4. Complexation Equilibria

A complex, as was noted previously, is a central atom or ion coordinated by ligands. Very often a complex with several ligands is formed in a stepwise manner, one ligand at a time, in a series of equilibrium reactions. Methods exist to measure the equilibrium constants and tables of data exist. In a solution then, all complexes may be present at various concentrations, depending on the equilibrium constant values. Determining the composition of such solutions requires solution of several simultaneous equilibria.

Complexation may dramatically affect the solubility of a metal compound. This can be employed to good effect in leaching and other aspects of hydro metallurgy. Treatment of complexation equilibria and effects on metal compound solubility will be introduced here.

Complexes may be formed with just one type of ligand, or with different types. For example,

\[\ce{Ni^{2+}(aq) + NH3(aq) = [Ni(NH3)]^{2+}}\tag{356}\]

\[\ce{[Ni(NH3)]^{2+}(aq) + NH3(aq) = [Ni(NH3)2]^{2+}}\tag{357}\]

\[\ce{[Ni(NH3)2]^{2+}(aq) + NH3(aq) = [Ni(NH3)3]^{2+}}\tag{358}\]

\[\ce{[Ni(NH3)3]^{2+}(aq) + NH3(aq) = [Ni(NH3)4]^{2+}}\tag{359}\]

\[\ce{[Ni(NH3)4]^{2+}(aq) + NH3(aq) = [Ni(NH3)5]^{2+}}\tag{360}\]

\[\ce{[Ni(NH3)5]^{2+}(aq) + NH3(aq) = [Ni(NH3)6]^{2+}}\tag{361}\]

(The Ni⁺² is 6-coordinate throughout. Water occupies coordination sites that NH₃ does not.) Each reaction adds one more NH₃ ligand so these are stepwise reactions. The equilibrium constants are called stepwise formation constants, i.e.

\[\ce{K_1 = \frac{[Ni(NH3)]^{2+}}{[Ni^{2+}][NH3]}}\tag{362}\]

\[\ce{K_2 = \frac{[Ni(NH3)2]^{2+}}{[Ni(NH3)]^{2+}[NH3]}}\tag{363}\]

\[\ce{K_i = \frac{[Ni(NH3)_i]^{2+}}{[Ni(NH3)_{i-1}]^{2+}[NH3]}}\tag{364}\]

The formation reactions can also be written in cumulative form:

\[\ce{Ni^{2+} + NH3 = [Ni(NH3)]^{2+}}\tag{365}\]

\[\ce{Ni^{2+} + 2NH3 = [Ni(NH3)2]^{2+}}\tag{366}\]

\[\ce{Ni^{2+} + 3NH3 = [Ni(NH3)3]^{2+} \;\text{etc.}}\tag{367}\]

For reactions written this way the equilibrium constants are called cumulative formation constants:

\[\ce{\beta_1 = \frac{[Ni(NH3)]^{2+}}{[Ni^{2+}][NH3]}}\tag{368}\]

\[\ce{\beta_2 = \frac{[Ni(NH3)2]^{2+}}{[Ni^{2+}][NH3]^2}}\tag{369}\]

\[\ce{\beta_3 = \frac{[Ni(NH3)3]^{2+}}{[Ni^{2+}][NH3]^3} \;\text{etc.}}\tag{370}\]

\[\ce{\beta_i = \frac{[Ni(NH3)_i]^{2+}}{[Ni^{2+}][NH3]^i}}\tag{371}\]

It can be easily shown that,

\[\ce{\beta_i = K_1 \times K_2 \times \dots \times K_i}\tag{372}\]

For instance, adding reaction (356) to reaction (358):

\[\ce{Ni^{2+}(aq) + NH3(aq) = [Ni(NH3)]^{2+} \qquad K_1}\tag{373}\]

\[\ce{[Ni(NH3)]^{2+}(aq) + NH3(aq) = [Ni(NH3)2]^{2+} \qquad K_2}\tag{374}\]

\[\ce{[Ni(NH3)2]^{2+}(aq) + NH3(aq) = [Ni(NH3)3]^{2+} \qquad K_3}\tag{375}\]

\[\ce{Ni^{2+} + 3NH3 = [Ni(NH3)3]^{2+} \qquad \beta_3 = K_1 K_2 K_3}\tag{376}\]

