Definition
Common ion effect
Suppression of ionisation of a weak electrolyte by adding a strong electrolyte with a common ion. Example: adding NaCl to NH₄OH suppresses NH₄⁺.
-- NCERT Class 11 Chemistry, Ch. 6, p. 201Common Ion Effect
8 MCQs3 revision cards9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 27 Sep 2026
Common Ion Effect, explained for NEET
The common ion effect is one of the most directly tested consequences of equilibrium in NEET chemistry — and the trap is almost always the same: students forget to raise ion concentrations to their stoichiometric powers when an external source supplies a common ion.
What is the common ion effect? When a soluble salt that shares an ion with a sparingly soluble salt is added to the solution, the concentration of that shared (common) ion increases. By Le Chatelier's principle, the dissolution equilibrium shifts backward, and the solubility of the sparingly soluble salt decreases. NCERT Class 11 Chemistry Chapter 6 (Equilibrium), page 201, defines this as the suppression of ionisation or solubility caused by the presence of a common ion from a second electrolyte.
The NEET trap — stoichiometric powers in K_sp. Consider CaF₂ dissolving: CaF₂ ⇌ Ca²⁺ + 2F⁻. If solubility is s, then [Ca²⁺] = s and [F⁻] = 2s, giving K_sp = s·(2s)² = 4s³. Now add NaF (a soluble fluoride). The F⁻ concentration is no longer just 2s — it becomes (2s + C_NaF). Many students write K_sp = [Ca²⁺][F⁻] without the square on fluoride, and then apply the common ion incorrectly.
How NEET tests this: a problem gives K_sp of a non-1:1 salt and asks for its solubility in the presence of a common ion source. The correct distractor exploits the stoichiometric-power mistake — the answer you get by ignoring the exponent lands on one of the wrong options.
Watch out: the common ion effect applies only when the added ion is genuinely common to the equilibrium. Adding a non-common salt (say NaCl to a CaF₂ solution) does not suppress CaF₂ solubility via this mechanism.
Can you answer these Common Ion Effect MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
The solubility product of PbCl₂ is 1.6 × 10⁻⁵. What is the molar solubility of PbCl₂ in pure water?
Show answer and why every option is right or wrong
Answer: A. PbCl₂ ⇌ Pb²⁺ + 2Cl⁻. If solubility = s, then [Pb²⁺] = s, [Cl⁻] = 2s. K_sp = s(2s)² = 4s³. So s = (K_sp/4)^(1/3) = (1.6 × 10⁻⁵/4)^(1/3) = (4.0 × 10⁻⁶)^(1/3) ≈ 1.59 × 10⁻² mol/L ≈ 1.6 × 10⁻² mol/L. (NCERT Class 11 Chemistry Chapter 6, page 205.)
Why B is wrong: B. This value results from incorrectly using K_sp = s³ (forgetting the coefficient 4 from the stoichiometry of Cl⁻): s = (1.6 × 10⁻⁵)^(1/3) ≈ 2.5 × 10⁻² mol/L. (trap: stoichiometric power error)
Why C is wrong: C. This value comes from writing K_sp = s · (2s) = 2s² instead of s · (2s)² = 4s³ — dropping the square on the chloride term: s = √(8.0 × 10⁻⁶) ≈ 2.8 × 10⁻³ mol/L. (trap: stoichiometric power error)
Why D is wrong: D. This results from solving K_sp = s² (treating PbCl₂ as a 1:1 salt): s = √(1.6 × 10⁻⁵) = 4.0 × 10⁻³ mol/L. (trap: ignoring stoichiometry entirely)
MCQ 2Easy RecallPractice
What happens to the solubility of AgCl (K_sp = 1.8 × 10⁻¹⁰) when it is dissolved in 0.10 M NaCl solution instead of pure water?
Show answer and why every option is right or wrong
Answer: A. NaCl supplies Cl⁻, which is common to the AgCl equilibrium (AgCl ⇌ Ag⁺ + Cl⁻). The increased [Cl⁻] shifts the equilibrium left, decreasing AgCl solubility. K_sp remains constant but solubility drops. (NCERT Class 11 Chemistry Chapter 6, page 206.)
