3. Sparingly Soluble Salts

\In Part 1 of the chemistry review the rules for solubility of salts were given. In this section we will quantify the solubility of salts using equilibrium constant concepts already developed.

1. The Solubility Product, Ksp

Any salt (a compound containing ions) that dissolves in water it dissociates into its constituent ions, so long as those ions persist in solution (O^2-, for instance, does not; it forms OH^– quantitatively). For example, silver chloride is an “insoluble” salt (by insoluble we mean sparingly soluble). The dissolution reactions can be written as:

\[ \ce{AgCl_{(s)} = Ag^+_{(aq)} + Cl^-_{(aq)}} \tag{260}\]

The equilibrium constant for this reaction is given by:

\[ K_{sp} = \ce{[Ag^+][Cl^-]} = 1.8 \times 10^{-10} \tag{261}\]

The activity of a pure solid is 1 and does not enter into the expression. This type of equilibrium constant is called the solubility product, hence Ksp. Some other examples are,

\[ \ce{[Fe2(SO4)3]\cdot5H2O_{(s)} = 2Fe^{3+}_{(aq)} + 3SO4^{2-}_{(aq)} + 5H2O_{(l)}} \tag{262}\]

\[ K_{sp} = \ce{[Fe^{3+}]^2[SO4^{2-}]^3} \tag{263}\]

\[ \ce{Zn2[Fe(CN)6]_{(s)} = 2Zn^{2+}_{(aq)} + [Fe(CN)6]^{4-}_{(aq)}} \tag{264}\]

\[ K_{sp} = \ce{[Zn^{2+}]^2[Fe(CN)6]^{4-}} = 2.1 \times 10^{-16} \tag{265}\]

The [Fe(CN)₆]₄^– complex is stable and persists in solution, hence Ksp refers to the dissolution of the salts into the indicated ions. A Ksp may be greater than 1 if a salt is quite soluble, e.g. NaCl.

1. Calculations using Ksp.

A compilation of Ksp data is provided in Table 13. Examples of the basic types of calculations that can be done are provided below.

I. Calculate the solubility of lead sulfate in water at 25°C.

PbSO₄ has a Ksp of 6.3 x 10^-7.

\[ \ce{PbSO4_{(s)} = Pb^{2+}_{(aq)} + SO4^{2-}_{(aq)}} \tag{266}\]

\[ K_{sp} = \ce{[Pb^{2+}][SO4^{2-}]} \tag{267}\]

Let [Pb⁺²] = x = [SO₄^2-]. Then,

\[ K_{sp} = x^2 \quad \text{and} \quad x = 7.94 \times 10^{-4}\ \text{M} \tag{268}\]

This is equal to only 0.165 g Pb⁺²/L. This makes sulfate medium inadequate for lead hydrometallurgy.

II. In order to remove residual lead from an aqueous solution, excess sulfate might be added (e.g. as Na2SO4). What will the concentration of lead be in solution if we add 2 M SO42-?

Addition of excess sulfate will act to drive the reaction toward the right. To start with we have 0.0008 M sulfate in solution, some of which will be removed as PbSO4,

\[ \ce{PbSO4_{(s)} = Pb^{2+}_{(aq)} + SO4^{2-}_{(aq)}} \tag{269}\]

Initial 0.0008 2.0008
Final 0.0008 – x 2.0008 – x

This results in a quadratic that is easy to solve, but since 0.0008 << 2 it is easier to

