The molar solubility of silver sulphate is `1.5xx10^(-2)mol L^(-1)`. The solubility product of the salt will be

Options

A. `2.25xx10^(-4)`
B. `1.4xx10^(-5)`
C. `1.7xx10^(-6)`
D. `3.0xx10^(-3)`

Answer

B

Explanation

Solubility equilibrium equation: `Ag_(2)SO_(4)(s)hArr2Ag^(+)(aq.)+SO_(4)^(2-)(aq.)`
1 mol of `Ag_(2)SO_(4)` produces 2 mol of `Ag^(+)` and 1 mol of `SO_(4)^(2-)` in solution. Therefore, when `1.5xx10^(-2)` mol `Ag_(2)SO_(4)` is dissolved in 1 L of solution, the concentrations are
`C_(Ag^(+))= 2(1.5xx10^(-2)M)`
`= 3.0xx10^(-2)M`
`C_(SO_(4)^(2-))= 1.5xx10^(-2)M`
Solubility product:
`K_(sp)=C_(Ag^(+))^(2)C_(SO_(4)^(2-))`
`= (3.0xx10^(-2))^(2)(1.5xx10^(-2))`
`= 1.4xx10^(-5)`