Answer : The concentration of [tex]Pb^{2+}[/tex] ion is 0.0375 M.
Explanation :
The balanced equilibrium reaction will be:
[tex]Pb^{2+}+2Cl^-\rightleftharpoons PbCl_2[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Pb^{2+}][Cl^-]^2[/tex]
Now put all the given values in this expression, we get:
[tex]2.40\times 10^{-4}=[Pb^{2+}]\times (0.0800)^2[/tex]
[tex][Pb^{2+}]=0.0375M[/tex]
Therefore, the concentration of [tex]Pb^{2+}[/tex] ion is 0.0375 M.
