Common ion effect — explain with example of AgCl precipitation

Question

Explain the common ion effect with the example of AgCl precipitation. How does adding NaCl to a saturated solution of AgCl affect the solubility of AgCl?

(NCERT Class 11, Chapter 7 — Equilibrium)

Solution — Step by Step

AgCl is a sparingly soluble salt. In water, it establishes an equilibrium:

\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)

The solubility product: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

At 25°C, $K_{sp}$ of AgCl $= 1.8 \times 10^{-10}$ .

NaCl is a strong electrolyte — it dissociates completely:

\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

The $\text{Cl}^-$ from NaCl is the common ion (same ion that AgCl also produces). Adding NaCl increases $[\text{Cl}^-]$ in the solution.

$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ is a constant at a given temperature.

When $[\text{Cl}^-]$ increases (due to added NaCl), the product $[\text{Ag}^+][\text{Cl}^-]$ would momentarily exceed $K_{sp}$ . To restore equilibrium, the system shifts backward — more AgCl precipitates out.

This means $[\text{Ag}^+]$ decreases, and the solubility of AgCl decreases.

Suppose we add NaCl so that $[\text{Cl}^-] = 0.1$ M.

K_{sp} = [\text{Ag}^+] \times 0.1 = 1.8 \times 10^{-10}

[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}

Without NaCl, $[\text{Ag}^+] = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ M.

The solubility dropped from $1.34 \times 10^{-5}$ M to $1.8 \times 10^{-9}$ M — a reduction by a factor of about 7400. The common ion effect is dramatic.

Why This Works

The common ion effect is a direct application of Le Chatelier’s principle to solubility equilibria. When you add an ion that’s already part of the equilibrium, you’re pushing the equilibrium backward, forcing more solid to precipitate.

This is why in qualitative analysis (salt analysis), we add HCl to precipitate group I cations (Ag⁺, Pb²⁺) — the excess Cl⁻ ensures maximum precipitation. It’s also why washing a precipitate with a solution containing a common ion (not pure water) minimises losses.

Alternative Method — Think of it as ion product vs Ksp

If $Q_{sp} = [\text{Ag}^+][\text{Cl}^-] > K_{sp}$ , precipitation occurs.

Adding NaCl increases $[\text{Cl}^-]$ , pushing $Q_{sp}$ above $K_{sp}$ , triggering precipitation until $Q_{sp}$ returns to $K_{sp}$ .

For NEET, the common ion effect can be summarised in one line: adding a common ion to a saturated solution decreases the solubility of the sparingly soluble salt. This principle applies to all salts, not just AgCl. Questions often use PbCl₂ with HCl or BaSO₄ with Na₂SO₄.

Common Mistake

Students sometimes think that adding NaCl increases the solubility of AgCl because “more ions are in solution.” This is exactly backwards. The common ion ( $\text{Cl}^-$ ) suppresses the dissolution of AgCl. More $\text{Cl}^-$ in solution means less AgCl can dissolve. Don’t confuse the total ion concentration in solution with the solubility of AgCl specifically.

Question

Solution — Step by Step

Why This Works

Alternative Method — Think of it as ion product vs Ksp

Common Mistake

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