Ionic Equilibrium – pH: NEET MCQs with Answers

15 free questions on Ionic Equilibrium – pH for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: pH, Buffer, Indicators and Salt Hydrolysis: pH scale, hydrolysis of salts and pH of their solutions, , the solubility of sparingly soluble salts and solubility products, buffer solutions.

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Question 1

Previous-year question
An acidic buffer can be prepared by mixing: [NEET 2023 (Manipur)]
  1. a weak acid and its salt with a strong base
  2. equal volumes of equimolar solutions of a weak acid and a weak base (the pK_a of the acid = the pK_b of the base)
  3. a strong acid and its salt with a strong base
  4. a strong acid and its salt with a weak base
Show answer
Answer: (A) a weak acid and its salt with a strong base
Explanation: An acidic buffer consists of a weak acid and its salt with a strong base, e.g. CH₃COOH + CH₃COONa. Equimolar weak acid and weak base with pK_a = pK_b give a neutral solution (pH = 7), not an acidic buffer, and a strong acid is fully ionised so it cannot form a buffer.

Question 2

Previous-year question
Which of the following salts gives a basic aqueous solution? [NEET 2020 (COVID-19)]
  1. Ammonium chloride
  2. Ammonium sulphate
  3. Ammonium nitrate
  4. Sodium acetate
Show answer
Answer: (D) Sodium acetate
Explanation: Sodium acetate is a salt of a weak acid (CH₃COOH) and a strong base (NaOH); acetate ion hydrolyses (CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻) to make the solution basic. Ammonium chloride, sulphate and nitrate are salts of a weak base with strong acids, so their solutions are acidic.

Question 3

Previous-year question
What will be the pH of a 0.01 M aqueous NaOH solution? [NEET (Odisha) 2019]
  1. 7.01
  2. 2
  3. 12
  4. 9
Show answer
Answer: (C) 12
Explanation: NaOH is a strong base and ionises completely, so [OH⁻] = 0.01 = 10⁻² M. pOH = 2, hence pH = 14 − 2 = 12.

Question 4

Previous-year question
Which of the following will make a basic buffer? [NEET 2019]
  1. 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH₃COOH
  2. 100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M NaOH
  3. 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH
  4. 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Show answer
Answer: (C) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH
Explanation: 10 mmol HCl neutralises 10 mmol of the 20 mmol NH₄OH, leaving 10 mmol NH₄OH with 10 mmol NH₄Cl, a weak base plus its conjugate acid, i.e. a basic buffer. The NaOH + CH₃COOH mixture leaves excess strong base, and the other two mixtures are exact neutralisations.

Question 5

Previous-year question
The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations: (i) 60 mL M/10 HCl + 40 mL M/10 NaOH (ii) 55 mL M/10 HCl + 45 mL M/10 NaOH (iii) 75 mL M/5 HCl + 25 mL M/5 NaOH (iv) 100 mL M/10 HCl + 100 mL M/10 NaOH The pH of which one of them will be equal to 1? [NEET 2018]
  1. (ii)
  2. (i)
  3. (iii)
  4. (iv)
Show answer
Answer: (C) (iii)
Explanation: In (iii) the excess acid = (75 − 25) × (1/5) = 10 mmol in a total of 100 mL, so [H⁺] = 0.1 M and pH = 1. For (i) the excess is 2 mmol in 100 mL (pH 1.7), for (ii) 1 mmol in 100 mL (pH 2), and (iv) is an exact neutralisation (pH 7).

Question 6

Previous-year question
Which of the following can form a buffer solution? [NEET 2017]
  1. aq. NH₃ + NH₄Cl
  2. KOH + HNO₃
  3. NaOH + HCl
  4. KI + KOH
Show answer
Answer: (A) aq. NH₃ + NH₄Cl
Explanation: Aqueous ammonia is a weak base and NH₄Cl supplies its conjugate acid NH₄⁺, so together they form a basic buffer. KOH + HNO₃ and NaOH + HCl are strong acid–strong base pairs giving neutral salts, and KI + KOH is a salt plus a strong base.

Question 7

Previous-year question
Which one of the following pairs of solutions is NOT an acidic buffer? [AIPMT 2015]
  1. HClO₄ and NaClO₄
  2. CH₃COOH and CH₃COONa
  3. H₂CO₃ and Na₂CO₃
  4. H₃PO₄ and Na₃PO₄
Show answer
Answer: (A) HClO₄ and NaClO₄
Explanation: HClO₄ is one of the strongest known acids, so it is completely ionised and its conjugate base ClO₄⁻ has no tendency to accept protons. Such a pair cannot resist pH change. The other three each combine a weak acid with its own conjugate base.
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Question 8

Previous-year question
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? [AIPMT 2015]
  1. 12.65
  2. 2.0
  3. 7.0
  4. 1.04
Show answer
Answer: (A) 12.65
Explanation: Taking 1 L of each, moles of OH⁻ = 0.1 and moles of H⁺ = 0.01, so 0.09 mol of OH⁻ survives in 2 L, giving [OH⁻] = 0.045 M. Then pOH = −log(0.045) = 1.35 and pH = 14 − 1.35 = 12.65.

Question 9

Previous-year question
Which of the following salts will give the highest pH in water? [AIPMT 2014]
  1. KCl
  2. NaCl
  3. Na₂CO₃
  4. CuSO₄
Show answer
Answer: (C) Na₂CO₃
Explanation: Na₂CO₃ is the salt of the strong base NaOH and the weak acid H₂CO₃. Its carbonate ion hydrolyses, CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻, making the solution distinctly alkaline. KCl and NaCl are neutral, while CuSO₄ is acidic through cationic hydrolysis.

