Chapter 3: Ionic Equilibria
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The pH of a solution is defined as the negative base-10 logarithm of the molar concentration of hydrogen ions (or hydronium ions) in the solution.
Mathematically: $pH = -\log_{10}[H^+]$ or $pH = -\log_{10}[H_3O^+]$
Answer: A pair of an acid and a base which differ from each other by a single proton ($H^+$) is called a conjugate acid-base pair.
Example: In the reaction $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$, the pair ($NH_3$, $NH_4^+$) is a conjugate base-acid pair.
Answer: $PbI_2$ dissociates as: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.
If the solubility is $S$, then $[Pb^{2+}] = S$ and $[I^-] = 2S$.
$$K_{sp} = [Pb^{2+}][I^-]^2 = (S)(2S)^2 = 4S^3$$
Answer: The degree of dissociation ($\alpha$) of an electrolyte is defined as the fraction of the total number of moles of the electrolyte that dissociate into its constituent ions at equilibrium.
2 Mark Questions (Short Answer-I)
Answer:
The ionic product of water ($K_w$) is given by:
$$K_w = [H^+][OH^-]$$
At 298 K ($25^\circ\text{C}$), the value of $K_w$ is $1.0 \times 10^{-14}$.
Therefore, $[H^+][OH^-] = 10^{-14}$
Taking the base-10 logarithm on both sides:
$$\log_{10}([H^+][OH^-]) = \log_{10}(10^{-14})$$
$$\log_{10}[H^+] + \log_{10}[OH^-] = -14 \log_{10}(10)$$
Multiplying the entire equation by $-1$:
$$(-\log_{10}[H^+]) + (-\log_{10}[OH^-]) = 14$$
Since $pH = -\log_{10}[H^+]$ and $pOH = -\log_{10}[OH^-]$,
$$pH + pOH = 14$$
Answer:
| Property | Strong Electrolyte | Weak Electrolyte |
|---|---|---|
| 1. Ionization | Ionizes completely or almost completely in aqueous solution. | Ionizes only to a small extent (partially) in aqueous solution. |
| 2. Degree of Dissociation ($\alpha$) | $\alpha \approx 1$ | $\alpha \ll 1$ |
| 3. Equilibrium | Reaction goes to completion; no equilibrium exists between ions and unionized molecules. | A dynamic equilibrium exists between the ions and the unionized molecules. |
| 4. Examples | Strong acids ($HCl, HNO_3$), Strong bases ($NaOH, KOH$), Salts ($NaCl$). | Weak acids ($CH_3COOH$), Weak bases ($NH_4OH$). |
Answer:
An acidic buffer is a solution which maintains its pH practically constant upon the addition of a small amount of strong acid or strong base, and its pH value is less than 7.
It is prepared by mixing a weak acid and its salt formed with a strong base.
Example: A mixture of Acetic acid ($CH_3COOH$, a weak acid) and Sodium acetate ($CH_3COONa$, its salt with a strong base).
3 Mark Questions (Short Answer-II / Numericals)
Answer:
Statement: Ostwald's dilution law states that the degree of dissociation of a weak electrolyte is inversely proportional to the square root of its concentration (or directly proportional to the square root of the volume containing 1 mole of the electrolyte).
Derivation for a weak acid ($HA$):
1. Consider $1 \text{ mole}$ of a weak acid $HA$ dissolved in volume $V$ liters of solution. Let $\alpha$ be its degree of dissociation at equilibrium. Concentration $c = \frac{1}{V} \text{ mol/L}$.
| $HA(aq)$ | $\rightleftharpoons$ | $H^+(aq)$ | $+$ | $A^-(aq)$ | |
|---|---|---|---|---|---|
| Initial moles | $1$ | $0$ | $0$ | ||
| Moles at eqm | $1 - \alpha$ | $\alpha$ | $\alpha$ | ||
| Conc. at eqm ($c$) | $\frac{1 - \alpha}{V} = c(1-\alpha)$ | $\frac{\alpha}{V} = c\alpha$ | $\frac{\alpha}{V} = c\alpha$ |
2. By the Law of Mass Action, the acid dissociation constant ($K_a$) is:
$$K_a = \frac{[H^+][A^-]}{[HA]}$$
3. Substituting the equilibrium concentrations:
$$K_a = \frac{(c\alpha) \times (c\alpha)}{c(1 - \alpha)}$$
$$K_a = \frac{c^2\alpha^2}{c(1 - \alpha)} = \frac{c\alpha^2}{1 - \alpha}$$
4. For a very weak acid, the degree of dissociation $\alpha$ is very small compared to 1 ($\alpha \ll 1$). Therefore, $(1 - \alpha) \approx 1$.
