Solveeit Logo
Login

Question

Chemistry Question on Buffer Solutions

Solubility product of silver bromide is 5.0×10−135.0 \times 10^{-13}. The quantity of potassium bromide (molar mass taken as 120 g of mol−1mol^{-1}) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

A

1.2×10−10 g1.2 \times 10^{-10}\, g

B

1.2×10−9 g1.2 \times 10^{-9}\, g

C

1.2×10−8 g1.2 \times 10^{-8}\, g

D

6.2×10−5 g6.2 \times 10^{-5}\, g

Answer

1.2×10−9 g1.2 \times 10^{-9}\, g

Explanation

Solution

Ag++Br−<=>AgBr {Ag+ + Br^{-} <=> AgBr} Precipitation starts when ionic product just exceeds solubility product Ksp=[Ag+][Br−]K_{sp} = \left[Ag^{+}\right]\left[Br^{-}\right] [Br−]=Ksp[Ag+]=5×10−130.05=10−11\left[Br^{-}\right] = \frac{K_{sp}}{\left[Ag^{+}\right]} = \frac{5\times10^{-13}}{0.05} = 10^{-11} i.e., precipitation just starts when 10−1110^{-11} moles of KBr is added to 1L of AgNO3AgNO_3 solution. No. of moles of KBr to be added =10−11= 10^{-11} ∴\therefore weight of KBr to be added =10−11×120= 10^{-11} \times 120 =1.2×10−9 g= 1.2 \times 10^{-9}\, g