Write the K_sp expression for lead(II) chloride, PbCl_2. [1 mark]

The solubility of Mg(OH)_2 is 1.2 × 10^-4 mol dm^-3. Calculate its K_sp. [3 marks]

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How is the solubility of a sparingly soluble ionic compound described and predicted?

Define and write expressions for the solubility product Ksp, calculate solubility from Ksp and vice versa, predict precipitation by comparing the ionic product with Ksp, and explain the common ion effect

A focused answer to the H2 Chemistry learning outcome on solubility equilibria. Defining and writing Ksp, converting between Ksp and molar solubility, using the ionic product to predict precipitation, and explaining the common ion effect quantitatively.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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What this dot point is asking
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What this dot point is asking

SEAB wants you to define the solubility product $K_{sp}$ , write its expression for any sparingly soluble salt, convert between $K_{sp}$ and molar solubility, predict whether a precipitate forms by comparing the ionic product with $K_{sp}$ , and explain the common ion effect. These solubility calculations are a compact, high-value Paper 2 topic and connect to qualitative analysis.

The answer

The solubility product

For a sparingly soluble ionic solid in contact with its saturated solution, a dynamic equilibrium exists:

\text{A}_x\text{B}_y(s) \rightleftharpoons x\text{A}^{y+}(aq) + y\text{B}^{x-}(aq)

The solubility product is the product of the ion concentrations, each raised to its stoichiometric power, at saturation:

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

The solid does not appear (it is a pure solid). $K_{sp}$ is constant at a fixed temperature.

Writing expressions

Salt	Dissolution	$K_{sp}$ expression
$\text{AgCl}$	$\text{Ag}^+ + \text{Cl}^-$	$[\text{Ag}^+][\text{Cl}^-]$
$\text{CaF}_2$	$\text{Ca}^{2+} + 2\text{F}^-$	$[\text{Ca}^{2+}][\text{F}^-]^2$
$\text{Mg(OH)}_2$	$\text{Mg}^{2+} + 2\text{OH}^-$	$[\text{Mg}^{2+}][\text{OH}^-]^2$

Linking $K_{sp}$ to solubility

Let molar solubility be $s$ . The ion concentrations follow the stoichiometry:

For AgCl (1:1): $K_{sp} = s^2$ , so $s = \sqrt{K_{sp}}$ .
For $\text{CaF}_2$ (1:2): $[\text{Ca}^{2+}] = s$ , $[\text{F}^-] = 2s$ , so $K_{sp} = s(2s)^2 = 4s^3$ , giving $s = \sqrt[3]{K_{sp}/4}$ .

Always set up the stoichiometric relationship before solving.

Predicting precipitation

Compute the ionic product $Q$ using the actual concentrations when two solutions are mixed (after dilution), and compare with $K_{sp}$ :

$Q < K_{sp}$ : unsaturated, no precipitate.
$Q = K_{sp}$ : exactly saturated.
$Q > K_{sp}$ : supersaturated, a precipitate forms until $Q = K_{sp}$ .

Remember to account for the dilution that happens on mixing.

The common ion effect

Adding an ion already present in the equilibrium (a common ion) decreases the solubility of the salt. Because $K_{sp}$ is fixed, raising one ion's concentration forces the other ion's concentration down, so less solid dissolves. This is a direct application of Le Chatelier to the solubility equilibrium, and it underlies selective precipitation in qualitative analysis.

Predicting precipitation on mixing two solutions

Two solutions are mixed: $50\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\text{CaCl}_2$ and $50\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\text{Na}_2\text{SO}_4$ . Will a precipitate of calcium sulfate form? ( $K_{sp}(\text{CaSO}_4) = 9.1 \times 10^{-6}\ \text{mol}^2\ \text{dm}^{-6}$ .)

Step 1: Recalculate concentrations after mixing

Mixing equal volumes doubles the total volume from $50\ \text{cm}^3$ to $100\ \text{cm}^3$ , so each concentration is halved. This dilution step is essential before computing $Q$ .

[\text{Ca}^{2+}] = 0.0100\ \text{mol dm}^{-3}, \qquad [\text{SO}_4^{2-}] = 0.0100\ \text{mol dm}^{-3}

Step 2: Calculate the ionic product $Q$

The ionic product $Q$ has the same form as $K_{sp}$ but uses the actual concentrations after mixing, not the equilibrium concentrations. For $\text{CaSO}_4$ the dissolution gives one $\text{Ca}^{2+}$ and one $\text{SO}_4^{2-}$ , so the expression is simply the product of the two ion concentrations.

