Ksp from a solubility in g dm⁻³

Worked example

The solubility of a salt MX₂ (M = 376.0 g mol⁻¹) in water is 0.509 g dm⁻³. Calculate its solubility product, Ksp.

1. 1. Mass concentration: solve for Molar solubility

   $c = \dfrac{\rho}{M}$

   = 0.001354 mol dm⁻³

2. 2. Ksp and solubility (MₓAᵧ): solve for Ksp = 4s³

   $K_\mathrm{sp} = x^x y^y s^{x+y}$

   = 9.923 × 10⁻⁹

Answer: Ksp = 9.92 × 10⁻⁹

Method

Convert to a molar solubility: s = (g dm⁻³) ÷ M. Each formula unit gives one M²⁺ and two X⁻, so Ksp = [M²⁺][X⁻]² = s(2s)² = 4s³ (with concentrations in mol dm⁻³).

Common mistakes

• MX₂ ⇌ M²⁺ + 2X⁻: [X⁻] = 2s, so Ksp = s × (2s)² = 4s³, not s².

Formulas used

• Mass concentration

  $\rho = cM$

• Ksp and solubility (MₓAᵧ)

  $K_\mathrm{sp} = (xs)^x(ys)^y$

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