# Utilization of α-Values for the Metal Ion

There is one other aspect to this problem we have not yet considered.  That is whether calcium can complex with the hydroxide ion and whether this complexation is significant enough to alter any of the results we have seen before regarding complexation of Ca2+ by E4-.  Looking in the table of formation constants indicates that calcium ions can complex with hydroxide according to the following equation.  Since only one Kf value is listed, it is only a one-step process.  We might also notice that it’s a fairly small association constant, so that we might anticipate that this reaction would never represent that much of an interference in the complexation of Ca2+ with E4-.

$\ce{Ca^2+ + OH- \leftrightarrow Ca(OH)+} \hspace{60px} \mathrm{K_f = 1.99×10}$

We can couple this process into the overall scheme as shown below:

\begin{align} &\ce{Ca^2+} \hspace{25px} + &&\ce{E^4-} \hspace{25px}\leftrightarrow \hspace{25px}\ce{CaE^2-} \\ \ce{OH-} &\:\Updownarrow &&\:\Updownarrow\\ &\ce{Ca(OH)+} &&\ce{HE^3-} \\ & &&\:\Updownarrow\\ & &&\ce{H2E^2-} \\ & &&\:\Updownarrow\\ & &&\ce{H3E-}\\ & &&\:\Updownarrow\\ & &&\ce{H4E} \end{align}

The approach we will use is analogous to that employed with the protonation of E4–.  If we know the concentration of ligand, it is possible to calculate α-values for the uncomplexed metal ion and the metal-ligand species.  In this case of hydroxide ion, the concentration is known and fixed provided the pH is known and fixed.  With other ligands, we may need to assess whether the initial ligand concentration we are provided remains fixed.  In some cases, the ligand will complex with the metal and this causes the concentration to drop from its initial value, changing the α-values that were calculated.

For the situation in this problem, we need to calculate $$\mathrm{α_{Ca^{2+}}}$$.  We do this by setting up a ratio of Ca2+ to the total of other calcium species.  There is one important thing to realize in setting up this ratio.  We only want to look at the distribution of calcium species in the set of reactions involving complexation with hydroxide ion.  We therefore do not include CaE2– as a term in the ratio.

$\mathrm{α_{Ca^{2+}} = \dfrac{[Ca^{2+}]}{[Ca(OH)^+] + [Ca^{2+}]}}$

The next step is to take the reciprocal, and divide the equation into a series of separate terms.

$\mathrm{\dfrac{1}{α_{Ca^{2+}}} = \dfrac{[Ca(OH)^+]}{[Ca^{2+}]}+\dfrac{[Ca^{2+}]}{[Ca^{2+}]}}$

We can now use the Kf expression to substitute in for the first term in this equation.  Using Kf for complexation of Ca2+ with hydroxide, we get as follows:

$\ce{Ca^2+ + OH- \leftrightarrow Ca(OH)+} \hspace{60px} \mathrm{K_f = 1.99×10}$

$\mathrm{K_f =\dfrac{[Ca(OH)^+]}{[Ca^{2+}][OH^-]}}$

Rearrange the Kf expression as follows:

$\mathrm{\dfrac{[Ca(OH)^+]}{[Ca^{2+}]} = K_f[OH^-]}$

Substitute this into the 1/$$\mathrm{α_{Ca^{2+}}}$$ expression to get:

$\mathrm{\dfrac{1}{α_{Ca^{2+}}} = K_f[OH^-] + 1}$

What we see is that the fraction of calcium that exists as Ca2+ only depends on the Kf value and the concentration of ligand (hydroxide in this case).  We could also write the following expression:

$\mathrm{[Ca^{2+}] = α_{Ca^{2+}}[Ca]_{TOT}}$

Remember, [Ca]TOT  = [Ca2+] + [Ca(OH)+] in this expression.

We can substitute this into our original Kf expression for the complexation of Ca2+ with E4–, just as we did previously to account for the protonation of EDTA as a function of pH.

