JEE Chemistry Ionic Equilibrium MCQs Set 25

Multiple Choice Questions (MCQs) for JEE Chemistry: Ionic Equilibrium

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Multiple Choice Questions

Question 1: Which of the following are polyprotic Arrhenius acids: H\(_3\)PO\(_2\), H\(_3\)PO\(_3\), H\(_3\)BO\(_3\), HCOOH, (COOH)\(_2\)?
(a) H\(_3\)PO\(_3\) and (COOH)\(_2\) only
(b) H\(_3\)PO\(_2\) and HCOOH only
(c) H\(_3\)PO\(_3\), H\(_3\)BO\(_3\) and (COOH)\(_2\)
(d) H\(_3\)PO\(_2\), H\(_3\)BO\(_3\) and HCOOH
Show Answer & Explanation

Answer: (a) H\(_3\)PO\(_3\) and (COOH)\(_2\) only

Explanation:
1. H\(_3\)PO\(_3\) has two ionisable O-H groups and oxalic acid has two COOH groups.
2. H\(_3\)PO\(_2\) and HCOOH are monoprotic, and H\(_3\)BO\(_3\) is not an Arrhenius acid (it accepts OH\(^-\)).

Teacher's Note:
1. Count replaceable hydrogens that are bonded to oxygen.

Question 2: Classify H\(_2\)PO\(_2^-\), H\(_2\)PO\(_3^-\), H\(_2\)PO\(_4^-\), HPO\(_3^{2-}\), HPO\(_4^{2-}\), NH\(_4^+\) and CH\(_3\)COOH\(_2^+\) by the protonic concept. Which statement is correct?
(a) H\(_2\)PO\(_2^-\) and HPO\(_3^{2-}\) are amphiprotic
(b) H\(_2\)PO\(_3^-\), H\(_2\)PO\(_4^-\) and HPO\(_4^{2-}\) are amphiprotic; H\(_2\)PO\(_2^-\) and HPO\(_3^{2-}\) are bases only; NH\(_4^+\) and CH\(_3\)COOH\(_2^+\) are acids only
(c) NH\(_4^+\) is amphiprotic
(d) all of them are amphiprotic
Show Answer & Explanation

Answer: (b) H\(_2\)PO\(_3^-\), H\(_2\)PO\(_4^-\) and HPO\(_4^{2-}\) are amphiprotic; H\(_2\)PO\(_2^-\) and HPO\(_3^{2-}\) are bases only; NH\(_4^+\) and CH\(_3\)COOH\(_2^+\) are acids only

Explanation:
1. H\(_3\)PO\(_2\) has one acidic H and H\(_3\)PO\(_3\) two, so H\(_2\)PO\(_2^-\) and HPO\(_3^{2-}\) have no acidic hydrogen and can only accept protons.
2. H\(_2\)PO\(_3^-\), H\(_2\)PO\(_4^-\) and HPO\(_4^{2-}\) still carry acidic H and have lone pairs.
3. NH\(_4^+\) and protonated acetic acid have no lone pair to accept H\(^+\).

Teacher's Note:
1. An amphiprotic species needs both an acidic proton and a basic site.

Question 3: At \(-50^\circ\)C liquid NH\(_3\) has an ionic product of \(10^{-30}\). The number of amide (NH\(_2^-\)) ions per mm\(^3\) of pure liquid NH\(_3\) is (\(N_A = 6 \times 10^{23}\)):
(a) 600
(b) 6
(c) 0.6
(d) 6 x 10\(^{5}\)
Show Answer & Explanation

Answer: (a) 600

Explanation:
1. \([NH_2^-] = \sqrt{10^{-30}} = 10^{-15}\) mol/L.
2. 1 mm\(^3\) = \(10^{-6}\) L, so ions = \(10^{-15} \times 10^{-6} \times 6 \times 10^{23} = 600\).

Teacher's Note:
1. Convert litres to mm\(^3\) carefully.

