Question

Calculate the simultaneous solubility of $AgSCN$ and $AgBr$.

$\left(K_{sp}\right(AgSCN)=1.1\times {10}^{-12}$, $K_{sp}\left(AgBr\right)=5\times {10}^{-13})$

• A. $4\times {10}^{-7}mol/LAgBr$; $9\times {10}^{-7}mol/LAgSCN$
• B. $9\times {10}^{-7}mol/LAgBr$; $4\times {10}^{-7}mol/LAgSCN$
• C. $8\times {10}^{-7}mol/LAgBr$; $18\times {10}^{-7}mol/LAgSCN$
• D. $18\times {10}^{-7}mol/LAgBr$; $8\times {10}^{-7}mol/LAgSCN$

Answer 1

The correct option is A $4\times {10}^{-7}mol/LAgBr$; $9\times {10}^{-7}mol/LAgSCN$

Let $aM$ be the solubility of $AgSCN$ and $bM$ be the solubility of $AgBr$.
AgSCN$\rightleftharpoons \underset{a}{A{g}^{+}}+\underset{b}{CN{S}^{-}}$
$AgBr\rightleftharpoons \underset{b}{A{g}^{+}}+\underset{b}{B{r}^{-}}$
$\left[A{g}^{+}\right]=(a+b)$
$\left[CN{S}^{-}\right]=a$
$\left[B{r}^{-}\right]=b$

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

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

Dividing equation (i) by equation (ii), we get
$\frac{a}{b}=\frac{1\times {10}^{-12}}{5\times {10}^{-13}}=2$
$a=2b$ or $b=\frac{a}{2}$
Substittue the value of (a) in (i),
$(a+b)a={10}^{-12}$
$(2a+b)2b={10}^{-12}$
$4b^2+2b^2={10}^{-12}$
$b=4.08\times {10}^{-7}$
This is the solubility of $AgBr$.
Substitute the value $b=\frac{a}{2}$ in equation (i),
$(a+b)a={10}^{-12}$
$(a+\frac{a}{2})a={10}^{-12}$
$3a^2=2\times {10}^{-12}$
$a=8.16\times {10}^{-7}M$
This is the solubility of $AgSCN$.