The solubility of `BaSO_4` in water is `2.42 xx 10^(-3) gL^(-1)` at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of `BaSO_4 = 233 g "mol"^(-1)`)

Options

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

Answer

B

Explanation

Solubility of `BaSO_4 ("mol" L^(-1) ) = (2.42 xx 10^(-3))/(233) = S`
`K_(sp) " of " BaSO_4 = (s)^2`
` = ((2.42 xx 10^(-3))^2)/(233))`
` = 1.078 xx 10^(-10) "mol"^2 L^(-2)`
`1.08 xx 10^(-10) "mol"^2 L`