Question 4: The \(K_b\) of OCl\(^-\) is \(4 \times 10^{-10}\). The \(K_a\) and \(pK_a\) of HOCl are:
(a) \(2.5 \times 10^{-5}\) and 9.4
(b) \(4 \times 10^{-10}\) and 9.4
(c) \(2.5 \times 10^{-5}\) and 4.6
(d) \(4 \times 10^{-24}\) and 23.4
Show Answer & Explanation

Answer: (c) \(2.5 \times 10^{-5}\) and 4.6

Explanation:
1. \(K_a = K_w/K_b = 10^{-14}/(4 \times 10^{-10}) = 2.5 \times 10^{-5}\).
2. \(pK_a = 5 - \log 2.5 = 4.6\).

Teacher's Note:
1. For a conjugate pair, \(K_aK_b = K_w\).

Question 5: The pH values of (i) 0.001 M HNO\(_3\), (ii) 0.005 M H\(_2\)SO\(_4\), (iii) 0.01 M KOH, (iv) \(10^{-8}\) M NaOH and (v) 0.0008 M Ba(OH)\(_2\) are respectively:
(a) 3, 2, 12, 8, 10.8
(b) 3, 2.3, 12, 8, 11.9
(c) 3, 2, 2, 6, 2.8
(d) 3, 2, 12, 7.02, 11.2
Show Answer & Explanation

Answer: (d) 3, 2, 12, 7.02, 11.2

Explanation:
1. (i) pH = 3. (ii) \([H^+] = 0.01\), pH = 2. (iii) pOH = 2, pH = 12.
2. (iv) The OH\(^-\) of water cannot be ignored: \([OH^-] = 1.05 \times 10^{-7}\), so pH = 7.02.
3. (v) \([OH^-] = 1.6 \times 10^{-3}\), pOH = 2.8, pH = 11.2.

Teacher's Note:
1. A very dilute base cannot give a pH above 7.02 here.

Question 6: The pH of 2.21 g of a strong base TOH (molar mass 221) dissolved in water to give 2 L of solution is:
(a) 2.3
(b) 11.7
(c) 12.0
(d) 10.7
Show Answer & Explanation

Answer: (b) 11.7

Explanation:
1. Moles = 0.01, so \([OH^-] = 0.005\) M.
2. pOH = 2.3, pH = 11.7.

Teacher's Note:
1. Find molarity before the pOH.

Question 7: 1 mL of 12 M HCl is diluted to 1 L. The pH is:
(a) 3.0
(b) 1.92
(c) 2.08
(d) 1.0
Show Answer & Explanation

Answer: (b) 1.92

Explanation:
1. Concentration = 12 x 1/1000 = 0.012 M.
2. pH = 2 - log 1.2 = 1.92.

Teacher's Note:
1. Dilution factor 1000.

Question 8: The pH of the solution obtained by mixing 10 mL of 0.2 M HCl and 40 mL of 0.1 M H\(_2\)SO\(_4\) is:
(a) 0.7
(b) 0.3
(c) 1.0
(d) 1.3
Show Answer & Explanation

Answer: (a) 0.7

Explanation:
1. H\(^+\) = 2 + 8 = 10 mmol in 50 mL, so [H\(^+\)] = 0.2 M.
2. pH = 0.7.

Teacher's Note:
1. Add millimoles of H\(^+\) from both acids.

Question 9: The pH of a 500 mL solution of 1 M weak base BOH (\(K_b = 2.5 \times 10^{-5}\)) is:
(a) 2.3
(b) 9.3
(c) 12.0
(d) 11.7
Show Answer & Explanation

Answer: (d) 11.7

Explanation:
1. Molarity is 1 M (volume does not matter).
2. \([OH^-] = \sqrt{2.5 \times 10^{-5}} = 5 \times 10^{-3}\) M, so pOH = 2.3 and pH = 11.7.

Teacher's Note:
1. Volume is irrelevant once molarity is known.

