Question

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. $1.6\times {10}^{-9}M$

Answer 1

The correct option is D $1.6\times {10}^{-9}M$

$$\begin{array}{cccc}NaCl\left(aq\right)\to & N{a}^{+}\left(aq\right)& +& C{l}^{-}\left(aq\right)\\ 0.1M& 0& & 0\\ 0& 0.1M& & 0.1+S\end{array}$$

$$\begin{array}{cccc}AgCl\left(s\right)\rightleftharpoons & A{g}^{+}\left(aq\right)& +& C{l}^{-}\left(aq\right)\\ a& 0& & 0\\ a-S& S& & S+0.1\end{array}$$

$K_{sp}=1.6\times {10}^{-10}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=S(0.1+S)$

$\because K_{sp}$ is small, S is neglected with respect to 0.1 M

$1.6\times {10}^{-10}=S\times 0.1$

$S=1.6\times {10}^{-9}M$