JEE Chemistry Equilibrium MCQs Set 19

Multiple Choice Questions (MCQs) for JEE Chemistry: Equilibrium

Access targeted multiple-choice questions for Equilibrium designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Equilibrium MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: What is the pH of 0.01 M H\(_2\)S? (\(K_{a1} = 9 \times 10^{-8}\), \(K_{a2} = 1.2 \times 10^{-13}\))
(a) 7.00
(b) 3.52
(c) 4.52
(d) 5.52
Show Answer & Explanation

Answer: (c) 4.52

Explanation:
1. The second ionisation is negligible, so \([H^+] = \sqrt{K_{a1}C} = \sqrt{9 \times 10^{-8} \times 0.01} = 3 \times 10^{-5}\) M.
2. \(pH = 5 - \log 3 = 4.52\).

Teacher's Note:
1. Treat a polyprotic acid as a monoprotic acid when K\(_{a2}\) is much smaller than K\(_{a1}\).

Question 2: In 0.01 M H\(_2\)S (\(K_{a1} = 9 \times 10^{-8}\), \(K_{a2} = 1.2 \times 10^{-13}\)), the concentration of S\(^{2-}\) is:
(a) \(3 \times 10^{-5}\) M
(b) \(1.2 \times 10^{-15}\) M
(c) \(9 \times 10^{-8}\) M
(d) \(1.2 \times 10^{-13}\) M
Show Answer & Explanation

Answer: (d) \(1.2 \times 10^{-13}\) M

Explanation:
1. \([HS^-] = [H^+] = 3 \times 10^{-5}\) M.
2. \([S^{2-}] = K_{a2}[HS^-]/[H^+] = K_{a2} = 1.2 \times 10^{-13}\) M.

Teacher's Note:
1. In a pure solution of a diprotic acid, [A\(^{2-}\)] is approximately K\(_{a2}\).

Question 3: The pH of a 0.74 g/L solution of propane-1,2-diamine (M = 74; \(pK_{b1} = 4.18\), \(pK_{b2} = 7.39\)) is nearly:
(a) 3.1
(b) 10.9
(c) 9.8
(d) 11.9
Show Answer & Explanation

Answer: (b) 10.9

Explanation:
1. C = 0.74/74 = 0.01 M.
2. \(pOH = \frac12(pK_{b1} - \log C) = \frac12(4.18 + 2) = 3.09\).
3. \(pH = 14 - 3.09 = 10.91\).

Teacher's Note:
1. Only the first ionisation of the base matters here.

Question 4: 0.01 mol of HCl gas is passed into 100 mL of 0.05 M formic acid (\(K_a = 1.8 \times 10^{-4}\)). The change in pH is about:
(a) no change
(b) a decrease of 0.5
(c) a decrease of 1.5
(d) an increase of 1.5
Show Answer & Explanation

Answer: (c) a decrease of 1.5

Explanation:
1. Initially \(pH = \frac12(3.745 + 1.301) = 2.52\).
2. HCl gives 0.1 M H\(^+\), so the pH becomes 1.00.
3. The pH falls by about 1.5 units.

Teacher's Note:
1. The strong acid suppresses the ionisation of the weak acid.

Question 5: After passing HCl as in the previous case (0.1 M H\(^+\)), the concentration of HCOO\(^-\) in 0.05 M formic acid (\(K_a = 1.8 \times 10^{-4}\)) is:
(a) \(9 \times 10^{-3}\) M
(b) \(9 \times 10^{-5}\) M
(c) \(1.8 \times 10^{-4}\) M
(d) \(3 \times 10^{-3}\) M
Show Answer & Explanation

Answer: (b) \(9 \times 10^{-5}\) M

Explanation:
1. \([HCOO^-] = K_a[HCOOH]/[H^+]\).
2. \(= 1.8 \times 10^{-4} \times 0.05/0.1 = 9 \times 10^{-5}\) M.

