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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
A
$1.6 \times 10^{-11}$ M
B
zero
C
$1.26 \times 10^{-5}$ M
D
$1.6 \times 10^{-9}$ M
Explanation
$S = K_{sp}/[Cl^-]$.
Detailed Solution
$NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq)$ 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
$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ 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