Solubility with a common ion

Worked example

Ksp(AgCl) = 1.7 × 10⁻¹⁰. Calculate the solubility of silver chloride, in mol dm⁻³, in 0.017 mol dm⁻³ sodium chloride solution.

1. 1. Equilibrium from initial concentrations (ICE): solve for Solubility of AgCl

   $\text{Solve } K = \prod_i \left(\frac{[\mathrm{X}_i]_0 + \nu_i\,\xi}{c^\circ}\right)^{\nu_i} \text{ for } \xi$

   = 1.000 × 10⁻⁸ mol dm⁻³

Answer: s(AgCl) = 1.00 × 10⁻⁸ mol dm⁻³

Method

AgCl(s) ⇌ Ag⁺ + Cl⁻ with [Ag⁺] = s and [Cl⁻] = c + s. Then Ksp = s(c + s). Because s is tiny compared with c, s ≈ Ksp ÷ c: far less than the √Ksp in pure water.

Common mistakes

• That is the solubility in pure water. The chloride already in solution suppresses dissolving: s ≈ Ksp ÷ [Cl⁻].

Formulas used

• Equilibrium from initial concentrations (ICE)

  $K = \prod_i \left(\frac{[\mathrm{X}_i]_0 + \nu_i\,\xi}{c^\circ}\right)^{\nu_i}$

More equilibrium questions

• pH of a weak acid
• Buffer pH (Henderson–Hasselbalch)
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• pH of benzoic acid from a weighed sample
• H₂ + I₂ ⇌ 2HI: equilibrium amounts