Table 13. Table of Ksp data at 25°C.
Compound Ksp Compound Ksp
Carbonates Oxides 3
Ag2CO3 8.1  10-12 Ag2O (= 2Ag+ + 2OH-) 3.8  10-16
CaCO3 (calcite) 1 4.5  10-9 CuO (= Cu+2 + 2OH-) 2.2 x 10-21
CaCO3 (aragonite) 6.0 x 10-9 Cu2O (= 2Cu+ + 2OH-) 4 x 10-30
CuCO3 2.3  10-10 HgO (= Hg+2 + 2OH-) 3.6  10-26
Cu2CO3(OH)2 1.66 x 10-34 NiO (= Ni+2 + 2OH-) 1.6 x 10-16
FeCO3 2.1  10-11 PbO 1 (yellow; = Pb+2 + 2OH-) 8 x 10-16
MgCO3 3.5  10-8 PbO (red; = Pb+2 + 2OH-) 5 x 10-16
PbCO3 7.4 x 10-14 ZnO (= Zn+2 + 2OH-) 3.7 x 10-17
ZnCO3 1.0  10-10 Sulfates
Chlorides Ag2SO4 1.5 x 10-5
AgCl 1.78 x 10-10 BaSO4 1.1 x 10-10
CuCl (= Cu+ + Cl-) 2 1.9 x 10-7 CaSO4 2.4 x 10-5
Hg2Cl2 (= Hg2+2 + 2Cl-) 1.2 x 10-18 CaSO4·2H2O 4.4 x 10-5
PbCl2 1.7 x 10-5 KFe3(SO4)2(OH)6 4 2.8 x 10-99
Cyanides PbSO4 6.3 x 10-7
AgCN 2.2 x 10-16 Sulfides 5
CuCN 3.0 x 10-20 Ag2S 8 x 10-51
Zn(CN)2 3.2 x 10-16 CdS 1 x 10-27
Fluorides CoS 1 5 x 10-22
CaF2 3.9 x 10-11 CuS 8 x 10-37
LiF 1.7 x 10-3 Cu2S 3 x 10-49
PbF2 3.6 x 10-8 FeS 8 x 10-19
Hydroxides HgS (black) 1 2 x 10-53
Ca(OH)2 6.5 x 10-6 MnS (green) 1 3 x 10-14
Mg(OH)2 7.1  10-12 NiS 1 1 x 10-21
Ni(OH)2 6 x 10-16 PbS 3 x 10-28
Cu(OH)2 4.8 x 10-20 ZnS 1 2 x 10-25
Fe(OH)3 1.6 x 10-39
Al(OH)3 ( crystalline form) 3 x 10-34
Co(OH)2 2 x 10-16
Co(OH)3 1 x 10-43
Pb(OH)2 1.2 x 10-20

1. Two or more different crystalline forms exist, each with its own Ksp. Generally the
   form most relevant in hydrometallurgy is listed.
2. Cu+ is highly unstable in air; forms Cu⁺².
3. All oxides: MO + H₂O = M⁺² + 2OH^–, M₂O + H₂O = 2M⁺ + 2OH^–, etc.
4. KFe₃(SO₄)₂(OH)₆, ammonium jarosite, is among the least soluble of the many
   M(I)Fe₃(SO₄)₂(OH)₆ jarosites. Ksp = [K+][Fe⁺³]³[SO₄^2-]²[OH-]⁶; solubility of the
   compound in water would be ~2 x 10^-9 M.
5. Ksp data for sulfides may have exhibit considerable variation among different
   sources; measurements are difficult. Calculated Ksp data based on ΔG°_f data may
   also suffer from large uncertainties in ΔG°_f for S^2-.

assume that [SO₄^2-] ≈ 2 and solve for [Pb⁺²]:

\[\ce{2y = 6.3 \times 10^{-7} \;\Rightarrow\; y = 3.15 \times 10^{-7}}\tag{270}\]

This is equal to 6.5 x 10^-5 g/L Pb⁺². The error in this calculation as a result of our assumption can be seen to be negligible by recalculating Ksp:

\[\ce{x = 0.0008 – 3.15 \times 10^{-7} = 7.99685 \times 10^{-4}}\tag{271}\]

\[\ce{K_{sp} = 3.15 \times 10^{-7} \times (2.0008 – 7.99685 \times 10^{-4}) = 6.30 \times 10^{-7}}\tag{272}\]

Note: too that sulfate is a weak base, so some of it could hydrolyze and form HSO₄^–. Pb(HSO₄)₂ in turn also happens to be quite insoluble. Thus the actual solubility of PbSO₄ might differ from the simple value we have just calculated. However, since Kb is very small (9.8 x 10^-13) the extent of subsequent reactions will be small and will have a small effect on the solubility. However, if the pH was lower than ~4, then the effects of bisulfate involvement would start to become important. (pKa (HSO₄^–) = 1.99; at pH <4 [HSO₄^–]/[SO₄^2-] starts to exceed 1%.)

