO4, CH3COONH4, H3PO4.

Answer:
CH3COONH4 is a strong electrolyte since in aqueous solution it dissociates completely. Sparingly soluble salts AgCl, CuSO4 are also strong electrolytes.
In simple words: Strong electrolytes fully break apart into ions when dissolved in water, making them good conductors of electricity. From the given options, CH3COONH4, AgCl, and CuSO4 are considered strong electrolytes due to their complete or significant dissociation.

🎯 Exam Tip: Understand the definition of a strong electrolyte – complete dissociation in solution. For salts, solubility also plays a role in their effective concentration as electrolytes.

Use your brain power (Textbook Page No. 49)

Question 1. All Bronsted bases are also Lewis bases, but all Bronsted acids are not Lewis acids. Explain.

Answer:
NH3 is a Bronsted base since it can accept a proton while it is also a Lewis base since it has a lone pair of electrons to donate.
In simple words: Bronsted bases accept protons, which requires them to have a lone pair of electrons, making them electron-pair donors (Lewis bases). However, Bronsted acids donate protons but don't necessarily accept electron pairs, so they are not always Lewis acids.

🎯 Exam Tip: Distinguish between Bronsted-Lowry and Lewis acid-base theories. Remember that the Lewis definition is broader. A good example like NH3 helps illustrate the point.

Use your brain power (Textbook Page No. 53)

Question 1. Suppose that pH of monobasic and dibasic acid is the same. Does this mean that the molar concentrations of both acids are identical ?

Answer:
Even if monobasic acid and dibasic acid give same pH, their molar concentrations are different. One mole of monobasic acid like HCl gives 1 mol of H⁺ while one mole of dibasic acid gives 2 mol of H⁺ in solution. Hence the concentration of dibasic acid will be half of the concentration of monobasic acid. For example, for same pH. [Monobasic acid] = [Dibasic acid]/2
In simple words: No, if a monobasic acid and a dibasic acid have the same pH, their molar concentrations are not identical because a dibasic acid releases twice as many H+ ions per mole, requiring a lower molar concentration to achieve the same pH as a monobasic acid.

🎯 Exam Tip: When comparing acids, consider their basicity (number of dissociable protons). For the same pH, a dibasic acid will have a lower molar concentration than a monobasic acid because it produces more H+ ions per molecule.

Question 2. How does pH of pure water vary with temperature ? Explain.

Answer:
Since the increase in temperature, increases the dissociation of water, its pH decreases.
In simple words: As temperature increases, pure water dissociates more into H+ and OH- ions. Although the water remains neutral (equal concentrations of H+ and OH-), the increased [H+] leads to a lower pH value.

🎯 Exam Tip: Remember that the autoionization of water is an endothermic process. Higher temperatures shift the equilibrium towards products, increasing [H+] and [OH-], thus decreasing pH (making it more acidic numerically, but still neutral overall).

Can you tell? (Textbook Page No. 54)

Question 1. Why
(i) an aqueous solution of NH4Cl is acidic.
(ii) while that of HCOOK basic ?

