Solubility product of silver bromide is $5.0 x 10^{- 13}$ . 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:

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

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

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

$6.2 \times 10^{- 5} g$

answer is B.

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Detailed Solution

$\begin{array}{l} Molarity = \frac{n_{KBr}}{V_{Solution} (L)} \\ 1 \times 10^{- 11} = \frac{w_{KBr} / 120}{1} (Mol . wt. of KBr = 120) \end{array}$

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

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