Some formation constants are shown in the table below. Finally, it can be seen that,

\[\ce{\log \beta_i = \log K_1 + \log K_2 + \dots + \log K_i}\tag{377}\]

1. Effects of Complexation on Solubility of Minerals.

The effects of complexation can sometimes be estimated by assuming that the most stable complex is the only one that forms. This is not strictly correct, but it greatly simplifies the calculations and may afford a quick estimate of how complexation affects solubility of minerals. The crucial consideration is the size of the cumulative formation constants. (More thorough speciation calculation procedures are illustrated in the following section.)

Consider the example of CuO solubility in ammonia solution and buffered ammonia/ammonium solutions. (This example is treated in more detail in the following section. The results here can be compared with the speciation calculations there.) The cumulative formation constants (as log βi) for [Cu(NH₃)_i]⁺² are: 4.3, 7.9, 10.9, 13.1 and 12.6 (i = 1-5, respectively). The [Cu(NH₃)₄]⁺² complex is the most stable, and the gap between β₄ and β₃ is 2.2, i.e. the tetraammine complex is >100 times more stable than the triammine complex. The pentaammine complex is somewhat less stable. Assume that the main complex formed is [Cu(NH₃)₄]⁺².

The two reactions of interest are dissolution of CuO in an ammonia solution (1 M), and dissolution of CuO in a buffered ammonia (0.5 M) plus ammonium sulfate (0.25 M) solution. (i) Determine the equilibrium constants for the reactions, (ii) the solubility of CuO (expressed as mol/L Cu(II) in solution) in these solutions and (iii) estimate the pH of each solution.

In the first case the reaction is:

\[\ce{CuO(s) + 4NH3 + H2O = [Cu(NH3)4]^{2+} + 2OH^-}\tag{378}\]

First we need an equilibrium constant. This can be obtained in two ways. The ΔG°_f for the reaction can be calculated if we have suitable ΔG°_f data for the chemical

Table 14. Stepwise formation constants for selected complexes at 25°C (unless otherwise noted). In some cases log10 of the cumulative formation constants are listed (βi).
Ligand Metal ion log K1 log K2 log K3 log K4 log K5 log K6
Ammonia Ag+ 3.37 3.84
NH3 Co+2 1a 1.99 1.51 0.93 0.64 0.06 -0.74
Co+3 1a,2 7.3 6.7 6.1 5.6 5.05 4.41
Cu+ 1b,2 5.93 4.93
Cu+2 4.30 3.60 3.0 2.2 -0.5
Ni+2 2.36 1.90 1.55 1.23 0.85 0.42
Zn+2 1a 2.18 2.25 2.31 1.96
Chloride Ag+ 3.22 1.78
Cl- Au+3 2 β4 26.0
Bi+3 3 2.43 2.0 1.35 0.43
Cu+ 2 3.5 1.44
Fe+3 1.48 0.65 -1.0
Pb+2 1.10 1.16 -0.40 -1.05
Zn+2 0.43 0.18 -0.08 -0.33
Cyanide Ag+ 4 β2 19.85
CN- Au+ 2 β2 36.6
Cu+ 2 4 β2 24.0 4.59 1.70
Fe+2 β5 27.4 8.0
Zn+2 5.4 5.7 4.9 3.6
Hydroxide Al+3 8.98 β4 32.43
OH- Cu+2 6.66
Fe+2 4.5
Fe+3 10.95 10.74
Ni+2 3.08
Pb+2 6.9 4 β3 13.95
Zn+2 4.3 4 β3 14.23
Sulfate Cu+2 2.36
SO42- Fe+2 2.30
Fe+3 4.04 1.34
UO22+ 3.14 0.94
Zn+2 2.38
Carbonate UO22+ β2 15.04 3.61
CO32-

1. (a) At 30°C. (b). At 18°C.
2. Not stable in aqueous solution (although Cu+ can exist at low concentrations).
3. Only in very strongly acidic solution; BiO⁺ predominant otherwise.
4. Insoluble; see table of Ksp data.

