Common Ion Effect — Why Solubility Decreases

medium CBSE JEE-MAIN • JEE Main 2024 • 4 min read
Tags Equilibrium

Question

When excess NaCl is added to a saturated solution of AgCl, the solubility of AgCl decreases. Explain this using Le Chatelier’s Principle and calculate the solubility of AgCl in 0.1 M NaCl solution.

Given: KspK_{sp} of AgCl = 1.8×10−101.8 \times 10^{-10}

Solution — Step by Step

AgCl dissolves as:

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

The KspK_{sp} expression is Ksp=[Ag+][Cl−]K_{sp} = [\text{Ag}^+][\text{Cl}^-]. In pure water, if solubility is ss, then [Ag+]=[Cl−]=s[\text{Ag}^+] = [\text{Cl}^-] = s.

NaCl is fully dissociated: NaCl→Na++Cl−\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

So adding 0.1 M NaCl dumps 0.1 M Cl−\text{Cl}^- into the solution before any AgCl dissolves. By Le Chatelier’s Principle, the extra Cl−\text{Cl}^- shifts the equilibrium left, suppressing AgCl dissolution. This is the common ion effect — Cl−\text{Cl}^- is common to both NaCl and AgCl.

Let ss = solubility of AgCl in 0.1 M NaCl.

Ag+\text{Ag}^+Cl−\text{Cl}^-
Initial00.1
Change+s+s+s+s
Equilibriumss0.1+s0.1 + s

The KspK_{sp} equation becomes: s(0.1+s)=1.8×10−10s(0.1 + s) = 1.8 \times 10^{-10}

Since KspK_{sp} is very small, s≪0.1s \ll 0.1, so (0.1+s)≈0.1(0.1 + s) \approx 0.1.

s×0.1=1.8×10−10s \times 0.1 = 1.8 \times 10^{-10} s=1.8×10−100.1=1.8×10−9 mol/Ls = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \text{ mol/L}

In pure water: s=Ksp=1.8×10−10≈1.34×10−5s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} mol/L

In 0.1 M NaCl: s=1.8×10−9s = 1.8 \times 10^{-9} mol/L

The solubility dropped by a factor of roughly 7400. That’s a massive suppression from just 0.1 M NaCl.

Final Answer: Solubility of AgCl in 0.1 M NaCl = 1.8×10−91.8 \times 10^{-9} mol/L

Why This Works

KspK_{sp} is a constant at a given temperature — the product [Ag+][Cl−][\text{Ag}^+][\text{Cl}^-] must always equal 1.8×10−101.8 \times 10^{-10}. When we introduce extra Cl−\text{Cl}^- from NaCl, the ion product immediately exceeds KspK_{sp}. The system responds by precipitating AgCl until equilibrium is restored.

This is just Le Chatelier at work: stress the equilibrium by adding a product, and the reaction shifts to consume it. The “common ion” just means one of the ions in the KspK_{sp} expression is also contributed by another soluble salt.

The practical consequence is significant — this principle is used in gravimetric analysis in labs. Adding excess Cl−\text{Cl}^- ensures nearly complete precipitation of Ag+\text{Ag}^+, which is exactly what you want when trying to measure silver quantitatively.

Alternative Method — Using the Ion Product QspQ_{sp}

Instead of ICE, think in terms of reaction quotient QQ.

Before AgCl dissolves: Q=[Ag+][Cl−]=(0)(0.1)=0Q = [\text{Ag}^+][\text{Cl}^-] = (0)(0.1) = 0. As AgCl dissolves, [Ag+][\text{Ag}^+] rises. The system reaches equilibrium when Q=KspQ = K_{sp}, i.e., when [Ag+]×0.1=1.8×10−10[\text{Ag}^+] \times 0.1 = 1.8 \times 10^{-10}.

This directly gives [Ag+]=1.8×10−9[\text{Ag}^+] = 1.8 \times 10^{-9} M, which equals the solubility ss. Same answer, different framing — and in JEE, this Q-based thinking often cuts solution time.

The small-s approximation is valid when s≪s \ll [common ion]. Always verify: here 1.8×10−9≪0.11.8 \times 10^{-9} \ll 0.1. ✓ If the approximation fails (s > 5% of common ion concentration), you’d need the quadratic. For AgCl with typical NaCl concentrations, it always holds.

Common Mistake

The classic error: students write Ksp=s×s=s2K_{sp} = s \times s = s^2 even when a common ion is present. They ignore the initial 0.1 M Cl−\text{Cl}^- and get s=1.34×10−5s = 1.34 \times 10^{-5} — the pure water answer. This appeared directly in JEE Main 2024 and many students dropped marks here. The moment you see “dissolved in 0.1 M NaCl” (or any solution containing Ag+\text{Ag}^+ or Cl−\text{Cl}^-), the equilibrium concentration of that ion is NOT simply ss — it’s ss plus whatever was already there.

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