JEE Chemistry Equilibrium MCQs Set 20

Answer: (b) All four

Explanation:
1. Salts of a weak acid with a weak base, and amphiprotic anions, have pH given by an average of pK values.
2. The concentration cancels out in all four cases.

Teacher's Note:
1. Each gives \(pH = \frac12(pK_1 + pK_2)\)-type expressions.

Answer: (b) \(2 \times 10^{-3}\) and 9.0

Explanation:
1. \(K_h = K_w/K_a = 2 \times 10^{-8}\).
2. \(h = \sqrt{K_h/C} = \sqrt{2 \times 10^{-8}/0.005} = 2 \times 10^{-3}\).
3. \([OH^-] = Ch = 10^{-5}\) M, pH = 9.0.

Teacher's Note:
1. Chromate is the conjugate base of HCrO\(_4^-\).

Answer: (d) 10,000

Explanation:
1. In pure H\(_3\)A, \([HA^{2-}] \approx K_{a2} = 10^{-8}\) M.
2. With 0.1 M H\(^+\), \([H_2A^-] = K_{a1}(0.1)/0.1 = 10^{-5}\) and \([HA^{2-}] = K_{a2}[H_2A^-]/[H^+] = 10^{-12}\) M.
3. Ratio = \(10^{-8}/10^{-12} = 10^4\).

Teacher's Note:
1. The strong acid suppresses every ionisation step.

Answer: (a) 4

Explanation:
1. When step 1 is complete, \([H^+] = [H_2A^-]\) and \([A^{3-}] = K_{a2}K_{a3}/[H^+]\).
2. \([H^+] = 10^{-21}/10^{-17} = 10^{-4}\) M.
3. pH = 4.

Teacher's Note:
1. K\(_{a2}K_{a3}\) is the overall constant for H\(_2\)A\(^-\) to A\(^{3-}\).

Answer: (a) 0.11 M

Explanation:
1. Let x be the H\(^+\) from the weak acid: \((0.05 + x)x/(0.2 - x) = 0.05\).
2. \(x^2 + 0.1x - 0.01 = 0\) gives \(x = 0.0618\).
3. \([H^+] = 0.05 + 0.0618 = 0.112\) M.

Teacher's Note:
1. The weak acid is not suppressed enough to ignore its ionisation, so solve the quadratic.

Answer: (d) 9 : 20

Explanation:
1. Both anions share the same [H\(^+\)] and equal acid concentrations.
2. The ratio = \(K_{a1} : K_{a2} = 1.8 : 4 = 9 : 20\).

Teacher's Note:
1. The ratio equals the ratio of the dissociation constants.

Answer: (b) 1.0

Explanation:
1. \(K_h = K_w/K_{a3} = 10^{-14}/(4 \times 10^{-13}) = 0.025\).
2. \(x^2/(0.5 - x) = 0.025\) gives \(x = 0.1\) M.
3. pOH = 1.0.

Teacher's Note:
1. h is not small, so use the quadratic formula.

Answer: (d) I, II and IV only

Explanation:
1. Common-ion effect from 0.1 M OH\(^-\) suppresses NH\(_3\) ionisation (I true).
2. Before: \([OH^-] = 1.41 \times 10^{-3}\), pH 11.15; after: pH 13, a change of 1.85 (II true).
3. \([OH^-] \approx 0.1\) M, not 0.2 (III false).
4. K\(_b\) depends only on temperature (IV true).

Teacher's Note:
1. Equilibrium constants change only with temperature.

Answer: (a) 0.01%

Explanation:
1. \(K_h = 10^{-14}/10^{-5} = 10^{-9}\).
2. \(h = \sqrt{K_h/C} = \sqrt{10^{-8}} = 10^{-4}\).
3. Percentage = 0.01%.

Teacher's Note:
1. h is small, so the approximation \(h = \sqrt{K_h/C}\) is valid.

Answer: (c) 8

Explanation:
1. \(pH = \frac12(pK_w + pK_a + \log C) = \frac12(14 + 4 - 2) = 8\).

Teacher's Note:
1. Salts of weak acids are basic.

Answer: (c) 6

Explanation:
1. Diprotic (two ionisable H): H\(_2\)SO\(_4\), H\(_3\)PO\(_3\), H\(_2\)CO\(_3\), H\(_2\)S\(_2\)O\(_7\), H\(_2\)CrO\(_4\), H\(_2\)SO\(_3\).
2. H\(_3\)PO\(_4\) is triprotic, H\(_3\)BO\(_3\) acts as a monoprotic Lewis acid and H\(_3\)PO\(_2\) is monoprotic.

Teacher's Note:
1. Phosphorous acid has only two P-OH groups.