Teacher's Note:
1. Common-ion effect lowers the anion concentration.

Question 6: Equal volumes of 0.02 M KOH and 0.2 M NH\(_3\) (\(K_b = 2 \times 10^{-5}\)) are mixed. In the final solution, the pH, \([NH_4^+]\) and % dissociation of NH\(_3\) are respectively:
(a) 11.3, \(2 \times 10^{-4}\) M, 0.2%
(b) 12, \(2 \times 10^{-5}\) M, 0.02%
(c) 12, \(2 \times 10^{-4}\) M, 0.2%
(d) 12, \(2 \times 10^{-3}\) M, 2%
Show Answer & Explanation

Answer: (c) 12, \(2 \times 10^{-4}\) M, 0.2%

Explanation:
1. After mixing, [OH\(^-\)] from KOH = 0.01 M (pH 12) and [NH\(_3\)] = 0.1 M.
2. \([NH_4^+] = K_b[NH_3]/[OH^-] = 2 \times 10^{-5} \times 0.1/0.01 = 2 \times 10^{-4}\) M.
3. Degree of dissociation = \(2 \times 10^{-4}/0.1 = 0.2\%\).

Teacher's Note:
1. The strong base fixes [OH\(^-\)].

Question 7: A solution contains 0.1 M HCOOH (\(K_a = 2.4 \times 10^{-4}\)) and 0.1 M HOCN (\(K_a = 4 \times 10^{-4}\)). The values of \([H^+]\), \([HCOO^-]\) and \([OCN^-]\) are:
(a) \(8 \times 10^{-3}\), \(5 \times 10^{-3}\), \(3 \times 10^{-3}\) M
(b) \(3 \times 10^{-3}\), \(3 \times 10^{-3}\), \(3 \times 10^{-3}\) M
(c) \(4 \times 10^{-3}\), \(2 \times 10^{-3}\), \(2 \times 10^{-3}\) M
(d) \(8 \times 10^{-3}\), \(3 \times 10^{-3}\), \(5 \times 10^{-3}\) M
Show Answer & Explanation

Answer: (d) \(8 \times 10^{-3}\), \(3 \times 10^{-3}\), \(5 \times 10^{-3}\) M

Explanation:
1. \([H^+] = \sqrt{C_1K_1 + C_2K_2} = \sqrt{2.4 \times 10^{-5} + 4 \times 10^{-5}} = 8 \times 10^{-3}\) M.
2. \([HCOO^-] = 2.4 \times 10^{-5}/8 \times 10^{-3} = 3 \times 10^{-3}\) M.
3. \([OCN^-] = 4 \times 10^{-5}/8 \times 10^{-3} = 5 \times 10^{-3}\) M.

Teacher's Note:
1. The ratio of the anion concentrations equals the ratio of the K\(_a\) values: 3 : 5.

Question 8: The pH of 0.1 M Na\(_3\)A, the salt of a tribasic acid H\(_3\)A with \(K_{a1}, K_{a2}, K_{a3} = 10^{-4}, 10^{-7}, 10^{-9}\) (only first hydrolysis counted), is:
(a) 12
(b) 11
(c) 13
(d) 9
Show Answer & Explanation

Answer: (b) 11

Explanation:
1. \(A^{3-}\) is the conjugate base of the third step, so use \(pK_{a3} = 9\).
2. \(pH = \frac12(pK_w + pK_{a3} + \log C) = \frac12(14 + 9 - 1) = 11\).

Teacher's Note:
1. Use the K\(_a\) of the acid that is the conjugate of the ion.

Question 9: The pH of M/20 potassium hydrogen phthalate (KHC\(_8\)H\(_4\)O\(_4\)) solution (\(pK_{a1} = 2.94\), \(pK_{a2} = 5.44\)) is:
(a) 2.94
(b) 5.44
(c) 4.19
(d) 8.38
Show Answer & Explanation

Answer: (c) 4.19

Explanation:
1. HP\(^-\) is amphiprotic.
2. \(pH = \frac12(pK_{a1} + pK_{a2}) = \frac12(2.94 + 5.44) = 4.19\).

