Chapter 3: Ionic Equilibria Mock Test
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
- Use of logarithmic tables is allowed. Calculators are not permitted.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
The relationship between pH and pOH for an aqueous solution at 298 K is:
(A) $pH \times pOH = 14$(B) $pH - pOH = 14$(C) $pH + pOH = 14$(D) $pH / pOH = 14$
-
What is the conjugate base of the hydronium ion ($H_3O^+$)?
(A) $H^+$(B) $OH^-$(C) $H_2O$(D) $O^{2-}$
-
The solubility product ($K_{sp}$) expression for $Al(OH)_3$ is:
(A) $27S^4$(B) $4S^3$(C) $108S^5$(D) $S^2$
-
Which of the following mixtures forms an acidic buffer?
(A) $HCl + NaCl$(B) $NaOH + NaCl$(C) $CH_3COOH + CH_3COONa$(D) $NH_4OH + NH_4Cl$
Q2. Answer the following questions in one sentence: [3 Marks]
- Define: Degree of dissociation ($\alpha$).
- State Ostwald's dilution law.
- What is the pH of a $10^{-3}$ M $HCl$ solution?
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Derive the relationship: $pH + pOH = 14$ at 298 K.
- Distinguish between strong electrolytes and weak electrolytes. (Any 2 points).
- Identify the Lewis acids and Lewis bases from the following: $NH_3, BF_3, AlCl_3, H_2O$.
- Define a buffer solution. Write any two applications of buffer solutions.
- Write the solubility product expression ($K_{sp}$) and its relation with solubility ($S$) for $Ag_2CrO_4$.
SECTION C
Attempt any TWO of the following: [6 Marks]
- Derive the expression for the dissociation constant ($K_a$) of a weak acid based on Ostwald's dilution law.
- Calculate the pH of a buffer solution containing 0.05 M $CH_3COOH$ and 0.01 M $CH_3COONa$. ($K_a$ for acetic acid = $1.8 \times 10^{-5}$).
- The solubility of $AgCl$ in water is $1.06 \times 10^{-5} \text{ mol L}^{-1}$ at 298 K. Calculate its solubility product ($K_{sp}$).
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) A weak monobasic acid is 0.04% dissociated in a 0.025 M solution. Calculate the pH of the solution. [3 Marks]
(b) What is the common ion effect? [1 Mark] - (a) Explain the salt hydrolysis of a salt of strong acid and strong base with a suitable example. Why is its aqueous solution neutral? [2 Marks]
(b) Derive the relationship between solubility ($S$) and solubility product ($K_{sp}$) for $Bi_2S_3$. [2 Marks]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (C) $pH + pOH = 14$ [1 Mark for correct option]
2. (C) $H_2O$ [1 Mark. Conjugate base is formed by removing an $H^+$ ion]
3. (A) $27S^4$ [1 Mark. $Al(OH)_3 \rightarrow Al^{3+} + 3OH^-$, so $K_{sp} = (S)(3S)^3 = 27S^4$]
4. (C) $CH_3COOH + CH_3COONa$ [1 Mark for correct option]
Q2. Very Short Answers:
1. Degree of dissociation ($\alpha$):
It is defined as the fraction of the total number of moles of an electrolyte that dissociate into its constituent ions at equilibrium. [1 Mark for correct definition]
2. Ostwald's dilution law:
It states that the degree of dissociation of a weak electrolyte is inversely proportional to the square root of its concentration. [1 Mark for correct statement]
3. pH of $10^{-3}$ M $HCl$:
$HCl$ is a strong acid, so $[H^+] = 10^{-3}$ M. $pH = -\log_{10}(10^{-3}) = 3$. [1 Mark for correct answer]
SECTION B [8 Marks]
Q3. Derivation of $pH + pOH = 14$:
The ionic product of water is $K_w = [H^+][OH^-]$. At 298 K, $K_w = 1.0 \times 10^{-14}$. [1/2 Mark]
Therefore, $[H^+][OH^-] = 10^{-14}$. Taking $\log_{10}$ on both sides:
$\log_{10}[H^+] + \log_{10}[OH^-] = \log_{10}(10^{-14}) = -14$ [1 Mark]
Multiplying by -1: $(-\log_{10}[H^+]) + (-\log_{10}[OH^-]) = 14$
Since $pH = -\log_{10}[H^+]$ and $pOH = -\log_{10}[OH^-]$, we get $pH + pOH = 14$. [1/2 Mark]
Q4. Strong vs Weak Electrolytes:
| Strong Electrolytes | Weak Electrolytes |
|---|---|
| Ionize completely in aqueous solution ($\alpha \approx 1$). | Ionize only partially in aqueous solution ($\alpha \ll 1$). |
| No equilibrium exists between ions and unionized molecules. | A dynamic equilibrium exists between ions and unionized molecules. |
[1 Mark for each point of distinction. Total 2 Marks]
Q5. Lewis Acids and Bases:
A Lewis acid is an electron-pair acceptor. A Lewis base is an electron-pair donor.
