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The common ion effect

Covers AS91392

What it is

A common ion is an ion that is already present in the solution and is also one of the ions of the sparingly soluble solid.
Adding a common ion decreases the solubility of that solid.

Why it happens

Consider a saturated solution of silver chloride:

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

Adding sodium chloride solution increases $[\text{Cl}^-]$ — a product of this equilibrium.
By Le Châtelier's principle the equilibrium shifts to oppose the increase, so it moves left.
More solid forms, and less of it dissolves.

The solubility decreases, but $K_s$ itself is unchanged.

$K_s$ does not change

This distinction is worth stating carefully, because questions test it.
$K_s$ is a constant at a given temperature. Adding a common ion does not alter it.
What changes is the balance between the two ion concentrations. One goes up, so the other must come down to keep their product equal to $K_s$ .

Calculating solubility with a common ion

Write the $K_s$ expression as usual.
Set the common ion's concentration equal to that supplied by the added salt — the solid contributes so little that its own contribution is negligible.
Let the other ion's concentration equal $s$ .
Solve for $s$ .
This is much easier than the pure-water case, because there is no square or cube root to take.

The size of the effect

For $\text{AgCl}$ , $K_s = 1.8 \times 10^{-10}$ :

Solvent	Working	Solubility
pure water	$s = \sqrt{K_s}$	$1.34 \times 10^{-5}$ mol L⁻¹
0.10 mol L⁻¹ NaCl	$s = K_s / 0.10$	$1.8 \times 10^{-9}$ mol L⁻¹

That is a reduction by a factor of about 7400. The effect is large, which is why it is used deliberately.

Where it is used

Washing a precipitate with a dilute solution of one of its own ions, rather than pure water, minimises the amount that redissolves and is lost.
In gravimetric analysis, a slight excess of the precipitating reagent is added for exactly this reason.

Tips and Tricks

Say which ion is common and which direction the equilibrium shifts. "Adding NaCl increases $[\text{Cl}^-]$ , so the equilibrium shifts left to oppose the increase, and less solid dissolves."

$K_s$ does not change. If your answer says the common ion lowers $K_s$ , it is wrong. $K_s$ is a constant at a given temperature; only the solubility changes.

In the calculation, neglect the ion supplied by the solid itself. The solid is sparingly soluble, so its contribution to the common ion is tiny compared with the added salt — say so explicitly.

Check the magnitude of your answer. The solubility with a common ion should be much smaller than in pure water. If it comes out larger, you have divided the wrong way.

A salt with no ion in common — potassium nitrate, say — has essentially no effect on solubility. Questions often include one as a distractor.

Worked ExampleComparing solubility in water and in a common ion solution

Calculate the solubility of silver chloride

(a) in pure water, and (b) in $0.10$ mol L⁻¹ sodium chloride solution,

and explain the difference. $K_s(\text{AgCl}) = 1.8 \times 10^{-10}$

Part (a) — in pure water

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \qquad K_s = [\text{Ag}^+][\text{Cl}^-]$

Both ions come only from the dissolving solid, so $[\text{Ag}^+] = [\text{Cl}^-] = s$ :

$K_s = s^2 \quad \Rightarrow \quad s = \sqrt{1.8 \times 10^{-10}}$

Part (b) — in 0.10 mol L⁻¹ NaCl

Step 1 — Identify the common ion. Sodium chloride supplies chloride ions, which are also produced by the dissolving AgCl. So $\text{Cl}^-$ is the common ion.

Step 2 — Set the chloride concentration. The chloride comes from two sources: the added NaCl at 0.10 mol L⁻¹, and the dissolving AgCl at $s$ .

But part (a) showed $s$ is only about $10^{-5}$ in pure water, and the common ion will make it smaller still. So the AgCl contribution is utterly negligible beside 0.10:

$[\text{Cl}^-] \approx 0.10 \text{ mol L}^{-1}$

Step 3 — Set the silver concentration. Silver ions come only from the dissolving AgCl, so

$[\text{Ag}^+] = s$

Step 4 — Solve. No root is needed this time.

$K_s = [\text{Ag}^+][\text{Cl}^-] = s \times 0.10 = 1.8 \times 10^{-10}$

$s = \frac{1.8 \times 10^{-10}}{0.10}$

Step 5 — Explain the difference

$\frac{1.34 \times 10^{-5}}{1.8 \times 10^{-9}} \approx 7400$

Silver chloride is about 7400 times less soluble in 0.10 mol L⁻¹ NaCl than in pure water.

Why. The sodium chloride supplies a large concentration of $\text{Cl}^-$ , which is a product of the dissolving equilibrium:

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

By Le Châtelier's principle the equilibrium shifts left to oppose the increase in chloride, so more solid forms and less dissolves.

Note that $K_s$ has not changed — it is still $1.8 \times 10^{-10}$ . What has changed is the balance: $[\text{Cl}^-]$ is now about $10^4$ times larger than before, so $[\text{Ag}^+]$ must be about $10^4$ times smaller for their product to remain constant.

A practical check on the assumption. We assumed the AgCl contributed negligibly to $[\text{Cl}^-]$ . The answer confirms it: the solid contributes only $1.8 \times 10^{-9}$ mol L⁻¹ of chloride against the added 0.10 — smaller by a factor of $5 \times 10^7$ . The approximation was more than safe.

