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^(-14) "mol"^(2)L^(-2)`
D. `1.08xx10^(-8) "mol"^(2)L^(-2)`

Answer

A

Explanation

Solubility (S) `=2.42xx10^(-3) g L^(-1)`
`=(2.42xx10^(-3))/(233) "mol" L^(-1)`
`=0.0104xx10^(-3) "mol" L^(-1)`
`=1.04xx10^(-5) "mol" L^(-1)`
`{:(BaSO_(4) ,hArr,Ba^(2+),+,SO_(4)^(2-)),(,,S,,S):}`
`K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=SxxS=S^(2)`
`=(1.04xx10^(-5) "mol" L^(-1))^(2)`
`=1.08xx10^(-10) "mol"^(2)L^(-2)`