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Common ion effect on solubility
Concepts tested here
- common-ion-agcl-solubility
All Questions
2016 Phase II 1 question
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The solubility of AgCl(s) with solubility product $1.6 \times 10^{-10}$ in 0.1 M NaCl solution would be
$S = K_{sp}/[Cl^-]$.
$NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq)$ gives $[Cl^-] = 0.1$ M
$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ with S and S + 0.1
$K_{sp} = 1.6\times10^{-10} = [Ag^+][Cl^-] = S(0.1 + S)$
$\because K_{sp}$ is small, S is neglected with respect to 0.1 M: $1.6\times10^{-10} = S\times0.1$
$S = 1.6\times10^{-9}$ M