Note: The values of formation constants depend on ionic strength, which is a measure of the effects of ionic charge and concentration. Suffice it to say for now that formation constants vary somewhat with concentration of ions in solution. Data in the tables may involve differing ionic strength solutions.

species. A ΔG°_f for [Cu(NH₃)₄]⁺² might not be easy to find. Data are readily available for the other species. We can calculate ΔG°_f from the formation reaction:

\[\ce{Cu^{2+}(aq) + 4NH3(aq) = [Cu(NH3)4]^{2+}\qquad \beta_4 = 10^{13.1} = 1.259 \times 10^{13}}\tag{379}\]

\[\Delta G^\circ= -RT \ln \beta_4 = -8.314 \times 298.15 \times \ln(1.259 \times 10^{13}) = -74{,}771\ \text{J·mol}^{-1} = -74.77\ \text{kJ·mol}^{-1}\tag{380}\]

Other data (in kJ/mol):

ΔG°_f (OH-) = -157.24
ΔG°_f (H2O) = -237.15
ΔG°_f (NH3 aq) = -26.5
ΔG°_f (CuO) = -129.7
ΔG°_f (Cu+2) = 65.49

For the complexation reaction (169),

\[\Delta G^\circ
= \Delta G_f^\circ\!\left([Cu(NH3)_4]^{2+}\right)
– \Delta G_f^\circ(Cu^{2+})
– 4\,\Delta G_f^\circ(NH_3)\tag{381}\]

\[\Delta G_f^\circ\!\left([Cu(NH3)_4]^{2+}\right)
= -74.77 + 65.49 + 4(-26.5)
= -115.28\ \text{kJ·mol}^{-1}\tag{382}\]

For the dissolution of CuO in ammonia solution:

\[\Delta G^\circ
= 2\,\Delta G_f^\circ(\ce{OH^-})
+ \Delta G_f^\circ\!\left([Cu(NH3)_4]^{2+}\right)
– \Delta G_f^\circ(\ce{H2O})
– 4\,\Delta G_f^\circ(\ce{NH3})
– \Delta G_f^\circ(\ce{CuO})\tag{383}\]

\[
= 2(-157.24) + (-115.28) – (-237.15) – 4(-26.5) – (-129.7)
= 43.1\ \text{kJ·mol}^{-1}\tag{384}
\]

\[\ce{K = e^{-\Delta G^\circ/(RT)}
= e^{-83600/(8.314 \times 298.15)}
= 2.8 \times 10^{-8}}\tag{385}\]

Another way to find the equilibrium constant is to recognize that the reaction is made up of two reactions; first CuO dissolving to form Cu⁺² and OH^–, then Cu⁺² complexing with ammonia:

\[\ce{CuO + H2O = Cu^{2+} + 2OH^- \qquad K_{sp} = 2.2 \times 10^{-21}}\tag{386}\]

\[\ce{Cu^{2+} + 4NH3 = [Cu(NH3)4]^{2+} \qquad \beta_4 = 1.259 \times 10^{13}}\tag{387}\]

\[\ce{CuO + H2O + 4NH3 = [Cu(NH3)4]^{2+} + 2OH^-}\tag{388}\]

\[
K = K_{sp}\,\beta_4
= (2.2 \times 10^{-21})(1.259 \times 10^{13})
= 2.8 \times 10^{-8}\tag{389}
\]

The solubility of CuO can be determined now. Let x = the concentration of Cu(II) as [[Cu(NH₃)₄]⁺²]. The final equilibrium concentrations are given by:

\[\ce{CuO + H2O + 4NH3 = [Cu(NH3)4]^{2+} + 2OH^-}\tag{390}\]

Initial 1
Final 1 – 4x x 2x

\[\frac{x(2x)^2}{(1 – 4x)^4}
= K
= 2.8 \times 10^{-8}\tag{391}\]

Solve for x = 0.00189 M (0.12 g Cu/L)

\[\text{pH}
= 14 – \log[\ce{OH^-}]
= 14 – \log(2x)
= 11.6\tag{391}\]

The assumption that the only complex that forms is [Cu(NH₃)₄]⁺² is probably not very good in this case since there is much left over ammonia, and [Cu(NH₃)₅]⁺² is likely to be significant.

