Question

Solubility product of silver bromide is $5.0\times {10}^{-13}$. How much of potassium bromide $\left(120{\text{g mol}}^{-1}\right)$ has to be added to $1\text{L}$ of $0.05M$ solution of silver nitrate to start the precipitation of $AgBr$ ?

• A. $1.2\times {10}^{-10}\text{g}$
• B. $5.0\times {10}^{-8}\text{g}$
• C. $6.0\times {10}^{-5}\text{g}$
• D. $1.2\times {10}^{-9}\text{g}$

Answer 1

The correct option is D $1.2\times {10}^{-9}\text{g}$

$\underset{0.05M}{AgN{O}_3}\rightleftharpoons \underset{0.05M}{A{g}^{+}}+\underset{0.05M}{N{O}_3^{-}}$
$\underset{xM}{KBr}\rightleftharpoons \underset{xM}{K^{+}}+\underset{xM}{B{r}^{-}}$
$AgBr\left(s\right)$ is precipitated if
$\left[A{g}^{+}\right]\left[B{r}^{-}\right]>K_{sp}$
$\left(0.05\right)\left(x\right)>5\times {10}^{-13}$
$\therefore x=\frac{5\times {10}^{-13}}{0.05}=1\times {10}^{-11}{\text{mol L}}^{-1}$
$\therefore KBr$ to be added, $=1\times {10}^{11}{\text{mol L}}^{-1}=1\times {10}^{-11}\times 120{\text{g mol}}^{-1}$
$=1.2\times {10}^{-9}\text{g}$