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Ionization, pH, common-ion effect and salt hydrolysis: NEET MCQs with answers

73 questions on this topic of Equilibrium (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The ionization of acetic acid (CH₃COOH) is suppressed by the addition of sodium acetate (CH₃COONa). This phenomenon is governed by:

  1. Ostwald's dilution law
  2. Hess's law of constant heat summation
  3. Henry's law
  4. Le Chatelier's principle and common-ion effect
Show answer and explanation

Answer: (D) Le Chatelier's principle and common-ion effect

Adding CH₃COONa supplies common acetate ions (CH₃COO⁻). According to Le Chatelier's principle, the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts to the left, suppressing the ionization of the weak acid.

medium

2. The pH of a 0.005 M aqueous solution of NaOH at 25 °C (complete dissociation) is:

  1. 10.7
  2. 11.7
  3. 2.3
  4. 3
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Answer: (B) 11.7

[OH⁻] = 0.005 M, pOH = 2.3, so pH = 14 − 2.3 = 11.7.

hard

3. The hydrogen-ion concentration of a 0.1 M acetic acid solution (Ka = 1.8 × 10⁻⁵) is approximately:

  1. 1.8 × 10⁻⁶ mol/L
  2. 0.1 mol/L
  3. 0.0134 mol/L
  4. 0.00134 mol/L
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Answer: (D) 0.00134 mol/L

For a weak acid with small α, [H⁺] = √(Ka C) = √(1.8 × 10⁻⁵ × 0.1) = 0.00134 mol/L.

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4. The pH of a 0.01 M aqueous solution of NaOH at 25 °C (complete dissociation) is:

  1. 11
  2. 24
  3. 2
  4. 12
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Answer: (D) 12

[OH⁻] = 0.01 M, pOH = 2, so pH = 14 − 2 = 12.

easy

5. Which of the following salts undergoes both anionic and cationic hydrolysis, and produces an aqueous solution whose pH is practically independent of its concentration?

  1. NH₄Cl
  2. CH₃COONH₄
  3. CH₃COONa
  4. Na₂SO₄
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Answer: (B) CH₃COONH₄

Ammonium acetate (CH₃COONH₄) is formed from a weak acid (CH₃COOH) and a weak base (NH₄OH). Both ions hydrolyze simultaneously. The pH of such a salt solution is given by pH = 7 + 0.5(pKa - pKb), which is completely independent of the salt concentration C.

hard

6. At 25 °C, what is the hydroxyl ion concentration [OH⁻] in a 0.01 M aqueous solution of aniline hydrochloride (C₆H₅NH₃⁺Cl⁻)? (Given: Kb for aniline = 4.0 × 10⁻¹⁰, Kw = 1.0 × 10⁻¹⁴)

  1. 5.0 × 10⁻¹² M
  2. 5.0 × 10⁻⁴ M
  3. 2.0 × 10⁻³ M
  4. 2.0 × 10⁻¹¹ M
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Answer: (D) 2.0 × 10⁻¹¹ M

Aniline hydrochloride is a salt of a weak base and strong acid. For cationic hydrolysis: [H⁺] = √(Kh · C) = √((Kw · C) / Kb) = √((1.0 × 10⁻¹⁴ × 10⁻²) / (4.0 × 10⁻¹⁰)) = √(2.5 × 10⁻⁷) = √(25 × 10⁻⁸) = 5.0 × 10⁻⁴ M. Then, [OH⁻] = Kw / [H⁺] = (1.0 × 10⁻¹⁴) / (5.0 × 10⁻⁴) = 2.0 × 10⁻¹¹ M.

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7. The pH of a 0.1 M aqueous solution of NaOH at 25 °C (complete dissociation) is:

  1. 12
  2. 26
  3. 1
  4. 13
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Answer: (D) 13

[OH⁻] = 0.1 M, pOH = 1, so pH = 14 − 1 = 13.

medium

8. An aqueous solution of a weak base BOH has a concentration of 0.2 M and is 2.0% ionized at 25 °C. What is the value of the base dissociation constant (Kb)?

  1. 4.0 × 10⁻⁵
  2. 8.0 × 10⁻⁵
  3. 1.6 × 10⁻⁴
  4. 8.0 × 10⁻⁴
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Answer: (B) 8.0 × 10⁻⁵

The degree of ionization α = 2.0% = 0.02. For a weak base, Kb ≈ c × α² = (0.2)(0.02)² = 0.2 × 4.0 × 10⁻⁴ = 8.0 × 10⁻⁵.

easy

9. Which of the following salts dissolves in water to give a solution with pH > 7 due to anionic hydrolysis?

  1. NH₄NO₃
  2. FeCl₃
  3. NaCl
  4. KNO₂
Show answer and explanation

Answer: (D) KNO₂

KNO₂ is a salt formed from a strong base (KOH) and a weak acid (HNO₂). Its nitrite ion (NO₂⁻) undergoes anionic hydrolysis (NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻) releasing OH⁻ ions, making the aqueous solution alkaline (pH > 7).

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10. The pH of a 0.5 M aqueous solution of HCl at 25 °C (complete dissociation) is:

  1. 0.301
  2. 1.3
  3. 1
  4. 13.7
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Answer: (A) 0.301

HCl is a strong acid, so [H⁺] = 0.5 M and pH = −log(0.5) = 0.301.

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11. The pH of a 0.001 M aqueous solution of NaOH at 25 °C (complete dissociation) is:

  1. 22
  2. 11
  3. 3
  4. 10
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Answer: (B) 11

[OH⁻] = 0.001 M, pOH = 3, so pH = 14 − 3 = 11.

easy

12. For a polyprotic acid H₃PO₄, the three stepwise ionization constants are Ka₁, Ka₂, and Ka₃. Which of the following relationships correctly describes their relative magnitudes?

  1. Ka₁ = Ka₂ = Ka₃
  2. Ka₁ < Ka₂ < Ka₃
  3. Ka₂ > Ka₁ > Ka₃
  4. Ka₁ > Ka₂ > Ka₃
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Answer: (D) Ka₁ > Ka₂ > Ka₃

It is much more difficult to remove a positively charged proton from an increasingly negatively charged anion (H₂PO₄⁻ and HPO₄²⁻) due to electrostatic attraction. Consequently, Ka₁ >> Ka₂ >> Ka₃.

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