Question

$K_{sp}ofAgCl$ is $=2.8\times {10}^{-10}$ at ${25}^{\circ }$ C. Calculate solubility of AgCl in 0.1 (M) NaCl

• A. $2.9\times {10}^{-5}moleli{t}^{-1}$
• B. $2.6\times {10}^{-8}moleli{t}^{-1}$
• C. $2.8\times {10}^{-6}moleli{t}^{-1}$
• D. $2.8\times {10}^{-9}moleli{t}^{-1}$

Answer 1

The correct option is D $2.8\times {10}^{-9}moleli{t}^{-1}$

$AgCl\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$
s s
$KCl\left(s\right)\to K^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$
0.1 0.1
$\therefore K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$=S(0.1+S)$
$\Rightarrow S=2.8\times {10}^{-9}moleli{t}^{-1}$