JEE Chemistry Equilibrium MCQs Set 23

Download JEE MCQs for JEE Chemistry: Equilibrium

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Chapter-wise Objective Questions: Equilibrium

Access the complete set of multiple-choice questions for Equilibrium below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The pH of a soft drink is 3.82. Its hydrogen ion concentration in mol dm\(^{-3}\) is:
(a) \(1.96 \times 10^{-3}\)
(b) \(1.5 \times 10^{-4}\)
(c) \(1.96 \times 10^{-2}\)
(d) \(1.96 \times 10^{-1}\)
Show Answer & Explanation

Answer: (b) \(1.5 \times 10^{-4}\)

Explanation:
1. \([H^+] = 10^{-3.82} = 1.51 \times 10^{-4}\) M.

Teacher's Note:
1. \([H^+] = 10^{-pH}\).

Question 2: The colour change of an acid-base indicator is due to the formation of:
(a) ionic bond
(b) covalent bond
(c) benzoid structure
(d) quinoid structure
Show Answer & Explanation

Answer: (d) quinoid structure

Explanation:
1. Indicators such as phenolphthalein and methyl orange change between benzenoid and quinonoid resonance structures as the pH changes, altering the absorbed light.

Teacher's Note:
1. Ionisation changes the conjugation of the dye.

Question 3: The hydrolysis constant of a salt of a strong acid and a weak base is:
(a) \(K_h = K_w/K_b\)
(b) \(K_h = K_b/K_w\)
(c) \(K_h = K_w/(K_aK_b)\)
(d) \(K_h = K_w/K_a\)
Show Answer & Explanation

Answer: (a) \(K_h = K_w/K_b\)

Explanation:
1. The cation is the conjugate acid of the weak base, so its constant is \(K_a = K_w/K_b\).
2. That equals \(K_h\).

Teacher's Note:
1. Only the cation hydrolyses here.

Question 4: The pH of \(10^{-6}\) M KOH is nearest to:
(a) 8.0
(b) 6.0
(c) 7.0
(d) 9.0
Show Answer & Explanation

Answer: (a) 8.0

Explanation:
1. The OH\(^-\) from KOH (\(10^{-6}\) M) adds to that from water, giving \([OH^-] \approx 1.0 \times 10^{-6}\) M (more exactly \(1.05 \times 10^{-6}\) M).
2. pOH is nearly 6, so pH is nearly 8.

Teacher's Note:
1. A basic solution cannot have a pH below 7.

Question 5: H\(_2\)S is passed into acidified ZnSO\(_4\) solution but ZnS does not precipitate because:
(a) the ionic product does not exceed \(K_{sp}\) of ZnS
(b) \(K_{sp}\) of ZnS decreases
(c) \(K_{sp}\) of ZnS increases
(d) ZnS is soluble in water
Show Answer & Explanation

Answer: (a) the ionic product does not exceed \(K_{sp}\) of ZnS

Explanation:
1. Acid suppresses the ionisation of H\(_2\)S, so [S\(^{2-}\)] is very low.
2. The ionic product stays below \(K_{sp}\).
3. \(K_{sp}\) itself depends only on temperature.

Teacher's Note:
1. The ionic product, not \(K_{sp}\), changes.

Question 6: The \(pK_a\) of acetic acid is 4.74. This implies that:
(a) at pH 4.74, half of the acetic acid molecules are dissociated
(b) dissociation is maximum at pH 4.74
(c) the pH of 1 N acetic acid is 4.74
(d) dissociation is minimum at pH 4.74
Show Answer & Explanation

Answer: (a) at pH 4.74, half of the acetic acid molecules are dissociated

Explanation:
1. At \(pH = pK_a\), \([A^-] = [HA]\), so 50% is dissociated.

Teacher's Note:
1. Henderson's equation gives equal amounts when \(pH = pK_a\).

Question 7: The minimum volume of water required to dissolve 1.9 mg of AgCl (molar mass 143.5, \(K_{sp} = 1.8 \times 10^{-10}\)) is about:
(a) 2 L
(b) 10 L
(c) 1 L
(d) 20 L
Show Answer & Explanation

Answer: (c) 1 L

Explanation:
1. \(s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}\) M.
2. 1.9 mg = \(1.32 \times 10^{-5}\) mol, so volume is about 1 L.

