The solubility of a springly soluble salt `AB_(2)` in water is `1.0xx10^(-5) mol L^(-1)`. Its solubility product is:

Options

A. `4 xx 10^(-15)`
B. `4 xx 10^(-10)`
C. `1 xx 10^(-15)`
D. ` 1 xx 10^(-10)`

Answer

A

Explanation

`AB_(2)hArr A^(2+)+2B^(-)`
At eqm. ` 1.0 xx 10^(-5)" "2.0 xx 10^(-5)`
`K_(sp)=[A^(2+)][B^(-)]^(2)`
`=(1 xx 10^(-5))(2 xx 10^(-5))^(2)=4 xx 10^(-15)`