Answer: a) 0.0144mol/L
b) [tex]1.19\times 10^{-5}[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
The equation for the ionization of the [tex]PbCl_2[/tex] is given as:
We are given:
Solubility of [tex]PbCl_2[/tex] = [tex]\frac{2.0g}{0.5L}=4g/L[/tex]
Molar Solubility of [tex]PbCl_2[/tex] = [tex]\frac{4g/L}{278.1g/mol}=0.0144mol/L[/tex]
1 mole of [tex]PbCl_2[/tex] gives 1 mole of [tex]Pb^{2+}[/tex] and 2 moles of [tex]Cl^-[/tex] ions
Solubility product of [tex]PbCl_2[/tex] = [tex][Pb^{2+}][Cl^-]^2[/tex]
[tex]K_{sp}=[0.0144][2\times 0.0144]^2[/tex]
[tex]K_{sp}=1.19\times 10^{-5}[/tex]
Thus the solubility product constant is [tex]1.19\times 10^{-5}[/tex]
