Ionic Equilibrium – Solubility Product and Common Ion Effect: NEET MCQs with Answers

15 free questions on Ionic Equilibrium – Solubility Product and Common Ion Effect 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: Solubility Product and Common Ion Effect: ionization of water, common ion effect,

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Question 1

Previous-year question
What is the ratio of the solubility of AgCl in 0.1 M KCl solution to its solubility in water? (Given: solubility product of AgCl = 10⁻¹⁰) [Re-NEET 2024]
  1. 10⁻⁴
  2. 10⁻⁶
  3. 10⁻⁹
  4. 10⁻⁵
Show answer
Answer: (A) 10⁻⁴
Explanation: In water, s₁ = √Kₛₚ = √(10⁻¹⁰) = 10⁻⁵ M. In 0.1 M KCl, s₂ = Kₛₚ/[Cl⁻] = 10⁻¹⁰/0.1 = 10⁻⁹ M. Ratio s₂/s₁ = 10⁻⁹/10⁻⁵ = 10⁻⁴.

Question 2

Previous-year question
BaSO₄ has a solubility product of 1.5 × 10⁻⁹ in water. Find its molar solubility in a 0.1 M solution of Ba(NO₃)₂. [NEET 2022 (Overseas)]
  1. 0.5 × 10⁻⁸ M
  2. 1.5 × 10⁻⁸ M
  3. 1.0 × 10⁻⁸ M
  4. 2.0 × 10⁻⁸ M
Show answer
Answer: (B) 1.5 × 10⁻⁸ M
Explanation: Ba(NO₃)₂ supplies the common ion Ba²⁺, [Ba²⁺] ≈ 0.1 M. Kₛₚ = [Ba²⁺][SO₄²⁻] = 0.1 × s, so s = 1.5 × 10⁻⁹/0.1 = 1.5 × 10⁻⁸ M.

Question 3

Previous-year question
Silver iodide is used for cloud seeding to produce rain: AgI(s) ⇌ Ag⁺(aq) + I⁻(aq); Kₛₚ = 8.5 × 10⁻⁷. When AgNO₃ and KI are mixed so that [Ag⁺] = 0.010 M and [I⁻] = 0.015 M, will AgI be precipitated? [NEET 2021 (Paper-2)]
  1. Yes
  2. No
  3. Can't say
  4. Depends on [NO₃⁻] and [K⁺]
Show answer
Answer: (A) Yes
Explanation: Ionic product Q = [Ag⁺][I⁻] = 0.010 × 0.015 = 1.5 × 10⁻⁴. Since Q (1.5 × 10⁻⁴) > Kₛₚ (8.5 × 10⁻⁷), the solution is supersaturated and AgI will precipitate. NO₃⁻ and K⁺ are spectator ions and do not affect this.

Question 4

Previous-year question
A salt of the AB type has a solubility product of 4 × 10⁻⁸. What is the molarity of its saturated (standard) solution? [NEET 2020 (COVID-19)]
  1. 2 × 10⁻⁴ mol/L
  2. 16 × 10⁻¹⁶ mol/L
  3. 2 × 10⁻¹⁶ mol/L
  4. 4 × 10⁻⁴ mol/L
Show answer
Answer: (A) 2 × 10⁻⁴ mol/L
Explanation: For AB ⇌ A⁺ + B⁻, Kₛₚ = s². s = √(4 × 10⁻⁸) = 2 × 10⁻⁴ mol/L, which is the molarity of its saturated solution.

Question 5

Previous-year question
Calculate the molar solubility of CaF₂ (Kₛₚ = 5.3 × 10⁻¹¹) in a 0.1 M solution of NaF. [NEET (Odisha) 2019]
  1. 5.3 × 10⁻¹¹ mol L⁻¹
  2. 5.3 × 10⁻⁸ mol L⁻¹
  3. 5.3 × 10⁻⁹ mol L⁻¹
  4. 5.3 × 10⁻¹⁰ mol L⁻¹
Show answer
Answer: (C) 5.3 × 10⁻⁹ mol L⁻¹
Explanation: CaF₂ ⇌ Ca²⁺ + 2F⁻; the common ion F⁻ from NaF gives [F⁻] ≈ 0.1 M. Kₛₚ = s[F⁻]² = s × (0.1)², so s = 5.3 × 10⁻¹¹/10⁻² = 5.3 × 10⁻⁹ mol L⁻¹.