The second leaching reaction is:

\[\ce{CuO(s) + 2NH3 + 2NH4^+ = [Cu(NH3)4]^{2+} + H2O}\tag{392}\]

Additional data: ΔG°_f (NH₄⁺) = -79.31 kJ/mol

The equilibrium constant can be found from:

\[\Delta G^\circ
= \Delta G_f^\circ(\ce{H2O})
+ \Delta G_f^\circ\!\left([Cu(NH3)_4]^{2+}\right)
– 2\,\Delta G_f^\circ(\ce{NH4^+})
– 2\,\Delta G_f^\circ(\ce{NH3})
– \Delta G_f^\circ(\ce{CuO})\tag{393}\]

\[
\Delta G^\circ
= -237.15 + (-115.28) – 2(-79.31) – 2(-26.5) – (-129.7)
= -11.11\ \text{kJ·mol}^{-1}\tag{394}
\]

\[\ce{K
= e^{11{,}110/(8.314 \times 298.15)}
= 88.4}\tag{395}\]

The alternative method is to break the overall reaction into parts for which equilibrium constants are known: CuO dissolves as per Ksp, NH₄⁺ reacts with OH^– to form water and NH₃ and finally, NH₃ complexes Cu⁺².

\[\ce{CuO + H2O = Cu^{2+} + 2OH^- \qquad K_{sp} = 2.2 \times 10^{-21}}\tag{396}\]

\[\ce{2NH4^+ = 2NH3 + 2H^+ \qquad K_{a2} = (5.75 \times 10^{-10})^2 = 3.306 \times 10^{-19}}\tag{397}\]

\[\ce{2H^+ + 2OH^- = 2H2O \qquad 1/K_w^2 = 1 \times 10^{28}}\tag{398}\]

\[\ce{Cu^{2+} + 4NH3 = [Cu(NH3)4]^{2+} \qquad \beta_4 = 1.259 \times 10^{13}}\tag{399}\]

The overall reaction and equilibrium constant are:

\[\ce{CuO + 2NH3 + 2NH4^+ = [Cu(NH3)4]^{2+} + H2O}\tag{400}\]

\[K = \frac{K_{sp}\,K_{a2}\,\beta_4}{K_w^2}= 91.6\tag{401}\]

(The discrepancy is minor. A change in ΔG°_f (NH₄⁺) from -79.31 to -79.27 would make the values identical.) Now the equilibrium [Cu(II)] can be estimated. Let the x = [Cu(II)]. The initial concentration of ammonium is 2 x [(NH₄)₂SO₄] = 0.5 M.

\[\ce{CuO + 2NH3 + 2NH4^+ = [Cu(NH3)4]^{2+} + H2O}\tag{402}\]

Initial 0.5 0.5
Final 0.5 – 2x 0.5 – 2x x

\[\frac{x}{(0.5 – 2x)^4}= K= 88.4\tag{403}\]

\[x = 0.149\ \text{M}\quad (9.45\ \text{g Cu·L}^{-1})\tag{404}\]

This is considerably greater than we had with ammonia alone. The pH is given by:

\[
\text{pH}
= \text{p}K_a(\ce{NH4^+})
+ \log\!\left(\frac{[\ce{NH3}]}{[\ce{NH4^+}]}\right)
\tag{405}
\]

\[
= -\log(5.75 \times 10^{-10})
+ \log\!\left(\frac{0.5 – 2x}{0.5 – 2x}\right)
= 9.24
\tag{406}
\]

Note that the pH in this case is much lower than in the case with ammonia alone (11.6). The solubility of CuO is greatly enhanced by using a buffered solution of ammonia and ammonium. Without NH₄⁺, CuO dissolution increases the [OH-] and raises the pH. The very low Ksp for CuO prevents this from occurring to a significant degree. When we add ammonium, we now have a weak acid that can neutralize that base, keeping the pH lower. Note too that now we have combined the effects of acid-base, solubility and complexation equilibria. Further elaboration is provided in the speciation calculation for the Cu⁺²-NH₃ system below.