Answer:
(i) Ammonium chloride (NH4Cl) is a salt of a strong acid and a weak base. In aqueous solution, it undergoes hydrolysis:
NH4Cl\(_{\text{(aq)}}\) + H\(_2\)O\(_{\text{(l)}}\) \( \rightleftharpoons \) NH4OH\(_{\text{(aq)}}\) + HCl\(_{\text{(aq)}}\)
The strong acid (HCl) completely dissociates, releasing H\(_3\)O\(_\text{+}\) ions, while the weak base (NH4OH) remains largely undissociated. Thus, the concentration of H\(_\text{3}\)O\(_\text{+}\) ions is greater than OH\(_\text{-}\) ions (\( [\text{H}_\text{3}\text{O}^\text{+}] > [\text{OH}^\text{-}] \)), making the solution acidic.
(ii) HCOOK (potassium formate) is a salt of a weak acid HCOOH (formic acid) and a strong base KOH. In aqueous solution, it undergoes hydrolysis:
HCOOK\(_{\text{(aq)}}\) + H\(_2\)O\(_{\text{(l)}}\) \( \rightleftharpoons \) HCOOH\(_{\text{(aq)}}\) + KOH\(_{\text{(aq)}}\)
The strong base (KOH) completely dissociates, releasing OH\(_\text{-}\) ions, while the weak acid (HCOOH) remains largely undissociated. Thus, the concentration of OH\(_\text{-}\) ions is greater than H\(_\text{3}\)O\(_{\text{+}}\) ions (\( [\text{OH}^\text{-}] > [\text{H}_\text{3}\text{O}^\text{+}] \)), making the solution basic.
In simple words: NH4Cl solution is acidic because its ammonium ion hydrolyzes to produce H+ ions, overwhelming the small amount of OH- from water. HCOOK solution is basic because its formate ion hydrolyzes to produce OH- ions.

🎯 Exam Tip: For hydrolysis questions, identify the strong/weak acid and base components of the salt. The stronger component's influence (through its conjugate) will determine the pH of the solution. Write balanced hydrolysis equations.

Can you think? (Textbook Page No. 56)

Question 1. Home made jams and jellies without any added chemical preservative additives spoil in a few days whereas commercial jams and jellies have a long shelf life. Explain. What role does added sodium benzoate play ?

Answer:
Sodium benzoate added to jams and jellies in commercial products maintains the pH constant and acts as a preservative. Hence jams and jellies are not spoiled for a very long time unlike homemade products.
In simple words: Sodium benzoate acts as a preservative in commercial jams and jellies by creating and maintaining a specific pH level, which inhibits microbial growth and spoilage, thus extending their shelf life significantly compared to homemade versions without such additives.

🎯 Exam Tip: When discussing food preservation, relate chemical additives to their specific function, such as maintaining pH (buffering action) and inhibiting microbial growth, for a complete answer.

Can you tell? (Textbook Page No. 56)

Question 1. It is enough to add a few mL of a buffer solution to maintain its pH. Which property of buffer is used here ?

Answer:
The important property of reserve acidity and reserve basicity of a buffer solution is used to maintain constant pH. Weak acid or weak base along with ions (cations or anions) from salt react with excess of added acid (H+) or base [OH¯] and maintain pH constant.
In simple words: The property of a buffer solution used here is its ability to resist changes in pH upon the addition of small amounts of acid or base, due to its components neutralizing the added ions.

🎯 Exam Tip: Focus on the "reserve acidity" and "reserve basicity" concept of buffers. Explain how the weak acid/base components and their conjugate react to neutralize added H+ or OH- ions, maintaining pH.

Use your brain power (Textbook Page No. 59)

Question 1. What is the relationship between molar solubility and solubility product for salts given below :
(i) Ag2CrO4
(ii) Ca3(PO4)2
(iii) Cr(OH)3.

Answer:
(i) Ag2CrO4\(_{\text{(s)}}\) \( \rightleftharpoons \) 2Ag\(_{\text{(aq)}}^\text{+}\) + CrO\(_\text{4(aq)}^\text{2-}\)
Here, x=2, y=1
K\(_{\text{sp}}\) = [Ag\(_{\text{+}}\)]\(^2\) × [CrO\(_\text{4}^\text{2-}\)]
\( \implies \) K\(_{\text{sp}}\) = x\(^{\text{x}}\) y\(^{\text{y}}\) (S)\(^{\text{x+y}}\)
\( \implies \) K\(_{\text{sp}}\) = \( 2^2 \times 1^1 \times \text{S}^{2+1} \)
\( \implies \) K\(_{\text{sp}}\) = \( 4\text{S}^3 \)
\( \implies \) S = \( \left(\frac{\text{K}_\text{sp}}{4}\right)^{1/3} \) mol dm\(^{-3}\)

