Question

A saturated solution of pbcl2 is found to contain 2.88 x 10-2 m chloride ions. what is the correct value of the solubility product, ksp

Answer 1

The solubility equilibrium of PbCl
`_(2)`:

`PbCl_(2)(aq) <---> Pb^(2+)(aq) + 2 Cl^(-)(aq)`

`K_(sp)=[Pb^(2+)][Cl^(-)]^(2)`

`[Cl^(-)] = 2.88 * 10^(-2) M`

`[Pb^(2+)]=([Cl^(-)])/(2) = (2.88 * 10^(-2))/(2)=1.44 *10^(-2)`

`K_(sp)=[Pb^(2+)][Cl^(-)]^(2)`

=
`(1.44 * 10^(-2))(2.88*10^(-2))^(2)`

=
`1.19 * 10^(-5)`

So, the corrected solubility product will be
`1.19 * 10^(-5)`