Why B is wrong: B. More ions in solution does not increase solubility of a sparingly soluble salt; the equilibrium shifts backward when a common ion is added. (trap: confusing ionic strength with solubility)
Why C is wrong: C. While K_sp is indeed constant at a given temperature, solubility is NOT the same as K_sp. Solubility changes when the equilibrium shifts due to a common ion. (trap: conflating K_sp constancy with solubility constancy)
Why D is wrong: D. NaCl being a strong electrolyte means it fully dissociates, which actually ensures maximum supply of the common ion Cl⁻ — this suppresses AgCl solubility, not increases it. (trap: misapplying electrolyte strength)
MCQ 3Direct ApplicationPractice
The K_sp of BaSO₄ is 1.1 × 10⁻¹⁰. Its molar solubility in 0.10 M Na₂SO₄ solution is approximately:
Show answer and why every option is right or wrong
Answer: C. BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. In 0.10 M Na₂SO₄, [SO₄²⁻] ≈ 0.10 M (common ion dominates). K_sp = [Ba²⁺][SO₄²⁻] = s × 0.10. So s = 1.1 × 10⁻¹⁰ / 0.10 = 1.1 × 10⁻⁹ mol/L. This is drastically lower than solubility in pure water (~1.05 × 10⁻⁵ mol/L). (NCERT Class 11 Chemistry Chapter 6, page 206.)
Why A is wrong: A. This is the solubility of BaSO₄ in pure water (√K_sp ≈ 1.05 × 10⁻⁵). The question asks for solubility in 0.10 M Na₂SO₄, where the common ion effect drastically reduces solubility. (trap: ignoring the common ion entirely)
Why B is wrong: B. This error comes from dividing K_sp by 1.0 × 10⁻⁵ (the pure-water solubility) instead of 0.10 M. (trap: circular substitution)
Why D is wrong: D. This results from treating BaSO₄ as a 1:2 salt in pure water, K_sp = 4s³, so s = (1.1 × 10⁻¹⁰/4)^(1/3) ≈ 3.0 × 10⁻⁴: the wrong stoichiometry, and the common ion ignored as well. (trap: stoichiometric power error)
MCQ 4Direct ApplicationPractice
For the sparingly soluble salt Ag₂CrO₄ (K_sp = 1.12 × 10⁻¹²), the correct expression for K_sp in terms of molar solubility s is:
Show answer and why every option is right or wrong
Answer: B. Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻. If solubility = s, then [Ag⁺] = 2s and [CrO₄²⁻] = s. K_sp = (2s)² · s = 4s³. (NCERT Class 11 Chemistry Chapter 6, page 205.)
Why A is wrong: A. This treats Ag₂CrO₄ as a 1:1 salt (K_sp = [Ag⁺][CrO₄²⁻] = s·s). The stoichiometry gives 2 mol Ag⁺ per mole dissolved. (trap: stoichiometric power error)
Why C is wrong: C. This uses (2s) · s² = 2s³, swapping which ion gets the coefficient. Ag⁺ has coefficient 2 (squared), CrO₄²⁻ has coefficient 1. (trap: assigning the power to the wrong ion)
Why D is wrong: D. This is the K_sp expression for a 3:1 salt like M₃X → 3M + X, giving (3s)³ · s = 27s⁴. Ag₂CrO₄ is 2:1, not 3:1. (trap: wrong stoichiometry)
MCQ 5Easy RecallPractice
Which of the following statements about the common ion effect is correct?
Show answer and why every option is right or wrong
Answer: D. Adding a common ion shifts the dissolution equilibrium backward, decreasing solubility. K_sp itself remains constant at a given temperature — only the solubility changes. (NCERT Class 11 Chemistry Chapter 6, page 206.)
Why A is wrong: A. The common ion effect suppresses ionisation of a weak electrolyte (shifts equilibrium left), it does not increase it. (trap: direction of equilibrium shift reversed)
Why B is wrong: B. The common ion effect operates in aqueous solutions. It is a direct consequence of equilibrium principles in water. (trap: no basis in NCERT)
Why C is wrong: C. K_sp is an equilibrium constant and depends only on temperature, not on the concentrations of added ions. The common ion changes solubility, not K_sp. (trap: confusing K_sp with solubility)
MCQ 6CalculationPractice
CaF₂ has K_sp = 3.9 × 10⁻¹¹. What is the molar solubility of CaF₂ in a solution that already contains 0.010 M NaF?