$\ce{Ca^2+ + E^4- \leftrightarrow CaE^2-}$

$\mathrm{K_f =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E^{4-}]}}$

$\mathrm{K_f =\dfrac{[CaE^{2-}]}{(α_{Ca^{2+}}[Ca]_{TOT})(α_{E^{4-}}[E]_{TOT})}}$

$\mathrm{(K_f)(α_{Ca^{2+}})(α_{E^{4-}}) =\dfrac{[CaE^{2-}]}{[Ca]_{TOT}[E]_{TOT}}}$

This provides a conditional constant $$\mathrm{(K_f)(α_{Ca^{2+}})(α_{E^{4-}})}$$ that incorporates both conditions that are present: protonation of the E4– and complexation of Ca2+ by hydroxide ion.  This conditional constant is essentially the equilibrium constant for the following reaction:

$\ce{[Ca]_{TOT} + [E]_{TOT} \leftrightarrow CaE^2-}$

What we then need to do is examine the magnitude of this conditional constant to assess whether the complexation of calcium with EDTA will occur.

Table 7 is a compilation of $$\mathrm{α_{E^{4-}}}$$, $$\mathrm{α_{Ca^{2+}}}$$, conditional constants, and extent of reaction for this entire process as a function of pH.

Table 7.  Conditional constants for the complexation of Ca2+ with E4-.

pH $$\mathrm{α_{E^{4-}}}$$ $$\mathrm{α_{Ca^{2+}}}$$ $$\mathrm{(K_f)(α_{E^{4-}})(α_{Ca^{2+}})}$$ Extent of reaction
1 3.66×10-18 1 1.83×10-7 Very small
2 2.00×10-14 1 1.00×10-3 Fairly small
3 1.61×10-11 1 0.805 Intermediate
4 2.48×10-9 1 1.24×102 Intermediate
5 2.47×10-7 1 1.24×104 Close to completion
6 1.67×10-5 1 8.35×106 Completion
7 3.89×10-4 1 1.95×107 Completion
8 4.47×10-3 1 2.24×108 Completion
9 4.36×10-2 1 2.18×109 Completion
10 0.314 0.998 1.57×1010 Completion
11 0.820 0.980 4.02×1010 Completion
12 0.979 0.834 4.09×1010 Completion
13 0.998 0.334 1.67×1010 Completion

First, let’s consider the situation at pH 2.  The $$\mathrm{α_{Ca^{2+}}}$$ value is 1, which means that essentially none of the Ca2+ is complexed with the hydroxide ion.  This makes sense since there is a very low level of hydroxide ion at pH 2 ([OH-] = 10-12) and because the Kf value for calcium complexation with hydroxide is not that large.  The complexation of calcium by hydroxide has no significant effect on the system at this pH.  As the pH becomes more basic, notice how $$\mathrm{α_{Ca^{2+}}}$$ eventually falls below 1.  This means that some of the calcium ion will complex with hydroxide.  But if we examine the overall conditional constant, we also see that there is so much E4– available at the more basic pH values, that complexation with hydroxide is never sufficient enough to overcome the complexation of Ca2+ with the E4–.  It would take a much larger Kf value for complexation of Ca2+ with hydroxide for this reaction to compete with the reaction with E4–.

Calcium ion in tap water forms an insoluble precipitate with soap molecules and prevents the formation of lots of suds.  Because it’s hard to get suds when a high concentration of calcium ion is present, the water is referred to as “hard water”.  The classic procedure for analyzing the calcium concentration in hard water is to perform a titration with EDTA.  The solution is buffered at a pH of 10 to ensure that there is complete complexation of the calcium with the EDTA.  The conditional constants in the table above show the reason why a pH of 10 is used.

One last thing we need to consider is how we would handle a metal complex in which there were multiple formation constants.  For example, if we look up the complexation of Cd2+ with hydroxide, we see that there are four steps in the process and that the Kf values are larger than the one with Ca2+.  If we had substituted Cd2+ for Ca2+ in the problem above, the competing complexation of Cd2+ with hydroxide might have had more of an influence on the complexation of Cd2+ with E4–.  Of course, we also need to examine the complexation of Cd2+ with E4–, which has a Kf value of 3.16×1016 from the table.

