The solubility of `BaSO_(4)` in water is `2.33xx10^(-3) g//litre`. Its solubility product will be (molecular weight of `BaSO_(4)=233`)

Options

A. `1 xx 10^(-5)`
B. `1 xx 10^(-10)`
C. `1 xx 10^(-15)`
D. `1 xx 10^(-20)`

Answer

B

Explanation

The solubility of `BaSO_(4)` in g/litre is given `2.33 xx 10^(-3)` `because` in mole/litre. `n = (W)/(m.wt) = 1 xx 10^(-5) = (2.33 xx 10^(-3))/(233)`
Because `BaSO_(4)` is a compound
`K_(sp) = S^(2) = [1 xx 10^(-5)]^(2) = 1 xx 10^(-10)`.