Show answer and why every option is right or wrong
Answer: D. CaF₂ ⇌ Ca²⁺ + 2F⁻. Let solubility = s. [Ca²⁺] = s, [F⁻] = 2s + 0.010 ≈ 0.010 M (since s is very small compared to 0.010). K_sp = s × (0.010)² = s × 1.0 × 10⁻⁴. So s = 3.9 × 10⁻¹¹ / 1.0 × 10⁻⁴ = 3.9 × 10⁻⁷ mol/L. (NCERT Class 11 Chemistry Chapter 6, page 206.)
Why A is wrong: A. This comes from taking [F⁻] = 2 × 0.010 = 0.020 M because the formula has two F, but NaF supplies 0.010 M F⁻ and the 2 applies only to F⁻ released by CaF₂: s = 3.9 × 10⁻¹¹/(0.020)² ≈ 9.8 × 10⁻⁸. (trap: doubling the common-ion concentration)
Why B is wrong: B. This results from using [F⁻] = 0.10 M instead of 0.010 M (misreading the NaF concentration by a factor of 10), or from forgetting to square [F⁻] = 0.010 M: both give s = 3.9 × 10⁻¹¹/1.0 × 10⁻² = 3.9 × 10⁻⁹. (trap: decimal error in common ion concentration)
Why C is wrong: C. This is the solubility in pure water: s = (K_sp/4)^(1/3) = (3.9 × 10⁻¹¹/4)^(1/3) ≈ 2.1 × 10⁻⁴. The question asks for solubility in 0.010 M NaF, not pure water. (trap: ignoring the common ion)
MCQ 7Easy RecallPractice
If the molar solubility of Mg(OH)₂ in pure water is s, the correct K_sp expression is:
Show answer and why every option is right or wrong
Answer: C. Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻. [Mg²⁺] = s, [OH⁻] = 2s. K_sp = s · (2s)² = 4s³. This is a standard 1:2 salt stoichiometry. (NCERT Class 11 Chemistry Chapter 6, page 205.)
Why A is wrong: A. This treats Mg(OH)₂ as a 1:1 salt. Two moles of OH⁻ are produced per mole dissolved, so [OH⁻] = 2s, not s. (trap: stoichiometric power error)
Why B is wrong: B. This incorrectly computes (2)(s)(s) = 2s² — the coefficient 2 must be inside the square: (2s)² = 4s². (trap: pulling the coefficient outside the exponent)
Why D is wrong: D. This uses s · s · s = s³, ignoring that two OH⁻ ions are produced and that [OH⁻] = 2s. (trap: stoichiometric power error)
MCQ 8CalculationPractice
The K_sp of PbI₂ is 8.0 × 10⁻⁹. When PbI₂ is dissolved in 0.10 M KI solution, the solubility of PbI₂ is approximately:
Show answer and why every option is right or wrong
Answer: B. PbI₂ ⇌ Pb²⁺ + 2I⁻. In 0.10 M KI, [I⁻] ≈ 0.10 M (common ion dominates). K_sp = s × (0.10)² = s × 0.010. So s = 8.0 × 10⁻⁹ / 0.010 = 8.0 × 10⁻⁷ mol/L. (NCERT Class 11 Chemistry Chapter 6, page 206.)
Why A is wrong: A. This is the solubility in pure water: s = (K_sp/4)^(1/3) = (2.0 × 10⁻⁹)^(1/3) ≈ 1.26 × 10⁻³. The question asks for solubility in 0.10 M KI. (trap: ignoring the common ion)
Why C is wrong: C. This comes from using [I⁻] = 0.10 without squaring it: s = K_sp / 0.10 = 8.0 × 10⁻⁸... then a decimal slip to 10⁻⁵. (trap: stoichiometric power error combined with arithmetic error)
Why D is wrong: D. This results from taking the cube root of K_sp directly (s = (8.0 × 10⁻⁹)^(1/3) ≈ 2.0 × 10⁻³) without the factor of 4 or the common ion. (trap: ignoring both stoichiometry and common ion)
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Common Ion Effect: quick recall before you leave
How do you solve a Common Ion Effect question? A worked example
Pattern: Compute solubility from K_sp with the common ion effect (PYQ pattern NEET pattern: solubility ksp, observed in NEET 2021 and 2023).
