Question

The solubility product of AgI at 25${}^o$C is $1.0\times {10}^{-16}mo{l}^2{L}^{-2}$. The solubility of AgI in ${10}^{-4}$ N solution of KI at ${25}^{o}C$ is approximately (in mol L${}^{-1}$) :

• A. $1.0\times {10}^{-16}$
• B. $1.0\times {10}^{-12}$
• C. $1.0\times {10}^{-10}$
• D. $1.0\times {10}^{-8}$

Answer 1

The correct option is B $1.0\times {10}^{-12}$

$AgI\rightleftharpoons {Ag}^{+}+I^{-}$

$KI\rightleftharpoons K^{+}+I^{-}$ $AgI\rightleftharpoons {Ag}^{+}+I^{-}$

${10}^{-4}$ ${10}^{-4}$ $x$ $x+{10}^{-4}$

$\nearrow$ negligible

Given, $K_{sp}=x(x+{10}^{-4})=x^2+x\left({10}^{-4}\right)={10}^{-4}x$

$K_{sp}=\left({10}^{-4}\right)x$

$x=\frac{K_{sp}}{{10}^{-4}}=\frac{{10}^{-16}}{{10}^{-4}}=1.0\times {10}^{-12}$

$\Rightarrow$ Solubility of $AgI$ is $1.0\times {10}^{-12}mol{L}^{-1}$.

Option B is correct.