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Acid/base indicator, oxalic-acid/KMnO4 and Mohr-salt/KMnO4 titrations: NEET MCQs with answers

89 questions on this topic of Principles Related to Practical Chemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In the titration of acidified KMnO₄ solution with Mohr's salt, the equivalent weight of KMnO₄ (molecular mass = M) is:

  1. M / 3
  2. M / 5
  3. M / 1
  4. M / 2
Show answer and explanation

Answer: (B) M / 5

In an acidic medium, the permanganate ion undergoes reduction according to: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The change in oxidation state of manganese is from +7 to +2 (n-factor = 5). Therefore, Equivalent weight = Molecular mass / 5 = M / 5.

medium

2. An acid-base titration of a strong acid (HCl) with a weak base (NH₄OH) is carried out. The pH at the equivalence point and the most suitable indicator for this titration are respectively:

  1. pH > 7, Phenolphthalein
  2. pH < 7, Methyl orange
  3. pH = 7, Bromothymol blue
  4. pH < 7, Phenolphthalein
Show answer and explanation

Answer: (B) pH < 7, Methyl orange

At the equivalence point of a strong acid and weak base titration, NH₄⁺ hydrolyses to give an acidic solution (pH < 7). The steep change in pH occurs in the acidic region (around 4 to 6.5), making methyl orange (pH range 3.1-4.4) the appropriate indicator.

hard

3. 15 mL of 0.1 M oxalic acid is titrated in acid medium and requires 20 mL of KMnO₄ (2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O). The molarity of the KMnO₄ is:

  1. 0.075 M
  2. 0.015 M
  3. 0.03 M
  4. 0.188 M
Show answer and explanation

Answer: (C) 0.03 M

Moles of MnO₄⁻ = (2/5) × moles of oxalate: M = (2/5) × 0.1 × 15/20 = 0.03 M.

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4. Given below are two statements: Statement I: During the titration of oxalic acid with KMnO₄, the pink colour imparted by the first drop of KMnO₄ takes a noticeable time to disappear. Statement II: The reaction is autocatalysed by the Mn²⁺ ions formed as a product of the redox process. In light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Statement I is correct because initially no Mn²⁺ ions are present and the activation energy is high, so the first few drops decolourise very slowly. Statement II is also correct because the Mn²⁺ produced acts as an autocatalyst, greatly speeding up the subsequent decolourisation.

easy

5. Which of the following acid-base indicators shows a pink colour in basic medium and remains colourless in acidic medium?

  1. Phenolphthalein
  2. Methyl orange
  3. Methyl red
  4. Bromothymol blue
Show answer and explanation

Answer: (A) Phenolphthalein

Phenolphthalein is a synthetic indicator with a working pH range of approximately 8.3 to 10.0. It is colourless in acidic solutions and turns pink in alkaline solutions due to quinonoid structure formation.

hard

6. 10 mL of 0.1 M oxalic acid is titrated in acid medium and requires 12 mL of KMnO₄ (2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O). The molarity of the KMnO₄ is:

  1. 0.0833 M
  2. 0.0333 M
  3. 0.208 M
  4. 0.0167 M
Show answer and explanation

Answer: (B) 0.0333 M

Moles of MnO₄⁻ = (2/5) × moles of oxalate: M = (2/5) × 0.1 × 10/12 = 0.0333 M.

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7. 25 mL of 0.05 M oxalic acid (H₂C₂O₄) is exactly neutralised by 34 mL of NaOH solution. The molarity of the NaOH is:

  1. 0.0735 M
  2. 0.136 M
  3. 0.068 M
  4. 0.0368 M
Show answer and explanation

Answer: (A) 0.0735 M

Oxalic acid is diprotic: 2 × M₁V₁ = M₂V₂, so M₂ = 2 × 0.05 × 25/34 = 0.0735 M.

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8. 15 mL of 0.05 M oxalic acid (H₂C₂O₄) is exactly neutralised by 14 mL of NaOH solution. The molarity of the NaOH is:

  1. 0.0467 M
  2. 0.0536 M
  3. 0.107 M
  4. 0.0933 M
Show answer and explanation

Answer: (C) 0.107 M

Oxalic acid is diprotic: 2 × M₁V₁ = M₂V₂, so M₂ = 2 × 0.05 × 15/14 = 0.107 M.

medium

9. 15 mL of 0.1 M oxalic acid (H₂C₂O₄) is exactly neutralised by 22 mL of NaOH solution. The molarity of the NaOH is:

  1. 0.0682 M
  2. 0.147 M
  3. 0.136 M
  4. 0.293 M
Show answer and explanation

Answer: (C) 0.136 M

Oxalic acid is diprotic: 2 × M₁V₁ = M₂V₂, so M₂ = 2 × 0.1 × 15/22 = 0.136 M.

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10. 20 mL of 0.1 M oxalic acid (H₂C₂O₄) is exactly neutralised by 10 mL of NaOH solution. The molarity of the NaOH is:

  1. 0.2 M
  2. 0.05 M
  3. 0.4 M
  4. 0.1 M
Show answer and explanation

Answer: (C) 0.4 M

Oxalic acid is diprotic: 2 × M₁V₁ = M₂V₂, so M₂ = 2 × 0.1 × 20/10 = 0.4 M.

hard

11. 10 mL of 0.05 M oxalic acid is titrated in acid medium and requires 12 mL of KMnO₄ (2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O). The molarity of the KMnO₄ is:

  1. 0.104 M
  2. 0.0417 M
  3. 0.00833 M
  4. 0.0167 M
Show answer and explanation

Answer: (D) 0.0167 M

Moles of MnO₄⁻ = (2/5) × moles of oxalate: M = (2/5) × 0.05 × 10/12 = 0.0167 M.

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12. 10 mL of 0.05 M oxalic acid (H₂C₂O₄) is exactly neutralised by 22 mL of NaOH solution. The molarity of the NaOH is:

  1. 0.0455 M
  2. 0.0227 M
  3. 0.11 M
  4. 0.22 M
Show answer and explanation

Answer: (A) 0.0455 M

Oxalic acid is diprotic: 2 × M₁V₁ = M₂V₂, so M₂ = 2 × 0.05 × 10/22 = 0.0455 M.

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