Question

Solubility product of silver bromide is 5.0 × 10–13 . This 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 :

• A. 1.2 × 10–10 g
• B. 1.2 × 10–9 g
• C. 6.2 × 10–5 g
• D. 5.0 × 10–8 g

Answer 1

Answer: (B)

1.2 × 10–9 g

Answer 2

Ksp = [Ag+] [Br–] = 5.0 × 10–13

[Ag+] = 0.05 M

[0.05] [Br–] = 5.0 × 10–13

[Br–] = $\frac { 5.0 \times 10 ^ { - 13 } } { 0.05 }$ = 1 × 10–11 M

moles of KBr = M × V = 1 × 10–11 × 1 = 1 × 10–11

weight of KBr = 1 × 10–11 × 120 = 1.2 × 10–9 g