Solubility equilibria and Ksp: TCE Chemistry (Tasmania)

Syllabus dot point

“Write solubility product expressions, calculate Ksp and solubility, and predict precipitation.”

9 min readUpdated 2 Jun 2026

Quick answer

A sparingly soluble salt sits in equilibrium with its dissolved ions. The solubility product $K_{sp}$ is the product of the ion concentrations, each raised to its coefficient (the solid is omitted). Relate $K_{sp}$ to molar solubility $s$ by writing each ion concentration in terms of $s$ . To predict precipitation, calculate the ionic product $Q$ from the mixed concentrations and compare it with $K_{sp}$ : precipitate forms when $Q > K_{sp}$ .

What this dot point is asking

TASC expects you to write $K_{sp}$ expressions, link $K_{sp}$ to molar solubility (including salts with coefficients greater than one), apply the common ion effect, and predict precipitation.

The dissolution equilibrium

When an ionic solid is only slightly soluble, a saturated solution reaches equilibrium with the undissolved solid:

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

The pure solid is omitted from the expression, so the equilibrium constant is the product of the dissolved ion concentrations.

Solubility product

For $\text{A}_x\text{B}_y(s) \rightleftharpoons x\text{A}^{m+}(aq) + y\text{B}^{n-}(aq)$ :

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

A larger

K_{sp}

means a more soluble salt. The value depends only on temperature.

Relating Ksp to solubility

Molar solubility $s$ is the moles of salt that dissolve per litre to give a saturated solution. From the dissolution equation, write each ion concentration in terms of $s$ , then substitute into the $K_{sp}$ expression. The coefficients matter: for a $1:1$ salt $K_{sp} = s^2$ , but for a $1:2$ salt $K_{sp} = 4s^3$ .

Solubility of a 1:1 salt from Ksp

Calculate the molar solubility of silver chloride in pure water at $25\ ^\circ\text{C}$ , given $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ .

Step 1: Write the dissolution equilibrium

When AgCl dissolves it releases one silver ion and one chloride ion for every formula unit that enters solution. Writing $s$ for the number of moles that dissolve per litre, the equilibrium and the resulting ion concentrations are:

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

[\text{Ag}^+] = s \quad \text{and} \quad [\text{Cl}^-] = s.

Step 2: Write the $K_{sp}$ expression in terms of $s$

The solid AgCl is omitted from the equilibrium expression because its concentration does not change. Substituting the ion concentrations in terms of $s$ gives a simple relationship:

K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \times s = s^2.

Step 3: Solve for the molar solubility

Set the expression equal to the known $K_{sp}$ value and take the square root. The result is the moles of AgCl that dissolve per litre of solution:

s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}\ \text{mol L}^{-1}.

This tiny value confirms that silver chloride is sparingly soluble: only about $1.3 \times 10^{-5}\ \text{mol}$ dissolves in each litre of pure water.

Final answer: Molar solubility $s = 1.3 \times 10^{-5}\ \text{mol L}^{-1}$

Predicting precipitation

When two solutions are mixed, calculate the ionic product $Q$ using the mixed concentrations (account for dilution on mixing) and compare with $K_{sp}$ :

$Q > K_{sp}$ : the solution is supersaturated, so a precipitate forms.
$Q = K_{sp}$ : the solution is exactly saturated; no further change.
$Q < K_{sp}$ : the solution is unsaturated, so no precipitate forms.

The common ion effect

Adding a soluble salt that shares an ion with the sparingly soluble salt pushes the dissolution equilibrium left (Le Chatelier), reducing solubility. For example, $\text{AgCl}$ is less soluble in sodium chloride solution than in pure water because the added $\text{Cl}^-$ suppresses dissolution.

Selective precipitation

When a solution contains two ions that both form sparingly soluble salts with the same added reagent, the one with the smaller $K_{sp}$ (relative to its concentration) precipitates first. By adding the reagent gradually, the two ions can be separated, a technique called selective or fractional precipitation, used in qualitative analysis to identify ions and in water treatment to remove specific contaminants. The ion that requires the lower added concentration to reach $Q = K_{sp}$ comes out of solution first.

Factors affecting solubility

Solubility depends on temperature: most ionic solids dissolve more readily as temperature rises, so $K_{sp}$ values are quoted at a stated temperature (usually $25\ ^\circ\text{C}$ ). The common ion effect lowers solubility when a shared ion is already present. Solubility can also change with pH when the anion is basic: a salt such as calcium carbonate dissolves more in acidic solution because $\text{H}^+$ removes carbonate ions (as $\text{CO}_2$ and water), pulling the dissolution equilibrium to the right. This is the chemistry behind acid attack on limestone and the formation of caves.

Forgetting dilution when mixing solutions

When two solutions are combined, each is diluted by the other, so the concentrations fall before you calculate $Q$ . Using the original undiluted concentrations overestimates $Q$ and wrongly predicts a precipitate. Always recalculate concentrations for the new total volume first. A second slip is omitting the power: for $\text{PbI}_2$ the $[\text{I}^-]$ term must be squared, and $[\text{I}^-] = 2s$ , so $K_{sp} = 4s^3$ , not $s^2$ .

In the exam, write the dissolution equation, build $K_{sp}$ with the correct powers, express ion concentrations in terms of $s$ (watch the coefficients), and compare $Q$ with $K_{sp}$ after accounting for dilution on mixing.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Lead(II) iodide,

\text{PbI}_2

, is a sparingly soluble salt with

K_{sp} = 7.1 \times 10^{-9}

25\ ^\circ\text{C}

. (a) Write the dissolution equation and the

K_{sp}

expression. (b) Calculate the molar solubility of

\text{PbI}_2

in pure water at

25\ ^\circ\text{C}

Show worked answer →

(a) $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)$ , so $K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2$ . (1 mark)

(b) If $s$ moles dissolve per litre, $[\text{Pb}^{2+}] = s$ and $[\text{I}^-] = 2s$ . So $K_{sp} = (s)(2s)^2 = 4s^3$ . (1 mark)

s^3 = \frac{K_{sp}}{4} = \frac{7.1 \times 10^{-9}}{4} = 1.78 \times 10^{-9}, \qquad s = \sqrt[3]{1.78 \times 10^{-9}} = 1.2 \times 10^{-3}\ \text{mol L}^{-1}.

(2 marks)

Practice3 marks

Equal volumes of

0.0020\ \text{mol L}^{-1}

silver nitrate and

0.0020\ \text{mol L}^{-1}

sodium chloride are mixed. For

\text{AgCl}

K_{sp} = 1.8 \times 10^{-10}

. Determine, with a calculation, whether a precipitate of silver chloride forms.

Show worked answer →

Mixing equal volumes halves each concentration: $[\text{Ag}^+] = [\text{Cl}^-] = 0.0010\ \text{mol L}^{-1}$ . (1 mark)

Calculate the ionic product: $Q = [\text{Ag}^+][\text{Cl}^-] = (1.0 \times 10^{-3})(1.0 \times 10^{-3}) = 1.0 \times 10^{-6}$ . (1 mark)

$Q\ (1.0 \times 10^{-6}) > K_{sp}\ (1.8 \times 10^{-10})$ , so the solution is supersaturated and a precipitate of $\text{AgCl}$ forms. (1 mark)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC