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

Answer: (D)

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

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
a 0 0
a-S S S+0.1
$k_{sp}=1.6\times {10}^{-10}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$=S(0.1+S)$

$k_{sp}$ is value seems to be very small

S value can be ignored, with respect to 0.1 M

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

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

Hence option D is correct.