- Home
- Standard 11
- Chemistry
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
$1.2 \times 10^{-10}\,g$
$1.2 \times 10^{-9}\,g$
$6.2 \times 10^{-5}\,g$
$5.0 \times 10^{-8}\,g$
Solution
$A g B r \rightleftharpoons A g^{+}+B r$
$K_{s p}=\left[A g^{+}\right]\left[B r^{-}\right]$
For precipitation to occur
lonic product $>$ Solubility product
$\left[B r^{-}\right]=\frac{K_{m}}{| A g^{+1}}=\frac{5 \times 10^{-13}}{0.05}=10^{-11}$
i.e., precipitation just starts when $10^{-11}$
moles of $K B r$ is added to $1\, \ell \,A g N O_{3}$ solution
$\therefore$ Number of moles of $B r^{-}$ needed
from $K B r=10^{-11}$
$\therefore$ Mass of $K B r=10^{-11} \times 120$
$=1.2 \times 10^{-9} g$
