Sodium chromate solution is gradually added to a mixture containing `0.05 M Pb^(2+)` ions and `0.10 M Ba^(2+)` ions. The concentration of the ion precipitating first when the second ion begins to form a percipitate is [Note: `K_(sp)` of `BaCrO_(4)=2.4xx10^(-10)` and `K_(sp)` of `PbCrO_(4)=1.8xx10^(-14)`]

Options

A. `7.5xx10^(-6)`
B. `2.5xx10^(-5)`
C. `8.2xx10^(-3)`
D. `5.0xx10^(-4)`

Answer

A

Explanation

`2nd` ion precipitated `=Ba^(2+)`
`[CrO_(4)^(-2)]=(K_(sp)(BaCrO_(4)))/([Ba^(2+)])=(2.4xx10^(-10))/(10^(-1))=2.4xx10^(-9)`
At this concentration of `[CrO_(4)^(2-)] ions`,
`[Pb^(2+)]=(1.8xx10^(-14))/(2.4xx10^(-9))=0.75xx10^(-5)=7.5xx10^(-6)`