JEE Chemistry Equilibrium MCQs Set 22

Practice MCQs for JEE Chemistry Equilibrium

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Multiple Choice Questions

Question 1: The degree of ionisation of water at a certain temperature is \(1.8 \times 10^{-9}\). The \(K_a\) of water (taking [H\(_2\)O] = 55.55 M) is:
(a) \(10^{-14}\)
(b) \(1.8 \times 10^{-16}\)
(c) \(1.8 \times 10^{-12}\)
(d) \(10^{-7}\)
Show Answer & Explanation

Answer: (b) \(1.8 \times 10^{-16}\)

Explanation:
1. \(K_a = C\alpha^2/(1-\alpha) \approx 55.55 \times (1.8 \times 10^{-9})^2\).
2. \(= 1.8 \times 10^{-16}\).

Teacher's Note:
1. Here \(K_a = K_w/55.55\), not \(K_w\) itself.

Question 2: The pH of a \(10^{-2}\) M H\(_2\)SO\(_4\) solution prepared in \(10^{-3}\) M Na\(_2\)SO\(_4\) solution is nearly (\(\log 2 = 0.3\)):
(a) 2.3
(b) 1.7
(c) 2.0
(d) 7.0
Show Answer & Explanation

Answer: (b) 1.7

Explanation:
1. Treating H\(_2\)SO\(_4\) as fully ionised, \([H^+] = 2 \times 10^{-2}\) M.
2. Na\(_2\)SO\(_4\) is a strong electrolyte and exerts no common-ion effect on the strong acid.
3. \(pH = 2 - \log 2 = 1.7\).

Teacher's Note:
1. Count both protons of H\(_2\)SO\(_4\) here.

Question 3: 100 mL of a strong acid of pH 1 is mixed with 100 mL of another strong acid of pH 2. The pH of the mixture is nearly (\(\log 11 = 1.04\)):
(a) 1.26
(b) 0.96
(c) 1.50
(d) 1.74
Show Answer & Explanation

Answer: (a) 1.26

Explanation:
1. Total H\(^+\) = 0.1 x 0.1 + 0.1 x 0.01 = 0.011 mol in 0.2 L.
2. \([H^+] = 0.055\) M, so \(pH = 2 - \log 5.5 = 1.26\).

Teacher's Note:
1. Add moles of H\(^+\), then divide by the total volume.

Question 4: \(V_1\) mL of 0.1 M HNO\(_3\) is mixed with \(V_2\) mL of 0.1 M Sr(OH)\(_2\). The final solution is neutral when:
(a) \(2V_1 = V_2\)
(b) \(V_1 = V_2\)
(c) \(V_1 = 2V_2\)
(d) \(V_1 > 2V_2\) only
Show Answer & Explanation

Answer: (c) \(V_1 = 2V_2\)

Explanation:
1. HNO\(_3\) supplies 0.1\(V_1\) mmol H\(^+\); Sr(OH)\(_2\) supplies 0.2\(V_2\) mmol OH\(^-\).
2. Neutrality needs \(0.1V_1 = 0.2V_2\), i.e. \(V_1 = 2V_2\).

Teacher's Note:
1. Sr(OH)\(_2\) is diacidic, so each mole gives two OH\(^-\).

Question 5: The dissociation constants of two bases B\(_1\)OH and B\(_2\)OH are \(2.56 \times 10^{-4}\) and \(1.6 \times 10^{-5}\). The ratio of their relative strengths (proportional to the degree of dissociation at the same concentration) is:
(a) 4 : 1
(b) 1 : 4
(c) 16 : 1
(d) 1 : 16
Show Answer & Explanation

Answer: (a) 4 : 1

Explanation:
1. At equal concentration \(\alpha \propto \sqrt{K_b}\).
2. \(\sqrt{2.56 \times 10^{-4}} : \sqrt{1.6 \times 10^{-5}} = 0.016 : 0.004 = 4 : 1\).

Teacher's Note:
1. Relative strength goes as the square root of K, not K itself.

