If the molar concentration of `Cl^(-)` ions in saturated aqueous solution of `PbCl_(2)` is `3.2xx10^(-2)mol" "L^(-1)` at a certain temperature, then find the solubility product of `PbCl_(2)` at that temperature.

Explanation

In saturated aqueous solution, `PbCl_(2)` remains in the following equilibrium: `PbCl_(2)(s)hArrPb^(2+)(aq)+2Cl^(-)(aq)`
`therefore`Solublity product of `PbCl_(2),K_(sp)=[Pb^(2+)]xx[Cl^(-)]^(2)`
As given in the question, `[Cl^(-)]=3.2xx10^(-2)mol*L^(-1)`
`therefore [Pb^(2+)]=(1)/(2)[Cl^(-)]=1.6xx10^(-2)mol*L^(-1)`
[`because` 1 molecule of `PbCl_(2)` dissociates into one `Pb^(2+)` and two `Cl^(-)` ions]
`therefore ` Solubility product of `PbCl_(2)=[Pb^(2+)][Cl^(-)]^(2)`
`=(1.6xx10^(-2))xx(3.2xx10^(-2))^(2)=1.63xx10^(-5)`.