III. Hydrated lime is calcium hydroxide, Ca(OH)₂. It is insoluble and important as a base in hydrometallurgy; Ksp = 6.5 x 10^-6. What is the pH of a saturated Ca(OH)₂ solution?

Let the concentration of Ca(OH)₂ that dissolves = x, which also equals [Ca⁺²]. Two OH- go into solution per Ca⁺²:

\[\ce{Ca(OH)2 = Ca^{2+} + 2OH^-}\tag{273}\]

\[\ce{K_{sp} = x(2x)^2 = 4x^3 = 6.5 \times 10^{-6}}\tag{274}\]

\[\ce{x = 0.0118\ M;\ [OH^-] = 0.0236\ M}\tag{275}\]

\[\ce{pH = 14 – pOH = 12.4}\tag{276}\]

This is the maximum pH we can attain with Ca(OH)₂. If we need a higher pH then a different base would be needed (say NaOH).

IV. The magnitude of Ksp may sometimes create a fall sense of the solubility of a salt, unless the exponents are considered. Consider the example of KFe₃(SO₄)₂(OH)₆ for which Ksp = 2.8 × 10^-99. This is incredibly small. But the expression for Ksp in terms of [Fe⁺³] = x is:

\[\ce{K_{sp} = \left(\frac{1}{3}x\right)(x^3)\left(\frac{2}{3}x\right)^2(2x)^6 = 2.8 \times 10^{-99}}\tag{277}\]

Solving gives x = 5 x 10^-9 M Fe⁺³, which is on the order of Ksp 1/12.

2. The Special Case of Metal Oxides.

A metal oxide, by the above rules, should dissociate into a metal cation and oxide anions. However, O^2- does not survive in aqueous solution. It is completely converted to OH^–. Therefore we write:

\[\ce{MO_{(s)} = M^{2+}_{(aq)} + O^{2-}}\tag{278}\]

\[\ce{O^{2-} + H2O_{(l)} = 2OH^-_{(aq)}}\tag{279}\]

\[\ce{MO_{(s)} + H2O_{(l)} = M^{2+} + 2OH^- \quad K_{sp} = [M^{2+}][OH^-]^2}\tag{280}\]

Calculations are otherwise done in the same way as shown above.

3. Considerations for Metal hydroxides and Metal Oxides.

The solubility product for all metal oxides is defined as above. However, there are additional complexities. Many metal cations possess oxide and hydroxide salts. A good example is lime (CaO) and hydrated, or “slaked” lime, i.e. Ca(OH)₂.

\[\ce{CaO_{(s)} + H2O_{(l)} = Ca^{2+}_{(aq)} + 2OH^-_{(aq)}}\tag{281}\]

\[\ce{K_{sp} = [Ca^{2+}][OH^-]^2}\tag{282}\]

and,

\[\ce{Ca(OH)2_{(s)} = Ca^{2+}_{(aq)} + 2OH^-_{(aq)}}\tag{283}\]

\[\ce{K_{sp} = [Ca^{2+}][OH^-]^2}\tag{284}\]

The two expressions are the same. Therefore it is crucially important that the salt be indicated; the two Ksp values will differ. Tables of Ksp data may list Ca(OH)₂, but not CaO. Why? We can calculate Ksp for CaO from thermodynamic data. The standard free energy change (ΔG°) for the reaction is needed. For reaction (119),

\[\ce{\Delta G^\circ = 2\Delta G_f^\circ(OH^-) + \Delta G_f^\circ(Ca^{2+}) – \Delta G_f^\circ(H2O) – \Delta G_f^\circ(CaO)}\tag{285}\]

\[\ce{= 2 \times (-157.24) + (-553.58) – (-237.15) – (-604.03) = -26.88\ kJ/mol}\tag{286}\]

