Question

Calculate simultaneous solubility of $AgCNS$ and $AgBr$ in a solution of water.

[Given: $K_{sp}$ of $AgCNS=1\times {10}^{-12}$, $K_{sp}$ of $AgBr=5\times {10}^{-13}]$.

Answer 1

Let solubility of $AgCNS$ and $AgBr$ in $a$ and $b$ respectively.
$AgCNS\rightleftharpoons {Ag}^{+}+CN{S}^{-}$
$a$ $a$

$AgBr\rightleftharpoons {Ag}^{+}+{Br}^{-}$
$b$ $b$
$\left[{Ag}^{+}\right]=a+b,\left[CN{S}^{-}\right]=a$ and $\left[{Br}^{-}\right]=b$

$K_{sp}AgCNS=\left[{Ag}^{+}\right]\left[CN{S}^{-}\right]=a(a+b)$
$1\times {10}^{-12}=a(a+b)...\left(i\right)$

$K_{sp}AgBr=\left[{Ag}^{+}\right]\left[{Br}^{-}\right]=b(a+b)$
$5\times {10}^{-13}=b(a+b)...\left(ii\right)$

Dividing eq. (i) by (ii)
$\frac{1\times {10}^{-12}}{5\times {10}^{-13}}=\frac{a}{b}$
$2=\frac{a}{b}$ or $a=2b$
Putting the value of $a$ in eq. (i)
$6b^2=1\times {10}^{-12}$

$b=4.08\times {10}^{-7}mol{L}^{-1}$

$a=2\times 4.08\times {10}^{-7}=8.16\times {10}^{-7}mol{L}^{-1}$