The `K_(sp)` of `AgCrO_(4)` is `1.1xx10^(-12)` at `298K`. The solubility (in mol `L^(-1)`) of `Ag_(2)CrO_(4)` in a `0.1M AgNO_(3)` solution is

Options

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

Answer

A

Explanation

Let us assume that the solubility of `Ag_(2)CrO_(4)` in `0.1M AgNO_(3)` is S mol `L^(-1)`:
`{:(Ag_(2)CrO_(4)(s)hArr2Ag^(+)(aq.)+CrO_(4)^(2-)(aq.)),("Initial (M)" " 0.1 0"),("Change (M)" " 0.1+2S S"),(bar("Equilibrium (M) 0.1+2S S ")):}`
`K_(sp)= C_(Ag^(+))^(2)C_(CrO_(4)^(2-))`
`(1.1xx10^(-12))=(0.1+2S)^(2)(S)`
Since `K_(sp)` of `Ag_(2)CrO_(4)` is very low, we can neglect 2S contribution relative to `0.1` to have
`1.1xx10^(-12)=(0.1)^(2)S`
or `S= (1.1xx10^(-12))/(10^(-2))`
`=1.1xx10^(-10)`