The solubility product of `BaCrO_(4)` is `2.4xx10^(-10)M^(2)`. The maximum concentration of `Ba(N0_(3))_(2)` possible without precipitation in a `6xx10^(-4)` M `K_(2)CrO_(4)` solution is :

Options

A. `4xx10^(-7) M`
B. `1.2 xx 10^(-10)` M
C. `6xx10^(-4)` M
D. `3xx10^(-4)M`.

Answer

A

Explanation

`[Ba^(2+)][CrO_(4)^(2-)]=K_(sp)`
`[Ba^(2+)]=K_(sp)//[CrO_(4)^(2-)]`
`[CrO_(4)^(2-)]=[K_(2)CrO_(4)]=6xx10^(-4)M`
`=(2.4 xx 10^(-10))/(6 xx 10^(-4))=4 xx 10^(-7)M`