Solubility product

CIE A-Level Chemistry· Unit 18: Further chemical equilibria· 20 min read

1. Definition and Ksp Expressions★★☆☆☆⏱ 5 min

📘 Definition

Solubility Product

$K_{s p}$

For the dissociation equilibrium $A_{x} B_{y} (s) ⇌ x A^{y +} (a q) + y B^{x -} (a q)$ , $K_{s p}$ is the product of equilibrium ion concentrations, each raised to their stoichiometric power. Undissolved solid has an activity of 1, so it is excluded from the expression.

Example:

For AgCl: $K_{s p} = [A g^{+}] [C l^{-}]$

A higher $K_{s p}$ indicates a more soluble sparingly soluble salt. $K_{s p}$ is only affected by temperature, and remains constant at a given temperature regardless of other ions in solution.

📐 Worked Example

Write the correct $K_{s p}$ expression for calcium fluoride, $C a F_{2}$ .

1
1. Write the balanced dissociation equation for solid calcium fluoride:
2
$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$
3
1. Write the product of ion concentrations, raised to their stoichiometric powers, exclude the solid:
4
$K_{sp} = [Ca^{2+}][F^-]^2$

2. Interconverting Ksp and Molar Solubility★★★☆☆⏱ 6 min

Molar solubility ( $s$ ) is the maximum moles of salt that dissolve in 1 dm³ of solution. We can relate $s$ to $K_{s p}$ using the stoichiometry of the dissociation reaction.

Write the balanced dissociation equation
Express equilibrium ion concentrations in terms of $s$
Substitute into the $K_{s p}$ expression and solve for the unknown

📐 Worked Example

The $K_{s p}$ of AgCl is $1.8 \times 1 0^{- 10}$ at 25°C. Calculate the molar solubility of AgCl in pure water.

1
Let $s$ = molar solubility of AgCl. Write the dissociation:
2
$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$
3
1 mole of AgCl produces 1 mole of each ion, so:
4
$[Ag^+] = s, \quad [Cl^-] = s$
5
Substitute into Ksp:
6
$K_{sp} = s \times s = s^2$
7
Solve for $s$ :
8
$s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} \text{ mol dm}^{-3}$

📐 Worked Example

The molar solubility of $C a F_{2}$ is $2.1 \times 1 0^{- 4}$ mol dm⁻³. Calculate $K_{s p}$ of $C a F_{2}$ .

1
1 mole of $C a F_{2}$ produces 1 mole $C a^{2 +}$ and 2 moles $F^{-}$ :
2
$[Ca^{2+}] = s, \quad [F^-] = 2s$
3
Substitute into Ksp expression:
4
$K_{sp} = (s)(2s)^2 = 4s^3$
5
Plug in $s = 2.1 \times 1 0^{- 4}$ :
6
$K_{sp} = 4(2.1 \times 10^{-4})^3 = 3.7 \times 10^{-11}$

3. Predicting Precipitation★★★☆☆⏱ 5 min

To predict if a precipitate forms when two solutions are mixed, we calculate the ion product $Q_{s p}$ , which uses initial ion concentrations after mixing, then compare it to $K_{s p}$ .

📘 Definition

Ion Product

$Q_{s p}$

Product of initial ion concentrations after mixing, raised to their stoichiometric powers, used to test for precipitation.

$Q_{s p} > K_{s p}$ : Solution is supersaturated, precipitation occurs
$Q_{s p} = K_{s p}$ : Solution is saturated, no precipitation
$Q_{s p} < K_{s p}$ : Solution is unsaturated, no precipitation

📐 Worked Example

Equal volumes of 0.002 mol dm⁻³ $A g N O_{3}$ and 0.001 mol dm⁻³ NaCl are mixed. Will AgCl precipitate? $K_{s p} (A g C l) = 1.8 \times 1 0^{- 10}$ .

1
Mixing equal volumes halves all concentrations:
2
$[Ag^+] = \frac{0.002}{2} = 0.001 \text{ mol dm}^{-3}, \quad [Cl^-] = \frac{0.001}{2} = 0.0005 \text{ mol dm}^{-3}$
3
Calculate $Q_{s p}$ :
4
$Q_{sp} = (0.001)(0.0005) = 5 \times 10^{-7}$
5
Compare $Q_{s p}$ to $K_{s p}$ : $5 \times 1 0^{- 7} > 1.8 \times 1 0^{- 10}$ , so AgCl precipitation will occur.

4. Common Ion Effect on Solubility★★★★☆⏱ 4 min

Adding a soluble salt that shares a common ion with a sparingly soluble salt reduces the solubility of the sparingly soluble salt. This follows Le Chatelier's principle: adding product ion shifts equilibrium left to form more solid. $K_{s p}$ does not change, only solubility decreases.

