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Solubility products and buffer solutions: NEET MCQs with answers

80 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. Which of the following salts produces an alkaline solution when dissolved in pure water?

  1. NH₄Cl
  2. CH₃COONa
  3. NaCl
  4. (NH₄)₂SO₄
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Answer: (B) CH₃COONa

CH₃COONa is a salt of a weak acid (CH₃COOH) and a strong base (NaOH). Its anion undergoes hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, generating excess OH⁻ ions and yielding a basic (alkaline) solution with pH > 7.

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2. A buffer solution is prepared by mixing 50 mL of 0.2 M HCN with 50 mL of 0.1 M NaCN. Given that Ka for HCN is 4.0 × 10⁻¹⁰, what is the pH of the resulting mixture? (Take log 2 = 0.30)

  1. 9.70
  2. 10.00
  3. 8.80
  4. 9.10
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Answer: (D) 9.10

First find pKa: pKa = -log(4.0 × 10⁻¹⁰) = 10 - 2 log 2 = 10 - 0.60 = 9.40. Millimoles of HCN = 50 × 0.2 = 10 mmol, millimoles of NaCN = 50 × 0.1 = 5 mmol. Using the Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]) = 9.40 + log(5/10) = 9.40 - log 2 = 9.40 - 0.30 = 9.10.

hard

3. How many moles of solid sodium acetate (CH₃COONa) must be added to 500 mL of a 0.20 M acetic acid solution to prepare a buffer with pH = 5.04? (Given: Ka of CH₃COOH = 1.8 × 10⁻⁵; pKa = 4.74)

  1. 0.40 mol
  2. 0.10 mol
  3. 0.05 mol
  4. 0.20 mol
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Answer: (D) 0.20 mol

Using Henderson's equation: pH = pKa + log([Salt]/[Acid]) ⇒ 5.04 = 4.74 + log([Salt]/0.20) ⇒ log([Salt]/0.20) = 0.30 ⇒ [Salt]/0.20 = 10⁰°³⁰ = 2.0. Thus, [Salt] = 0.40 M. The number of moles required in 500 mL (0.5 L) is moles = Molarity × Volume = 0.40 mol L⁻¹ × 0.5 L = 0.20 mol.

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4. A buffer solution is prepared by mixing 200 mL of 0.1 M NH₄Cl and 100 mL of 0.1 M NH₄OH. Given pKb for NH₄OH is 4.74, what is the pH of the resulting solution? (Take log 2 = 0.30)

  1. 4.44
  2. 9.56
  3. 8.96
  4. 5.04
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Answer: (C) 8.96

Millimoles of NH₄OH = 100 × 0.1 = 10 mmol; Millimoles of NH₄Cl (salt) = 200 × 0.1 = 20 mmol. Using Henderson-Hasselbalch equation: pOH = pKb + log([Salt]/[Base]) = 4.74 + log(20/10) = 4.74 + 0.30 = 5.04. Therefore, pH = 14 - pOH = 14 - 5.04 = 8.96.

easy

5. The solubility product (Ksp) of a sparingly soluble salt A₂B₃ in water is related to its molar solubility 's' by which of the following expressions?

  1. Ksp = 108s⁵
  2. Ksp = 36s⁵
  3. Ksp = 6s⁵
  4. Ksp = 108s⁴
Show answer and explanation

Answer: (A) Ksp = 108s⁵

For A₂B₃(s) ⇌ 2A³⁺(aq) + 3B²⁻(aq), if solubility is 's', then [A³⁺] = 2s and [B²⁻] = 3s. Therefore, Ksp = [A³⁺]²[B²⁻]³ = (2s)²(3s)³ = 4s² × 27s³ = 108s⁵.

hard

6. At what minimum pH will Fe(OH)₃ begin to precipitate from a 0.001 M FeCl₃ solution? (Given: Ksp of Fe(OH)₃ = 1.0 × 10⁻³⁸)

  1. 11.67
  2. 3.45
  3. 2.33
  4. 4.25
Show answer and explanation

Answer: (C) 2.33

Precipitation begins when [Fe³⁺][OH⁻]³ = Ksp. Given [Fe³⁺] = 10⁻³ M, [OH⁻]³ = 1.0 × 10⁻³⁸ / 10⁻³ = 10⁻³⁵, so [OH⁻] = 10⁻¹¹·⁶⁷ M. Thus, pOH = 11.67, and pH = 14 - 11.67 = 2.33.

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7. The molar solubility of BaSO₄ in water is 0.003 mol/L. Its solubility product is:

  1. 1.8 × 10⁻⁵
  2. 9 × 10⁻⁶
  3. 3.6 × 10⁻⁵
  4. 0.003
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Answer: (B) 9 × 10⁻⁶

For a 1:1 salt, Ksp = s² = (0.003)² = 9 × 10⁻⁶.

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8. An acetic acid–sodium acetate buffer has [salt]/[acid] = 0.5. Taking pKa of acetic acid as 4.74, the pH of the buffer is:

  1. 4.44
  2. 9.56
  3. 2.37
  4. 5.04
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Answer: (A) 4.44

Henderson equation: pH = pKa + log([salt]/[acid]) = 4.74 + log 0.5 = 4.44.

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9. The molar solubility of AgCl in water is 0.001 mol/L. Its solubility product is:

  1. 2 × 10⁻⁶
  2. 1 × 10⁻⁶
  3. 0.001
  4. 4 × 10⁻⁶
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Answer: (B) 1 × 10⁻⁶

For a 1:1 salt, Ksp = s² = (0.001)² = 1 × 10⁻⁶.

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10. The molar solubility of BaSO₄ in water is 3 × 10⁻⁵ mol/L. Its solubility product is:

  1. 9 × 10⁻¹⁰
  2. 3 × 10⁻⁵
  3. 1.8 × 10⁻⁹
  4. 3.6 × 10⁻⁹
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Answer: (A) 9 × 10⁻¹⁰

For a 1:1 salt, Ksp = s² = (3 × 10⁻⁵)² = 9 × 10⁻¹⁰.

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11. The solubility of a sparingly soluble hydroxide M(OH)₂ in pure water gives a saturated solution with pH = 9.0. What is the value of Ksp for M(OH)₂?

  1. 5.0 × 10⁻¹⁶
  2. 1.0 × 10⁻¹⁵
  3. 2.0 × 10⁻¹⁵
  4. 5.0 × 10⁻¹¹
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Answer: (A) 5.0 × 10⁻¹⁶

At pH = 9.0, pOH = 14 - 9.0 = 5.0, so [OH⁻] = 10⁻⁵ M. For M(OH)₂ ⇌ M²⁺ + 2OH⁻, [M²⁺] = [OH⁻]/2 = 0.5 × 10⁻⁵ M = 5.0 × 10⁻⁶ M. Therefore, Ksp = [M²⁺][OH⁻]² = (5.0 × 10⁻⁶)(10⁻⁵)² = 5.0 × 10⁻¹⁶.

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12. The molar solubility of Ag₂CrO₄ in water is 5 × 10⁻⁴ mol/L. Its solubility product is:

  1. 2.5 × 10⁻⁷
  2. 1.25 × 10⁻¹⁰
  3. 0.002
  4. 5 × 10⁻¹⁰
Show answer and explanation

Answer: (D) 5 × 10⁻¹⁰

For a 1:2 or 2:1 salt, Ksp = 4s³ = 4 × (5 × 10⁻⁴)³ = 5 × 10⁻¹⁰.

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