Solubility product of silver bromide is 5.0 × 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

Question

Solubility product of silver bromide is 5.0 × 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 × 10-10 g (B) 1.2 × 10-9 g (C) 1.2 × 10-8 g (D) 6.2 × 10-5 g

Tardigrade Admin · Accepted Answer

Ag+ + Br- leftharpoons AgBr Precipitation starts when ionic product just exceeds solubility product Ksp = [Ag+][Br-] [Br-] = (Ksp/[Ag+]) = (5×10-13/0.05) = 10-11 i.e., precipitation just starts when 10-11 moles of KBr is added to 1L of AgNO3 solution. No. of moles of KBr to be added = 10-11 ∴ weight of KBr to be added = 10-11 × 120 = 1.2 × 10-9 g

Q. Solubility product of silver bromide is $5.0 \times 1 0^{- 13}$ . The quantity of potassium bromide (molar mass taken as $120 g$ of $m o l^{- 1}$ ) to be added to $1$ litre of $0.05 M$ solution of silver nitrate to start the precipitation of $A g B r$ is

$1.2 \times 1 0^{- 10} g$

$1.2 \times 1 0^{- 9} g$

$1.2 \times 1 0^{- 8} g$

$6.2 \times 1 0^{- 5} g$

$5.0 \times 1 0^{- 8} g$

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

1.2×10−10g

1.2×10−9g

1.2×10−8g

6.2×10−5g