Question 6

Previous-year question
Find the solubility of Ni(OH)₂ in 0.1 M NaOH, given that the ionic product of Ni(OH)₂ is 2 × 10⁻¹⁵. [NEET 2020]
  1. 2 × 10⁻⁸ M
  2. 1 × 10⁻¹³ M
  3. 1 × 10⁸ M
  4. 2 × 10⁻¹³ M
Show answer
Answer: (D) 2 × 10⁻¹³ M
Explanation: NaOH supplies so much OH⁻ that the contribution from Ni(OH)₂ is negligible, so [OH⁻] = 0.1 M. Then s = Ksp/[OH⁻]² = (2 × 10⁻¹⁵)/(0.1)² = 2 × 10⁻¹³ M — a dramatic suppression by the common-ion effect.

Question 7

Previous-year question
The pH of a saturated solution of Ca(OH)₂ is 9. The solubility product Ksp of Ca(OH)₂ is [NEET 2019]
  1. 0.5 × 10⁻¹⁵
  2. 0.25 × 10⁻¹⁰
  3. 0.125 × 10⁻¹⁵
  4. 0.5 × 10⁻¹⁰
Show answer
Answer: (A) 0.5 × 10⁻¹⁵
Explanation: pH 9 gives pOH = 5, so [OH⁻] = 1 × 10⁻⁵ M and s = ½(10⁻⁵) = 5 × 10⁻⁶ M. Then Ksp = 4s³ = 4(5 × 10⁻⁶)³ = 5.0 × 10⁻¹⁶ = 0.5 × 10⁻¹⁵.
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Question 8

Previous-year question
The solubility of BaSO₄ in water is 2.42 × 10⁻³ g L⁻¹ at 298 K. Its solubility product Ksp is (molar mass of BaSO₄ = 233 g mol⁻¹) [NEET 2018]
  1. 1.08 × 10⁻¹⁰ mol²L⁻²
  2. 1.08 × 10⁻¹² mol²L⁻²
  3. 1.08 × 10⁻⁸ mol²L⁻²
  4. 1.08 × 10⁻¹⁴ mol²L⁻²
Show answer
Answer: (A) 1.08 × 10⁻¹⁰ mol²L⁻²
Explanation: Molar solubility s = (2.42 × 10⁻³)/233 = 1.04 × 10⁻⁵ mol L⁻¹. For the 1 : 1 salt BaSO₄, Ksp = s² = (1.04 × 10⁻⁵)² = 1.08 × 10⁻¹⁰ mol² L⁻².

Question 9

Previous-year question
The concentration of Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10⁻⁴ mol L⁻¹. The solubility product of Ag₂C₂O₄ is [NEET 2017]
  1. 2.66 × 10⁻¹²
  2. 4.5 × 10⁻¹¹
  3. 5.3 × 10⁻¹²
  4. 2.42 × 10⁻⁸
Show answer
Answer: (C) 5.3 × 10⁻¹²
Explanation: Each formula unit gives two Ag⁺, so s = ½(2.2 × 10⁻⁴) = 1.1 × 10⁻⁴ M. Then Ksp = [Ag⁺]²[C₂O₄²⁻] = 4s³ = 4(1.1 × 10⁻⁴)³ = 5.3 × 10⁻¹².

Question 10

Previous-year question
MY and NY₃, two nearly insoluble salts, have the same Ksp of 6.2 × 10⁻¹³ at room temperature. Which statement is true about MY and NY₃? [NEET 2016]
  1. The molar solubilities of MY and NY₃ in water are identical
  2. The molar solubility of MY in water is less than that of NY₃
  3. The salts MY and NY₃ are more soluble in 0.5 M KY than in pure water
  4. Adding the salt KY to solutions of MY and NY₃ will have no effect on their solubilities
Show answer
Answer: (B) The molar solubility of MY in water is less than that of NY₃
Explanation: For MY, Ksp = s², so s = 7.9 × 10⁻⁷ M. For NY₃, Ksp = 27s⁴, so s = (6.2 × 10⁻¹³/27)¹ᐟ⁴ ≈ 3.9 × 10⁻⁴ M. Equal Ksp therefore does not mean equal solubility — MY is much the less soluble. Adding KY supplies the common ion Y⁻ and would lower both solubilities.