Teacher's Note:
1. The pH of an amphiprotic ion does not depend on its concentration.

Question 10: \(K_{b1}\) of N\(_2\)H\(_4\) is \(4 \times 10^{-6}\). The acid dissociation constants of N\(_2\)H\(_5^+\) and N\(_2\)H\(_6^{2+}\) are respectively:
(a) \(2.5 \times 10^{-9}\); \(4 \times 10^{-6}\)
(b) data insufficient; \(2.5 \times 10^{-8}\)
(c) data insufficient; \(4 \times 10^{-6}\)
(d) \(2.5 \times 10^{-9}\); data insufficient
Show Answer & Explanation

Answer: (d) \(2.5 \times 10^{-9}\); data insufficient

Explanation:
1. \(K_a(N_2H_5^+) = K_w/K_{b1} = 10^{-14}/(4 \times 10^{-6}) = 2.5 \times 10^{-9}\).
2. \(K_a(N_2H_6^{2+}) = K_w/K_{b2}\), and \(K_{b2}\) is not given.

Teacher's Note:
1. Each acid K\(_a\) pairs with a different base K\(_b\) of the sequence.

Question 11: In 0.1 M H\(_3\)PO\(_4\) (\(K_{a1} = 10^{-3}\), \(K_{a2} = 10^{-7}\), \(K_{a3} = 10^{-12}\)), the approximate pH is:
(a) 5
(b) 2
(c) 1
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. The quadratic \(x^2/(0.1 - x) = 10^{-3}\) gives \(x = 0.0095 \approx 0.01\) M.
2. \(pH \approx 2\).

Teacher's Note:
1. Only the first step contributes; check that x is less than 10% of C (here it is just under).

Question 12: In the same 0.1 M H\(_3\)PO\(_4\) solution (\([H^+] \approx 0.01\) M, \([H_2PO_4^-] \approx 0.01\) M), the concentration of PO\(_4^{3-}\) is:
(a) \(10^{-15}\) M
(b) \(10^{-12}\) M
(c) \(10^{-20}\) M
(d) \(10^{-17}\) M
Show Answer & Explanation

Answer: (d) \(10^{-17}\) M

Explanation:
1. \([HPO_4^{2-}] = K_{a2} = 10^{-7}\) M.
2. \([PO_4^{3-}] = K_{a3}[HPO_4^{2-}]/[H^+] = 10^{-12} \times 10^{-7}/10^{-2} = 10^{-17}\) M.

Teacher's Note:
1. Each successive ion is smaller by a large factor.

Question 13: Ionisation from the second step is negligible in aqueous solutions of weak polyacidic bases because:
(a) the second step is not possible in water
(b) only the common ion effect operates
(c) \(K_{b2} \ll K_{b1}\) and also the OH\(^-\) from the first step suppresses the second ionisation
(d) only \(K_{b2} \ll K_{b1}\)
Show Answer & Explanation

Answer: (c) \(K_{b2} \ll K_{b1}\) and also the OH\(^-\) from the first step suppresses the second ionisation

Explanation:
1. Each later step has a far smaller constant.
2. The OH\(^-\) already produced in step 1 is a common ion for step 2 and shifts it back.
3. Both reasons apply.

Teacher's Note:
1. The same argument holds for polyprotic acids.

Question 14: The degree of dissociation of 0.01 M acetic acid (\(K_a = 1.69 \times 10^{-5}\)) in the presence of 0.01 M HCl is:
(a) \(1.69 \times 10^{-5}\)
(b) \(2.9 \times 10^{-2}\)
(c) \(1.69 \times 10^{-3}\)
(d) \(1.69 \times 10^{-7}\)
Show Answer & Explanation

Answer: (c) \(1.69 \times 10^{-3}\)

Explanation:
1. \([H^+] \approx 0.01\) M from HCl.
2. \(\alpha = K_a/[H^+] = 1.69 \times 10^{-5}/0.01 = 1.69 \times 10^{-3}\).

