The solubility equilibrium of PbCl[tex] _{2} [/tex]:
[tex] PbCl_{2}(aq) <---> Pb^{2+}(aq) + 2 Cl^{-}(aq)[/tex]
[tex] K_{sp}=[Pb^{2+}][Cl^{-}]^{2} [/tex]
[tex] [Cl^{-}] = 2.88 * 10^{-2} M [/tex]
[tex] [Pb^{2+}]=\frac{[Cl^{-}]}{2} = \frac{2.88 * 10^{-2}}{2}=1.44 *10^{-2} [/tex]
[tex] K_{sp}=[Pb^{2+}][Cl^{-}]^{2} [/tex]
= [tex] (1.44 * 10^{-2})(2.88*10^{-2})^{2} [/tex]
= [tex] 1.19 * 10^{-5} [/tex]
So, the corrected solubility product will be [tex] 1.19 * 10^{-5} [/tex]