📐 Worked Example

Calculate the solubility of AgCl in 0.10 mol dm⁻³ NaCl solution. $K_{s p} (A g C l) = 1.8 \times 1 0^{- 10}$ .

1
Let $s$ = solubility of AgCl. All $C l^{-}$ from NaCl is 0.10 mol dm⁻³. $s$ is very small, so $[C l^{-}] \approx 0.10$ :
2
$[Ag^+] = s, \quad [Cl^-] = 0.10 + s \approx 0.10$
3
Substitute into $K_{s p}$ :
4
$K_{sp} = s(0.10) = 1.8 \times 10^{-10}$
5
Solve for $s$ :
6
$s = 1.8 \times 10^{-9} \text{ mol dm}^{-3}$
7
This is far lower than the solubility in pure water ( $1.3 \times 1 0^{- 5}$ mol dm⁻³), matching the expected common ion effect.

5. Common Pitfalls

Wrong move:

Writing Ksp for CaF₂ as $[C a^{2 +}] [F^{-}]$ (forgetting to raise $[F^{-}]$ to the power of 2)

Why:

Stoichiometric coefficients from the balanced dissociation equation must be used as exponents

Correct move:

Always write the full balanced dissociation equation first before writing the Ksp expression

Wrong move:

Concluding precipitation occurs when $Q_{s p} < K_{s p}$

Why:

$Q_{s p} < K_{s p}$ means the solution can still dissolve more solid, so no precipitation occurs

Correct move:

Memorize: precipitation only occurs when $Q_{s p} > K_{s p}$

Wrong move:

Changing Ksp value when a common ion is added

Why:

Ksp is an equilibrium constant, it only changes with temperature

Correct move:

Ksp remains constant; only the solubility of the sparingly soluble salt changes

Wrong move:

Forgetting to dilute concentrations after mixing two solutions before calculating Qsp

Why:

Mixing increases total volume, so initial ion concentrations are lower than in the starting solutions

Correct move:

Always recalculate concentrations after mixing using $c_{1} V_{1} = c_{2} V_{2}$ before calculating Qsp

Wrong move:

Writing $[F^{-}] = s$ for CaF₂ instead of $2 s$

Why:

Each mole of dissolved CaF₂ produces 2 moles of fluoride ions

Correct move:

Relate ion concentration to solubility using the reaction stoichiometry, check the dissociation equation

6. Quick Reference Cheatsheet

Compound type	Dissociation	Ksp expression	Ksp-s relation
AB (1:1)	$A B ⇌ A^{+} + B^{-}$	$K_{s p} = [A^{+}] [B^{-}]$	$K_{s p} = s^{2}, s = K_{s p}$
AB₂ (1:2)	$A B_{2} ⇌ A^{2 +} + 2 B^{-}$	$K_{s p} = [A^{2 +}] [B^{-}]^{2}$	$K_{s p} = 4 s^{3}, s = 3 \frac{K _{s p}}{4}$
A₂B (2:1)	$A_{2} B ⇌ 2 A^{+} + B^{2 -}$	$K_{s p} = [A^{+}]^{2} [B^{2 -}]$	$K_{s p} = 4 s^{3}, s = 3 \frac{K _{s p}}{4}$
AB₃ (1:3)	$A B_{3} ⇌ A^{3 +} + 3 B^{-}$	$K_{s p} = [A^{3 +}] [B^{-}]^{3}$	$K_{s p} = 27 s^{4}, s = 4 \frac{K _{s p}}{27}$
Precipitation rule			$Q > K_{s p}$ : precipitate; $Q \leq K_{s p}$ : no precipitate

7. Frequently Asked

Do I need to include units for Ksp?

CIE accepts both dimensionless Ksp and units based on stoichiometry. Always check the question prompt; if units are requested, calculate them from ion concentration units.

When does the value of Ksp change?

Ksp, like all equilibrium constants, only changes with temperature. Adding other ions or changing concentration does not alter Ksp.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 22
Calculate Ksp from solubility data
2023 · 13
Predict precipitation from ion concentrations
2021 · 21
Common ion effect solubility calculation

Going deeper

syllabusCIE 9701 A-Level Chemistry SyllabusCheck unit 18 for latest assessment requirements

What's Next

Solubility product is a core application of equilibrium principles to ionic systems, and it appears frequently in both multiple choice and structured exam questions. Mastery of Ksp calculations underpins topics like selective precipitation of salts, qualitative analysis of metal ions, and pH calculations for sparingly soluble hydroxides. You will next explore the common ion effect in more depth, and extend these ionic equilibria concepts to acid-base buffers and titration curves. Ksp also connects to thermodynamics, as it can be used to calculate Gibbs free energy change for dissolution reactions.