The common ion effect

Covers AS91392

What it is

A common ion is an ion that is already present in the solution and is also one of the ions of the sparingly soluble solid.
Adding a common ion decreases the solubility of that solid.

Why it happens

Consider a saturated solution of silver chloride:

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

Adding sodium chloride solution increases $[\text{Cl}^-]$ — a product of this equilibrium.
By Le Châtelier's principle the equilibrium shifts to oppose the increase, so it moves left.
More solid forms, and less of it dissolves.

The solubility decreases, but $K_s$ itself is unchanged.

$K_s$ does not change

This distinction is worth stating carefully, because questions test it.
$K_s$ is a constant at a given temperature. Adding a common ion does not alter it.
What changes is the balance between the two ion concentrations. One goes up, so the other must come down to keep their product equal to $K_s$ .

Calculating solubility with a common ion

Write the $K_s$ expression as usual.
Set the common ion's concentration equal to that supplied by the added salt — the solid contributes so little that its own contribution is negligible.
Let the other ion's concentration equal $s$ .
Solve for $s$ .
This is much easier than the pure-water case, because there is no square or cube root to take.

The size of the effect

For $\text{AgCl}$ , $K_s = 1.8 \times 10^{-10}$ :

Solvent	Working	Solubility
pure water	$s = \sqrt{K_s}$	$1.34 \times 10^{-5}$ mol L⁻¹
0.10 mol L⁻¹ NaCl	$s = K_s / 0.10$	$1.8 \times 10^{-9}$ mol L⁻¹

That is a reduction by a factor of about 7400. The effect is large, which is why it is used deliberately.

Where it is used

Washing a precipitate with a dilute solution of one of its own ions, rather than pure water, minimises the amount that redissolves and is lost.
In gravimetric analysis, a slight excess of the precipitating reagent is added for exactly this reason.

Tips and Tricks

$K_s$ does not change. If your answer says the common ion lowers $K_s$ , it is wrong. $K_s$ is a constant at a given temperature; only the solubility changes.

In the calculation, neglect the ion supplied by the solid itself. The solid is sparingly soluble, so its contribution to the common ion is tiny compared with the added salt — say so explicitly.

Check the magnitude of your answer. The solubility with a common ion should be much smaller than in pure water. If it comes out larger, you have divided the wrong way.

A salt with no ion in common — potassium nitrate, say — has essentially no effect on solubility. Questions often include one as a distractor.

Worked ExampleComparing solubility in water and in a common ion solution

Calculate the solubility of silver chloride

(a) in pure water, and (b) in $0.10$ mol L⁻¹ sodium chloride solution,

and explain the difference. $K_s(\text{AgCl}) = 1.8 \times 10^{-10}$

Part (a) — in pure water

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \qquad K_s = [\text{Ag}^+][\text{Cl}^-]$

Both ions come only from the dissolving solid, so $[\text{Ag}^+] = [\text{Cl}^-] = s$ :

$K_s = s^2 \quad \Rightarrow \quad s = \sqrt{1.8 \times 10^{-10}}$

Part (b) — in 0.10 mol L⁻¹ NaCl

Step 1 — Identify the common ion. Sodium chloride supplies chloride ions, which are also produced by the dissolving AgCl. So $\text{Cl}^-$ is the common ion.

Step 2 — Set the chloride concentration. The chloride comes from two sources: the added NaCl at 0.10 mol L⁻¹, and the dissolving AgCl at $s$ .

But part (a) showed $s$ is only about $10^{-5}$ in pure water, and the common ion will make it smaller still. So the AgCl contribution is utterly negligible beside 0.10:

$[\text{Cl}^-] \approx 0.10 \text{ mol L}^{-1}$

Step 3 — Set the silver concentration. Silver ions come only from the dissolving AgCl, so

$[\text{Ag}^+] = s$

Step 4 — Solve. No root is needed this time.

$K_s = [\text{Ag}^+][\text{Cl}^-] = s \times 0.10 = 1.8 \times 10^{-10}$

$s = \frac{1.8 \times 10^{-10}}{0.10}$

Step 5 — Explain the difference

$\frac{1.34 \times 10^{-5}}{1.8 \times 10^{-9}} \approx 7400$

Silver chloride is about 7400 times less soluble in 0.10 mol L⁻¹ NaCl than in pure water.

Why. The sodium chloride supplies a large concentration of $\text{Cl}^-$ , which is a product of the dissolving equilibrium:

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

By Le Châtelier's principle the equilibrium shifts left to oppose the increase in chloride, so more solid forms and less dissolves.

The common ion effect

What it is

Why it happens

KsK_sKs​ does not change

Calculating solubility with a common ion

The size of the effect

Where it is used

Part (a) — in pure water

Part (b) — in 0.10 mol L−1 NaCl

Step 5 — Explain the difference

The common ion effect

What it is

Why it happens

KsK_sKs​ does not change

Calculating solubility with a common ion

The size of the effect

Where it is used

Part (a) — in pure water

Part (b) — in 0.10 mol L−1 NaCl

Step 5 — Explain the difference

$K_s$ does not change

Part (b) — in 0.10 mol L⁻¹ NaCl

$K_s$ does not change

Part (b) — in 0.10 mol L⁻¹ NaCl