(ii) Ca3(PO4)2\(_{\text{(s)}}\) \( \rightleftharpoons \) 3Ca\(_{\text{(aq)}}^{\text{2+}}\) + 2PO\(_\text{4(aq)}^\text{3-}\)
Here, x=3, y=2
K\(_{\text{sp}}\) = [Ca\(^{\text{2+}}\)]\(^3\) × [PO\(_\text{4}^\text{3-}\)]\(^2\)
\( \implies \) K\(_{\text{sp}}\) = x\(^{\text{x}}\) y\(^{\text{y}}\) (S)\(^{\text{x+y}}\)
\( \implies \) K\(_{\text{sp}}\) = \( 3^3 \times 2^2 \times \text{S}^{3+2} \)
\( \implies \) K\(_{\text{sp}}\) = \( 108 \text{S}^5 \)
\( \implies \) S = \( \left(\frac{\text{K}_\text{sp}}{108}\right)^{1/5} \) mol dm\(^{-3}\)

(iii) Cr (OH)3\(_{\text{(s)}}\) \( \rightleftharpoons \) Cr\(_{\text{(aq)}}^{\text{3+}}\) + 3OH\(_{\text{(aq)}}^\text{-}\)
Here, x=1, y=3
K\(_{\text{sp}}\) = [Cr\(^{\text{3+}}\)]\(^1\) × [OH\(^{\text{-}}\)]\(^3\)
\( \implies \) K\(_{\text{sp}}\) = x\(^{\text{x}}\) y\(^{\text{y}}\) (S)\(^{\text{x+y}}\)
\( \implies \) K\(_{\text{sp}}\) = \( 1^1 \times 3^3 \times \text{S}^{1+3} \)
\( \implies \) K\(_{\text{sp}}\) = \( 27 \text{S}^4 \)
\( \implies \) S = \( \left(\frac{\text{K}_\text{sp}}{27}\right)^{1/4} \) mol dm\(^{-3}\)
In simple words: For sparingly soluble salts, the solubility product (Ksp) is related to the molar solubility (S) by considering the stoichiometry of dissociation. If a salt MxAy dissociates into xM+ and yA- ions, then Ksp = x^x * y^y * S^(x+y).

🎯 Exam Tip: Accurately write the dissociation equation for each salt, identify the stoichiometric coefficients (x and y), and correctly apply the formula Ksp = x^x * y^y * S^(x+y). Algebraic manipulation to express S in terms of Ksp is also important.

Can you tell? (Textbook Page No. 60)

Question 1. How is the ionization of NH4OH suppressed by addition of NH4Cl to the solution of NH4OH ?

Answer:
Ionisation of NH4OH is represented as follows :
NH4OH\(_{\text{(aq)}}\) \( \rightleftharpoons \) NH\(_\text{4(aq)}^\text{+}\) + OH\(_{\text{(aq)}}^\text{-}\)
It has ionisation constant, K\(_{\text{b}}\) = \( \frac{[\text{NH}_\text{4}^\text{+}][\text{OH}^\text{-}]}{[\text{NH}_\text{4}\text{OH}]} \)
K\(_{\text{b}}\) has constant value at constant temperature. When strong electrolyte NH4Cl is added to its solution, it dissociates completely.
NH4Cl\(_{\text{(aq)}}\) \( \implies \) NH\(_\text{4(aq)}^\text{+}\) + Cl\(_{\text{(aq)}}^\text{-}\)
Due to common ion NH\(_\text{4}^\text{+}\), by Le Chatelier's principle, the equilibrium of NH4OH is shifted from right to left, suppressing the ionisation of NH4OH.
In simple words: The addition of NH4Cl to an NH4OH solution introduces a common ion, NH4+, into the system. According to Le Chatelier's principle, this increase in product concentration shifts the weak base's equilibrium to the left, reducing the dissociation of NH4OH.

🎯 Exam Tip: This question tests Le Chatelier's principle and the common ion effect. Clearly state the common ion, explain how it affects the equilibrium of the weak electrolyte, and identify the resulting shift in equilibrium.