Question

Solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \,g$ of $mol^{-1}$) to be added to $1$ litre of $0.05 \,M$ 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. $1.2 \times 10^{-8}\, g$
• D. $6.2 \times 10^{-5}\, g$
• E. $5.0 \times 10^{-8}\, g$

Answer 1

Answer: (B)

$Ag+ + Br^{-} \rightleftharpoons AgBr$
Precipitation starts when ionic product just exceeds solubility product
$K_{sp} = \left[Ag^{+}\right]\left[Br^{-}\right]$
$\left[Br^{-}\right] = \frac{K_{sp}}{\left[Ag^{+}\right]} = \frac{5\times10^{-13}}{0.05} = 10^{-11}$
i.e., precipitation just starts when $10^{-11}$ moles of $KBr$ is added to $1L$ of $AgNO_3$ solution.
No. of moles of $KBr$ to be added $= 10^{-11}$
$\therefore $ weight of $KBr$ to be added $= 10^{-11} \times 120$
$= 1.2 \times 10^{-9}\, g$