Then using,

\[\ce{\Delta G^\circ = -RT \ln K}\tag{287}\]

\[\ce{K = e^{-\Delta G^\circ / RT}}\tag{288}\]

\[\ce{K = e^{-1000\ J/kJ \times (-26.88\ kJ/mol)\,/\, (8.314\ J/(mol\ K) \times 298.15\ K)} = 5.12 \times 10^{4}}\tag{289}\]

This is a large number, but CaO s is only sparingly soluble. We have an apparent contradiction. The explanation is that while CaO would appear to be very soluble, Ca(OH)₂ is not; Ksp = 6.5 x 10^-6. Thus most of the CaO that would dissolve to form Ca⁺² and OH^– will reprecipitate as Ca(OH)₂. Hence CaO in water reacts to form Ca(OH)₂ s:

\[\ce{CaO + H2O = Ca^{2+} + 2OH^- \quad K = 5.12 \times 10^{4}}\tag{290}\]

\[\ce{Ca^{2+} + 2OH^- = Ca(OH)2_{(s)} \quad K’ = \frac{1}{K_{sp}} = 1.54 \times 10^{5}}\tag{291}\]

\[\ce{CaO + H2O = Ca(OH)2}\tag{292}\]

\[\ce{K” = K’ K_{sp} = 5.12 \times 10^{4} \times 1.54 \times 10^{5} = 7.9 \times 10^{9}}\tag{293}\]

This reaction is very favourable and it readily occurs at room temperature; CaO in the presence of water converts to Ca(OH)₂. The hydroxide Ca(OH)₂ is called slaked lime, for the process by which water and lime are mixed to form it. Hence in this case it makes sense to speak of Ksp only for Ca(OH)₂, not CaO. Note that K” has no expression; all the chemicals involved in the reaction are pure solids and liquid water.

The extent of similar reactions for other oxides and hydroxides depends on the relative stabilities of the two. As another example consider Cu(OH)₂ and CuO.

\[\ce{Cu(OH)2_{(s)} = Cu^{2+} + 2OH^- \quad K_{sp}(\text{Cu(OH)2}) = 2.2 \times 10^{-20}}\tag{294}\]

\[\ce{CuO_{(s)} + H2O_{(l)} = Cu^{2+} + 2OH^- \quad K_{sp}(\text{CuO}) = 4.4 \times 10^{-26}}\tag{295}\]

The reaction,

\[\ce{CuO + H2O = Cu(OH)2}\tag{296}\]

can be obtained by adding the reverse of reaction (294) to reaction (295). Then the equilibrium constant is given by:

\[\ce{K = \frac{K_{sp}(\text{CuO})}{K_{sp}(\text{Cu(OH)2})}
= \frac{4.4 \times 10^{-26}}{2.2 \times 10^{-20}}
= 2 \times 10^{-6}}\tag{297}\]

In this case CuO is much more stable than Cu(OH)₂. However, Cu(OH)₂ can survive in the presence of water for some time at room temperature, despite its instability with respect to CuO. (Cu(OH)₂ converts to CuO rapidly at temperatures a little above room temperature, but slowly at room temperature.) Therefore we can reasonably employ Ksp values for both species. Thus knowledge of the chemistry of the oxides and hydroxides is important when seeking to use Ksp data.

As a final illustration, ferric ion can precipitate form sulfate solutions in numerous different forms, depending on conditions. There are three oxide-hydroxide solids: Fe(OH)₃, Fe(O)OH and Fe₂O₃. The last one is the most stable, but the other two can form at lower temperatures; Fe₂O₃ forms at and above 100°C. In addition there are numerous basic iron(III) sulfates, such as the jarosites, e.g. M(I)Fe₃(SO₄)₂(OH)₆, where M(I) can be a number of +1 cations, including Na+, K+, NH₄+, H₃O+, etc. All are sparingly soluble (despite being sulfate and alkali metal/ammonium, etc. salts) and several are important in the disposal of unwanted iron. Understanding the conditions under which these form and how they interconvert is very important in controlling iron rejection processes (removal from hydrometallurgical process solutions).

