TASC TCE 2023-style practice: Lead(II) iodide, PbI_2, is a sparingly soluble salt with K_sp = 7.1 × 10^-9 at 25\ ^C. (a) Write the dissolution equation and the K_sp expression. (b) Calculate the molar solubility of PbI_2 in pure water at 25\ ^C.

(a) PbI_2(s) Pb^2+(aq) + 2I^-(aq), so K_sp = [Pb^2+][I^-]^2. (1 mark) (b) If s moles dissolve per litre, [Pb^2+] = s and [I^-] = 2s. So K_sp = (s)(2s)^2 = 4s^3. (1 mark) $s^3 = K_sp4 = 7.1 × 10^-94 = 1.78 × 10^-9, s = [3]1.78 × 10^-9 = 1.2 × 10^-3\ mol L^-1.$ (2 marks)

TASC TCE 2021-style practice: Equal volumes of 0.0020\ mol L^-1 silver nitrate and 0.0020\ mol L^-1 sodium chloride are mixed. For AgCl, K_sp = 1.8 × 10^-10. Determine, with a calculation, whether a precipitate of silver chloride forms.

Mixing equal volumes halves each concentration: [Ag^+] = [Cl^-] = 0.0010\ mol L^-1. (1 mark) Calculate the ionic product: Q = [Ag^+][Cl^-] = (1.0 × 10^-3)(1.0 × 10^-3) = 1.0 × 10^-6. (1 mark) Q\ (1.0 × 10^-6) > K_sp\ (1.8 × 10^-10), so the solution is supersaturated and a precipitate of AgCl forms. (1 mark)

§-Syllabus dot point

TASChemistrySyllabus dot point

How do we describe the equilibrium of a sparingly soluble salt?

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

The solubility product Ksp, relating Ksp to molar solubility, the common ion effect, and predicting whether a precipitate forms using the ionic product Q, with fully worked TASC-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

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.

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.

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 practice questions

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

TCE 20234 marksLead(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)

TCE 20213 marksEqual 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)