Answer:
The solubility product of lead(II) chloride is [tex]1.61\times 10^{-5}[/tex].
Explanation:
Concentration of lead (II) ions = [tex][Pb^{2+}]=0.0159 M[/tex]
Concentration of chloride ion = [tex][Cl^-]=0.0318 M[/tex]
[tex]PbCl_2(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)[/tex]
The expression of a solubility product will be given as:
[tex]K_{sp}=[Pb^{2+}][Cl^-]^2[/tex]
[tex]=0.0159 M\times (0.0318 M)^2=1.61\times 10^{-5}[/tex]
The solubility product of lead(II) chloride is [tex]1.61\times 10^{-5}[/tex].
