Solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \mathrm{~g} \mathrm{~of} \mathrm{~mol}^{-1}$) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

(B) $1.2 \times 10^{-9} \mathrm{~g}$

(A) $1.2 \times 10^{-10} \mathrm{~g}$

(C) $6.2 \times 10^{-5} \mathrm{~g}$

(D) $5.0 \times 10^{-8} \mathrm{~g}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Equilibrium

Solubility product of silver bromide is

5.0 \times 1 0^{- 13}

. The quantity of potassium bromide (molar mass taken as

120 g of mol^{- 1}

) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

Answer: (B)

Silver bromide begins to precipitate when

[A g^{+}] [B r^{-}]

reaches

K_{s p}

. With

[A g^{+}] = 0.05

M the bromide needed is

[B r^{-}] = \frac{5.0 \times 1 0 ^{- 13}}{0.05} = 1.0 \times 1 0^{- 11}

M. In one litre that is

1.0 \times 1 0^{- 11}

mol of bromide, supplied by the same number of moles of KBr, whose mass is

1.0 \times 1 0^{- 11} \times 120 = 1.2 \times 1 0^{- 9}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option