Question 6: The pH of a solution containing 0.2 M acetic acid (\(K_a = 1.75 \times 10^{-5}\)) and 0.1 M benzoic acid (\(K_a = 6.5 \times 10^{-5}\)) is:
(a) 5.3
(b) 2.65
(c) 2.5
(d) 2.4
Show Answer & Explanation

Answer: (c) 2.5

Explanation:
1. \([H^+] = \sqrt{0.2 \times 1.75 \times 10^{-5} + 0.1 \times 6.5 \times 10^{-5}} = \sqrt{10^{-5}} = 3.16 \times 10^{-3}\) M.
2. \(pH = 2.5\).

Teacher's Note:
1. Add the \(CK_a\) contributions of the two weak acids.

Question 7: For a weak triprotic acid H\(_3\)A (\(K_{a1} = 10^{-5}\), \(K_{a2} = 10^{-9}\), \(K_{a3} = 10^{-13}\)) at 0.1 M, \(pX\) where \(X = [A^{3-}]/[HA^{2-}]\) is:
(a) 10
(b) 8
(c) 11
(d) 9
Show Answer & Explanation

Answer: (a) 10

Explanation:
1. \([H^+] \approx \sqrt{K_{a1}C} = 10^{-3}\) M.
2. \(X = K_{a3}/[H^+] = 10^{-13}/10^{-3} = 10^{-10}\), so \(pX = 10\).

Teacher's Note:
1. From the third equilibrium, \([A^{3-}]/[HA^{2-}] = K_{a3}/[H^+]\).

Question 8: FeCl\(_3\) dissolved in water gives a solution that is:
(a) basic
(b) acidic
(c) neutral
(d) amphoteric
Show Answer & Explanation

Answer: (b) acidic

Explanation:
1. Fe\(^{3+}\) is a highly charged cation that hydrolyses, releasing H\(^+\).
2. Cl\(^-\) does not hydrolyse, so the solution is acidic.

Teacher's Note:
1. Salts of a strong acid with a weak base are acidic.

Question 9: A 0.01 M solution of NaA (A\(^-\) from a weak acid HA) has a degree of hydrolysis of 0.01. The pH and the \(K_a\) of HA are respectively:
(a) 10 and \(10^{-6}\)
(b) 4 and \(10^{-8}\)
(c) 10 and \(10^{-8}\)
(d) 4 and \(10^{-6}\)
Show Answer & Explanation

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

Explanation:
1. \([OH^-] = Ch = 10^{-4}\) M, so pOH = 4 and pH = 10.
2. \(K_h = Ch^2 = 10^{-6}\), so \(K_a = K_w/K_h = 10^{-8}\).

Teacher's Note:
1. Hydrolysis of the anion makes the solution basic.

Question 10: 10 mL of M/5 CH\(_3\)COOH and 10 mL of M/5 NH\(_4\)OH are mixed (\(pK_a = 4.76\), \(pK_b = 4.74\)). The change in pH when each solution is diluted to double the volume is:
(a) -0.3
(b) +0.3
(c) +1
(d) no change
Show Answer & Explanation

Answer: (d) no change

Explanation:
1. The mixture is ammonium acetate, whose pH is \(7 + \frac12(pK_a - pK_b)\), independent of concentration.
2. So dilution causes no change in pH.

Teacher's Note:
1. The pH of a salt of a weak acid and a weak base does not depend on concentration.

Question 11: H\(_2\)A is a diprotic acid with \(K_{a1} = 10^{-7}\) and \(K_{a2} = 10^{-11}\). The solution with a pH closest to 9 is:
(a) 0.1 M Na\(_2\)A
(b) 0.1 M H\(_2\)A
(c) 0.1 M NaHA + 0.1 M Na\(_2\)A
(d) 0.1 M NaHA
Show Answer & Explanation

Answer: (d) 0.1 M NaHA

Explanation:
1. NaHA: \(pH = \frac12(7 + 11) = 9\).
2. H\(_2\)A is about 4, Na\(_2\)A about 12 and the NaHA + Na\(_2\)A buffer equals \(pK_{a2} = 11\).