Answer: (c) 3

Explanation:
1. Basic solutions come from salts of strong bases with weak acids: KCN, K\(_2\)CO\(_3\) and LiCN.
2. K\(_2\)SO\(_4\) and NaCl are neutral; Zn(NO\(_3\))\(_2\), FeCl\(_3\) and NH\(_4\)NO\(_3\) are acidic; ammonium oxalate is nearly neutral (slightly acidic).

Teacher's Note:
1. Red litmus turns blue only in basic solution.

Answer: (c) [H\(^+\)] and [HCO\(_3^-\)] are approximately equal

Explanation:
1. The first ionisation dominates: \([H^+] \approx [HCO_3^-] = \sqrt{4.2 \times 10^{-7} \times 0.034} = 1.2 \times 10^{-4}\) M.
2. \([CO_3^{2-}] \approx K_2 = 4.8 \times 10^{-11}\) M, which is much smaller.

Teacher's Note:
1. Look at the orders of magnitude for each species.

Answer: (a) \(3 \times 10^{-20}\) M

Explanation:
1. \([S^{2-}] = K_1K_2[H_2S]/[H^+]^2 = (10^{-7})(1.2 \times 10^{-13})(0.1)/(0.2)^2\).
2. \(= 3 \times 10^{-20}\) M.

Teacher's Note:
1. The overall constant K\(_1K_2\) applies to H\(_2\)S giving S\(^{2-}\).

Answer: (d) about 14.9 g

Explanation:
1. \(pK_{a2} = 10.328\). \(\log([CO_3^{2-}]/[HCO_3^-]) = 10 - 10.328\), so the ratio is 0.47.
2. HCO\(_3^-\) = 0.3 mol, so CO\(_3^{2-}\) = 0.141 mol.
3. Mass = \(0.141 \times 106 = 14.9\) g.

Teacher's Note:
1. Use the pK\(_a\) for the HCO\(_3^-\)/CO\(_3^{2-}\) pair, that is \(pK_{a2}\).

Answer: (a) 12.4

Explanation:
1. NaOH = 5 mmol, acid = 2.5 mmol. Excess NaOH = 2.5 mmol in 100 mL = 0.025 M.
2. \(pOH = 1.6\), \(pH = 12.4\).

Teacher's Note:
1. The acetate contributes negligibly beside the strong base.

Answer: (a) 4.7

Explanation:
1. NaOH = 2.5 mmol and acid = 5 mmol. Half the acid is neutralised.
2. Buffer with equal acid and salt: \(pH = pK_a = -\log(2 \times 10^{-5}) = 4.7\).

Teacher's Note:
1. Half-neutralisation gives \(pH = pK_a\).

Answer: (a) 9.3

Explanation:
1. NH\(_4\)OH = 5 mmol, HCl = 2.5 mmol; half the base is neutralised.
2. \(pOH = pK_b = 4.7\).
3. \(pH = 14 - 4.7 = 9.3\).

Teacher's Note:
1. At half neutralisation of a weak base, pOH = pK\(_b\).

Answer: (c) 5.3

Explanation:
1. The mixture is 0.05 M NH\(_4\)Cl.
2. \(pH = \frac12(pK_w - pK_b - \log C) = \frac12(14 - 4.7 + 1.3)\).
3. \(pH = 5.3\).

Teacher's Note:
1. Salt of a strong acid and weak base is acidic.

Answer: (c) 4.52 and 5.48

Explanation:
1. pH 3 for 0.1 M gives \(K_a = 10^{-6}/0.1 = 10^{-5}\), \(pK_a = 5\).
2. 1/4 neutralised: \(pH = 5 + \log(1/3) = 4.52\).
3. 3/4 neutralised: \(pH = 5 + \log 3 = 5.48\).

Teacher's Note:
1. Use the Henderson equation with salt : acid = x : (100 - x).

Answer: (c) 9.82

Explanation:
1. Base = 2.5 mmol; HCl = 1.25 mmol, which is half of the first neutralisation.
2. \(pOH = pK_{b1} = 4.18\).
3. \(pH = 9.82\).

Teacher's Note:
1. Half of the first equivalence gives pOH = pK\(_{b1}\).

Answer: (a) 8.21

Explanation:
1. At the first equivalence point the species is BH\(^+\) (amphiprotic).
2. \(pOH = \frac12(pK_{b1} + pK_{b2}) = \frac12(4.18 + 7.39) = 5.785\).
3. \(pH = 8.21\).

Teacher's Note:
1. The result is independent of concentration.

Answer: (d) 7.21

Explanation:
1. Total HCl = 3 mmol: the first step uses 2.5 mmol, the extra 0.5 mmol converts BH\(^+\) to BH\(_2^{2+}\), leaving 2.0 mmol BH\(^+\).
2. \(pOH = pK_{b2} + \log([BH_2^{2+}]/[BH^+]) = 7.39 + \log(0.5/2) = 6.79\).
3. \(pH = 7.21\).

Teacher's Note:
1. Use the second pair BH\(^+\)/BH\(_2^{2+}\).