Question 11

Previous-year question
The Ksp values of Ag₂CrO₄, AgCl, AgBr and AgI are respectively 1.1 × 10⁻¹², 1.8 × 10⁻¹⁰, 5.0 × 10⁻¹³ and 8.3 × 10⁻¹⁷. Which of these salts will precipitate last when AgNO₃ solution is added to a solution containing equal moles of NaCl, NaBr, NaI and Na₂CrO₄? [AIPMT 2015]
  1. AgCl
  2. AgBr
  3. Ag₂CrO₄
  4. AgI
Show answer
Answer: (C) Ag₂CrO₄
Explanation: The salt that precipitates last is the one needing the largest [Ag⁺]. For the 1 : 1 salts [Ag⁺] = Ksp/[X⁻], but for Ag₂CrO₄ it is √(Ksp/[CrO₄²⁻]). Taking the square root of a very small number gives a much larger value, so Ag₂CrO₄ requires the highest silver-ion concentration and appears last.

Question 12

Previous-year question
Using the Gibbs energy change ΔG° = +63.3 kJ for Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq), the Ksp of Ag₂CO₃ in water at 25 °C is (R = 8.314 J K⁻¹ mol⁻¹) [AIPMT 2014]
  1. 3.2 × 10⁻²⁶
  2. 8.0 × 10⁻¹²
  3. 2.9 × 10⁻³
  4. 7.9 × 10⁻²
Show answer
Answer: (B) 8.0 × 10⁻¹²
Explanation: ΔG° = −RT ln Ksp gives ln Ksp = −63300/(8.314 × 298) = −25.54, so Ksp = e⁻²⁵·⁵⁴ ≈ 8.0 × 10⁻¹². The large positive ΔG° correctly signals a very sparingly soluble salt.

Question 13

Previous-year question
The solubility of BaSO₄ in water is 2.42 × 10⁻³ g L⁻¹ at 298 K. The value of its solubility product (Ksp) will be (molar mass of BaSO₄ = 233 g mol⁻¹): [NEET 2018]
  1. 1.08 × 10⁻¹⁴ mol² L⁻²
  2. 1.08 × 10⁻¹² mol² L⁻²
  3. 1.08 × 10⁻¹⁰ mol² L⁻²
  4. 1.08 × 10⁻⁸ mol² L⁻²
Show answer
Answer: (C) 1.08 × 10⁻¹⁰ mol² L⁻²
Explanation: s = 2.42×10⁻³/233 = 1.04×10⁻⁵ M; Ksp = s² = (1.04×10⁻⁵)² = 1.08×10⁻¹⁰.

Question 14

Previous-year question
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10⁻⁴ mol L⁻¹. Solubility product of Ag₂C₂O₄ is: [2017]
  1. 2.66 × 10⁻¹²
  2. 4.5 × 10⁻¹¹
  3. 5.3 × 10⁻¹²
  4. 2.42 × 10⁻⁸
Show answer
Answer: (C) 5.3 × 10⁻¹²
Explanation: [Ag⁺] = 2s = 2.2×10⁻⁴ → s = 1.1×10⁻⁴. Ksp = (2s)²(s) = (2.2×10⁻⁴)² × 1.1×10⁻⁴ = 5.3×10⁻¹².

Question 15

Pure water at 25 °C has a molar concentration of about 55.5 M and [H⁺] = 1.0 × 10⁻⁷ M. Which value is closest to the fraction of water molecules that are ionised?
  1. 1.8 × 10⁻⁹
  2. 1.0 × 10⁻⁷
  3. 1.8 × 10⁻⁷
  4. 1.8 × 10⁻¹⁶
Show answer
Answer: (A) 1.8 × 10⁻⁹
Explanation: The fraction ionised equals the concentration ionised divided by the total concentration: 1.0 × 10⁻⁷ / 55.5 ≈ 1.8 × 10⁻⁹. This means only about two molecules in a billion are ionised, so the equilibrium lies far towards unionised water. For this reason the concentration of water is treated as constant in Kw.

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