Question 10: A 0.25 M solution of pyridinium chloride (C\(_5\)H\(_5\)NH\(^+\)Cl\(^-\)) has pH 2.75. The \(K_b\) of pyridine is nearest to:
(a) \(1.6 \times 10^{-9}\)
(b) \(1.8 \times 10^{-5}\)
(c) \(8 \times 10^{-10}\)
(d) \(3 \times 10^{-5}\)
Show Answer & Explanation

Answer: (c) \(8 \times 10^{-10}\)

Explanation:
1. \([H^+] = 1.78 \times 10^{-3}\) M, so \(K_h = [H^+]^2/C = 1.27 \times 10^{-5}\).
2. \(K_b = K_w/K_h = 10^{-14}/(1.27 \times 10^{-5}) = 7.9 \times 10^{-10}\), nearest to \(8 \times 10^{-10}\).

Teacher's Note:
1. Convert from \(K_h\) to \(K_b\).

Question 11: The percentage hydrolysis and the pH of 0.02 M CH\(_3\)COONH\(_4\) (\(K_a = K_b = 1.6 \times 10^{-5}\)) are:
(a) 6.25% and 7
(b) 1.25% and 7
(c) 0.625% and 9
(d) 0.625% and 7
Show Answer & Explanation

Answer: (d) 0.625% and 7

Explanation:
1. \(K_h = K_w/(K_aK_b) = 3.9 \times 10^{-5}\), so \(h/(1-h) = \sqrt{K_h} = 6.25 \times 10^{-3}\), h = 0.62%.
2. Since \(K_a = K_b\), \(pH = 7\).

Teacher's Note:
1. The pH of such a salt is independent of concentration.

Question 12: Select the incorrect statement.
(a) H\(_3\)PO\(_4\) is a tribasic acid
(b) Sr(OH)\(_2\) is a strong diacidic base
(c) H\(_3\)BO\(_3\) is not an Arrhenius acid
(d) NH\(_3\).H\(_2\)O is a strong monoacidic base
Show Answer & Explanation

Answer: (d) NH\(_3\).H\(_2\)O is a strong monoacidic base

Explanation:
1. Ammonia solution is a weak base.
2. The other statements are correct.

Teacher's Note:
1. Check strength versus acidity (number of OH\(^-\)).

Question 13: In \(HC_2O_4^- + PO_4^{3-} \rightleftharpoons HPO_4^{2-} + C_2O_4^{2-}\), the two Bronsted bases are:
(a) HC\(_2\)O\(_4^-\) and HPO\(_4^{2-}\)
(b) HPO\(_4^{2-}\) and C\(_2\)O\(_4^{2-}\)
(c) PO\(_4^{3-}\) and C\(_2\)O\(_4^{2-}\)
(d) HC\(_2\)O\(_4^-\) and PO\(_4^{3-}\)
Show Answer & Explanation

Answer: (c) PO\(_4^{3-}\) and C\(_2\)O\(_4^{2-}\)

Explanation:
1. Forward, PO\(_4^{3-}\) accepts H\(^+\) (base).
2. Reverse, C\(_2\)O\(_4^{2-}\) accepts H\(^+\) (base).

Teacher's Note:
1. Bases are proton acceptors in either direction.

Question 14: When HClO\(_4\) dissolves in HF solvent, \(HF + HClO_4 \rightleftharpoons ClO_4^- + H_2F^+\). Which is a conjugate acid-base pair?
(a) HF and H\(_2\)F\(^+\)
(b) HF and ClO\(_4^-\)
(c) HF and HClO\(_4\)
(d) HClO\(_4\) and H\(_2\)F\(^+\)
Show Answer & Explanation

Answer: (a) HF and H\(_2\)F\(^+\)

Explanation:
1. H\(_2\)F\(^+\) differs from HF by one proton.
2. Similarly HClO\(_4\) and ClO\(_4^-\) are a pair.

Teacher's Note:
1. A conjugate pair differs by exactly one H\(^+\).

Question 15: Which equation correctly describes boric acid in water?
(a) \(H_3BO_3 + H_2O \rightleftharpoons H_3O^+ + H_2BO_3^-\)
(b) \(H_3BO_3 + 2H_2O \rightleftharpoons 2H_3O^+ + HBO_3^{2-}\)
(c) \(H_3BO_3 + H_2O \rightleftharpoons B(OH)_4^- + H^+\)
(d) \(H_3BO_3 + 3H_2O \rightleftharpoons 3H_3O^+ + BO_3^{3-}\)
Show Answer & Explanation

Answer: (c) \(H_3BO_3 + H_2O \rightleftharpoons B(OH)_4^- + H^+\)

Explanation:
1. Boric acid is a Lewis acid: boron accepts OH\(^-\) from water to form the tetrahedral B(OH)\(_4^-\) and releases H\(^+\).

