Question

$\text{If}{K}_{sp}\left(AgCNS\right)=1\times {10}^{-12}\text{and}{K}_{sp}\left(AgBr\right)=5\times {10}^{-13}$ then the values of simultaneous solubility of $AgCNS$ and $AgBr$ in a solution of water will be:

• A. $8.16\times {10}^{-7},4.08\times {10}^{-7}$
• B. $4.08\times {10}^{-7},8.16\times {10}^{-7}$
• C. $8.16,4.08$
• D. $1\times {10}^{12},5\times {10}^{-13}$

Answer 1

The correct option is A $8.16\times {10}^{-7},4.08\times {10}^{-7}$

$AgCNS\rightleftharpoons \underset{x+y}{A{g}^{+}}+\underset{x}{CN{S}^{-}}$
$AgBr\rightleftharpoons \underset{x+y}{A{g}^{+}}+\underset{y}{B{r}^{+}}$
$\therefore \left[A{g}^{+}\right]=(x+y),\left[CN{S}^{-}\right]=x,\left[B{r}^{-}\right]=y$
$K_{sp}\left[AgCNS\right]=\left[A{g}^{+}\right]\left[CN{S}^{-}\right]=x(x+y)$
$\Rightarrow 1\times {10}^{-12}=x(x+y)......\left(1\right)$
$\text{and}{K}_{sp}\left[AgBr\right]=\left[A{g}^{+}\right]\left[B{r}^{-}\right]=y(x+y)$
$\Rightarrow 5\times {10}^{-13}=y(x+y).......\left(2\right)$
on solving equation (1) and (2) we get,
$x=8.16\times {10}^{-7}\text{mol/L}$
$y=4.08\times {10}^{-7}\text{mol/L}$