Teacher's Note:
1. Divide moles by molar solubility.

Question 8: 50.0 mL of 0.1 N HCl is titrated with 0.1 N NaOH. The pH after 49.9 mL of NaOH is added is approximately:
(a) 7.0
(b) 3.0
(c) 6.0
(d) 4.0
Show Answer & Explanation

Answer: (d) 4.0

Explanation:
1. HCl left = 0.1 mL x 0.1 N = 0.01 meq in 99.9 mL (about 100 mL), so \([H^+] \approx 10^{-4}\) M.
2. pH is about 4.0.

Teacher's Note:
1. Just before the equivalence point the pH rises steeply.

Question 9: Methanoic acid in its 0.20 M solution has a degree of dissociation of 0.032. Its dissociation constant is:
(a) \(1.6 \times 10^{-8}\)
(b) \(2.1 \times 10^{-4}\)
(c) \(2.1 \times 10^{-2}\)
(d) \(1.1 \times 10^{-6}\)
Show Answer & Explanation

Answer: (b) \(2.1 \times 10^{-4}\)

Explanation:
1. \(K_a = C\alpha^2/(1 - \alpha) = 0.2 \times (0.032)^2/0.968 = 2.1 \times 10^{-4}\).

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

Question 10: A buffer is made from 0.050 mol HA and 0.20 mol NaA in 500 mL (\(K_a = 4.5 \times 10^{-4}\), \(\log 4 = 0.6\)). Its pH is:
(a) 2.17
(b) 2.74
(c) 3.95
(d) 1.97
Show Answer & Explanation

Answer: (c) 3.95

Explanation:
1. \(pK_a = 3.35\).
2. \(pH = 3.35 + \log(0.20/0.050) = 3.35 + 0.60 = 3.95\).

Teacher's Note:
1. Volume cancels in the salt/acid ratio.

Question 11: Among the species formed when H\(_3\)PO\(_4\) dissolves in water, the conjugate acid of HPO\(_4^{2-}\) is:
(a) H\(_3\)PO\(_4\)
(b) H\(_3\)O\(^+\)
(c) PO\(_4^{3-}\)
(d) H\(_2\)PO\(_4^-\)
Show Answer & Explanation

Answer: (d) H\(_2\)PO\(_4^-\)

Explanation:
1. The conjugate acid is formed by adding one proton: HPO\(_4^{2-} + H^+ \rightarrow H_2PO_4^-\).

Teacher's Note:
1. Add one proton to get the conjugate acid.

Question 12: Which of these salts will not change the pH of water when dissolved?
(a) NH\(_4\)Cl
(b) NaCl
(c) KCN
(d) Na\(_2\)CO\(_3\)
Show Answer & Explanation

Answer: (b) NaCl

Explanation:
1. NaCl is the salt of a strong acid and a strong base and does not hydrolyse.
2. The others hydrolyse to give basic or acidic solutions.

Teacher's Note:
1. No hydrolysis means a neutral solution.

Question 13: Which indicator is suitable for the titration of acetic acid against NaOH?
(a) methyl orange (3.7)
(b) phenolphthalein (\(pK_{In} = 9.6\))
(c) chlorophenol red (6.1)
(d) bromocresol green (4.7)
Show Answer & Explanation

Answer: (b) phenolphthalein (\(pK_{In} = 9.6\))

Explanation:
1. At the equivalence point sodium acetate gives a basic solution (pH about 8.7).
2. The indicator should change near that pH, so phenolphthalein is suitable.

Teacher's Note:
1. Match the indicator range with the equivalence-point pH.

Question 14: Which compound added to water raises its pH above 7?
(a) AlCl\(_3\)
(b) HCl
(c) Na\(_2\)CO\(_3\)
(d) NaCl
Show Answer & Explanation

Answer: (c) Na\(_2\)CO\(_3\)

Explanation:
1. Carbonate hydrolyses (\(CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\)) and makes the solution basic.

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

Question 15: Which aqueous solution is the most acidic?
(a) Na\(_2\)HPO\(_4\)
(b) CH\(_3\)COONa
(c) Na\(_2\)CO\(_3\)
(d) NH\(_4\)Cl
Show Answer & Explanation

Answer: (d) NH\(_4\)Cl

Explanation:
1. NH\(_4\)Cl is a salt of a strong acid and a weak base, so it is acidic.
2. The other three are basic.

