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Inquiry Question 4: How does solubility relate to chemical equilibrium and the position of equilibrium?

Predict the solubility of an ionic substance by applying solubility equilibrium principles, and perform calculations involving the solubility product constant (Ksp) and the ionic product

A focused answer to the HSC Chemistry Module 5 dot point on the solubility product. Writing Ksp expressions, predicting precipitation using the ionic product Q vs Ksp, the common ion effect, and worked HSC calculations for solubility and precipitation.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to write Ksp expressions for sparingly soluble salts, calculate molar solubility, predict whether a precipitate will form by comparing the ionic product Q to Ksp, and understand the common ion effect. This dot point is examined as a 3-5 mark calculation question and as part of extended response questions on environmental chemistry.

The answer

The solubility product Ksp

For a sparingly soluble ionic compound, dissolution is an equilibrium between the solid and its ions in solution:

\text{M}_a\text{X}_{b(s)} \rightleftharpoons a\text{M}^{n+}_{(aq)} + b\text{X}^{m-}_{(aq)}

The solubility product constant Ksp is the equilibrium constant for this dissolution. The pure solid is excluded, so:

K_{sp} = [\text{M}^{n+}]^a [\text{X}^{m-}]^b

Examples:

Compound	Dissolution	Ksp expression
IMATH_9	IMATH_10	IMATH_11
IMATH_12	IMATH_13	IMATH_14
IMATH_15	IMATH_16	IMATH_17

Ksp values are small (typically $10^{-5}$ to $10^{-50}$ ), reflecting the fact that the compound is only slightly soluble.

The ionic product Q

For any solution (not necessarily at equilibrium), the same expression evaluated using current concentrations is the ionic product Q.

Compare Q to Ksp to predict precipitation:

If $Q < K_{sp}$ , the solution is unsaturated. More solid can dissolve. No precipitate forms.
If $Q = K_{sp}$ , the solution is saturated. The system is at equilibrium.
If $Q > K_{sp}$ , the solution is supersaturated. A precipitate forms until Q falls back to Ksp.

The common ion effect

Adding a common ion to a saturated solution decreases solubility. For example, adding NaCl to a saturated solution of AgCl increases $[Cl^-]$ , so $[Ag^+]$ must decrease to keep $Q = K_{sp}$ . Some AgCl precipitates.

This is a direct application of Le Chatelier's principle. The added ion shifts the dissolution equilibrium to the left (toward the solid).

Worked example

Calculate the molar solubility of lead(II) chloride, $PbCl_2$ , in pure water at 25°C. Ksp = $1.7 \times 10^{-5}$ .

Step 1: Dissolution equation.

PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}

Step 2: Ksp expression.

K_{sp} = [Pb^{2+}][Cl^-]^2

Step 3: Set up with molar solubility $s$ .

Each mole of $PbCl_2$ that dissolves gives 1 mole of $Pb^{2+}$ and 2 moles of $Cl^-$ .

$[Pb^{2+}] = s$ , $[Cl^-] = 2s$ .

Step 4: Substitute and solve.

K_{sp} = (s)(2s)^2 = 4s^3 = 1.7 \times 10^{-5}

s^3 = 4.25 \times 10^{-6}

s = (4.25 \times 10^{-6})^{1/3} = 1.62 \times 10^{-2} \text{ mol/L}

The molar solubility is $1.62 \times 10^{-2}$ mol/L (about 0.016 mol/L).

Compare to AgCl. Note that $PbCl_2$ has a larger Ksp than AgCl but the relationship to solubility is not direct (different stoichiometries). For $AB$ salts, $s = \sqrt{K_{sp}}$ . For $AB_2$ salts, $s = \sqrt[3]{K_{sp}/4}$ .

Worked example 2: the common ion effect

Calculate the molar solubility of $AgCl$ ( $K_{sp} = 1.8 \times 10^{-10}$ ) in 0.10 mol/L $NaCl$ , and compare with its solubility in pure water.

Step 1: Account for the common ion. $NaCl$ fully dissociates, so $[Cl^-]$ from the salt is 0.10 mol/L before any AgCl dissolves. Let $s'$ be the molar solubility of AgCl in this solution.

