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Simultaneous Solubility & Common Ion Effect | chemca

Ionic Equilibrium

Simultaneous Solubility

Calculating the solubility of mixed salts in the presence of a common ion.

By chemca Team • Updated Jan 2026

Simultaneous Solubility refers to the solubility of two or more sparingly soluble salts present in the same solution. When these salts share a common ion (e.g., $AgCl$ and $AgBr$ both have $Ag^+$), their solubilities decrease significantly compared to pure water due to the Common Ion Effect.

1. The Concept

Consider a solution containing two sparingly soluble salts, $AB$ and $AC$.

$AB \rightleftharpoons A^+ + B^-$
$AC \rightleftharpoons A^+ + C^-$

Here, $A^+$ is the common ion. The total concentration of $A^+$ in the solution is the sum of ions coming from both salts. This high concentration shifts both equilibria backward (Le Chatelier's Principle), reducing the solubility of both salts.

2. Mathematical Derivation

Let $S_1$ be the solubility of $AgCl$ and $S_2$ be the solubility of $AgBr$ in a mixture.
Given: $K_{sp}(AgCl) = K_1$ and $K_{sp}(AgBr) = K_2$.

Step 1: Write Equilibrium Expressions

1. $AgCl(s) \rightleftharpoons Ag^+ + Cl^-$

From $AgCl$: $[Ag^+] = S_1, \quad [Cl^-] = S_1$

2. $AgBr(s) \rightleftharpoons Ag^+ + Br^-$

From $AgBr$: $[Ag^+] = S_2, \quad [Br^-] = S_2$

Step 2: Total Ion Concentration

The common ion is $Ag^+$. $$ [Ag^+]_{total} = S_1 + S_2 $$

Step 3: Apply Solubility Product

$$ K_1 = [Ag^+][Cl^-] = (S_1 + S_2) S_1 \quad \dots(i) $$

$$ K_2 = [Ag^+][Br^-] = (S_1 + S_2) S_2 \quad \dots(ii) $$

Step 4: Solve for Solubility

Divide equation (i) by (ii):

$$ \frac{S_1}{S_2} = \frac{K_1}{K_2} $$

This implies that the ratio of solubilities in a simultaneous mixture is equal to the ratio of their solubility products (for salts of the same type, e.g., 1:1 electrolytes).

Using $S_1 = S_2 \frac{K_1}{K_2}$ substitute back into (i) to find exact values.

3. Solved Example

Problem: Calculate the simultaneous solubility of $AgCl$ and $AgSCN$.
Given $K_{sp}(AgCl) = 10^{-10}$ and $K_{sp}(AgSCN) = 10^{-12}$.

Solution:

Let solubilities be $x$ and $y$. Ratio $x/y = 10^{-10} / 10^{-12} = 100 \Rightarrow x = 100y$.

Equation for AgCl:

$$ 10^{-10} = (x + y)x \approx (100y + y)(100y) $$ $$ 10^{-10} \approx 101y \times 100y \approx 10^4 y^2 $$ $$ y^2 = 10^{-14} \Rightarrow y = 10^{-7} M $$ $$ x = 100y = 10^{-5} M $$

Result: $S_{AgCl} = 10^{-5}M$, $S_{AgSCN} = 10^{-7}M$.

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