The solubility of `Ag_(2)CrO_(4)` is `2 xx 10^(-2) "mol"//"lit"` its solubility product is

Options

A. `3.2 xx 10^(-5)`
B. `32 xx 10^(-8)`
C. `16 xx 10^(-8)`
D. `3.32 xx 10^(-10)`

Answer

A

Explanation

`Ag_(2)CrO_(4) rarr 2Ag^(+) + CrO_(4^(2-))`
`Ksp = (2S)^(2)S , = 4S^(3)`
`= 4 (2 xx 10^(-2))^(3) = 4 xx 8 xx 10^(-6)`
`= 3.2 xx 10^(-5)`