Question 6.4: Determine the value of the equilibrium constant for the reac...

This reaction can be broken down into three reactions. The first of these reactions is the solubility of \mathrm{PbCl}_{2}, described by its K_{\mathrm{sp}}

\mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)

and the second and third are the stepwise formation of \mathrm{PbCl}_{2} (aq), described by K_{1} and K_{2}

\begin{aligned} \mathrm{Pb}^{2+}(a q)+\mathrm{Cl}^{-}(a q) & \rightleftharpoons \mathrm{PbCl}^{+}(a q) \\ \mathrm{PbCl}^{+}(a q)+\mathrm{Cl}^{-}(a q) & \rightleftharpoons \mathrm{PbCl}_{2}(a q) \end{aligned}

Using values for K_{\mathrm{sp}}, K_{1}, and K_{2} from Appendices 3 \mathrm{~A} and 3 \mathrm{C}, we find the equilibrium constant to be

K=K_{\text {sp }} \times K_{1} \times K_{2}=\left(1.7 \times 10^{-5}\right)(38.9)(1.62)=1.1 \times 10^{-3}