The relevant equilibria for Cd2+ in this case are as follows:

\begin{align} &\ce{Cd^2+ + OH- \leftrightarrow Cd(OH)+} &&\mathrm{K_{f1}}\\ &\ce{Cd(OH)+ + OH- \leftrightarrow Cd(OH)2} &&\mathrm{K_{f2}}\\ &\ce{Cd(OH)2 + OH- \leftrightarrow Cd(OH)3-} &&\mathrm{K_{f3}}\\ &\ce{Cd(OH)3- + OH- \leftrightarrow Cd(OH)4^2-} &&\mathrm{K_{f4}} \end{align}

The evaluation of $$\mathrm{α_{Cd^{2+}}}$$ would involve the initial equation shown below:

$\mathrm{α_{Cd^{2+}} =\dfrac{[Cd^{2+}]}{[Cd(OH)_4^{2-}] + [Cd(OH)_3^-] + [Cd(OH)_2] + [Cd(OH)^+] + [Cd^{2+}]}}$

Taking the reciprocal leads to the following terms:

$\mathrm{ \dfrac{1}{α_{Cd^{2+}}} =\dfrac{[Cd(OH)_4^{2-}]}{[Cd^{2+}]} + \dfrac{[Cd(OH)_3^-]}{[Cd^{2+}]} + \dfrac{[Cd(OH)_2]}{[Cd^{2+}]} + \dfrac{[Cd(OH)^+]}{[Cd^{2+}]} + \dfrac{[Cd^{2+}]}{[Cd^{2+}]} }$

Using the Kf expressions for the complexation of Cd2+ with hydroxide, each ratio can be evaluated in terms of Kf values and [OH-], leading to the following equation.

$\mathrm{\dfrac{1}{α_{Cd^{2+}}} =K_{f1} K_{f2} K_{f3} K_{f4}[OH^-]^4 + K_{f1} K_{f2} K_{f3}[OH^-]^3 + K_{f1} K_{f2}[OH^-]^2 + K_{f1}[OH^-] + 1}$

$\mathrm{(K_f)(α_{Cd^{2+}})(α_{E^{4-}}) =\dfrac{[CdE^{2-}]}{[Cd]_{TOT}[E]_{TOT}}}$

Evaluation of a similar set of conditional constants over the entire pH range for the complexation of Cd2+ with E4- leads to the set of data in Table 8.

Table 8.  Conditional constants for the complexation of Cd2+ with E4-.

pH $$\mathrm{α_{E^{4-}}}$$ $$\mathrm{α_{Cd^{2+}}}$$ $$\mathrm{(K_f)(α_{Cd^{2+}})(α_{E^{4-}})}$$ Extent of reaction
1 3.66×10-18 1 1.20×10-1 Intermediate
2 2.00×10-14 1 6.32×102 Intermediate
3 1.61×10-11 1 5.09×105 Close to completion
4 2.48×10-9 1 7.84×107 Completion
5 2.47×10-7 1 7.81×109 Completion
6 1.67×10-5 1 5.28×1011 Completion
7 3.89×10-4 0.998 1.23×1013 Completion
8 4.47×10-3 0.980 1.38×1014 Completion
9 4.36×10-2 0.830 1.14×1015 Completion
10 0.314 0.284 2.81×1015 Completion
11 0.820 1.09×10-2 2.82×1014 Completion
12 0.979 2.84×10-5 8.79×1011 Completion
13 0.998 8.30×10-9 2.62×108 Completion

If we compare this data to that for Ca2+, we see that a much higher proportion of the Cd2+ is complexed with hydroxide ion at the more basic pH values.  The complexation with hydroxide is sufficient enough at pH 12 and 13 to significantly lower the conditional constant compared to the maximum at pH 10.  Nevertheless, the complexation of Cd2+ by the EDTA is still complete at pH 12 and 13 because of such a high formation constant.

What we see for a species like Cd2+ is some optimum pH for complexation with EDTA.  At low pH, protonation of the EDTA reduces the extent of complexation.  At high pH, complexation of the Cd2+ with hydroxide competes with the EDTA to some extent.  If we wanted to perform an analysis of Cd2+ using EDTA, we would buffer the solution at the pH that produces the maximum conditional constant, which is at a pH of 10.