Question

Solubility product of silver bromide is $5.0\times {10}^{-13}$. The quantity of potassium bromide (molar mass is 120 gmol${}^{-1})$ to be added to 1 litre of $0.05M$ solution of silver nitrate to start the precipitation of $AgBr$ is:

• A. $1.2\times {10}^{-10}g$
• B. $1.2\times {10}^{-9}g$
• C. $6.2\times {10}^{-5}g$
• D. $5.0\times {10}^{-8}g$

Answer 1

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

Given, $K_{sp}\left(AgBr\right)=5.0\times {10}^{-13},\left[A{g}^{+}\right]=0.05M$.
Molar mass of $KBr$ = 120 gmol${}^{-1}$

$A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)\rightleftharpoons AgBr\left(s\right)$
Precipitation starts when ionic product just exceeds solubility product.

$K_{sp}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$

putting the values,
$\Rightarrow \left[B{r}^{-}\right]=\frac{K_{sp}}{\left[A{g}^{+}\right]}=\frac{5\times {10}^{-13}}{0.05}={10}^{-11}$

Precipitation starts when ${10}^{-11}$ moles of $KBr$ is added to 1L of $AgN{O}_3$ solution.

Number of moles of $KBr$ to be added $={10}^{-11}$

Weight of $KBr=\text{number of moles}\times \text{molar mass}$

putting the values,
$={10}^{-11}\times 120=1.2\times {10}^{-9}g$