Question

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

$KBr + AgNO _{3}\to AgBr + KNO _{3}$

Given, $\left[ AgNO _{3}\right]=0.05 M ;\left[ Ag ^{+}\right]=\left[ NO _{3}^{-}\right]=0.05 \,M$

$\left[ Ag ^{+}\right]\left[ Br ^{-}\right]=\left(K_{s p}\right)_{ AgBr }$

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

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

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

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

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

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

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