The `K_(s)` of `Ag_(2)CrO_(4)` is `1.1xx10^(-12)` at `298K`. The solubility in mol per litre of `Ag_(2)CrO_(4)` in `0.1M.AgNO_(3)` solution is

Options

A. `1.1xx10^(-11)`
B. `1.1xx10^(-10)`
C. `1.1xx10^(-12)`
D. `1.1xx10^(-9)`

Answer

B

Explanation

`underset(s)(Ag_(2)CrO_(4))hArrunderset(2s+0.1)(2Ag^(+))+underset(s)(CrO_(4)^(2-))`
(`0.1` from `AgNO_(3)=0.1` as .s. is negligible compared to `10^(-12)` )
`K_(s)=[Ag^(+)]^(2)[CrO_(4)^(2-)]`
`1.1xx10^(-12)=(0.1)^(2)(s)`
`s=1.1xx10^(-10)M`