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

Options

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

Answer

A

Explanation

`K_(sp)=[Ba^(2+)][CrO_(4)^(2-)]`
`2.4xx10^(-10)=[Ba^(2+)](6xx10^(-4))`
`therefore [Ba^(2+)]=(2.4xx10^(-10))/(6xx10^(-4))`
`=4.0xx10^(-7)M`.