	IMATH_46	IMATH_47
From IMATH_48	0	0.10
From dissolved IMATH_49	IMATH_50	IMATH_51
Equilibrium	IMATH_52	IMATH_53

Step 2: Substitute into Ksp.

K_{sp} = [Ag^+][Cl^-] = s'(0.10 + s') = 1.8 \times 10^{-10}

Step 3: Approximate. Because Ksp is tiny and $[Cl^-]$ from NaCl is much larger than $s'$ , $0.10 + s' \approx 0.10$ .

s' \approx \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ mol/L}

Step 4: Compare. In pure water $s = 1.34 \times 10^{-5}$ mol/L; in 0.10 mol/L NaCl, $s' = 1.8 \times 10^{-9}$ mol/L. The solubility falls by a factor of ~7400. Adding the common ion drives the equilibrium to the left (toward solid), exactly as Le Chatelier's principle predicts.

Check the approximation. $s' / 0.10 = 1.8 \times 10^{-8}$ , far below 5%. Valid.

Common traps

Forgetting the stoichiometric exponent in Ksp. For $PbCl_2$ , $[Cl^-]$ is squared. Lose this and the calculation collapses.

Forgetting the stoichiometric multiplier in molar solubility. $[Cl^-] = 2s$ , not $s$ .

Comparing Ksp values directly across different stoichiometries. The salt with the larger Ksp is not necessarily more soluble. Convert to molar solubility before comparing.

Forgetting dilution when mixing solutions. When two solutions are combined, the concentrations of each ion are diluted by the dilution factor (volume mixed / total volume). Most "will a precipitate form" questions test this.

Calling Q a constant. Q changes as the system approaches equilibrium. Only Ksp is constant (at a given temperature).

In one sentence

The solubility product Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic solid, expressed as the product of ion concentrations raised to their stoichiometric coefficients, and a precipitate forms whenever the ionic product Q exceeds Ksp.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2020 HSC4 marksThe Ksp of silver chloride (AgCl) at 25°C is 1.8 × 10⁻¹⁰. Calculate the molar solubility of silver chloride in pure water at 25°C.

Show worked answer →

A 4 mark answer needs the dissolution equation, the Ksp expression, the algebraic solution, and the final value with units.

Step 1: Dissolution equation.

AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}

Step 2: Ksp expression. Pure solid AgCl is excluded.

K_{sp} = [Ag^+][Cl^-]

Step 3: Set up using molar solubility $s$ .

Each mole of AgCl that dissolves gives 1 mole of $Ag^+$ and 1 mole of $Cl^-$ , so $[Ag^+] = [Cl^-] = s$ .

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

Step 4: Solve.

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

Markers reward (1) the balanced dissolution equation with state symbols, (2) the correct Ksp expression (excluding the solid), (3) clear use of molar solubility $s$ with stoichiometry, (4) the answer in mol/L to appropriate significant figures.

2019 HSC3 marksEqual volumes of 0.0010 mol/L silver nitrate and 0.0020 mol/L sodium chloride are mixed. The Ksp of AgCl is 1.8 × 10⁻¹⁰. Determine, with calculation, whether a precipitate will form.

Show worked answer →

Mixing equal volumes halves each concentration.

After mixing:
$[Ag^+] = 0.0010 / 2 = 5.0 \times 10^{-4}$ mol/L.
$[Cl^-] = 0.0020 / 2 = 1.0 \times 10^{-3}$ mol/L.

Calculate the ionic product Q.

Q = [Ag^+][Cl^-] = (5.0 \times 10^{-4})(1.0 \times 10^{-3}) = 5.0 \times 10^{-7}

Compare to Ksp.

$Q = 5.0 \times 10^{-7}$ is much greater than $K_{sp} = 1.8 \times 10^{-10}$ .

Since $Q > K_{sp}$ , the solution is supersaturated and a precipitate of AgCl will form.

Markers reward (1) the dilution step (halving concentrations after mixing), (2) calculating Q correctly, (3) the explicit Q vs Ksp comparison and the conclusion.