# Solubility product

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u18-solubility-product/

This module covers the solubility product constant ($K_{sp}$) for sparingly soluble ionic equilibria. You will learn to interconvert Ksp and solubility, predict precipitation, and apply the common ion effect, all core exam topics.

**Prerequisites:** [Chemical equilibrium and equilibrium constants](https://www.owlsprep.com/study/cie-9701-u11-equilibrium/); [Balancing ionic equations](https://www.owlsprep.com/study/cie-9701-u02-ionic-equations/)

## Learning objectives

- Define solubility product ($K_{sp}$) and write correct expressions for sparingly soluble ionic compounds
- Interconvert $K_{sp}$ and molar solubility for ionic salts
- Predict precipitation reactions by comparing $Q_{sp}$ to $K_{sp}$
- Calculate solubility of salts in solutions containing common ions

## Definition and Ksp Expressions

**Solubility Product** — For the dissociation equilibrium $A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$, $K_{sp}$ 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.

*Notation:* $K_{sp}$

*Example:* For AgCl: $K_{sp} = [Ag^+][Cl^-]$

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

**Worked example:** Write the correct $K_{sp}$ expression for calcium fluoride, $CaF_2$.

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

## Interconverting Ksp and Molar Solubility

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

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

**Worked example:** The $K_{sp}$ of AgCl is $1.8 \times 10^{-10}$ 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 $CaF_2$ is $2.1 \times 10^{-4}$ mol dm⁻³. Calculate $K_{sp}$ of $CaF_2$.

1. 1 mole of $CaF_2$ produces 1 mole $Ca^{2+}$ 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 $s = 2.1 \times 10^{-4}$:
6. $$K_{sp} = 4(2.1 \times 10^{-4})^3 = 3.7 \times 10^{-11}$$

## Predicting Precipitation

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

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

*Notation:* $Q_{sp}$

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

**Worked example:** Equal volumes of 0.002 mol dm⁻³ $AgNO_3$ and 0.001 mol dm⁻³ NaCl are mixed. Will AgCl precipitate? $K_{sp}(AgCl) = 1.8 \times 10^{-10}$.

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 $Q_{sp}$:
4. $$Q_{sp} = (0.001)(0.0005) = 5 \times 10^{-7}$$
5. Compare $Q_{sp}$ to $K_{sp}$: $5 \times 10^{-7} > 1.8 \times 10^{-10}$, so AgCl precipitation will occur.

## Common Ion Effect on Solubility

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. $K_{sp}$ does not change, only solubility decreases.

**Worked example:** Calculate the solubility of AgCl in 0.10 mol dm⁻³ NaCl solution. $K_{sp}(AgCl) = 1.8 \times 10^{-10}$.

1. Let $s$ = solubility of AgCl. All $Cl^-$ from NaCl is 0.10 mol dm⁻³. $s$ is very small, so $[Cl^-] \approx 0.10$:
2. $$[Ag^+] = s, \quad [Cl^-] = 0.10 + s \approx 0.10$$
3. Substitute into $K_{sp}$:
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 ($1.3 \times 10^{-5}$ mol dm⁻³), matching the expected common ion effect.

## Common pitfalls

- **Wrong:** Writing Ksp for CaF₂ as $[Ca^{2+}][F^-]$ (forgetting to raise $[F^-]$ to the power of 2)
  - Why it fails: Stoichiometric coefficients from the balanced dissociation equation must be used as exponents
  - Correct: Always write the full balanced dissociation equation first before writing the Ksp expression
- **Wrong:** Concluding precipitation occurs when $Q_{sp} < K_{sp}$
  - Why it fails: $Q_{sp} < K_{sp}$ means the solution can still dissolve more solid, so no precipitation occurs
  - Correct: Memorize: precipitation only occurs when $Q_{sp} > K_{sp}$
- **Wrong:** Changing Ksp value when a common ion is added
  - Why it fails: Ksp is an equilibrium constant, it only changes with temperature
  - Correct: Ksp remains constant; only the solubility of the sparingly soluble salt changes
- **Wrong:** Forgetting to dilute concentrations after mixing two solutions before calculating Qsp
  - Why it fails: Mixing increases total volume, so initial ion concentrations are lower than in the starting solutions
  - Correct: Always recalculate concentrations after mixing using $c_1V_1 = c_2V_2$ before calculating Qsp
- **Wrong:** Writing $[F^-] = s$ for CaF₂ instead of $2s$
  - Why it fails: Each mole of dissolved CaF₂ produces 2 moles of fluoride ions
  - Correct: Relate ion concentration to solubility using the reaction stoichiometry, check the dissociation equation

## Cheatsheet

| Compound type | Dissociation | Ksp expression | Ksp-s relation |
| --- | --- | --- | --- |
| AB (1:1) | $AB \rightleftharpoons A^+ + B^-$ | $K_{sp} = [A^+][B^-]$ | $K_{sp} = s^2, s = \sqrt{K_{sp}}$ |
| AB₂ (1:2) | $AB_2 \rightleftharpoons A^{2+} + 2B^-$ | $K_{sp} = [A^{2+}][B^-]^2$ | $K_{sp} = 4s^3, s = \sqrt[3]{\frac{K_{sp}}{4}}$ |
| A₂B (2:1) | $A_2B \rightleftharpoons 2A^+ + B^{2-}$ | $K_{sp} = [A^+]^2[B^{2-}]$ | $K_{sp} = 4s^3, s = \sqrt[3]{\frac{K_{sp}}{4}}$ |
| AB₃ (1:3) | $AB_3 \rightleftharpoons A^{3+} + 3B^-$ | $K_{sp} = [A^{3+}][B^-]^3$ | $K_{sp} = 27s^4, s = \sqrt[4]{\frac{K_{sp}}{27}}$ |
| Precipitation rule | - | - | $Q > K_{sp}$: precipitate; $Q \leq K_{sp}$: no precipitate |

## 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.

- [pH titration curves and indicators](https://www.owlsprep.com/study/cie-9701-u18-ph-titration-curves-and-indicators/)
- [Further electrochemistry](https://www.owlsprep.com/study/cie-9701-u19-overview/)
- [Standard electrode potentials](https://www.owlsprep.com/study/cie-9701-u19-standard-electrode-potentials/)

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