Question 10

Previous-year question
Following solutions were prepared by mixing NaOH and HCl: (a) 60 mL M/10 HCl + 40 mL M/10 NaOH (b) 55 mL M/10 HCl + 45 mL M/10 NaOH (c) 75 mL M/5 HCl + 25 mL M/5 NaOH (d) 100 mL M/10 HCl + 100 mL M/10 NaOH. pH of which one will be equal to 1? [NEET 2018]
  1. d
  2. a
  3. b
  4. c
Show answer
Answer: (D) c
Explanation: For (c): H⁺ = 75×0.2 = 15 mmol, OH⁻ = 25×0.2 = 5 mmol; excess H⁺ = 10 mmol in 100 mL = 0.1 M → pH = 1.

Question 11

A 0.040 M aqueous solution of anilinium chloride, C₆H₅NH₃⁺Cl⁻, has pH 3.0 at 298 K. Taking the extent of hydrolysis as small, what is Kb of aniline? (Kw = 1.0 × 10⁻¹⁴)
  1. 2.5 × 10⁻⁹
  2. 4.0 × 10⁻¹²
  3. 4.0 × 10⁻¹⁰
  4. 2.5 × 10⁻⁵
Show answer
Answer: (C) 4.0 × 10⁻¹⁰
Explanation: The cation hydrolyses as a weak acid: C₆H₅NH₃⁺ + H₂O ⇌ C₆H₅NH₂ + H₃O⁺. With [H₃O⁺] = 1.0 × 10⁻³ M, Ka = (1.0 × 10⁻³)²/0.040 = 2.5 × 10⁻⁵. Since KaKb = Kw for a conjugate pair, Kb = 1.0 × 10⁻¹⁴/2.5 × 10⁻⁵ = 4.0 × 10⁻¹⁰.

Question 12

A patient passes 2.0 L of urine a day. If the urine pH falls from 6.0 to 5.0 with the volume unchanged, how does the daily loss of free H⁺ change, counted from pH alone?
  1. From 1.2 × 10⁻⁵ to 1.0 × 10⁻⁵ gram equivalents
  2. From 2.0 × 10⁻⁶ to 2.0 × 10⁻⁵ gram equivalents
  3. From 2.0 × 10⁻⁵ to 2.0 × 10⁻⁶ gram equivalents
  4. From 1.0 × 10⁻⁶ to 1.0 × 10⁻⁵ gram equivalents
Show answer
Answer: (B) From 2.0 × 10⁻⁶ to 2.0 × 10⁻⁵ gram equivalents
Explanation: At pH 6.0, [H⁺] = 1.0 × 10⁻⁶ M, so 2.0 L contains 2.0 × 10⁻⁶ mol of H⁺. At pH 5.0, [H⁺] = 1.0 × 10⁻⁵ M, so 2.0 L contains 2.0 × 10⁻⁵ mol. For H⁺, one mole is one gram equivalent, and a fall of one pH unit means a tenfold rise.

Question 13

A buffer is made from a weak acid HA (pKa = 4.70) and its sodium salt NaA. The sum [HA] + [A⁻] is 0.30 M and the pH of the buffer is 5.00. What is the concentration of A⁻? (log 2 = 0.30)
  1. 0.10 M
  2. 0.20 M
  3. 0.25 M
  4. 0.15 M
Show answer
Answer: (B) 0.20 M
Explanation: From the Henderson equation, 5.00 = 4.70 + log([A⁻]/[HA]), so the ratio [A⁻]/[HA] is 2. Sharing the total of 0.30 M in the ratio 2 : 1 gives [A⁻] = 0.20 M and [HA] = 0.10 M.

Question 14

One litre of a buffer contains 0.10 mol CH₃COOH and 0.10 mol CH₃COONa (pKa = 4.74). If 0.010 mol of HCl gas is dissolved in it with no volume change, what is the new pH? (log 0.82 = −0.087)
  1. 4.65
  2. 4.83
  3. 2.00
  4. 4.74
Show answer
Answer: (A) 4.65
Explanation: The added H⁺ is consumed by acetate: CH₃COO⁻ + H⁺ → CH₃COOH. This leaves 0.09 mol acetate and 0.11 mol acetic acid. Then pH = 4.74 + log(0.09/0.11) = 4.74 − 0.087 ≈ 4.65, a fall of only about 0.09 unit, whereas 0.01 M HCl alone would have a pH of 2.

Question 15

A few drops of dilute H₂SO₄ are added to the water when pure blue crystals of copper(II) sulphate are grown from a hot concentrated solution. What is the main purpose of this acid?
  1. To raise the pH of the solution so that crystals separate faster
  2. To suppress hydrolysis of Cu²⁺, which would otherwise give a turbid basic product
  3. To oxidise Cu²⁺ to a higher oxidation state that forms larger crystals
  4. To supply water of crystallisation for CuSO₄·5H₂O
Show answer
Answer: (B) To suppress hydrolysis of Cu²⁺, which would otherwise give a turbid basic product
Explanation: Cu²⁺ is the cation of a weak base and undergoes hydrolysis in water, giving insoluble basic species and H⁺. Added H⁺ shifts this equilibrium back towards free Cu²⁺, in line with Le Chatelier's principle. The solution therefore stays clear and pure crystals are obtained.

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