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 $1$ litre of $0.05M$ solution of silver nitrate to start the precipitation of $AgBr$ is

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

Answer 1

Answer: (C)

Given, ${\left(K_{sp}\right)}_{AgBr}=5.0\times 10^{-13}$

The required equation is,

$KBr+AgN{O}_3\to AgBr+KN{O}_3$

Given, $\left[AgN{O}_3\right]=0.05M;\left[A{g}^{+}\right]=\left[N{O}_3^{-}\right]=0.05M$

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

$\Rightarrow 0.05\times \left[B{r}^{-}\right]=5\times 10^{-13}$

$\Rightarrow \left[B{r}^{-}\right]=\frac{5\times 10^{-13}}{5\times 10^{-2}}=1\times 10^{-11}M$

$\because \left[K^{+}\right]=\left[B{r}^{-}\right]=[KBr]$

$\therefore [KBr]=1\times 10^{-11}M$

Molarity $=\frac{n_{KBr}}{V_{\text{Solution }}(L)}$

$1\times 10^{-11}=\frac{w_{KBr}/120}{1}$ ( Mol. wt. of $KBr=120)$

$\Rightarrow w_{KBr}=1\times 10^{-11}\times 120=120\times 10^{-11}=1.2\times 10^{-9}g$