Solubility product of silver bromide is 5.0 10^ -13 . The quantit…

MCQ

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
$1.2 \times 10^{-10}\,g$
✓
$1.2 \times 10^{-9}\,g$
C
$6.2 \times 10^{-5}\,g$
D
$5.0 \times 10^{-8}\,g$

✓

Answer

Correct option: B.

$1.2 \times 10^{-9}\,g$

b
$A g B r \rightleftharpoons A g^{+}+B r$

$K_{s p}=\left[A g^{+}\right]\left[B r^{-}\right]$

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 $K B r=10^{-11} \times 120$

$=1.2 \times 10^{-9} g$

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