Common Ion Effect on Solubility of Ionic Salts

The Common Ion Effect on Solubility of Ionic Salts

When a sparingly soluble salt is in equilibrium with its saturated solution, the product of the concentrations of its ions (raised to appropriate powers) equals $K_{sp}$. Le Chatelier's principle tells us what happens if we disturb this equilibrium by adding more of one of the ions.

If you increase the concentration of either ion — say, by adding a soluble salt that provides that same ion — the system responds by shifting the equilibrium to the left. Some of the salt precipitates until the ionic product $Q_{sp}$ once again equals $K_{sp}$. Conversely, if you decrease the concentration of one ion (by removing it through a reaction, for instance), more salt dissolves to restore the balance.

This principle applies even to highly soluble salts like sodium chloride, though with a practical difference. In concentrated solutions, the activities of ions (their effective concentrations) deviate significantly from molarities, so we must use activities in the $Q_{sp}$ expression rather than simple molar concentrations.

A Practical Demonstration: Purifying Sodium Chloride

Consider a saturated solution of NaCl. If you pass HCl gas through it, the concentration (more precisely, the activity) of chloride ions increases dramatically because HCl dissociates completely. The common ion $Cl^-$ shifts the equilibrium

to the left, precipitating solid NaCl. The sodium chloride obtained this way is remarkably pure — impurities like sodium sulphate and magnesium sulphate remain in solution because their concentrations never exceed their solubility products.

Gravimetric Estimation

The common ion effect is deliberately exploited in quantitative analysis. To precipitate a particular ion almost completely as a sparingly soluble salt, you add an excess of the precipitating agent (the common ion). This drives the solubility equilibrium so far to the left that the ion's concentration in solution becomes negligible. Silver ion is precipitated as AgCl, ferric ion as Fe(OH)$_3$ (or hydrated ferric oxide), and barium ion as BaSO$_4$ — all for gravimetric estimation.

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Quantitative Treatment

Problem 6.28 (below) works the numbers for exactly this situation: the molar solubility of $\text{Ni(OH)}_2$ ($K_{sp} = 2.0 \times 10^{-15}$) in 0.10 M NaOH comes out at just $2.0 \times 10^{-13}$ M — compared with about $7.9 \times 10^{-6}$ M in pure water, a suppression of some seven orders of magnitude by the common hydroxide ion.

Effect of pH on Solubility of Salts of Weak Acids

The solubility of salts whose anion is the conjugate base of a weak acid (like phosphates, carbonates, sulphides) is strongly pH-dependent. The reason is that at lower pH, the anion gets protonated, reducing its concentration in solution. To maintain $K_{sp} = Q_{sp}$, more salt must dissolve.

Derivation of the pH-Dependent Solubility Formula

Consider a sparingly soluble salt $MX$ where $X^-$ is the anion of a weak acid $HX$. The equilibria involved are:

1. Dissolution: $MX(s) \rightleftharpoons M^+(aq) + X^-(aq)$ with $K_{sp} = [M^+][X^-]$
2. Protonation of the anion: $X^-(aq) + H^+(aq) \rightleftharpoons HX(aq)$ with $K_a = \frac{[H^+][X^-]}{[HX]}$

Let $S$ be the solubility of the salt at a given pH. Then:

The total concentration of the anion in all forms (free $X^-$ plus protonated $HX$) is also $S$:

We need to express $[X^-]$ in terms of $S$, $K_a$, and $[H^+]$.

From the $K_a$ expression:

Substituting into the mass balance:

Define the fraction $f$ as the fraction of total anion that exists as free $X^-$:

Now substitute into the $K_{sp}$ expression:

Solving for $S$:

This is equation (6.46) in the textbook.

What This Tells Us …