Solubility product and the common ion effect

A saturated solution is an equilibrium

Shake some silver chloride with water. A tiny amount dissolves, and the rest stays at the bottom. The solution is saturated. Ions still leave the solid, and others settle back on it, at the same rate. This is a dynamic equilibrium:

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

The solubility product

The solid has a fixed concentration, so it is left out of the equilibrium constant. What remains is the solubility product:

• $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

For a general salt $\text{A}_x\text{B}_y$, each ion's concentration is raised to its number in the formula:

• $K_{sp} = [\text{A}^{y+}]^x [\text{B}^{x-}]^y$

So for $\text{CaF}_2$, $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$. For $\text{Bi}_2\text{S}_3$, $K_{sp} = [\text{Bi}^{3+}]^2[\text{S}^{2-}]^3$.

$K_{sp}$ depends only on temperature. A small $K_{sp}$ means a salt that hardly dissolves.

From Ksp to solubility

Let $s$ be the molar solubility, the moles of salt that dissolve in one litre of water.

• Type AB, such as AgCl: $[\text{A}^+] = s$ and $[\text{B}^-] = s$, so $K_{sp} = s^2$ and $s = \sqrt{K_{sp}}$.
• Type AB$_2$, such as $\text{CaF}_2$: $[\text{A}^{2+}] = s$ and $[\text{B}^-] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$.
• Type A$_2$B$_3$, such as $\text{Bi}_2\text{S}_3$: $[\text{A}] = 2s$ and $[\text{B}] = 3s$, so $K_{sp} = (2s)^2(3s)^3 = 108s^5$.

In general, for $\text{A}_x\text{B}_y$, $K_{sp} = x^x y^y s^{x+y}$.

Silver chloride. $K_{sp} = 1.8 \times 10^{-10}$:

• $s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}\ \text{mol/L}$

Calcium fluoride. $K_{sp} = 5.3 \times 10^{-11}$:

• $4s^3 = 5.3 \times 10^{-11}$, so $s = \sqrt[3]{1.33 \times 10^{-11}} = 2.37 \times 10^{-4}\ \text{mol/L}$

Notice that $\text{CaF}_2$ has the smaller $K_{sp}$ but the larger solubility. You can compare $K_{sp}$ values directly only for salts of the same type.

The common ion effect on solubility

Now drop silver chloride into $0.010\ \text{mol/L}$ sodium chloride instead of water. The chloride ions are already there. By Le Chatelier's principle, the equilibrium is pushed to the left, and much less AgCl dissolves.

Let the new solubility be $s$. Then $[\text{Ag}^+] = s$ and $[\text{Cl}^-] = 0.010 + s \approx 0.010\ \text{mol/L}$, since $s$ is tiny.

• $s = \dfrac{K_{sp}}{0.010} = \dfrac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8}\ \text{mol/L}$

That is about seven hundred times less than in pure water. Chemists use this to make a precipitate as complete as possible: they add a slight excess of the precipitating ion.

Will a precipitate form?

Mix two solutions, and work out the ionic product, $Q_{sp}$. It has the same form as $K_{sp}$, but uses the concentrations just after mixing.

• $Q_{sp} > K_{sp}$: the solution holds more ions than it can. A precipitate forms until $Q_{sp}$ falls to $K_{sp}$.
• $Q_{sp} = K_{sp}$: the solution is just saturated.
• $Q_{sp} < K_{sp}$: no precipitate. More solid could dissolve.

Suppose equal volumes of $2.0 \times 10^{-4}\ \text{mol/L}$ silver nitrate and sodium chloride are mixed. Mixing doubles the volume, so each ion is now $1.0 \times 10^{-4}\ \text{mol/L}$.

• $Q_{sp} = (1.0 \times 10^{-4})^2 = 1.0 \times 10^{-8}$, which is larger than $1.8 \times 10^{-10}$. Silver chloride precipitates.

The same idea separates ions in salt analysis. Sulphides of different metals precipitate at different acidities, because the acid controls how much sulphide ion is free.

Practise with new numbers

Try it in Python

The program finds the molar solubility of AgCl and of $\text{CaF}_2$ from their $K_{sp}$, and then of AgCl in salt water.

Common mistakes

• Writing $K_{sp} = s^2$ for every salt. For $\text{AB}_2$ it is $4s^3$, because one formula unit gives two B ions.
• Comparing $K_{sp}$ of different salt types. A smaller $K_{sp}$ need not mean a smaller solubility.
• Forgetting the dilution on mixing. Equal volumes halve each concentration.
• Adding the tiny $s$ to a large common-ion concentration. It changes nothing you can see, so leave it out.

Remember these

• $K_{sp}$ is the product of the ion concentrations in a saturated solution, each to the power of its number.
• AB: $K_{sp} = s^2$. $\text{AB}_2$: $K_{sp} = 4s^3$. $\text{A}_2\text{B}_3$: $K_{sp} = 108s^5$.
• A common ion lowers the solubility.
• $Q_{sp} > K_{sp}$: precipitate. $Q_{sp} < K_{sp}$: none.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write $K_{sp}$ for silver sulphate, $\text{Ag}_2\text{SO}_4$.

2. In terms of its molar solubility $s$, what is $K_{sp}$ of lead chloride, $\text{PbCl}_2$?

3. Why does barium sulphate dissolve less in sodium sulphate solution than in water?

4. After two solutions are mixed, $Q_{sp}$ is smaller than $K_{sp}$. Does a precipitate form?