Solubility product of `PbCl_(2)` at 298 K is `1.0xx10^(-6)`. At this temperature solubility of `PbCl_(2)` in moles per litre is :

Options

A. `(1.0xx10^(-6))^(1//2)`
B. `(1.0xx10^(-6))^(1//3)`
C. `(0.25xx10^(-6))^(1//3)`
D. `(0.25xx10^(-6))^(1//2)`

Answer

C

Explanation

Let solubility of `PbCl_(2)=x`
`K_(sp)=[Pb^(2+)][Cl^(-)]^(2)=(x)(2x)^(2)=4x^(3)`
`therefore x = ((1.0xx10^(-8))/(4))^(1//3)`
`=(0.25xx10^(-6))1//3`