Accounting for pH

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Page ID: 70571

Thomas Wenzel
Analytical Sciences Digital Library

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b) Calculate the solubility at pH 3.

Now you need to consider the protonation of phosphate that can occur. If we look back at the K_sp expression, we notice that it only contains lead ion and phosphate ion. Protonation of the phosphate will reduce the concentration of phosphate in solution, thereby causing more of the lead phosphate to dissolve based on Le Chatelier’s principle.

\[\begin{align}
\ce{Pb3(PO4)2 \leftrightarrow 3Pb^2+ +\: &2PO4^3- \\
&\:\:\:\Updownarrow\\
&HPO4^2- \\
&\:\:\:\Updownarrow\\
&H2PO4^- \\
&\:\:\:\Updownarrow\\
&H3PO4}
\end{align}\]

The problem with trying to solve this is that we do not know the concentration of phosphate (\(\ce{PO4^3-}\)) because it no longer relates directly to the amount of lead in solution. Once again, the way to approach solving this is to write two expressions for the solubility, one in terms of lead ion, the other in terms of phosphate species.

The situation for lead has not changed from part (a) of this problem, so we have the same expression for solubility for lead.

\[\mathrm{S =\dfrac{[Pb^{2+}]}{3} \hspace{30px} or \hspace{30px} [Pb^{2+}] = 3S}\]

For phosphate, we know that the only source of phosphate is by dissolution of lead phosphate. If there was a way for us to find the total amount of all phosphate species in solution, we could relate that back to the amount of lead phosphate that had to dissolve. This leads to the following expression relating the concentrations of phosphate species to solubility:

\[\mathrm{S =\dfrac{[H_3PO_4] + [H_2PO_4^-] + [HPO_4^{2-}] + [PO_4^{3-}]}{2} =\dfrac{[PO_4]_{TOT}}{2}}\]

\[\mathrm{[PO_4]_{TOT} = 2S}\]

But we also know the following:

\[\mathrm{[PO_4^{3-}] = α_{PO_4^{3-}}[PO_4]_{TOT}}\]

Since we were told the pH of this solution, we realize that we can evaluate the α-value and it’s a fixed number. We can then substitute in from the solubility expression above to get:

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

If we now substitute the terms for [Pb²⁺] and [\(\ce{PO4^3-}\)] back into the K_sp expression, we get the following equation:

\[\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = (3S)^3[α_{PO_4^{3-}}(2S)]^2}\]

\[\mathrm{K_{sp} = 108S^5(α_{PO_4^{3-}})^2 = 8.1\times 10^{-47}}\]

We now need to evaluate \(\mathrm{α_{PO_4^{3-}}}\) at a pH of 3. The form of the 1/α-value expression is as follows:

\[\mathrm{\dfrac{1}{α_{PO_4^{3-}}}=\dfrac{[H_3O^+]^3}{K_{a1} K_{a2} K_{a3}} + \dfrac{[H_3O^+]^2}{K_{a2} K_{a3}} + \dfrac{H_3O^+}{K_{a3}} + 1}\]

Substituting in for [H₃O⁺] and the K_a values gives an α-value of 2.315×10^-14 at a pH of 3. Putting this value into the K_sp expression above gives a final solubility of:

S = 6.75×10^-5

At this point, it would be worthwhile comparing the solubility in part (a) (no competing equilibria) to the solubility at pH 3.

(a) S = 2.37×10^-10

(b) S = 6.75×10^-5

Notice how the solubility is much higher at pH 3. This is reasonable since protonation of the phosphate ion was expected to increase the solubility. This trend points out an important aspect of the solubility of metal ions. Assuming that the anion of the solid is the anion of a weak acid, lowering the pH of the solution will cause a higher extent of protonation of the anion and increase the solubility of the solid.

In general, the solubility of sparingly soluble substances increases with the acidity of the water. It turns out that this is one of the principle concerns of acid rain. Acid rain into unbuffered natural waters raises the acidity (lowers the pH) of the water. The higher acidity causes solid metal salts and minerals in the lake or river bed to dissolve at higher levels. For example, there are lakes with poor buffering in which the impact of acid rain has increased the levels of dissolved aluminum ion (Al³⁺). Aluminum ion is known to form a highly insoluble complex with hydroxide ion [Al(OH)₃, K_sp = 2.20×10^-32]. Obviously the solubility of this complex is critically dependent on pH. At acidic pH values, it will dissolve because hydroxide is low. At neutral to basic pH, it will precipitate because the hydroxide level becomes high enough. Aluminum hydroxide is a very gelatinous solid that is sometimes used as a sticky flocculent in water treatment processes (undesirable impurities essentially stick to this material and slowly settle out with it). When the fish take the water through their gills (which are at a pH of 7.4) to remove the dissolved oxygen, the pH of the water increases and the aluminum ion now precipitates out as aluminum hydroxide. The gelatinous precipitate clogs up the gills of the fish and actually causes the fish to die of suffocation. The fish deaths that have occurred in some lakes heavily impacted by acid rain are attributable to this phenomenon.