Teacher's Note:
1. Without HCl, \(\alpha\) would be 0.041.

Question 15: 100 mL of \(10^{-2}\) M NaOH is added to 100 mL of 0.01 M triethylamine (\(K_b = 6.4 \times 10^{-5}\)). The change in pH relative to the triethylamine solution is:
(a) -0.8
(b) -1.1
(c) +1.1
(d) +0.8
Show Answer & Explanation

Answer: (d) +0.8

Explanation:
1. Original: \([OH^-] = \sqrt{6.4 \times 10^{-5} \times 0.01} = 8 \times 10^{-4}\), pOH 3.1.
2. After mixing, \([OH^-] = 5 \times 10^{-3}\) M from NaOH, pOH 2.3.
3. pH rises by 0.8.

Teacher's Note:
1. The strong base dominates the OH\(^-\) concentration.

Question 16: In the resulting solution of the previous question, the concentration of the triethylammonium ion is:
(a) \(K_b\) (numerically)
(b) \(100K_b\)
(c) \(200K_b\)
(d) \(10K_b\)
Show Answer & Explanation

Answer: (a) \(K_b\) (numerically)

Explanation:
1. \([BH^+] = K_b[B]/[OH^-]\).
2. With [B] = 0.005 M and [OH\(^-\)] = 0.005 M, the ratio is 1 and \([BH^+] = K_b\).

Teacher's Note:
1. [B] and [OH\(^-\)] have the same value here.

Question 17: In a 0.03 M HA (\(K_a = 3 \times 10^{-4}\)) and 0.1 M HB (\(K_a = 1 \times 10^{-10}\)) solution, which are correct? (I) \([H^+] = 3 \times 10^{-3}\) M (II) \([A^-] = 3 \times 10^{-3}\) M (III) \([B^-] = 3.33 \times 10^{-9}\) M
(a) I, II and III
(b) I and III only
(c) I and II only
(d) II and III only
Show Answer & Explanation

Answer: (a) I, II and III

Explanation:
1. HA gives \([H^+] = \sqrt{0.03 \times 3 \times 10^{-4}} = 3 \times 10^{-3}\) M; HB adds almost nothing.
2. \([A^-] = [H^+] = 3 \times 10^{-3}\) M.
3. \([B^-] = 10^{-10} \times 0.1/(3 \times 10^{-3}) = 3.33 \times 10^{-9}\) M.

Teacher's Note:
1. The stronger acid fixes [H\(^+\)] for both.

Question 18: Equal volumes of 0.1 M triethylamine (\(K_b = 6.4 \times 10^{-5}\)) and (4/45) M NH\(_4\)OH (\(K_b = 1.8 \times 10^{-5}\)) are mixed. The pH is:
(a) 11.45
(b) 11.3
(c) 10.3
(d) 12.3
Show Answer & Explanation

Answer: (b) 11.3

Explanation:
1. After mixing: 0.05 M and 2/45 M.
2. \([OH^-] = \sqrt{0.05 \times 6.4 \times 10^{-5} + (2/45) \times 1.8 \times 10^{-5}} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3}\).
3. \(pOH = 2.7\), \(pH = 11.3\).

Teacher's Note:
1. For a mixture of two weak bases add the contributions \(C K_b\).

Question 19: For BOH with \(K_b = 1.0 \times 10^{-12}\), the \([OH^-]\) in a 0.01 M solution is:
(a) \(2 \times 10^{-6}\) M
(b) \(\sqrt{2} \times 10^{-7}\) M
(c) \(10^{-5}\) M
(d) \(10^{-6}\) M
Show Answer & Explanation

Answer: (b) \(\sqrt{2} \times 10^{-7}\) M

Explanation:
1. \(\sqrt{K_bC} = 10^{-7}\) M is comparable to the \(10^{-7}\) M from water.
2. So \([OH^-]^2 = K_bC + K_w = 10^{-14} + 10^{-14}\).
3. \([OH^-] = \sqrt{2} \times 10^{-7}\) M.