Teacher's Note:
1. An amphiprotic ion has a pH midway between two pK values.

Question 12: Consider: S1: A mixture of 1 mol each of HCl, NaOH and CH\(_3\)COONH\(_4\) has pH 7 if \(pK_a\)(CH\(_3\)COOH) = \(pK_b\)(NH\(_3\)). S2: Methyl orange can be used in the titration of CH\(_3\)COOH with NaOH. S3: Water acts as an acid when ammonia dissolves in it. The truth values of S1, S2, S3 are:
(a) T F F
(b) F T T
(c) T T T
(d) T F T
Show Answer & Explanation

Answer: (d) T F T

Explanation:
1. S1 is true: HCl and NaOH cancel, leaving a neutral ammonium acetate solution.
2. S2 is false: the end point is basic, so phenolphthalein is needed.
3. S3 is true: water donates H\(^+\) to NH\(_3\).

Teacher's Note:
1. Indicator choice depends on the pH at the equivalence point.

Question 13: HCl gas is passed into 0.1 M 1-aminopropane (\(K_b = 5 \times 10^{-4}\)) until the pH is 9.7. The ratio \([PrNH_2]/[PrNH_3^+]\) is (\(\log 5 = 0.7\)):
(a) 0.1
(b) 10
(c) 0.25
(d) 4
Show Answer & Explanation

Answer: (a) 0.1

Explanation:
1. pOH = 4.3, so \([OH^-] = 5 \times 10^{-5}\) M.
2. \(K_b = [PrNH_3^+][OH^-]/[PrNH_2]\) gives \([PrNH_2]/[PrNH_3^+] = [OH^-]/K_b = 0.1\).

Teacher's Note:
1. Use the base constant with the known [OH\(^-\)].

Question 14: 8 g of a weak acid HX (molar mass 80, \(K_a = 10^{-4}\)) is dissolved in 100 mL water and titrated with 0.25 M NaOH. The pH at the equivalence point is (\(\log 2 = 0.3\)):
(a) 8.7
(b) 8.65
(c) 9.0
(d) 9.35
Show Answer & Explanation

Answer: (b) 8.65

Explanation:
1. HX = 0.1 mol needs 400 mL of NaOH; total volume 500 mL, so \([X^-] = 0.2\) M.
2. \(pH = \frac12(14 + 4 + \log 0.2) = \frac12(18 - 0.7) = 8.65\).

Teacher's Note:
1. At the equivalence point, find the salt's hydrolysis pH.

Question 15: 10 mL of M/5 CH\(_3\)COONa is titrated with M/5 HCl (\(pK_a\) of acetic acid = 4.76). The pH at the equivalence point is:
(a) 0.7
(b) 2.88
(c) 1.88
(d) 1.0
Show Answer & Explanation

Answer: (b) 2.88

Explanation:
1. At equivalence all acetate has become acetic acid: 0.002 mol in 20 mL, i.e. 0.1 M.
2. \(pH = \frac12(pK_a - \log C) = \frac12(4.76 + 1) = 2.88\).

Teacher's Note:
1. The product is a weak acid, not a salt.

Question 16: The minimum volume of water needed to dissolve 10 mg of CaCO\(_3\) (molar mass 100, \(K_{sp} = 2.5 \times 10^{-9}\)), neglecting hydrolysis, is:
(a) 2 L
(b) 0.02 L
(c) 20 L
(d) 0.2 L
Show Answer & Explanation

Answer: (a) 2 L

Explanation:
1. \(s = \sqrt{2.5 \times 10^{-9}} = 5 \times 10^{-5}\) M.
2. Moles = \(10^{-4}\), so volume = \(10^{-4}/(5 \times 10^{-5}) = 2\) L.

Teacher's Note:
1. Volume = moles divided by solubility.

