Chemistry / Unit 10 Chem 1916 worked example, 60 to 90 seconds
Molar Solubility with a Common Ion Present
The idea
Finding molar solubility with a common ion present means writing the ICE table with that ion's existing concentration already placed in the Initial row, substituting the equilibrium row into Ksp, and solving for x, usually applying the small-x approximation since the common ion suppresses solubility so heavily here that x barely moves it.
In practice
For AgCl, Ksp = 1.8 × 10⁻¹⁰, dissolving into 0.10 M NaCl: [Ag+] = x, [Cl−] ≈ 0.10 + x ≈ 0.10, so x(0.10) = 1.8 × 10⁻¹⁰, giving x ≈ 1.8 × 10⁻⁹ M, far below its solubility in plain pure water.
Where people slip
Omitting the pre-existing common-ion concentration from the Initial row treats the problem as if it were plain water, badly overstating the true solubility of the solid.
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