Teacher's Note:
1. Include water's contribution when the base is extremely weak.

Question 20: Equal volumes of 0.02 M KOH and 0.2 M of a weak diacidic base B (\(K_{b1} = 10^{-7}\), \(K_{b2} = 10^{-14}\)) are mixed. Which holds? (I) pH = 12 (II) \([BH^+] = 10^{-6}\) M (III) \([BH_2^{2+}] = 10^{-18}\) M
(a) II and III only
(b) I and II only
(c) I and III only
(d) I, II and III
Show Answer & Explanation

Answer: (d) I, II and III

Explanation:
1. After mixing, [OH\(^-\)] = 0.01 M (pH 12) and [B] = 0.1 M.
2. \([BH^+] = K_{b1}[B]/[OH^-] = 10^{-7} \times 0.1/0.01 = 10^{-6}\) M.
3. \([BH_2^{2+}] = K_{b2}[BH^+]/[OH^-] = 10^{-14} \times 10^{-6}/0.01 = 10^{-18}\) M.

Teacher's Note:
1. Apply the stepwise constants in order.

Question 21: The pH of the solution obtained by mixing 100 mL of 0.2 M CH\(_3\)COOH with 100 mL of 0.2 M NaOH (\(pK_a = 4.74\)) is:
(a) 8.87
(b) 9.10
(c) 4.74
(d) 8.57
Show Answer & Explanation

Answer: (a) 8.87

Explanation:
1. The mixture is 0.1 M sodium acetate.
2. \(pH = \frac12(pK_w + pK_a + \log C) = \frac12(14 + 4.74 - 1) = 8.87\).

Teacher's Note:
1. Acetate is a weak base, so the pH is above 7.

Question 22: For a BH\(_2\)Cl\(_2\) solution (chloride salt of a diacidic base with \(K_{b1}\) and \(K_{b2}\)), which is correct?
(a) \(K_{h1} = K_w/K_{b1}\), and the anion hydrolyses to give a basic solution
(b) \(K_{h1} = K_w/K_{b2}\), and the cation hydrolyses to give an acidic solution
(c) \(K_{h1} \ll K_{h2}\), and the solution is basic
(d) \(K_{h1} = K_{b1}/K_w\)
Show Answer & Explanation

Answer: (b) \(K_{h1} = K_w/K_{b2}\), and the cation hydrolyses to give an acidic solution

Explanation:
1. The cation \(BH_2^{2+}\) is the conjugate acid of \(BH^+\), so its hydrolysis constant is \(K_w/K_{b2}\).
2. Cl\(^-\) comes from a strong acid and does not hydrolyse.
3. The cation produces H\(_3\)O\(^+\), so the solution is acidic.

Teacher's Note:
1. The first hydrolysis of the cation corresponds to the last base ionisation step.

Question 23: The pH values of 0.1 M Na\(_2\)HPO\(_4\) and 0.2 M NaH\(_2\)PO\(_4\) (pK\(_a\) of H\(_3\)PO\(_4\): 2.2, 7.2, 12) are respectively:
(a) 4.4 and 9.3
(b) 4.7 and 9.6
(c) 9.3 and 4.4
(d) 9.6 and 4.7
Show Answer & Explanation

Answer: (d) 9.6 and 4.7

Explanation:
1. HPO\(_4^{2-}\): \(\frac12(pK_{a2} + pK_{a3}) = \frac12(7.2 + 12) = 9.6\).
2. H\(_2\)PO\(_4^-\): \(\frac12(pK_{a1} + pK_{a2}) = \frac12(2.2 + 7.2) = 4.7\).

Teacher's Note:
1. Amphiprotic ions use the average of two neighbouring pK\(_a\) values.

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