Q = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (0.0100)(0.0100) = 1.0 \times 10^{-4}\ \text{mol}^2\ \text{dm}^{-6}

Step 3: Compare $Q$ with $K_{sp}$

Since $Q = 1.0 \times 10^{-4}$ is greater than $K_{sp} = 9.1 \times 10^{-6}$ , the solution is supersaturated with respect to $\text{CaSO}_4$ . The system responds by precipitating solid until $Q$ falls back to $K_{sp}$ .

Final answer: Yes, a precipitate of $\text{CaSO}_4$ forms because $Q > K_{sp}$ .

Try it: Solubility product calculator - convert between $K_{sp}$ and solubility for any stoichiometry, and test whether mixing two solutions precipitates.

Examples in context

Example 1. Selective precipitation in qualitative analysis. When a mixture of chloride and iodide ions is treated with silver ions, silver iodide ( $K_{sp}$ very small) precipitates before silver chloride. By controlling the silver concentration, one ion can be removed selectively. This $K_{sp}$ reasoning underpins several separations met in the Analytical Techniques content area.

Example 2. Hard water and scale. Calcium carbonate has a small $K_{sp}$ , so it precipitates as scale when heated water concentrates the ions and shifts $Q$ above $K_{sp}$ . Adding carbonate ions (a common ion) to soften water deliberately precipitates calcium, an everyday application of the common ion effect that SEAB likes to set in context.

Try this

Q1. Write the $K_{sp}$ expression for lead(II) chloride, $\text{PbCl}_2$ . [1 mark]

Cue. $K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2$ .

Q2. The solubility of $\text{Mg(OH)}_2$ is $1.2 \times 10^{-4}$ mol dm $^{-3}$ . Calculate its $K_{sp}$ . [3 marks]

Cue. $[\text{Mg}^{2+}] = s$ , $[\text{OH}^-] = 2s$ ; $K_{sp} = s(2s)^2 = 4s^3 = 4(1.2\times10^{-4})^3 = 6.9 \times 10^{-12}$ mol $^3$ dm $^{-9}$ .

Q3. Explain, using $K_{sp}$ , why barium sulfate is safe to use as a medical "barium meal" despite barium ions being toxic. [2 marks]

Cue. $\text{BaSO}_4$ has an extremely small $K_{sp}$ , so almost no $\text{Ba}^{2+}$ dissolves; the ionic product stays below $K_{sp}$ and free barium ions in solution are negligible.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Specimen (9729)4 marksThe solubility product of silver chloride, AgCl, is 1.8 x 10^-10 mol squared per dm to the sixth at 298 K. Calculate the solubility of silver chloride in mol per dm cubed and in g per dm cubed. (Mr of AgCl = 143.4.)

Show worked answer →

Write the dissolution equilibrium and the Ksp expression.

$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ , so $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ .

Let solubility = s mol per dm cubed. Then [Ag+] = [Cl-] = s.

$K_{sp} = s^2 = 1.8 \times 10^{-10}$

$s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ mol per dm cubed.

In g per dm cubed: $1.34 \times 10^{-5} \times 143.4 = 1.92 \times 10^{-3}$ g per dm cubed.

Markers reward the Ksp expression, the 1:1 relationship s = [Ag+] = [Cl-], the molar solubility, and the conversion to g per dm cubed.

2023 (style)4 marksExplain, with reference to Ksp, why the solubility of silver chloride is much lower in 0.10 mol per dm cubed sodium chloride solution than in pure water.

Show worked answer →

Ksp is a constant at a fixed temperature: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ .

In pure water, [Ag+] = [Cl-] = s.

In 0.10 mol per dm cubed NaCl, there is already a high [Cl-] = 0.10 (a common ion).

To keep the product [Ag+][Cl-] equal to Ksp, [Ag+] must fall sharply: $[\text{Ag}^+] = K_{sp}/0.10 = 1.8 \times 10^{-9}$ mol per dm cubed.

So the amount of AgCl that dissolves (equal to [Ag+]) is far smaller than in pure water. This is the common ion effect.

Markers reward the constancy of Ksp, identifying the common chloride ion, the calculation of the reduced [Ag+], and naming the common ion effect.