Question

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

• A. $1.2 \times 10^{-10}\, g$
• B. $1.2 \times 10^{-9}\, g$
• C. $1.2 \times 10^{-8}\, g$
• D. $6.2 \times 10^{-5}\, g$

Answer 1

The Correct Option is B

$ {Ag+ + Br^{-} <=> AgBr}$ Precipitation starts when ionic product just exceeds solubility product $K_{sp} = \left[Ag^{+}\right]\left[Br^{-}\right]$ $\left[Br^{-}\right] = \frac{K_{sp}}{\left[Ag^{+}\right]} = \frac{5\times10^{-13}}{0.05} = 10^{-11}$ i.e., precipitation just starts when $10^{-11}$ moles of KBr is added to 1L of $AgNO_3$ solution. No. of moles of KBr to be added $= 10^{-11}$ $\therefore$ weight of KBr to be added $= 10^{-11} \times 120$ $= 1.2 \times 10^{-9}\, g$