Teacher's Note:
1. It is monobasic through Lewis behaviour.

Question 16: Which expression is NOT true?
(a) \([H^+] < \sqrt{K_w}\) and \([OH^-] > \sqrt{K_w}\) for an alkaline solution
(b) \([H^+] = [OH^-] = 10^{-7}\) M for a neutral solution at all temperatures
(c) \([H^+] = [OH^-] = \sqrt{K_w}\) for a neutral solution at all temperatures
(d) \([H^+] > \sqrt{K_w}\) and \([OH^-] < \sqrt{K_w}\) for an acidic solution
Show Answer & Explanation

Answer: (b) \([H^+] = [OH^-] = 10^{-7}\) M for a neutral solution at all temperatures

Explanation:
1. Neutral means \([H^+] = [OH^-] = \sqrt{K_w}\), which equals \(10^{-7}\) M only at 25\(^\circ\)C because \(K_w\) changes with temperature.

Teacher's Note:
1. Neutral does not always mean pH 7.

Question 17: \(K_w\) of water at 373 K is \(1 \times 10^{-12}\). Which statement is incorrect?
(a) pH + pOH = 12 for every aqueous solution at 373 K
(b) pH of water is 6 at 373 K
(c) Water is acidic at 373 K
(d) The ionisation of water has increased relative to 298 K
Show Answer & Explanation

Answer: (c) Water is acidic at 373 K

Explanation:
1. At 373 K, \([H^+] = [OH^-] = 10^{-6}\) M, so pH = pOH = 6 and water is still neutral.
2. So 'water is acidic' is the incorrect statement.

Teacher's Note:
1. Neutrality is defined by \([H^+] = [OH^-]\).

Question 18: In pure HCOOH liquid, \([HCOO^-] = 10^{-3}\) M at 27\(^\circ\)C. The self-ionisation constant \(K = [HCOOH_2^+][HCOO^-]\) is:
(a) \(10^{6}\)
(b) \(10^{-6}\)
(c) \(10^{3}\)
(d) \(10^{-3}\)
Show Answer & Explanation

Answer: (b) \(10^{-6}\)

Explanation:
1. \([HCOOH_2^+] = [HCOO^-] = 10^{-3}\) M.
2. \(K = 10^{-3} \times 10^{-3} = 10^{-6}\).

Teacher's Note:
1. Same idea as \(K_w\).

Question 19: How many moles of NaOH must be removed from one litre of its aqueous solution to change the pH from 12 to 11?
(a) 0.01
(b) 0.009
(c) 0.09
(d) 0.1
Show Answer & Explanation

Answer: (b) 0.009

Explanation:
1. At pH 12, [OH\(^-\)] = 0.01 M; at pH 11, 0.001 M.
2. Amount removed = 0.01 - 0.001 = 0.009 mol.

Teacher's Note:
1. Subtract the final from the initial moles.

Question 20: On adding 0.04 g of solid NaOH to 100 mL of M/200 Ba(OH)\(_2\) solution, the change in pH is:
(a) +0.7
(b) -0.3
(c) 0
(d) +0.3
Show Answer & Explanation

Answer: (d) +0.3

Explanation:
1. Initially \([OH^-] = 2/200 = 0.01\) M.
2. NaOH adds 0.001 mol in 0.1 L = 0.01 M, so \([OH^-] = 0.02\) M.
3. pH rises from 12 to 12.3.

Teacher's Note:
1. Doubling [OH\(^-\)] raises pH by 0.3.

Question 21: On mixing equal volumes of a strong acid HA solution and a strong base BOH solution, the pH of the resulting solution:
(a) may be less than 7 only
(b) may be greater than 7 only
(c) may be less than 7 or greater than 7
(d) will be 7
Show Answer & Explanation

Answer: (c) may be less than 7 or greater than 7

Explanation:
1. If the acid is more concentrated, the mixture is acidic; if the base is more concentrated, it is basic.
2. Only equal concentrations give pH 7.

