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Chemistry of Aqueous Solutions

Topic 5 of 6

Solubility and precipitation

Use Ksp and the ionic product with the correct dissolved-ion concentrations.

A-Level 9476 (2026-2027)

A saturated solution has a fixed product of free-ion concentrations

Ksp describes an equilibrium with an undissolved solid at a specified temperature.

For AgCl(s) ⇌ Ag+(aq) + Cl-(aq), Ksp = [Ag+][Cl-]. The pure solid is omitted. For CaF2(s) ⇌ Ca2+(aq) + 2F-(aq), Ksp = [Ca2+][F-]2. The exponent follows the dissolution coefficient.

Worked example

Convert Ksp into molar solubility

Find the molar solubility of AgCl with Ksp = 1.80 × 10-10, and CaF2 with Ksp = 3.20 × 10-11, in pure water under the simple dissolution model.

  1. For AgCl, let solubility be s. Both ion concentrations are s, so s2 = 1.80 × 10-10 and s = 1.34 × 10-5 mol dm-3.
  2. For CaF2, [Ca2+] = s and [F-] = 2s, so Ksp = s(2s)2 = 4s3.
  3. s = cube root(3.20 × 10-11 / 4) = 2.00 × 10-4 mol dm-3.
Answer

AgCl: 1.34 × 10-5 mol dm-3; CaF2: 2.00 × 10-4 mol dm-3. Comparing raw Ksp values alone is not a valid solubility ranking for different stoichiometries.

The ionic product Q has the same form as Ksp but uses the current free-ion concentrations. Q < Ksp means unsaturated; Q = Ksp means equilibrium saturation; Q > Ksp makes precipitation thermodynamically favoured. When mixing solutions, first calculate concentrations in the combined volume.

Worked example

Predict precipitation after mixing

Mix 10.0 cm3 of 0.00200 mol dm-3 AgNO3 with 10.0 cm3 of 0.00100 mol dm-3 NaCl. Does AgCl precipitate if Ksp = 1.80 × 10-10?

  1. Before precipitation, dilution gives [Ag+] = 0.00100 and [Cl-] = 0.000500 mol dm-3.
  2. Q = 0.00100 × 0.000500 = 5.00 × 10-7.
  3. This exceeds Ksp by a large factor; ions are removed into the solid until the equilibrium condition is approached.
Answer

AgCl precipitates. These diluted concentrations test the initial tendency; they are not the final dissolved concentrations after precipitation.