# Accounting for the Presence of Complexing Ligands

**c) Now you realize for the solution in part (b) that lead can form soluble hydroxide complexes. Incorporate these into the expression.**

The scheme below shows the total set of reactions that occur in this solution.

\[\begin{align}

\ce{Pb3(PO4)2}\hspace{25px} \leftrightarrow \hspace{25px} &\ce{3Pb^2+} \hspace{25px}+\: &&\ce{2PO4^3-} \\

\ce{&\:\:\:\Updownarrow &&\:\:\:\Updownarrow\\

&Pb(OH)+ &&HPO4^2- \\

&\:\:\:\Updownarrow &&\:\:\:\Updownarrow\\

&Pb(OH)2 &&H2PO4^- \\

&\:\:\:\Updownarrow &&\:\:\:\Updownarrow\\

&Pb(OH)3- &&H3PO4}

\end{align}\]

The approach in this case is going to be analogous to what we just did for the protonation of the phosphate ion. We know the concentration of hydroxide because the pH is known. This enables us to calculate an \(\mathrm{\alpha_{Pb^{2+}}}\) value and incorporate that into the K_{sp} expression.

The next step is to write two expressions for the solubility, one in terms of lead species, the other in terms of phosphate species.

The equation in terms of phosphate is identical to what was just done in part (b).

\[\mathrm{S = \dfrac{[PO_4]_{TOT}}{2} \hspace{60px} [PO_4]_{TOT} = 2S}\]

\[\mathrm{[PO_4^{3-}] = α_{PO_4^{3-}}[PO_4]_{TOT} = α_{PO_4^{3-}}(2S)}\]

The equation for lead is as follows:

\[\mathrm{S =\dfrac{[Pb^{2+}] + [Pb(OH)^+] + [Pb(OH)_2] + [Pb(OH)_3^-]}{3} =\dfrac{[Pb]_{TOT}}{3}}\]

\[\mathrm{[Pb]_{TOT} = 3S}\]

\[\mathrm{[Pb^{2+}] = \alpha_{Pb^{2+}} [Pb]_{TOT} = \alpha_{Pb^{2+}}(3S)}\]

Evaluation of \(\mathrm{\alpha_{Pb^{2+}}}\) is done by writing the ratio of Pb^{2+} over the total, taking the reciprocal so that there is a set of individual terms, and then using the K_{f} expressions for lead complexation with hydroxide to substitute in for each of the terms. The final equation for 1/\(\mathrm{\alpha_{Pb^{2+}}}\) is shown below.

\[\mathrm{\dfrac{1}{α_{Pb^{2+}}} = K_{f1} K_{f2} K_{f3}[OH^-]^3 + K_{f1} K_{f2}[OH^-]^2 + K_{f1}[OH^-] + 1}\]

Evaluation of \(\mathrm{\alpha_{Pb^{2+}}}\) at a pH of 3 gives a value of 0.999984. So very little of the lead actually complexes with hydroxide, which should not be that surprising given the small amount of hydroxide ion in solution at pH 3.

Above we have expressions for [Pb^{2+}] and [\(\ce{PO4^3-}\)] that are in terms of α-values and S. These can be substituted into the K_{sp} expression to give the following:

\[\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = [\alpha_{Pb^{2+}}(3S)]^3[\alpha_{PO_4^{3-}} (2S)]^2 = 8.1\times 10^{-47}}\]

\[\mathrm{8.1\times 10^{-47} = 108S^5(\alpha_{Pb^{2+}})^3(\alpha_{PO_4^{3-}})^2}\]

\[\mathrm{S = 6.75\times 10^{-5}}\]

If we compare this to the answer in part (b), it turns out that the two are the same. This means that so little lead complexes with the hydroxide ion at pH 3 that it does not lead to any increase in the solubility. If we were to make the solution more basic, complexation of lead by hydroxide would become more important. But also note that protonation of the phosphate would become less important, so the overall solubility is a balance between two processes that influence the solubility in opposite ways as a function of pH. What we might well observe for lead phosphate is that its solubility is smallest at some intermediate pH. At low pH, protonation of the phosphate increases the solubility. At high pH, complexation of lead with hydroxide increases the solubility. If we wanted to use precipitation of lead phosphate as a way to analyze lead (say by collecting the precipitate by filtration and weighing) or remove lead from a solution, we would need to perform a calculation over the entire pH range to find the best value for precipitation of the most amount of material.

**d) Revisit problem (a). What is the actual solubility of lead phosphate in unbuffered water given that other equilibria will simultaneously occur?**

This is a difficult situation because we know that hydroxide complexes of lead can form and that protonation of phosphate can occur, but it does not seem like we can use α-values because we really do not know the pH. The best approach might be to try some simplifying treatments to see if anything will work.

One thing we could do is assume that the pH of the water is 7, and that dissolving of the lead phosphate does not change it. If that were the case, we should evaluate \(\mathrm{α_{PO_4^{3-}}}\) at pH 7 to see what fraction of the phosphate stays in this form. Evaluation of \(\mathrm{α_{PO_4^{3-}}}\) at pH 7 gives a value of 1.62×10^{-6}. This means that only a small fraction of the phosphate species will exist as \(\ce{PO4^3-}\) and more of it will be protonated. The protonation has the possibility of changing the pH enough from 7 to make a difference. Similarly, if we evaluate \(\mathrm{α_{Pb^{2+}}}\) at a pH of 7 we get a value of 0.864, so some lead complexes as well. If we go ahead and plug in these values into the K_{sp} expression:

\[\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = [\alpha_{Pb^{2+}}(3S)]^3[\alpha_{PO_4^{3-}}(2S)]^2 = 8.1\times10^{-47}}\]

\[\mathrm{S = 7.49\times 10^{-6}}\]

This is a small number, but the problem is that it’s an appreciable number compared to the concentration of H_{3}O^{+} at a pH of 7. This means that the pH will probably change enough from 7 to make a difference in the solubility.

It turns out that we cannot make any simplifying assumptions in this case. In this event, we need to solve a series of simultaneous equations. If we write all the unknown species, you find that there are a total of ten for this solution.

\[\begin{align}

\ce{&[H3O+] &&[Pb^2+] &&[PO4^3- ]\\

&[OH- ] &&[Pb(OH)+] &&[HPO4^2- ]\\

& &&[Pb(OH)2] &&[H2PO4- ]\\

& &&[Pb(OH)3- ] &&[H3PO4]}

\end{align}\]

We may be able to eliminate some of these as insignificant, since it might be unlikely that we would get any significant levels of hydroxide complexes or protonation of phosphate besides the first species (Pb(OH)^{+} and \(\ce{HPO4^2-}\)). Even if that is the case, we would still need to solve a set of simultaneous equations.

What would be the ten equations? Eight of them are equilibrium constant expressions needed to describe the reactions taking place.

K_{sp} K_{a1} K_{a2} K_{a3} K_{f1} K_{f2} K_{f3} K_{w}

One is the mass balance, which involves the relationship between the two solubility expressions we can write for this solution.

\[\mathrm{S = \dfrac{[Pb]_{TOT}}{3} \hspace{60px} S = \dfrac{[PO_4]_{TOT}}{2}}\]

\[\mathrm{\dfrac{[Pb]_{TOT}}{3} = \dfrac{[PO_4]_{TOT}}{2}}\]

The final equation is the charge balance:

\[\ce{[H3O+] + 2[Pb^2+] + [Pb(OH)+]} = \ce{[OH- ] + [Pb(OH)3- ] + [H2PO4- ] + 2[HPO4^2- ] + 3[PO4^3- ] }\]

Next step? HAVE FUN!