Lewis Acids: $BF_3$, $AlCl_3$ (They are electron-deficient). [1 Mark]
Lewis Bases: $NH_3$, $H_2O$ (They have lone pairs of electrons to donate). [1 Mark]
Q6. Buffer Solution:
Definition: A solution which resists a drastic change in its pH upon the addition of a small amount of strong acid, strong base, or upon dilution. [1 Mark]
Applications: (1) In biochemical systems, blood is a natural buffer (pH ~7.4). (2) Used in agriculture to maintain soil pH. (3) Used in analytical chemistry to precipitate specific ions. [1 Mark for any two applications]
Q7. $K_{sp}$ for $Ag_2CrO_4$:
Dissociation: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$ [1/2 Mark]
Expression: $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$ [1/2 Mark]
Let solubility be $S$. Then $[Ag^+] = 2S$ and $[CrO_4^{2-}] = S$.
Relation: $K_{sp} = (2S)^2(S) = 4S^3$ [1 Mark]
SECTION C [6 Marks]
Q8. Ostwald's Dilution Law Derivation:
Consider a weak acid HA. Let initial concentration be $c$ mol/L and degree of dissociation be $\alpha$.
$HA \rightleftharpoons H^+ + A^-$
Eq. conc: $c(1-\alpha) \quad c\alpha \quad c\alpha$
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$ [1 Mark]
$K_a = \frac{c\alpha^2}{1-\alpha}$
For a very weak acid, $\alpha \ll 1$, so $(1-\alpha) \approx 1$. Hence, $K_a = c\alpha^2$. [1 Mark]
Q9. pH of Buffer Numerical:
Given: $[\text{Acid}] = 0.05 \text{ M}$, $[\text{Salt}] = 0.01 \text{ M}$, $K_a = 1.8 \times 10^{-5}$.
$pK_a = -\log_{10}(1.8 \times 10^{-5}) = 5 - \log_{10}(1.8) = 5 - 0.2553 = 4.7447$ [1 Mark]
Henderson-Hasselbalch equation: $pH = pK_a + \log_{10}\frac{[\text{Salt}]}{[\text{Acid}]}$ [1/2 Mark]
$pH = 4.7447 + \log_{10}\frac{0.01}{0.05} = 4.7447 + \log_{10}(1/5) = 4.7447 - \log_{10}5$ [1 Mark]
$pH = 4.7447 - 0.6990 = 4.0457$ [1/2 Mark for correct answer]
Q10. Solubility Product Numerical:
Given: $S = 1.06 \times 10^{-5} \text{ mol L}^{-1}$.
Reaction: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ [1/2 Mark]
Relation: $K_{sp} = [Ag^+][Cl^-] = S \times S = S^2$ [1 Mark]
Calculation: $K_{sp} = (1.06 \times 10^{-5})^2$ [1/2 Mark]
Answer: $K_{sp} = 1.1236 \times 10^{-10}$ [1 Mark]
SECTION D [4 Marks]
Q11. (a) pH of Weak Acid [3 Marks] (b) Common Ion Effect [1 Mark]
(a) pH calculation:
Given: \% diss = 0.04%, $c = 0.025 \text{ M}$.
$\alpha = \frac{0.04}{100} = 4 \times 10^{-4}$ [1 Mark]
$[H^+] = c \cdot \alpha = (0.025) \times (4 \times 10^{-4}) = 10^{-5} \text{ M}$ [1 Mark]
$pH = -\log_{10}[H^+] = -\log_{10}(10^{-5}) = 5$ [1 Mark]
(b) Common Ion Effect: The phenomenon where the degree of dissociation of a weak electrolyte is suppressed by the addition of a strong electrolyte containing a common ion. [1 Mark]
Q12. (a) Salt Hydrolysis [2 Marks] (b) $K_{sp}$ for $Bi_2S_3$ [2 Marks]
(a) Salt of Strong Acid and Strong Base:
Example: NaCl. It is a salt of strong acid (HCl) and strong base (NaOH). [1/2 Mark]
In aqueous solution, it dissociates completely into $Na^+$ and $Cl^-$ ions. Neither of these ions reacts with water molecules (they do not accept $OH^-$ or $H^+$ from water because the resulting NaOH and HCl would just dissociate completely again). Since no hydrolysis occurs, $[H^+] = [OH^-]$, keeping the solution neutral (pH = 7). [1.5 Marks for correct reasoning]
(b) $K_{sp}$ for $Bi_2S_3$:
Dissociation: $Bi_2S_3(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$ [1/2 Mark]
Let solubility be $S$. Then $[Bi^{3+}] = 2S$ and $[S^{2-}] = 3S$. [1/2 Mark]
$K_{sp} = [Bi^{3+}]^2[S^{2-}]^3 = (2S)^2(3S)^3$ [1/2 Mark]
$K_{sp} = (4S^2)(27S^3) = 108S^5$ [1/2 Mark]
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