- 1
Given
• K_sp(CaF₂) = 3.9 × 10⁻¹¹• NaF concentration = 0.050 mol/L (part b)
- 2
Required
• Molar solubility s of CaF₂ in pure water• Molar solubility s' of CaF₂ in 0.050 M NaF
- 3
Concept
CaF₂ is a sparingly soluble 1:2 salt. Its dissolution equilibrium is CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). The K_sp expression must reflect the stoichiometric coefficients as powers. When NaF is added, it supplies F⁻ (the common ion), shifting the equilibrium left and reducing CaF₂ solubility.
- 4
Formula
K_sp = [Ca²⁺] · [F⁻]²
In pure water: K_sp = s · (2s)² = 4s³
In 0.050 M NaF: K_sp = s' · (2s' + 0.050)² - 5
Substitution
(a) Pure water:
4s³ = 3.9 × 10⁻¹¹
s³ = 9.75 × 10⁻¹²
(b) In 0.050 M NaF:
Since s' will be very small (common ion suppresses it), 2s' << 0.050, so [F⁻] ≈ 0.050 M.
s' × (0.050)² = 3.9 × 10⁻¹¹
s' × 2.5 × 10⁻³ = 3.9 × 10⁻¹¹ - 6
Calculation
(a) s = (9.75 × 10⁻¹²)^(1/3)
Taking the cube root: 9.75 × 10⁻¹² = 9.75 × 10⁻¹². Cube root of 9.75 ≈ 2.14. Cube root of 10⁻¹² = 10⁻⁴.
s ≈ 2.1 × 10⁻⁴ mol/L
(b) s' = 3.9 × 10⁻¹¹ / 2.5 × 10⁻³ = 1.56 × 10⁻⁸ mol/L
Exact constants note: The stoichiometric coefficient 2 in "(2s)" and the factor 4 in "4s³" are exact counting numbers and do not limit significant figures. - 7
Final answer
• (a) Molar solubility in pure water: s ≈ 2.1 × 10⁻⁴ mol/L• (b) Molar solubility in 0.050 M NaF: s' ≈ 1.6 × 10⁻⁸ mol/L
The common ion effect reduces CaF₂ solubility by roughly four orders of magnitude — from ~10⁻⁴ to ~10⁻⁸ mol/L. - 8
Common trap
The high-frequency trap here is writing K_sp = s · (2s) = 2s² instead of s · (2s)² = 4s³. This stoichiometric power error gives the wrong solubility in pure water AND the wrong answer in the common ion calculation, because the F⁻ term must be squared.
A second common error is forgetting to replace [F⁻] with the external NaF concentration in part (b), effectively solving for pure-water solubility again. - 9
Similar NEET-style question
"The K_sp of Mg(OH)₂ is 5.6 × 10⁻¹². Calculate its molar solubility in 0.10 M NaOH." (Same pattern: 1:2 salt, common ion is OH⁻. Answer: s = K_sp / (0.10)² = 5.6 × 10⁻¹⁰ mol/L.)
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What to remember before solving Common Ion Effect questions
More in Equilibrium: 8 exam traps and mistakes · 5 formulas · 4 question patterns from its other lessons.
Common Ion Effect questions from past NEET papers
2 questions from NEET 2020. Answers verified against NTA official keys.
NEET 2020
11×10⁸ M
22×10⁻¹³ M
32×10⁻⁸ M
41×10⁻¹³ M
NTA Answer: Option 2(final)
NEET 2020
1NaCl, MgCl2 and CaCl2
2Both MgCl2 and CaCl2
3Only NaCl
4Only MgCl2
NTA Answer: Option 3(final)
Sources
NCERT refs: Class 11 Chemistry Chapter 6, p.201
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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