2. Speciation Calculations.

As previously we need one independent equation for every species to be accounted for. The principles have already been developed in previous sections.

I. A Copper-Ammonia Speciation Example.

Ammonia solution has been used for nickel and cobalt leaching, and considered for copper. Cupric ion forms five complexes with ammonia as shown in Table 14. The solubility of Cu(II) and other metal ions in ammonia solutions is limited by their tendency to precipitate as metal oxides or hydroxides. Ammonia in solution is, of course, associated with ammonium. Note again then that this ties together all three of the major types of equilibria we have considered in this review. We will assume that volatilization of ammonia,

\[\ce{NH3(aq) = NH3(g)}\tag{407}\]

is minor and can be ignored, though at higher temperatures this would not be a valid assumption.

Using speciation calculations we will determine the solubility of CuO in water, in ammonia solution (1 M) and in buffered ammonia/ammonium solutions (1 M total). This is the same problem considered in the preceding section. Now we also consider all the other complexes as well. We have the following 11 species to consider: Cu⁺², [Cu(NH₃)_i]⁺² (i = 1-5), NH₃, NH₄⁺, H⁺, OH^– and CuO. In calculating the maximum solubility, however, there won’t be any CuO in the system. For the sake of simplicity let M = Cu⁺² and L = NH₃ in this example.

For buffered solutions we will specify the pH, which takes care of [H+], and [OH-] is related to [H+] by Kw. We can write 5 complexation equilibria:

\[\ce{M + iL = ML_i}\tag{408}\]

\[
\beta_i = \frac{[\ce{ML_i}]}{[\ce{M}][\ce{L}]^i}
\tag{409}
\]

\[
[\ce{ML_i}] = \beta_i[\ce{M}][\ce{L}]^i
\tag{410}
\]

The concentration of Cu⁺² ion itself is governed by Ksp:

\[\ce{MO + H2O = M^{2+} + 2OH^-}
\qquad (\text{where } M = \ce{Cu^{2+}})\tag{411}\]

\[
K_{sp} = [\ce{M^{2+}}][\ce{OH^-}]^2
= 2.24 \times 10^{-21}
\tag{412}
\]

\[
[\ce{M^{2+}}]
= \frac{K_{sp}}{[\ce{OH^-}]^2}
\tag{413}
\]

Then [M] (uncomplexed Cu⁺²) can be found for a given pH. The [LH⁺] is given by:

\[
[\ce{LH^+}]
= \frac{[\ce{L}][\ce{H^+}]}{K_a}
\qquad
K_a = 5.75 \times 10^{-10}
\tag{414}
\]

Finally, we can write a mass balance on ammonia in all forms:

\[[\ce{L}]_T= [\ce{L}]+ [\ce{LH^+}]+ [\ce{ML}]+ 2[\ce{ML_2}]+ \dots+ 5[\ce{ML_5}]\tag{415}\]

Taking into account [H+] = 10^-pH gives:

\[
[\ce{L}]_T
= \frac{[\ce{L}][\ce{H^+}]}{K_a}
+ [\ce{L}]
+ \sum_{i=1}^{5} i\,\beta_i\,[\ce{M}][\ce{L}]^i
\tag{416}
\]

\[
[\ce{L}]_T
= \left(\frac{10^{-\text{pH}}}{K_a}+1\right)[\ce{L}]
+ [\ce{M}] \sum_{i=1}^{5} i\,\beta_i\,[\ce{L}]^i
\tag{417}
\]

We can solve this equation for [L] using approximate methods. Once we have this, the system is fully determined. The [MLi] can be found from equation (410). The sum + [M] is the total dissolved M(II) in all forms. This takes care of the speciation for NH₃/NH₄⁺ solutions at set pH values. A plot of the concentrations of the various species and the sum total [Cu(II)] is shown in the figure below.

Figure 4. Solubility of CuO in buffered ammonia solutions (1 M total) as a function of pH (25°C). Free [Cu+2] and [Cu(NH3)i]+2 (i = 1 , 2) are minor and omitted.

