The solubility of `BaSO_4` in water is ` 2.42 xx 10^(-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^(-14) " mol"^(2)L^(-2)`
B. `1.08xx10^(-12) " mol"^(2)L^(2)`
C. `1.08xx10^(-10) " mol"^(2)L^(-2)`
D. `1.08xx10^(-8) " mol"^(2)L^(-2)`

Answer

C

Explanation

For a general reaction,
`A_(x)B_(y) hArr xA^(y+)+yB^(x-)`
Solubility product `(K_(sp))=[A^(y+)]^(x)[B^(x-)]^(y)`
For `BaSO_(4)` (binary solute giving two ions)
`BaSO_(4)(s) hArr Ba^(2+)underset(S)((aq))+underset(S)(SO_(4)^(2-))(aq)`
`therefore K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=(S)(S)=S^(2) " " ["where, S=Solubility"]`
Given, `S=2.42xx10^(-3) gL^(-1)`
Molar mass of `BaSO_(4)=233 g "mol"^(-1)`
`therefore` Solubility of `BaSO_(4)`
`(S)=(2.42xx10^(-3))/(233) "mol" L^(-1)`
`=1.04xx10^(-5) "mol" L^(-1)`
On substituting the value of S in Eq. (i), we get
`K_(sp)=(1.04xx10^(-5) "mol" L^(-1))^(2)`
`=1.08xx10^(-10) "mol"^(2)L^(-2)`