$$K_a = c\alpha^2 \implies \alpha^2 = \frac{K_a}{c} \implies \alpha = \sqrt{\frac{K_a}{c}}$$
Answer:
Given Data:
Solubility of $AgCl$ ($S$) $= 1.06 \times 10^{-5} \text{ mol L}^{-1}$
Step 1: Write the dissociation equation
$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$
Step 2: Relate concentrations to solubility
Let the solubility of $AgCl$ be $S$ mol/L.
At equilibrium: $[Ag^+] = S$ and $[Cl^-] = S$
Step 3: Write the $K_{sp}$ expression and calculate
$$K_{sp} = [Ag^+][Cl^-]$$
$$K_{sp} = (S) \times (S) = S^2$$
$$K_{sp} = (1.06 \times 10^{-5})^2$$
$$K_{sp} = (1.06)^2 \times 10^{-10}$$
$$K_{sp} = 1.1236 \times 10^{-10}$$
The solubility product ($K_{sp}$) of $AgCl$ is $1.1236 \times 10^{-10}$.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark numerical combined with a 1-mark theory question).
(b) What is the common ion effect? [1 Mark] March 2015, Oct 2022
Answer (a): Numerical on Buffer Solution
Given Data:
Concentration of Acid $[CH_3COOH] = 0.05 \text{ M}$
Concentration of Salt $[CH_3COONa] = 0.01 \text{ M}$
Dissociation constant of acid $K_a = 1.8 \times 10^{-5}$
Step 1: Calculate $pK_a$
$pK_a = -\log_{10}(K_a)$
$pK_a = -\log_{10}(1.8 \times 10^{-5})$
$pK_a = 5 - \log_{10}(1.8) = 5 - 0.2553 = 4.7447$
Step 2: Apply Henderson-Hasselbalch Equation
$$pH = pK_a + \log_{10}\frac{[\text{Salt}]}{[\text{Acid}]}$$
$$pH = 4.7447 + \log_{10}\frac{0.01}{0.05}$$
$$pH = 4.7447 + \log_{10}\left(\frac{1}{5}\right)$$
$$pH = 4.7447 + (\log_{10} 1 - \log_{10} 5)$$
$$pH = 4.7447 + (0 - 0.6990)$$
$$pH = 4.7447 - 0.6990 = 4.0457$$
The pH of the buffer solution is 4.0457.
Answer (b): Theory
Common Ion Effect: The phenomenon in which the degree of dissociation of a weak electrolyte is suppressed (decreased) by the addition of a strong electrolyte containing a common ion is called the common ion effect. It is a direct consequence of Le Chatelier's principle.
(b) Write the conjugate acid of $NH_3$ and $H_2O$. [1 Mark] March 2018, March 2020
Answer (a): Numerical on pH of Weak Acid
Given Data:
Percentage dissociation $= 0.04\%$
Concentration ($c$) $= 0.025 \text{ M} = 2.5 \times 10^{-2} \text{ M}$
Step 1: Find Degree of Dissociation ($\alpha$)
$\alpha = \frac{\text{Percentage dissociation}}{100} = \frac{0.04}{100} = 0.0004 = 4 \times 10^{-4}$
Step 2: Calculate $[H^+]$ concentration
For a weak monobasic acid ($HA \rightleftharpoons H^+ + A^-$),
$[H^+] = c \times \alpha$
$[H^+] = (2.5 \times 10^{-2}) \times (4 \times 10^{-4}) = 10 \times 10^{-6} = 1.0 \times 10^{-5} \text{ M}$
Step 3: Calculate pH
$pH = -\log_{10}[H^+]$
$pH = -\log_{10}(1.0 \times 10^{-5})$
$pH = 5\log_{10}10 = 5 \times 1 = 5$
The pH of the solution is 5.
Answer (b): Theory
To find the conjugate acid, add one proton ($H^+$) to the base.
- The conjugate acid of $NH_3$ is $NH_4^+$ (Ammonium ion).
- The conjugate acid of $H_2O$ is $H_3O^+$ (Hydronium ion).
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