Kp, Ksp and Partition

For gas equilibria, partial pressure $p_A=x_A\times P_\text{total}$ , where mole fraction $x_A=n_A/n_\text{total}$ . $K_p$ uses partial pressures: for $\mathrm{N_2O_4\rightleftharpoons2NO_2}$ , $K_p=\frac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}}$ . Only temperature changes $K_p$ . The solubility product $K_{sp}$ for a sparingly soluble salt, e.g. $K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}]$ ; a precipitate forms if the ionic product exceeds $K_{sp}$ . Adding a common ion lowers solubility. The partition coefficient $K_{pc}=\frac{[X]_\text{solvent 1}}{[X]_\text{solvent 2}}$ describes how a solute distributes between two immiscible solvents.

Cambridge 9701 Paper 4Edexcel YCH11 WCH14AQA 7405 Paper 1≈ 50 minRevision 1 · updated 2026-10-30

Section 1 of 14

Learning objectives

Calculate mole fractions, partial pressures and $K_p$ .
Write and use $K_{sp}$ expressions.
Explain the common ion effect.
Use partition coefficients.

Pause and think

Why does adding NaCl to a saturated AgCl solution produce more white precipitate?

Section 2 of 14

Core idea

Start hereEquilibrium in gases and in precipitates

The Haber process is controlled with Kp; kidney stones form when Ksp is exceeded. Same equilibrium idea, different units.

The big idea

For gases, use partial pressures. For sparingly soluble salts, the product of ion concentrations at saturation is constant.

1. Kp

Worked exampleKp for N₂O₄ ⇌ 2NO₂

0.60 mol N₂O₄ and 0.40 mol NO₂ at 100 kPa. Find Kp.

Goal: Kp with units.
Strategy: Mole fractions → partial pressures → Kp.

$x_{\mathrm{N_2O_4}}=0.60,\ x_{\mathrm{NO_2}}=0.40$
Total 1.00 mol.
$p_{\mathrm{N_2O_4}}=60,\ p_{\mathrm{NO_2}}=40\ \text{kPa}$
$p=xP$ .
$K_p=\frac{40^2}{60}=26.7\ \text{kPa}$
Units: kPa²/kPa.

Answer26.7 kPa.

✓ CheckUnits cancel to kPa ✓.

2. Ksp and the common ion effect

Solubility from Ksp (AgCl)

$\mathrm{AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)}$
Saturated solution.
$K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}]=s^2$
Each dissolved AgCl gives one of each ion.
$s=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\ \text{mol dm}^{-3}$
Solve.

Key takeawayCommon ion In 0.100 mol dm⁻³ NaCl,

s=K_{sp}/0.100=1.8\times10^{-9}

: about 7000 times less soluble.

Summary

p = xP; Kp from partial pressures.
Only temperature changes K.
Ksp; precipitate when product > Ksp.
Partition coefficient for two solvents.

Mole fraction: Moles of a gas divided by total moles of gas.
Partial pressure: Pressure one gas would exert alone in the same volume.
Solubility product: Product of ion concentrations in a saturated solution, each raised to its stoichiometric power.

HistoryNernst again. Walther Nernst described the partition law in 1891, the same Nernst whose equation describes electrode potentials.

Fun factKidney stones. Calcium oxalate kidney stones form when [Ca²⁺][C₂O₄²⁻] in urine exceeds Ksp; drinking more water dilutes the ions below the limit.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in