Question

Solubility of $AgCl$ at $20{}^{o}C$ is $1.435\times {10}^{-3}g/L.$ The solubility product of $AgCl$ is

• A. $1\times {10}^{-5}$
• B. $1\times {10}^{-10}$
• C. $1.435\times {10}^{-5}$
• D. $108\times {10}^{-3}$

Answer 1

Answer: (B)

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$K_{sp}=S^2$
$(S=$ solubility in $mol/L)$
$S=\frac{1.435\times 10^{-3}g/L}{143.5}=1\times 10^{-10}mol/L$