The solubility of `BaSO_(4)` in water is `2.42xx10^(-3) gL^(-1)` at `298 K`. The value of its solubility product `(K_(sp))` will be (Given molar mass of `BaSO_(4)=233 g mol^(-1)`)

Options

A. `1.08xx10^(-10) mol^2 L^(-2)`
B. `1.08xx10^(-12) mol^2 L^(-2)`
C. `1.08xx10^(-8) mol^2 L^(-2)`
D. `1.08xx10^(-14) mol^2 L^(-2)`

Answer

A

Explanation

Solubility of `BaSO_4, s= (2.42xx10^(-3))/(233)(mol" "L^(-1))`
`= 1.04xx 10^(-5)(mol" "L^(-1))`
`BaSO_4(s) harr underset(s)(Ba^(2+))(aq)+underset(s)(SO_4^(2-))(aq)`
`K_(sp)=[Ba^(2+)][SO_4^(2-)]=s^2`
`=(1.04xx10^(-5))^2`
`=1.08xx10^(-10) mol^2 L^(-2)`