Common Ion Effect

8 MCQs1 revision card9-step worked example

Source: NCERT EquilibriumPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 7 (Equilibrium), page 32, 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.

Practice 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?

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?

MCQ 3Direct ApplicationPractice

The K_sp of BaSO₄ is 1.1 × 10⁻¹⁰. Its molar solubility in 0.10 M Na₂SO₄ solution is approximately:

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:

MCQ 5Easy RecallPractice

Which of the following statements about the common ion effect is correct?

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?

MCQ 7Easy RecallPractice

If the molar solubility of Mg(OH)₂ in pure water is *s*, the correct K_sp expression is:

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:

Quick recall before you leave

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.) ---

Before solving, remember these

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. 32

Formulas

Henderson-Hasselbalch (buffer)

pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

pH of acidic buffer in terms of conjugate base/acid concentrations. For basic buffer: pOH = pKb + log10([salt]/[base]).

Symbol	Quantity	SI Unit
pKa	-log Ka	-
[salt]	conjugate base conc	mol/L
[acid]	weak acid conc	mol/L

Valid when

Buffer (weak acid + conjugate base)
Concentrations not too dilute
Approximate (assumes negligible dissociation)

Ka, Kb, Kw relationship

K_{a} K_{b} = K_{w}

Stronger acid → weaker conjugate base, and vice versa. pKa + pKb = 14.

Symbol	Quantity	SI Unit
Ka	acid dissociation	-
Kb	base dissociation	-
Kw	water 10^-14	-

Valid when

Conjugate acid-base pair
25°C

Equilibrium constant K_p and K_c

K_{p} = K_{c} (R T)^{Δ n}

Convert between pressure-based and concentration-based equilibrium constants. T in K; R = 0.0821 L atm/mol/K (when K_p in atm).

Symbol	Quantity	SI Unit
K_p	pressure constant	-
K_c	concentration constant	-
Δn	mole change	-

Valid when

Gas-phase equilibrium
Same temperature

Solubility product

K_{sp} = [M^{+}]^{a} [X^{-}]^{b}

Equilibrium constant for sparingly soluble salt. Q < K_sp: dissolves; Q > K_sp: precipitates.

Symbol	Quantity	SI Unit
K_sp	solubility product	-

Valid when

Sparingly soluble salt
Saturated solution

pH and pOH

pH = - lo g [H^{+}]

Logarithmic acidity scale. Pure water at 25°C: pH = 7 = pOH.

Symbol	Quantity	SI Unit
[H+]	hydrogen ion conc	mol/L
[OH-]	hydroxide conc	mol/L

Valid when

Aqueous solution
Use Kw = 10^-14 at 25°C

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Catalyst does NOT shift equilibrium

Category: Overthinking

Student claims catalyst shifts equilibrium toward products. Catalyst speeds up forward AND reverse equally; equilibrium position unchanged.

When it triggers

Question lists catalyst addition among options for shifting equilibrium.

How to avoid

Catalyst lowers activation energy of BOTH forward and reverse reactions equally. Time to reach equilibrium decreases; equilibrium position is unchanged.

Trap

Δn sign in K_p ↔ K_c conversion

Category: Sign Convention

Δn = (mol gas product) - (mol gas reactant). Sign matters; K_p = K_c (RT)^Δn.

When it triggers

Convert K_p ↔ K_c for gas-phase reaction.

How to avoid

Count moles of gas only (ignore solids/liquids). Δn = product - reactant. If Δn = 0, K_p = K_c. If Δn = +1, K_p = K_c × RT.

Trap

K_sp stoichiometric powers

Category: Similar Terms

For salt M_aX_b: K_sp = [M⁺]^a [X⁻]^b. Student uses [M⁺][X⁻] regardless of stoichiometry.

When it triggers

K_sp problem with non-1:1 salt (e.g. CaF₂, Mg(OH)₂, Ag₂CrO₄).

How to avoid

Write dissolution: M_aX_b → aM + bX. Then K_sp = [M]^a · [X]^b. For CaF₂ ↔ Ca + 2F: K_sp = s · (2s)² = 4s³.

Mistake

Believes catalyst shifts equilibrium toward products.

Root cause: concept gap

Correction

Catalyst speeds up forward AND reverse equally. Equilibrium position is unchanged. Time to reach equilibrium decreases.

Mistake

Counts solid/liquid moles in Δn for K_p ↔ K_c conversion.

Root cause: concept gap

Correction

Δn = (moles GAS product) - (moles GAS reactant). Solids and liquids excluded.

Mistake

Uses K_sp = [M+][X-] regardless of salt stoichiometry.

Root cause: formula misuse

Correction

For M_aX_b: K_sp = [M]^a · [X]^b. CaF₂: K_sp = s · (2s)² = 4s³.

Mistake

Predicts wrong direction of shift on pressure change.

Root cause: concept gap

Correction

Increase P → shift toward FEWER moles of gas. Decrease P → shift toward MORE moles. Reactions with same number of moles of gas: P doesn't shift equilibrium.

Mistake

Treats weak acid as fully dissociated when computing pH.

Root cause: concept gap

Correction

Weak acid: only fraction ionises. [H+] ≈ √(Ka · C). For 0.1 M acetic acid (Ka ~10⁻⁵): [H+] ~10⁻³ (pH ~3), not 1 (pH=1).