4. Salts of Weak Bases.

Many salts of weak bases are insoluble, including sulfides, carbonates, fluorides, and even a few sulfates (like PbSO₄ and CaSO₄·2H₂O) The Ksp for these is defined exactly as above, e.g.

\[\ce{ZnCO3_{(s)} = Zn^{2+} + CO3^{2-}}\tag{298}\]

\[\ce{K_{sp} = [Zn^{2+}][CO3^{2-}] = 1.0 \times 10^{-10}}\tag{299}\]

\[\ce{Cu2(CO3)(OH)2_{(s)} = 2Cu^{2+} + CO3^{2-} + 2OH^-}\tag{300}\]

\[\ce{K_{sp} = [Cu^{2+}]^2[CO3^{2-}][OH^-]^2 = 1.66 \times 10^{-34}}\tag{301}\]

Note: Carbonate is a weak base. In water it will partially dissociate to form HCO₃^– and even a little CO₂. In some respects this is like the metal oxide situation. But, the key difference is that at least some of the CO₃^2- survives in solution, and so we write the Ksp expression for the ions that constitute the salt as written. For oxides no virtually no O^2- persists in solution, and so it makes no sense to include [O^2-] in the Ksp.

I. Consider the Ksp for Cu₂(CO₃)(OH)₂.

Its small magnitude suggests an extremely small solubility, but the exponents in the Ksp expression need to be considered as well. Let x = [Cu⁺²] that dissolves at equilibrium:

\[\ce{Cu2(CO3)(OH)2_{(s)} = 2Cu^{2+} + CO3^{2-} + 2OH^-}\tag{302}\]

Final concentrations: x 0.5x x

\[\ce{K_{sp} = (2x)\,(0.5x)\,(2x) \quad;\quad x = 2.01 \times 10^{-7}\ M}\tag{303}\]

The solubility is small; on the order of Ksp1/5. We can, in principle, address the dissociation reactions of carbonate and their effects on the solubility of the compound:

\[\ce{CO3^{2-} + H2O = HCO3^- + OH^- \quad K_{a1} = 4.47 \times 10^{-7}}\tag{304}\]

\[\ce{HCO3^- = OH^- + CO2_{(aq)} \quad K_{a2} = 4.68 \times 10^{-11}}\tag{305}\]

\[\ce{CO2_{(aq)} = CO2_{(g)} \quad K_3 = 29.4}\tag{306}\]

Such effects can have a significant effect on the solubility of a salt. The preceding example can be treated by consideration of the simultaneous equilibria as developed previously. (The CO₂ gas pressure would need to be fixed at some value that depends on the conditions, and the volume of the vessel would need to be known in order to accurately express a total carbonate mass balance.) For our purposes that exercise is academic. But, we can exploit the fact that carbonate is a weak base to leach the mineral.

The basic copper carbonate above is malachite. Another basic copper carbonate is azurite, Cu₃(CO₃)₂(OH)₂. Malachite and azurite show up in some copper ores. The sum of reactions (302), (304), (305), and (306) is,

\[\ce{Cu2(CO3)(OH)2 + H2O = 2Cu^{2+} + CO2_{(g)} + 4OH^-}\tag{307}\]

\[\ce{K = K_{sp}\,K_{a1}\,K_{a2}\,K_3 = 1.02 \times 10^{-49}}\tag{308}\]