Question 17: Four sparingly soluble salts have the same \(K_{sp}\) (less than 1): AB, CD\(_2\), E\(_2\)F\(_2\) (E\(_2^{2+}\) and F\(^-\)) and GH\(_3\). Which is most soluble?
(a) CD\(_2\)
(b) AB
(c) E\(_2\)F\(_2\)
(d) GH\(_3\)
Show Answer & Explanation

Answer: (d) GH\(_3\)

Explanation:
1. Solubilities are \(\sqrt{K}\), \((K/4)^{1/3}\) and \((K/27)^{1/4}\) for AB, AB\(_2\) and AB\(_3\) types.
2. For example with \(K = 10^{-10}\): \(10^{-5}\), \(2.9 \times 10^{-4}\), \(1.4 \times 10^{-3}\) mol/L.
3. GH\(_3\) is most soluble.

Teacher's Note:
1. For a given K, more ions per formula unit means higher solubility.

Question 18: \(K_{sp}\) of Ag\(_2\)CrO\(_4\) is \(8.64 \times 10^{-13}\). How many times is its solubility in pure water greater than in 0.6 M Na\(_2\)CrO\(_4\)?
(a) 1000
(b) 10
(c) 100
(d) 129
Show Answer & Explanation

Answer: (c) 100

Explanation:
1. In water \(s = (K/4)^{1/3} = 6 \times 10^{-5}\) M.
2. In 0.6 M chromate \([Ag^+] = \sqrt{K/0.6} = 1.2 \times 10^{-6}\), so \(s' = 6 \times 10^{-7}\) M.
3. Ratio = 100.

Teacher's Note:
1. Solubility equals half of [Ag\(^+\)] for Ag\(_2\)CrO\(_4\).

Question 19: \(K_{sp}\) of SrCO\(_3\) is \(10^{-10}\) and of ZnCO\(_3\) is \(1.5 \times 10^{-11}\). When both are dissolved together the ratio \([Sr^{2+}]/[Zn^{2+}]\) is:
(a) 10/3
(b) 3/10
(c) 3/20
(d) 20/3
Show Answer & Explanation

Answer: (d) 20/3

Explanation:
1. Both share the same [CO\(_3^{2-}\)], so \([Sr^{2+}]/[Zn^{2+}] = K_{sp}(SrCO_3)/K_{sp}(ZnCO_3)\).
2. \(= 10^{-10}/(1.5 \times 10^{-11}) = 20/3\).

Teacher's Note:
1. The common anion cancels in the ratio.

Question 20: The solubility of Fe(OH)\(_3\) is maximum in:
(a) 0.1 M KOH
(b) 0.1 M NaOH
(c) 0.1 M H\(_2\)SO\(_4\)
(d) 0.1 M HCl
Show Answer & Explanation

Answer: (c) 0.1 M H\(_2\)SO\(_4\)

Explanation:
1. Acid removes OH\(^-\) and shifts the equilibrium forward.
2. 0.1 M H\(_2\)SO\(_4\) gives 0.2 M H\(^+\), the highest, so it dissolves the most.

Teacher's Note:
1. Hydroxides dissolve in acid; bases add a common ion.

Question 21: Statement-1: AgCN is more soluble in aqueous KCN than in pure water. Statement-2: In KCN(aq) the complex [Ag(CN)\(_2\)]\(^-\) forms and the solubility equilibrium of AgCN shifts backward. Choose the correct option.
(a) Statement-1 is true and Statement-2 is false
(b) Both are true but Statement-2 does not explain Statement-1
(c) Statement-1 is false and Statement-2 is true
(d) Both are true and Statement-2 explains Statement-1
Show Answer & Explanation

Answer: (a) Statement-1 is true and Statement-2 is false

Explanation:
1. Complex formation removes Ag\(^+\), shifting the dissolution equilibrium forward and increasing the solubility.
2. So Statement-1 is true and Statement-2 is false.

Teacher's Note:
1. Removing an ion from solution drives dissolution.

Question 22: What is the effect of pH on the solubility of M(CN)\(_2\) (neglect hydrolysis of M\(^{2+}\))?
(a) Solubility increases both below and above pH 7
(b) Solubility increases as pH decreases
(c) It does not depend on pH
(d) Solubility decreases as pH decreases
Show Answer & Explanation

Answer: (b) Solubility increases as pH decreases

Explanation:
1. CN\(^-\) is the conjugate base of the weak acid HCN and combines with H\(^+\).
2. Lowering the pH therefore removes CN\(^-\) and increases the solubility.

Teacher's Note:
1. Salts of weak acids dissolve more in acidic solution.

Question 23: The solubility product of a salt MX is \(3 \times 10^{-10}\). Its solubility is:
(a) \(6.9 \times 10^{-10}\) mol dm\(^{-3}\)
(b) \(9 \times 10^{-20}\) mol dm\(^{-3}\)
(c) \(1.73 \times 10^{-5}\) mol dm\(^{-3}\)
(d) \(1.73 \times 10^{-10}\) mol dm\(^{-3}\)
Show Answer & Explanation

Answer: (c) \(1.73 \times 10^{-5}\) mol dm\(^{-3}\)

Explanation:
1. \(s = \sqrt{K_{sp}} = \sqrt{3 \times 10^{-10}} = 1.73 \times 10^{-5}\) mol dm\(^{-3}\).

Teacher's Note:
1. For an MX salt, \(s = \sqrt{K_{sp}}\).

Question 24: 40 cm\(^3\) of 0.1 M HCl is mixed with 10 cm\(^3\) of 0.45 M NaOH. The pH of the solution is:
(a) 12
(b) 5
(c) 10
(d) 8
Show Answer & Explanation

Answer: (a) 12

Explanation:
1. NaOH = 4.5 mmol and HCl = 4.0 mmol, leaving 0.5 mmol OH\(^-\) in 50 mL.
2. \([OH^-] = 0.01\) M, so pOH = 2 and pH = 12.

Teacher's Note:
1. Find the excess reagent first.

Question 25: In the reaction \(HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+\), which two species are Bronsted bases?
(a) CO\(_3^{2-}\) and H\(_3\)O\(^+\)
(b) CO\(_3^{2-}\) and H\(_2\)O
(c) HCO\(_3^-\) and H\(_3\)O\(^+\)
(d) HCO\(_3^-\) and CO\(_3^{2-}\)
Show Answer & Explanation

Answer: (b) CO\(_3^{2-}\) and H\(_2\)O

Explanation:
1. In the forward reaction H\(_2\)O accepts a proton (base).
2. In the reverse reaction CO\(_3^{2-}\) accepts a proton (base).

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

Question 26: The solubility of AgCl at 30\(^\circ\)C is \(1.56 \times 10^{-5}\) mol dm\(^{-3}\). Its solubility product is:
(a) \(2.43 \times 10^{-10}\)
(b) \(1.25 \times 10^{-10}\)
(c) \(5.90 \times 10^{-20}\)
(d) \(1.25 \times 10^{-5}\)
Show Answer & Explanation

Answer: (a) \(2.43 \times 10^{-10}\)

Explanation:
1. \(K_{sp} = s^2 = (1.56 \times 10^{-5})^2 = 2.43 \times 10^{-10}\).

Teacher's Note:
1. For AgCl, \(K_{sp} = s^2\).

Question 27: The pH of 0.1 M CH\(_3\)COOH (\(K_a = 1.80 \times 10^{-5}\)) is:
(a) 5.0
(b) 0.18
(c) 1.0
(d) 2.873
Show Answer & Explanation

Answer: (d) 2.873

Explanation:
1. \([H^+] = \sqrt{1.8 \times 10^{-5} \times 0.1} = 1.34 \times 10^{-3}\) M.
2. \(pH = 2.873\).

Teacher's Note:
1. Use \(\sqrt{K_aC}\) for a weak acid.

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