Teacher's Note:
1. The answer depends on which reagent is in excess.

Question 22: Which solution has a pH of exactly 1.0?
(a) 55 mL of M/10 HCl + 45 mL of M/10 NaOH
(b) 10 mL of M/10 HCl + 90 mL of M/10 NaOH
(c) 75 mL of M/5 HCl + 25 mL of M/5 NaOH
(d) 100 mL of M/10 HCl + 100 mL of M/10 NaOH
Show Answer & Explanation

Answer: (c) 75 mL of M/5 HCl + 25 mL of M/5 NaOH

Explanation:
1. 75 mL of 0.2 M HCl = 15 mmol and 25 mL of 0.2 M NaOH = 5 mmol.
2. 10 mmol H\(^+\) in 100 mL gives 0.1 M, pH = 1.
3. The other mixtures give pH 2, 7 and 12.

Teacher's Note:
1. Compute the excess H\(^+\) in each mixture.

Question 23: Which of 1 M, 0.1 M, 0.001 M and 0.0001 M NH\(_3\) has the maximum degree of ionisation?
(a) 0.1 M
(b) 0.0001 M
(c) 0.001 M
(d) 1 M
Show Answer & Explanation

Answer: (b) 0.0001 M

Explanation:
1. \(\alpha = \sqrt{K_b/C}\) rises as C falls.
2. The most dilute solution has the highest degree of ionisation.

Teacher's Note:
1. Ostwald's dilution law.

Question 24: A monobasic weak acid has a 0.1 M solution of pH 4.5. Its \(K_a\) is:
(a) \(10^{-10}\)
(b) \(10 \times 10^{-4}\)
(c) \(10 \times 10^{-6}\)
(d) \(10^{-8}\)
Show Answer & Explanation

Answer: (d) \(10^{-8}\)

Explanation:
1. \([H^+] = 10^{-4.5}\), so \([H^+]^2 = 10^{-9}\).
2. \(K_a = 10^{-9}/0.1 = 10^{-8}\).

Teacher's Note:
1. \(K_a = [H^+]^2/C\).

Question 25: An aqueous solution of NH\(_4\)Cl is ______ in nature due to the behaviour of the ______ ion.
(a) acidic; Cl\(^-\)
(b) acidic; NH\(_4^+\)
(c) alkaline; NH\(_4^+\)
(d) alkaline; Cl\(^-\)
Show Answer & Explanation

Answer: (b) acidic; NH\(_4^+\)

Explanation:
1. NH\(_4^+\) is the conjugate acid of a weak base and donates H\(^+\) to water.
2. Cl\(^-\) does not react with water.

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

Question 26: The chloride of a weak monoacidic base is hydrolysed 3% in its 0.1 M solution at 25\(^\circ\)C. The dissociation constant of the base is nearest to:
(a) \(1 \times 10^{-10}\)
(b) \(3.33 \times 10^{-10}\)
(c) \(1 \times 10^{-9}\)
(d) \(3.33 \times 10^{-9}\)
Show Answer & Explanation

Answer: (a) \(1 \times 10^{-10}\)

Explanation:
1. \(K_h = Ch^2 = 0.1 \times 9 \times 10^{-4} = 9 \times 10^{-5}\).
2. \(K_b = K_w/K_h = 1.1 \times 10^{-10}\), nearest to \(1 \times 10^{-10}\).

Teacher's Note:
1. Convert from \(K_h\) to \(K_b\).

Question 27: The degree of hydrolysis of a salt of a weak acid and weak base is 50% in 0.1 M solution. At 0.2 M the percentage hydrolysis is:
(a) none of these
(b) 50%
(c) 25%
(d) 100%
Show Answer & Explanation

Answer: (b) 50%

Explanation:
1. For a weak acid-weak base salt, \(h/(1-h) = \sqrt{K_h}\), independent of concentration.
2. So h stays at 50%.

Teacher's Note:
1. This type of salt has concentration-independent hydrolysis.

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