This reaction is obviously highly unfavourable, but it can be driven to the right simply by adding acid:

\[\ce{Cu2(CO3)(OH)2 + H2O = 2Cu^{2+} + CO2_{(g)} + 4OH^-}\tag{309}\]

\[\ce{4OH^- + 4H^+ = 4H2O \quad \frac{1}{K_w^4} = 1 \times 10^{56}}\tag{310}\]

\[\ce{Cu2(CO3)(OH)2 + 4H^+ = 2Cu^{2+} + CO2 + 3H2O}\tag{311}\]

\[\ce{K = 1.02 \times 10^{-49} \times 1 \times 10^{56} = 1 \times 10^{7}}\tag{312}\]

The latter reaction is very favourable, and even a buffered bisulfate/sulfate solution can easily leach the mineral. A number of minerals that are sparingly soluble salts of weak bases can be leached in this way, and this is relatively inexpensive.

II. Even many oxides/hydroxides are readily soluble in sulfuric acid solution, e.g. CuO, ZnO, NiO, Ni(OH)2, etc.

\[\ce{CuO_{(s)} + 2H^+ = Cu^{2+} + 2H2O}\tag{313}\]

\[\ce{K = \frac{K_{sp}}{K_w^2} = \frac{2.2 \times 10^{-21}}{1 \times 10^{-28}} = 2.2 \times 10^{7}}\tag{314}\]

III. Can covellite (CuS) in an ore be leached with strong acid?

What pH would be needed to build up a certain concentration of Cu⁺² in solution? Assume the solution H₂S concentration can be kept to 0.001 M (H₂S solubility in water is ~0.1 M under 1 atm pressure of H₂S) and that the Cu⁺² concentration needs to be 1 g/L (0.0157 M). We need to consider the equilibria,

\[\ce{CuS_{(s)} = Cu^{2+}_{(aq)} + S^{2-}_{(aq)}}\tag{315}\]

\[\ce{K_{sp} = [Cu^{2+}][S^{2-}] = 8 \times 10^{-37}}\tag{316}\]

\[\ce{S^{2-} + H^+ = HS^-}\tag{317}\]

\[\ce{K” = \frac{1}{K_{a2}} = \frac{1}{3.16 \times 10^{-19}}}\tag{318}\]

\[\ce{HS^- + H^+ = H2S_{(aq)}}\tag{319}\]

\[\ce{K’ = \frac{1}{K_{a1}} = \frac{1}{1.02 \times 10^{-7}}}\tag{320}\]

(S^2- will react with water quite strongly,

\[\ce{S^{2-} + H2O = HS^- + OH^-}\tag{321}\]

but we can also use the reverse of the dissociation of HS^–, reaction (317) to keep everything in terms of H⁺, and thus not have to involve OH^–, and similarly for HS^– reacting to form H₂S.)

Cupric ion will precipitate from HS^– as well as from S^2- solution, so in principle conversion right through to H₂S is what is needed. The sum of the above reactions gives,

\[\ce{CuS_{(s)} + 2H^+ = Cu^{2+} + H2S_{(aq)}}\tag{322}\]

\[\ce{K = \frac{[Cu^{2+}][H2S]}{[H^+]^2}}\tag{323}\]

\[\ce{K = K_{sp}\,K”\,K’ = \frac{K_{sp}}{K_{a2}\,K_{a1}} = 2.48 \times 10^{-11}}\tag{324}\]

Note how much smaller this K is than that for the reaction of CuO with acid. H₂S is modestly soluble in water and would escape into the gas phase. We are assuming that it can be kept in solution at only 0.001 M, which is quite low. We need a certain concentration of H⁺ to maintain this condition:

\[\ce{[H^+] = \frac{[Cu^{2+}][H2S]}{K}
= \left(\frac{0.0157 \times 0.001}{2.48 \times 10^{-11}}\right)^{0.5}
= 796\ \text{M};\quad pH = -2.9}\tag{325}\]

Such a high acid concentration is impossible; pure water has a concentration of 55.4 mol/L, and pure H₂SO₄ is only about 18 M. Clearly this cannot be achieved on the basis of concentration of acid in water. (However, some high concentration HCl + MgCl₂ solutions can achieve pH values approaching this pH, due to the extremely high activity of cations in such solutions. Such solutions are extremely aggressive and would be difficult to contain in vessels made with conventional materials.) In practice we oxidize sulfides in order to leach them.

