Question

Solubility product of silver bromide is $5.0\times {10}^{-13}$. The quantity of potassium bromide (molar mass taken as $120gmo{l}^{-1}$) to be added to $1litre$ 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 $1.2\times {10}^{-9}g$

$A{g}^{+}+B{r}^{-}\to AgBr$
Precipitation starts when ionic product just exceeds solubility product
$K_{sp}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$
$\left[B{r}^{-}\right]=\frac{K_{sp}}{\left[A{g}^{+}\right]}=\frac{5\times {10}^{-13}}{0.05}={10}^{-11}$
i.e., precipitation just starts when ${10}^{-11}$ moles of $KBr$ is added to 1 $L$ of $AgN{O}_3$ solution. No. of moles of $KBr$ to be added $={10}^{-11}$
Weight of $KBr$ to be added $={10}^{-11}\times 120$ $=1.2\times {10}^{-9}g$