Question

Solubility product of AgBr is 4.9 × 10^-13. What is its solubility?

Options

• 2.4 × 10^-7 mol dm^-3

• 4.9 × 10^-7 mol dm^-3

• 7.0 × 10^-7 mol dm^-3

• 3.2 × 10^-7 mol dm^-3

Answer 1

7.0 × 10^-7 mol dm^-3 

Explanation:

The equilibrium reaction of AgBr is

\[\ce{AgBr <=> Ag+ (aq) + Br- (aq)}\]

Molar solubility (S) of AgBr = `sqrt(4.9 xx 10^-13)` = 7.0 × 10^-7 mol dm^-3