Q. Solubility product of silver bromide is 5.0 $10^{13}$. The quantity of potassium bromide (molar mass taken as 120 g $mol^{1}$) to be added to 1 L of 0.05 M solution of silver nitrate to start the precipitation of $AgBr$ is
A) $1.2\ 10^{\ 10}$ g
B) $1.2\ \ 10^{\ 9}$ g
C) $6.2\ \ \ 10^{\ 5}$ g
D) $5.0\ \ 10^{\ 8}$ g
Explanation
(b)
$K_{sp} = \left\lbrack Ag^{+} \right\rbrack\left\lbrack Br^{-} \right\rbrack = 5.0 \times 10^{- 13}$
$\left\lbrack Ag^{+} \right\rbrack = 0.05$ M
$\left\lbrack Br^{-} \right\rbrack = \frac{5.0 \times 10^{- 13}}{0.05} = 1 \times 10^{- 11}$ M
Moles of $KBr = 1 \times 10^{- 11} \times 1 = 1 \times 10^{- 11}$
Weight of $KBr = 1 \times 10^{- 11} \times 120 = 1.2 \times 10^{- 9}$g
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.