The plot shows the maximum possible concentration of Cu(II) in the solutions containing a total of 1 M NH₃ (in all its forms) at 25°C as a function of pH. Clearly, near the NH₄⁺ buffer point (pKa = 9.24) we obtain the maximum solubility (0.151 M, 9.6 g Cu/L). Although the total solubility curve (heavy line in the figure) looks symmetrical, it is not quite. The maximum in solubility is actually at about pH 9.22, due to small contributions from other complexes. However, by far the main reaction is:

\[\ce{CuO(s) + 2NH4^+ + 2NH3 = [Cu(NH3)4]^{2+} + 2H2O}\tag{418}\]

If this were the only contributor the maximum in solubility would be precisely at pH = 9.24. Based on this stoichiommetry, a 1:1 mole ratio of NH₄⁺ and NH₃ is needed. At pH less than about 9.24 we would have an excess of NH₄⁺ over NH₃, and not enough NH₃ to optimally complex the Cu+2. Likewise at pH greater than about 9.24 we have not got enough NH₄⁺ to neutralize the base that would arise from CuO dissolution. Hence the peak is determined by the reaction stoichiommetry. For a reaction of the type,

\[
\ce{MO(s) + 2LH^+ + 4L -> [ML_6]^{2+} + H2O}
\]

and if β6 is substantially larger than any of the lower βi, then the peak in solubility would occur at a [NH₃]/[NH₄⁺] ratio of closer to 4/2, or pH 9.54 (= pKa + log2). Finally, if the βi values are not very different from each other (e.g. as for higher Ni⁺²-NH₃ complexes), then significant contributions to the solubility will arise from more than one complex, and this too will shift the peak pH.

Incidentally, the equilibrium [Cu(II)] is very close to the value we obtained when we assumed that [Cu(NH₃)₄]⁺² was the only complex formed (0.149 M), and with equimolar concentrations of ammonia and ammonium. This is the most stable complex, and the others contribute relatively little to the solubility. Even [Cu(NH₃)₅]⁺² contributes very little. There are two reasons. One is that much of the ammonia is used up to form [Cu(NH₃)₄]⁺². The amount available to form the pentaammine is relatively small, and it is somewhat less stable. The other is due to the stoichiommetries of the complexes. The optimum mole ratio of NH₃:NH₄⁺ is 2:2 for [Cu(NH₃)₄]⁺², but 3:2 for [Cu(NH₃)₅]⁺². This is why the pentaammine complex concentration peaks at pH > 9.24.

Finally, we need to calculate the maximum solubility of Cu⁺² in a solution of 1 M NH₃. As we might expect from the preceding discussion, this will be markedly less due to the lack of NH₄⁺. This is a more difficult problem. We only know the total [NH₃]. The pH is not known. Hence neither is [Cu⁺²] nor [NH₄⁺]. We have the βi expressions and we can write:

\[
[\ce{M^{2+}}]
= \frac{K_{sp}}{[\ce{OH^-}]^2}
= \frac{K_{sp}[\ce{H^+}]^2}{K_w^2}
\tag{419}
\]

The mass balance for [L]T is as above:

\[
[\ce{L}]_T
= \frac{[\ce{L}][\ce{H^+}]}{K_a}
+ [\ce{L}]
+ [\ce{M}] \sum_{i=1}^{5} i\,\beta_i\,[\ce{L}]^i
\tag{420}
\]

\[
[\ce{L}]_T
= \frac{[\ce{L}][\ce{H^+}]}{K_a}
+ [\ce{L}]
+ \frac{K_{sp}}{K_w^2}[\ce{H^+}]^2
\sum_{i=1}^{5} i\,\beta_i\,[\ce{L}]^i
\tag{421}
\]

This is a quadratic and can be solved for [H+] in terms of [L]:

\[
[\ce{H^+}]
= \frac{
-\frac{[\ce{L}]}{K_a}
+ \left(
\frac{[\ce{L}]^2}{K_a^2}
– \frac{4K_{sp}}{K_w^2}
\left(
\sum_{i=1}^{5} i\,\beta_i\,[\ce{L}]^i
\right)
([\ce{L}] – [\ce{L}]_T)
\right)^{0.5}
}{
\frac{2K_{sp}}{K_w^2}
\sum_{i=1}^{5} i\,\beta_i\,[\ce{L}]^i
}
\tag{422}
\]