Past Year Questions

15 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Among the following, choose the ones with equal number of atoms. A. 212 g of Na CO (s) [molar mass = 106 g] 2 3 B. 248 g of Na O(s) [molar mass = 62 g] 2 C. 240 g of NaOH(s) [molar mass = 40 g] D. 12 g of H (g) [molar mass = 2 g] 2 E. 220 g of CO (g) [molar mass = 44 g] 2 Choose the correct answer from the options given below :

1B, D, and E only

2A, B, and C only

3A, B, and D only

4B, C, and D only

NTA Answer: Option 3(final)

NEET 2025

Sugar ‘X’ A. is found in honey B. is a keto sugar C. exists in  and  – anomeric forms. D. Is laevorotatory. ‘X’ is :

1Sucrose

2D-Glucose

3D-Fructose

4Maltose

NTA Answer: Option 3(final)

NEET 2025

For the reaction A(g)  2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given : R = 0.0831 L atm mol–1 K–1] K for the reaction at 1000 K is P

10.021

283.1

32.077 × 105

40.033

NTA Answer: Option 4(final)

NEET 2024Revised key

C [A] = [B] = [C] = 2 × 10–3 M. Then, which of the following is correct?

1Reaction is at equilibrium.

2Reaction has a tendency to go in forward direction.

3Reaction has a tendency to go in backward direction.

4Reaction has gone to completion in forward direction.

NTA Answer: Option 3(revised_final)

NEET 2023

The relation between n , (n = the number of permissible values of magnetic quantum number (m)) for a m m given value of azimuthal quantum number (l), is

1l = 2n + 1

2n = 2l2 + 1 m m n –1

3n = l + 2

4l  m m 2

NTA Answer: Option 4(final)

NEET 2023

Amongst the given options which of the following molecules/ ion acts as a Lewis acid?

1H O

2BF 2 3

3OH–

4NH 3

NTA Answer: Option 2(final)

NEET 2023

Which complex compound is most stable?

1CoNH  NO    3 3 3 3

2CoCl en NO  2 2 3

3CoNH   SO   3 6 2 4 3

4CoNH  H OBrNO   3 4 2  3 2

NTA Answer: Option 2(final)

NEET 2022

Choose the correct statement:

1Both diamond and graphite are used as dry lubricants.

2Diamond and graphite have two dimensional network.

3Diamond is covalent and graphite is ionic.

4Diamond is sp3 hybridised and graphite is sp2 hybridized.

NTA Answer: Option 4(final)

NEET 2022

Which one is not correct mathematical equation for Dalton’s Law of partial pressure? Here p = total pressure of gaseous mixture

1p =p, where  = mole fraction of ith gas in gaseous mixture i i i i p = pressure of ith gas in pure state i

2p = p + p + p 1 2 3 RT RT RT

3p=n +n +n 1 V 2 V 3 V

4p = p, where p = partial pressure of ith gas i i i  = mole fraction of ith gas in gaseous mixture i

NTA Answer: Option 1(final)

NEET 2022

A 10.0 L flask contains 64 g of oxygen at 27ºC. (Assume O 2 gas is behaving ideally). The pressure inside the flask in bar is (Given R = 0.0831 L bar K–1 mol–1)

14.9

22.5

3498.6

449.8

NTA Answer: Option 1(final)

NEET 2022

- 19 - 3 O 2 ( g ) 2 O 3 ( g ) for the above reaction at 298 K, K is found to be 3.0 × 10–59. If the concentration of O at equilibrium is C 2 0.040 M then concentration of O in M is 3

11.2 × 1021

24.38 × 10–32

31.9 × 10–63

42.4 × 1031

NTA Answer: Option 2(final)

NEET 2022

Find the emf of the cell in which the following reaction takes place at 298 K - 20 - N (i s ) + 2 A g + ( 0 .0 0 1 M ) → N 2i + ( 0 .0 0 1 M ) + 2 A g ( s )  G iv e n th a t E c e ll = 1 0 .5 V , 2 .3 0 3 F R T = 0 .0 5 9 a t 2 9 8 K 

11.05 V

21.0385 V

31.385 V

40.9615 V Answer (NA)

NTA Answer: Option 5(final)

NEET 2021

BF is planar and electron deficient compound. 3 Hybridization and number of electrons around the central atom, respectively are : sp2 sp3

1and 8

2and 4 sp3 sp2

3and 6

4and 6

NTA Answer: Option 4(final)

NEET 2021

Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is :

1Beryllium chloride

2Calcium chloride

3Strontium chloride

4Magnesium chloride

NTA Answer: Option 1(final)

NEET 2021

Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O and 2 g H confined in 2 2 a total volume of one litre at 0°C is : mol–1K–1, [Given R = 0.082 L atm T = 273 K]

126.02

22.518

32.602

425.18

NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Convert between K_p and K_c using K_p = K_c (RT)^Δn. Identify Δn carefully.

Multi StepMedium

Common distractors

wrong delta n sign

Counts moles incorrectly

Predict equilibrium shift on changing concentration, P, T, or adding catalyst.

InterpretationMedium

Common distractors

treats catalyst as shifting

Believes catalyst shifts equilibrium

Compute pH of weak acid/base solution using Ka or Kb. Approximation: [H+] ≈ √(Ka·C).

Multi StepMedium

Common distractors

ignores weak vs strong

Treats weak acid as fully dissociated

Compute solubility from K_sp or vice versa. Apply common-ion effect when relevant.

Multi StepMedium

Common distractors

forgets stoichiometric coefficient power

Uses [M+]^1 [X-]^1 for M(X)_2 salt

Sources

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Common Ion Effect

Lesson

Practice MCQs

Quick recall before you leave

Solubility product (K_sp)

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Correction

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources