Solubility product of silver bromide is $5.0 \times 10^{- 13} M^{2}$ . 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

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$5.0 \times 10^{- 8} g$

$6.2 x 10^{- 5} g$

$1.2 x 10^{- 10} g$

$1.2 x 10^{- 9} g$

answer is C.

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Detailed Solution

$AgBr (s) ⇌ {Ag}^{+} (aq) + {Br}^{-} (aq); K_{sp} (AgBr) = [{Ag}^{+}] [{Br}^{-}] = 5.0 \times 10^{- 13} M^{2}$

It is given that $[{Ag}^{+}] = 0.05 M (from {AgNO}_{3} solution)$ Hence $[{Br}^{-}] = \frac{K_{sp} (AgBr)}{[{Ag}^{+}]} = \frac{5.0 \times 10^{- 13} M^{2}}{0.05 M} = 1 \times 10^{- 11} M$ Mass of KBr to be added $m = [KBr] M_{KBr} = (1 \times 10^{- 11} M) (120 g {mol}^{- 1}) = 1.2 \times 10^{- 9} {gL}^{- 1}$