Question 22.9: A solution is 0.100 M in Ag^+(aq) and 0.100 M in Pb^2+(aq). ......
A solution is 0.100 M in Ag^+(aq) and 0.100 M in Pb^{2+}(aq). Can these two ions be separated by precipitation with Cl^-(aq)? Assume that “separation” means that 99.0% of either ion is precipitated without precipitating the other ion.
The two solubility equilibrium equations and their respective solubility-product constants are
AgCl(s) ⇋ Ag^+(aq) + Cl^−(aq) \qquad K_{sp} = 1.8 × 10^{−10}\ M^2
PbCl_{2}(s) ⇋ Pb^{2+}(aq) + 2\ Cl^{−}(aq) \qquad K_{sp} = 1.5 × 10^{−5}\ M^3
The minimum values of [Cl^-] required to initiate precipitation of Ag^+(aq) and Pb^{2+}(aq) are given by solving each solubility-product expression for [Cl^-]. For Ag^+(aq), we have
K_{sp} = [Ag^+][Cl^−]so that
[Cl^–] =\frac{K_{sp}}{[Ag^+]} =\frac{1.8 × 10^{–10}\ M^2}{0.100\ M} =1.8 × 10^{–9}\ MFor Pb^{2+}(aq), we have
K_{sp} = [Pb^{2+}][Cl^{−}]^2so
[Cl^–] =\left(\frac{K_{sp}}{[Pb^{2+}]} \right) ^{1/2}=\left(\frac{1.5 × 10^{–5}\ M^3}{0.100\ M} \right) ^{1/2}=1.2 × 10^{–2}\ MBecause a much smaller concentration of Cl^-(aq) is required to precipitate the Ag^+(aq) ions than to precipitate the Pb^{2+}(aq) ions, the Ag^+(aq) will precipitate before the Pb^{2+}(aq).
If we precipitate 99.0% of the Ag^+(aq) according to the statement of the problem, then the concentration of Ag^+(aq) that would remain in solution is 1.0 % of 0.100 M, or 1.0 × 10^{-3}\ M. The concentration of Cl^-(aq) that will give this concentration of Ag^+(aq) is obtained from
[Cl^–] =\frac{K_{sp}}{[Ag^+]} =\frac{1.8 × 10^{–10}\ M^2}{1.0 × 10^{–3}\ M } =1.8 × 10^{–7}\ MTo determine whether Pb^{2+}(aq) will or will not precipitate with this concentration of Cl^-(aq), we calculate Q_{sp}:
Q_{sp} = [Pb^{2+}]_{0} [Cl^−]_{0}^{2} = (0.100\ M)(1.8 × 10^{−7}\ M)^2 = 3.2 × 10^{−15}\ M^3We see that Q_{sp} < K_{sp} for PbCl_{2}(s). Therefore, PbCl_{2}(s) will not precipitate, and the separation of the Ag^+(aq) and Pb^{2+}(aq) can be accomplished under these conditions.