Answer:
the concentration of Pb2+ ions is [tex][Pb^{2+}] } = 0.15 mol/L[/tex]
Explanation:
From the question we are told that
The concentration of [tex]Cl^{-}[/tex] is [tex]C_s = 0.0400 M[/tex]
The reaction is
[tex]Pb^{2+} + 2Cl^- ----> PbCl_2[/tex]
The solubility constant for this reaction is mathematically represented as
[tex]K_{sp} = [Pb^{2+}][Cl^{-}]^2[/tex]
Substituting values
[tex]2.4 0 * 10^{-4} = 0.0400^2 * [Pb^{2+}][/tex]
[tex][Pb^{2+}] } = \frac{2.4 0 * 10^{-4} }{0.0400^2 }[/tex]
[tex][Pb^{2+}] } = 0.15 mol/L[/tex]