Next we need one more equation. Charge balance is the only option:

\[
[\ce{H^+}]
+ [\ce{LH^+}]
+ 2[\ce{M}]
+ 2[\ce{M}] \sum_{i=1}^{5} \beta_i[\ce{L}]^i
= [\ce{OH^-}]
= \frac{K_w}{[\ce{H^+}]}
\tag{423}
\]

The 2 is because the Cu(II) species have a +2 charge. (This alone tells us that the total dissolved Cu(II) will be low). Incorporating [LH+] in terms of [L], [H+] and Ka and substituting Ksp [H+]²/Kw₂ for [M] gives:

\[
[\ce{H^+}]
+ \frac{[\ce{L}][\ce{H^+}]}{K_a}
+ \frac{2K_{sp}}{K_w^2}[\ce{H^+}]^2
\left(1 + \sum_{i=1}^{5} \beta_i[\ce{L}^{i-}]\right)
– \frac{K_w}{[\ce{H^+}]}
= 0
\tag{424}
\]

The procedure to solve this is to set a provisional value for [L] (it will be close to 1 since little Cu(II) goes into solution), calculate [H+], then calculate the function for equation (424). Vary [L] until it goes to zero. For 1 M NH₃ this yields,

[NH₃] = 0.9918 M (very close to 1)
[H⁺] = 1.82 x 10-12 (pH = 11.74)
[Cu(NH₃)_i⁺²] ( i = 0-5) = 1.19 x 10^-3 M = 0.076 g Cu/L

This differs significantly from the result obtained previously (0.12 g/L) when it was assumed that the only complex formed was [Cu(NH₃)₄]⁺². The discrepancy arises from ignoring the other equilibria in the system. In this case the free ammonia is substantial, so complexes other than [Cu(NH₃)₄]⁺² are significant.

The solubility of CuO in water without any ammonia is given by:

\[
K_{sp}
= x(2x)^2
= 4x^3
= 2.24 \times 10^{-21}
\tag{425}
\]

\[
x = 8.2 \times 10^{-8}\ \text{M}
= [\ce{Cu^{2+}}]
= 5 \times 10^{-6}\ \text{g Cu·L}^{-1}
\tag{426}
\]

Obviously this is negligible.

II. Example of AgCl Solubility as a Function of Chloride Concentration.

Silver is fairly rare. It may occur in concert with other metals, such as copper, nickel, cobalt, etc. If a chloride medium is used for leaching, Ag(I) may form silver chloride, AgCl. Silver chloride is sparingly soluble in water (Ksp = 1.78 x 10^-10). It is observed that when the chloride concentration in solution (e.g. as NaCl, MgCl₂, etc.) is increased, initially the solubility of AgCl decreases, then gradually increases at high chloride levels. The first effect we would expect is based on,

\[\ce{AgCl(s) = Ag^+ + Cl^-}\tag{427}\]

and Le Chatelier’s principle; increase [Cl-] and [Ag+] in solution will drop. The latter effect is due to complexation. The aqueous silver-chloride complexes are: AgCl_(aq) and [AgCl₂]^–. Note: We have to different forms or phases in which the compound AgCl exists; solid and solution. The first chloride complex of Ag⁺ is AgCl_(aq). This means we can write a reaction:

\[\ce{AgCl(s) = AgCl(aq)}\tag{428}\]

Interestingly the equilibrium constant for this reaction is,

\[
K = [\ce{AgCl}]
\tag{429}
\]

meaning that concentration of AgCl itself in solution is a constant.

