The Common Ion Effect — and Where It Is Actually Used
The common ion effect is Le Chatelier's principle applied to ionic equilibrium, and it is one of the few topics where a single idea explains a whole laboratory practical. It is why silver chloride is washed with dilute HCl rather than water, why a buffer works at all, why hydrogen sulphide precipitates one group of cations in acid and a different group in ammonia, and why soap is thrown out of solution with common salt. This page gives the definition, three fully worked calculations and the practical uses examiners ask about.
The definition
Nothing new is happening. Adding a product ion pushes the equilibrium backwards, exactly as Le Chatelier predicts. The point worth grasping is that K itself does not change — Ka and Ksp are constants at a fixed temperature. What changes is how the fixed product is split between the two ions. If one ion is forced up, the other must come down to keep the product equal to K.
Worked example 1 — solubility of AgCl in sodium chloride solution
Take Ksp(AgCl) = 1.8 × 10⁻¹⁰ at 298 K. Different data books quote slightly different values, so use the one printed in your paper; the method does not change.
(a) In pure water. AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), so if the solubility is s mol L⁻¹ then [Ag⁺] = s and [Cl⁻] = s.
Ksp = s × s = s²
s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol L⁻¹
Check: (1.34 × 10⁻⁵)² = 1.80 × 10⁻¹⁰ ✓
(b) In 0.10 M NaCl. NaCl is fully dissociated, so it delivers 0.10 mol L⁻¹ of chloride before any AgCl dissolves. Now [Cl⁻] = 0.10 + s. Because s will be tiny, 0.10 + s ≈ 0.10 — an approximation we will verify at the end.
Ksp = [Ag⁺][Cl⁻] = s × 0.10
s = 1.8 × 10⁻¹⁰ ÷ 0.10 = 1.8 × 10⁻⁹ mol L⁻¹
Verify the approximation: 0.10 + 1.8 × 10⁻⁹ = 0.100000002, so treating it as 0.10 was entirely safe.
(c) The size of the effect.
1.34 × 10⁻⁵ ÷ 1.8 × 10⁻⁹ = (1.34 ÷ 1.8) × 10⁴ = 0.744 × 10⁴ = 7.4 × 10³
AgCl is roughly 7 400 times less soluble in 0.10 M NaCl than in pure water. This is why a silver chloride precipitate is washed with very dilute HCl, not with distilled water — washing with water would dissolve part of the precipitate and spoil a gravimetric result.
Worked example 2 — a 1 : 2 salt, where the effect is even stronger
Take Ksp(Mg(OH)₂) = 5.6 × 10⁻¹².
(a) In pure water. Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻, so [Mg²⁺] = s and [OH⁻] = 2s.
Ksp = s × (2s)² = 4s³
s³ = 5.6 × 10⁻¹² ÷ 4 = 1.4 × 10⁻¹²
s = (1.4 × 10⁻¹²)^(1/3) = (1400 × 10⁻¹⁵)^(1/3) = 11.19 × 10⁻⁵ =
1.12 × 10⁻⁴ mol L⁻¹
Check: 4 × (1.119 × 10⁻⁴)³ = 4 × 1.40 × 10⁻¹² = 5.6 × 10⁻¹² ✓
(b) In 0.10 M NaOH. Here [OH⁻] ≈ 0.10 from the strong base.
Ksp = s × (0.10)² = s × 0.010
s = 5.6 × 10⁻¹² ÷ 0.010 = 5.6 × 10⁻¹⁰ mol L⁻¹
(c) Ratio: 1.12 × 10⁻⁴ ÷ 5.6 × 10⁻¹⁰ = 2.0 × 10⁵.
A 200 000-fold drop. The effect is larger here because the common ion is squared in the Ksp expression. Rule to remember: the higher the stoichiometric coefficient of the common ion, the more violently the solubility is suppressed.
Worked example 3 — suppressing a weak acid, and how a buffer is born
Take Ka(CH₃COOH) = 1.8 × 10⁻⁵.
(a) 0.10 M acetic acid alone.
[H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) =
1.34 × 10⁻³ mol L⁻¹
pH = −log(1.34 × 10⁻³) = 3 − 0.128 = 2.87
Degree of dissociation α = 1.34 × 10⁻³ ÷ 0.10 = 0.0134 = 1.34%
(b) The same acid, now also 0.10 M in sodium acetate. The salt is fully dissociated and supplies the common ion CH₃COO⁻.
[H⁺] = Ka × ([acid] ÷ [salt]) = 1.8 × 10⁻⁵ × (0.10 ÷ 0.10) =
1.8 × 10⁻⁵ mol L⁻¹
pH = −log(1.8 × 10⁻⁵) = 5 − 0.255 = 4.74
α = 1.8 × 10⁻⁵ ÷ 0.10 = 1.8 × 10⁻⁴ = 0.018%
(c) Compare. [H⁺] falls by 1.34 × 10⁻³ ÷ 1.8 × 10⁻⁵ = 74 times, and the acid is now about 75 times less dissociated. The pH has risen by almost two units without adding a drop of alkali.
That mixture of a weak acid and its salt is an acidic buffer, and the expression used in (b) is the Henderson–Hasselbalch equation in disguise. The common ion effect is not merely related to buffers — it is the mechanism by which they work.
Where the common ion effect is genuinely used
| Application | Common ion added | What it achieves |
|---|---|---|
| Group II qualitative analysis (H₂S in dilute HCl) | H⁺ from HCl | Suppresses ionisation of the weak acid H₂S, keeping [S²⁻] low so that only the least soluble sulphides — Cu, Pb, Cd, Bi and so on — precipitate |
| Group III (NH₄OH with NH₄Cl) | NH₄⁺ from NH₄Cl | Suppresses ionisation of NH₄OH, keeping [OH⁻] low so Fe(OH)₃, Al(OH)₃ and Cr(OH)₃ precipitate while Mg²⁺ and later groups stay in solution |
| Group IV (H₂S in ammoniacal solution) | Removal of H⁺ instead | Raises [S²⁻] so that the more soluble sulphides of Zn, Mn, Ni and Co now precipitate |
| Washing a precipitate | The precipitating ion, in dilute form | Prevents loss of the precipitate during washing in gravimetric analysis |
| Buffer solutions | Conjugate base or acid from the salt | Holds pH steady in blood, fermentation, plating baths and titration control |
| Salting out of soap | Na⁺ from saturated NaCl | Exceeds the solubility limit of sodium stearate so the soap separates from the glycerol and water |
| Purification of common salt | Cl⁻ from HCl gas | Passing HCl through saturated brine crystallises out pure NaCl and leaves the more soluble impurities behind |
The first three rows are the same reagent, H₂S, being made to behave in two different ways purely by controlling one common ion. If you understand that, you understand the logic of the whole qualitative-analysis scheme rather than memorising it.
Mistakes that cost marks
- Saying Ksp or Ka decreases. They are constants at a given temperature. What decreases is the solubility or the degree of dissociation. This is the single most common wording error in this topic.
- Forgetting to square the common ion. For Mg(OH)₂ in NaOH the expression is s × [OH⁻]², not s × [OH⁻]. Missing the square changes the answer by a factor of ten.
- Adding s to the common ion and then panicking about the cubic. Write [Cl⁻] = 0.10 + s, then justify 0.10 + s ≈ 0.10 in one line. Examiners award the justification.
- Assuming any added salt suppresses solubility. Adding a salt with no ion in common — say KNO₃ to an AgCl solution — actually increases solubility very slightly through ionic-strength effects. That is the salt effect, the opposite phenomenon, and it is not the common ion effect.
- Confusing suppression with complex formation. Adding a large excess of chloride to AgCl eventually dissolves it as [AgCl₂]⁻. The common ion effect describes the initial suppression only, and this exception is worth a line in a long answer.
- Using molarity of the salt instead of of the ion. 0.10 M CaCl₂ supplies 0.20 M chloride, not 0.10 M. Always multiply by the stoichiometry.
Where this is asked in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 11 | Ionic equilibrium — calculate solubility in the presence of a common ion; explain buffer action |
| CBSE / ICSE Class 12 | Salt analysis reasoning; why NH₄Cl is added before NH₄OH in Group III |
| JEE / NEET | Numerical solubility comparisons; selective precipitation problems |
| IIT-JAM / CUET-PG | Ksp with common ion, pH of buffers, fractional precipitation |
| GATE / CSIR-NET | Analytical separations, masking and selective precipitation, activity corrections |
Convert between Ksp and solubility without slips. The 1 : 1, 1 : 2 and 1 : 3 salts each need a different power of s, and that is where marks are lost. The Ksp ↔ Solubility tool does the conversion both ways so you can check the pure-water figure before you apply the common ion.
Open the Ksp ↔ Solubility Calculator →Ionic equilibrium rewards steady practice more than any other Class 11 chapter. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — see abcchemistry.in. For students in Delhi, Noida or Gurgaon who learn better one-to-one at home, delhihometutor.com arranges home tuition in the NCR.