Teacher's Note:
1. Only one of these is an acidic salt.

Question 16: 25 mL of 0.01 N NaOH is added to 50 mL of 0.01 N acetic acid (\(pK_a = 4.76\)). The pH is:
(a) 4.5
(b) 2.48
(c) 4.76
(d) 5.5
Show Answer & Explanation

Answer: (c) 4.76

Explanation:
1. Acid left = 0.25 mmol and acetate formed = 0.25 mmol (half neutralisation).
2. \(pH = pK_a = 4.76\).

Teacher's Note:
1. Half neutralisation gives pH equal to \(pK_a\).

Question 17: The pH of a solution made by mixing 150 cm\(^3\) of 0.10 M CH\(_3\)COONa and 250 cm\(^3\) of 0.10 M CH\(_3\)COOH (\(K_a = 1.8 \times 10^{-5}\)) is:
(a) 2.37
(b) 4.97
(c) 4.74
(d) 4.52
Show Answer & Explanation

Answer: (d) 4.52

Explanation:
1. \(pK_a = 4.74\).
2. \(pH = 4.74 + \log(15/25) = 4.74 - 0.22 = 4.52\).

Teacher's Note:
1. The volume factors cancel in the ratio.

Question 18: The pH of a 0.01 M solution of ammonium acetate can be changed by changing:
(a) the temperature
(b) the volume of the solution
(c) the concentration
(d) the pressure
Show Answer & Explanation

Answer: (a) the temperature

Explanation:
1. The pH of a weak acid-weak base salt is \(\frac12(pK_w + pK_a - pK_b)\), independent of concentration.
2. But \(K_w\), \(K_a\) and \(K_b\) vary with temperature, so the pH changes with temperature.

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

Question 19: The pH of a mixture of 100 mL of 0.25 N HCl and 100 mL of 0.125 N NaOH is:
(a) 1.20
(b) 0.90
(c) 0.42
(d) 0.70
Show Answer & Explanation

Answer: (a) 1.20

Explanation:
1. Acid left = 25 - 12.5 = 12.5 meq in 200 mL.
2. \([H^+] = 0.0625\) M, so pH = 1.20.

Teacher's Note:
1. Neutralise first, then dilute.

Question 20: CaF\(_2\) (molar mass 78) has a solubility of \(1.6 \times 10^{-2}\) g dm\(^{-3}\). Its solubility product is:
(a) \(4.20 \times 10^{-8}\)
(b) \(2.56 \times 10^{-4}\)
(c) \(3.45 \times 10^{-11}\)
(d) \(4.69 \times 10^{-9}\)
Show Answer & Explanation

Answer: (c) \(3.45 \times 10^{-11}\)

Explanation:
1. \(s = 1.6 \times 10^{-2}/78 = 2.05 \times 10^{-4}\) M.
2. \(K_{sp} = 4s^3 = 3.45 \times 10^{-11}\).

Teacher's Note:
1. Convert g/L to mol/L before using \(4s^3\).

Question 21: Which of these mixtures is a buffer? Mixture 1: 25 mL 0.10 M HNO\(_3\) + 25 mL 0.10 M NaNO\(_3\). Mixture 2: 25 mL 0.10 M CH\(_3\)COOH + 25 mL 0.10 M NaOH.
(a) 1 only
(b) neither 1 nor 2
(c) 2 only
(d) both 1 and 2
Show Answer & Explanation

Answer: (b) neither 1 nor 2

Explanation:
1. Mixture 1 contains a strong acid and its salt, which is not a buffer.
2. Mixture 2 gives only sodium acetate, since the acid is completely neutralised.
3. Neither is a buffer.

Teacher's Note:
1. A buffer needs excess weak acid or base.

Question 22: A 0.200 g sample of benzoic acid (molar mass 122.1) is titrated with 0.120 M Ba(OH)\(_2\). The volume needed to reach the equivalence point is:
(a) 13.6 mL
(b) 17.6 mL
(c) 35.2 mL
(d) 6.82 mL
Show Answer & Explanation

Answer: (d) 6.82 mL

Explanation:
1. Acid = 1.638 mmol.
2. Ba(OH)\(_2\) supplies 2 OH\(^-\), so mmol of base needed = 0.819, giving 0.819/0.120 = 6.82 mL.

Teacher's Note:
1. Ba(OH)\(_2\) is diacidic.

Question 23: Which indicator best suits titrating NaHCO\(_3\) with HCl (\(K_{a1}\) of H\(_2\)CO\(_3\) = \(2.5 \times 10^{-4}\), \(K_{a2} = 2.4 \times 10^{-8}\))?
(a) methyl orange (\(pK_{In} = 3.7\))
(b) phenolphthalein (9.3)
(c) bromothymol blue (7.0)
(d) alizarin yellow (12.5)
Show Answer & Explanation

Answer: (a) methyl orange (\(pK_{In} = 3.7\))

Explanation:
1. At the end point HCO\(_3^-\) has become H\(_2\)CO\(_3\) (\(pK_a = 3.6\)), giving an acidic solution of pH about 4.
2. Methyl orange changes in this range.

Teacher's Note:
1. The end point is acidic.

Question 24: Carbonic acid has \(K_1 = 4.2 \times 10^{-7}\) and \(K_2 = 4.7 \times 10^{-11}\). The 0.1 M solution with pH closest to 9 is:
(a) 0.1 M H\(_2\)CO\(_3\)
(b) 0.1 M NaHCO\(_3\) + 0.1 M Na\(_2\)CO\(_3\)
(c) 0.1 M Na\(_2\)CO\(_3\)
(d) 0.1 M NaHCO\(_3\)
Show Answer & Explanation

Answer: (d) 0.1 M NaHCO\(_3\)

Explanation:
1. NaHCO\(_3\): \(pH = \frac12(6.38 + 10.33) = 8.4\).
2. H\(_2\)CO\(_3\) is about 3.7, Na\(_2\)CO\(_3\) about 11.6 and the mixture 10.3.
3. 8.4 is closest to 9.

Teacher's Note:
1. Compare each solution's calculated pH.

Question 25: The strongest Bronsted base among these compounds is:
(a) NaNH\(_2\)
(b) NaOH
(c) NaF
(d) NaCH\(_3\)
Show Answer & Explanation

Answer: (d) NaCH\(_3\)

Explanation:
1. Basic strength follows the weakness of the conjugate acid.
2. CH\(_4\) is the weakest acid (largest pK) of CH\(_4\), H\(_2\)O, HF and NH\(_3\), so CH\(_3^-\) is the strongest base.

Teacher's Note:
1. Weaker conjugate acid means stronger base.

Question 26: For CaF\(_2\), \(K_{sp} = 1.6 \times 10^{-10}\). The moles that must dissolve in 2.0 L to form a saturated solution is:
(a) \(6.8 \times 10^{-4}\) mol
(b) \(2.6 \times 10^{-2}\) mol
(c) \(1.3 \times 10^{-3}\) mol
(d) \(3.4 \times 10^{-4}\) mol
Show Answer & Explanation

Answer: (a) \(6.8 \times 10^{-4}\) mol

Explanation:
1. \(s = (1.6 \times 10^{-10}/4)^{1/3} = 3.42 \times 10^{-4}\) M.
2. Moles in 2 L = \(6.8 \times 10^{-4}\).

Teacher's Note:
1. Multiply the molar solubility by the volume.

Question 27: A 0.10 M solution of NaZ has pH 8.90. The \(K_a\) of HZ is:
(a) \(1.6 \times 10^{-4}\)
(b) \(6.3 \times 10^{-11}\)
(c) \(6.3 \times 10^{-10}\)
(d) \(1.6 \times 10^{-5}\)
Show Answer & Explanation

Answer: (d) \(1.6 \times 10^{-5}\)

Explanation:
1. pOH = 5.10, so \([OH^-] = 7.9 \times 10^{-6}\) M.
2. \(K_h = [OH^-]^2/C = 6.3 \times 10^{-10}\).
3. \(K_a = K_w/K_h = 10^{-14}/(6.3 \times 10^{-10}) = 1.6 \times 10^{-5}\).

Teacher's Note:
1. Do not stop at K\(_h\); convert to K\(_a\).

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