Question

Calculate simultaneous solubility of silver thiocyanate $\left(AgSCN\right)$and silver bromide $\left(AgBr\right)$ in water respectively.
Given that $K_{sp}$ of silver thiocyanate = ${10}^{-12}$ and $K_{sp}$ of silver bromide $=5\times {10}^{-13}$ .

• A. $4.16\times {10}^{-7}M$ and $2.08\times {10}^{-7}M$
• B. $2.08\times {10}^{-12}M$ and $4.16\times {10}^{-12}M$
• C. $5.08\times {10}^{-9}M$ and $9.16\times {10}^{-9}M$
• D. $8.16\times {10}^{-7}M$ and $4.08\times {10}^{-7}M$

Answer 1

The correct option is D $8.16\times {10}^{-7}M$ and $4.08\times {10}^{-7}M$

Given, $K_{sp}\left(AgSCN\right)$ = ${10}^{-12}$ and $K_{sp}\left(AgBr\right)$ $=5\times {10}^{-13}$

Let the simultaneous solubility of $AgSCN$ be x and that of $AgBr$ is y.

Then:
$AgSCN\left(s\right)\rightleftharpoons \underset{x+y}{A{g}^{+}}\left(aq\right)+\underset{x}{SC{N}^{-}}\left(aq\right)$ $AgBr\left(s\right)\rightleftharpoons \underset{x+y}{A{g}^{+}}\left(aq\right)+\underset{y}{B{r}^{-}}\left(aq\right)$

$K_{sp}\left(AgSCN\right)={10}^{-12}=x(x+y)....\left(i\right)$
$K_{sp}\left(AgBr\right)=5\times {10}^{-13}=y(x+y)....\left(ii\right)$

Dividing $eq\left(i\right)$ by $eq\left(ii\right)$,
$\frac{x}{y}=2$
$x=2y$
Putting in $eq\left(ii\right)$
$3y^2=5\times {10}^{-13}$
$\therefore y=4.08\times {10}^{-7}M$ and $x=2y=8.16\times {10}^{-7}M.$

Hence, simultaneous solubility of silver thiocyanate and silver bromide in water is $8.16\times {10}^{-7}M.$ and $4.08\times {10}^{-7}M$respectively.