Consider the solubility of AgCl in solution as a function of solution [Cl^–], e.g. as NaCl. The species we have to account for in solution are: Ag⁺, AgCl aq, [AgCl₂]^–, Cl^– and Na⁺. We can specify the Cl^– concentration and see how AgCl_(s) solubility varies. The following are the relevant reactions:

\[\ce{AgCl(s) = Ag^+ + Cl^-}\tag{430}\]

\[
K_{sp}
= [\ce{Ag^+}][\ce{Cl^-}]
= 1.78 \times 10^{-10}
\tag{431}
\]

\[\ce{Ag^+ + Cl^- = AgCl(aq)}\tag{432}\]

\[
\beta_1
= \frac{[\ce{AgCl(aq)}]}{[\ce{Ag^+}][\ce{Cl^-}]}
= 1660
\tag{433}
\]

\[\ce{Ag^+ + 2Cl^- = AgCl2^-}\tag{434}\]

\[
\beta_2
= \frac{[\ce{AgCl2^-}]}{[\ce{Ag^+}][\ce{Cl^-}]^2}
= 1.00 \times 10^{5}
\tag{435}
\]

We have five species and three equations. One species (Cl^–) will be specified arbitrarily. That leaves four unknowns. But, sodium does not enter into any equation above. Sodium is simply a counterion, and we would be concerned about it only if we needed to use a charge balance equation. Having three undetermined species then and specifying [Cl^–] leaves us three unknowns, and we have three equations. Having a value for [Cl^–] means we know [Ag⁺]:

\[
\frac{[\ce{Ag^+}]}{[\ce{Cl^-}]}
= K_{sp}
\tag{436}
\]

This means that we are specifying the free, uncomplexed [Cl^–] at specific values. Knowing [Ag⁺] allows us to calculate [AgCl_(aq)] and [AgCl₂^–]:

\[
[\ce{AgCl(aq)}]
= \beta_1[\ce{Ag^+}][\ce{Cl^-}]
\tag{437}
\]

\[
[\ce{AgCl2^-}]
= \beta_2[\ce{Ag^+}][\ce{Cl^-}]^2
\tag{438}
\]

Chloride concentrations can be specified over a wide range of added NaCl. We need to calculate the added NaCl. The total chloride in solution is:

\[
[\ce{Cl^-}]_T
= [\ce{Cl^-}]
+ [\ce{AgCl(aq)}]
+ 2[\ce{AgCl2^-}]
\tag{439}
\]

The chloride due to AgCl dissolution to form the three solution species is:

\[
[\ce{Cl^-}]_{\text{from AgCl}}
= [\ce{Ag^+}]
+ [\ce{AgCl(aq)}]
+ [\ce{AgCl2^-}]
\tag{440}
\]

One chloride comes from every AgCl that dissociates to form Ag⁺ + Cl^–, one from every AgCl_(aq) that dissolves and one from every AgCl that forms AgCl₂^–. (All Ag⁺ entering solution brings along with it one Cl^– from AgCl_(s).) The added chloride then would be:

\[
[\ce{Cl^-}]_{\text{added}}
= [\ce{Cl^-}]_T
– [\ce{Cl^-}]_{\text{from AgCl}}
\tag{441}
\]

\[
[\ce{Cl^-}]_{\text{added}}
= [\ce{Cl^-}]
+ [\ce{AgCl2^-}]
– [\ce{Ag^+}]
= [\ce{NaCl}]_{\text{added}}
\tag{442}
\]

Then we can plot AgCl solubility and speciation as a function of added NaCl, including zero added, corresponding to AgCl in contact with water alone. Since the numbers are very small and span many orders of magnitude, it is more instructive to plot the log₁₀ of the concentrations versus log₁₀[NaCl]. This is shown in the figure below.

Below about 10^-6 M NaCl (log[NaCl] = -6) added chloride has no significant effect, as we might expect for such low levels. The [Ag⁺], [AgCl] and [AgCl₂^–] are virtually constant and Ag⁺ is by far the dominant species (therefore total [Ag(I)] and [Ag⁺] merge), and adequately approximated on the basis of the Ksp of AgCl alone. As added [NaCl] increases the [Ag⁺] goes down, as expected based on Ksp. A minimum in the solubility is seen at log[NaCl] = -2.5 (0.0032 M). Then the solubility of AgCl begins to increase due to formation of [AgCl₂]^–. In very strong chloride solution the solubility approaches 10^-4 M, or about 0.01 g/L. So at least partial leaching of Ag(I) in strong chloride solutions is a possibility.

Figure 6. Plots of concentrations of silver(I) species in solution as a function of concentration of added NaCl. Concentrations expressed as log10 values.