IV. As a final example, calcium sulfate dihydrate (gypsum) is CaSO₄·2H₂O; Ksp = 4.4 x 10^-5.

It is only sparingly soluble in water and is a source of considerable difficulty in hydrometallurgy operations. The use of calcium-containing bases (CaO, Ca(OH)₂ and CaCO₃) and the very common sulfate medium for hydrometallurgical processes often results in precipitation of gypsum in pipes, tanks etc. This causes a build-up of scale, which can block pipes and cause other problems. Often the onset of gypsum precipitation occurs due to pH changes and temperature changes. What is the solubility of gypsum in g/L Ca⁺² in water and in 0.2 M sulfuric acid? (Calcium bisulfate is quite soluble.)

\[\ce{CaSO4\cdot 2H2O = Ca^{2+} + SO4^{2-} + 2H2O}\tag{326}\]

\[\ce{K_{sp} = [Ca^{2+}][SO4^{2-}] = 4.4 \times 10^{-5}}\tag{327}\]

\[\ce{x^2 = 4.4 \times 10^{-5};\quad x = 6.6 \times 10^{-3}\ \text{M}}\tag{328}\]

Since the atomic weight of Ca is 40.08 g/mol this equals 0.266 g/L as Ca⁺². In the presence of 0.2 M H₂SO₄, we need to consider the additional equilibrium:

\[\ce{HSO4^- = H^+ + SO4^{2-}}\tag{329}\]

\[\ce{K_{a2} = \frac{[H^+][SO4^{2-}]}{[HSO4^-]} = 0.0102}\tag{330}\]

(Ka₁ for H₂SO₄ = large; virtually fully dissociates.) There are two sources of sulfate in the final solution, one from CaSO₄·2H₂O dissolution and one from HSO₄^– dissociation. Let the sulfate arising from HSO₄^– dissociation = x and let y = [SO₄^2-] arising from gypsum dissolution.

\[\ce{HSO4^- = H^+ + SO4^{2-}}\tag{331}\]

Initial Co Co
Co – x Co +x x (for HSO4- dissociation only)

\[\ce{CaSO4\cdot 2H2O = Ca^{2+} + SO4^{2-} + 2H2O}\tag{332}\]

y y (for gypsum dissolution only)

Final [SO₄^2-] = x + y

\[\ce{K_{sp} = y(x + y)}\tag{333}\]

\[\ce{K_a = \frac{(C_0 + x)(x + y)}{(C_0 – x)}}\tag{334}\]

\[\ce{K_{sp} = xy + y^2}\tag{335}\]

\[\ce{y^2 + xy – K_{sp} = 0}\tag{336}\]

\[\ce{y = \frac{-x + \sqrt{x^2 + 4K_{sp}}}{2}}\tag{337}\]

From equation (334),

\[\ce{x^2 + (C_0 + y + K_a)x + C_0y – K_aC_0 = 0}\tag{338}\]

Set a tentative value for x and calculate y. Obtain a value for equation (338) based on x and the calculated y. Vary x until equation (338) returns zero. For 0.2 M H₂SO₄ this results in:

x = 0.005187 and y = 0.00453

[H⁺] = 0.205 M, [HSO₄^–] = 0.195, [SO₄^2-] = 0.00972 and [Ca⁺²] = 0.00453

The solubility of CaSO₄·2H₂O is now less than in the absence of acid and precipitation would be expected. Variations in pH can result in increased and decreased solubility of gypsum.

The other way to approach this problem is by the speciation approach developed earlier for H₃PO₄. There are five species: Ca⁺², SO₄^2-, HSO₄^–, H⁺ and OH^–. Three equilibrium expressions are available for three reactions: Ksp for CaSO₄·2H₂O, Ka for HSO₄^– (H₂SO₄ → H⁺ + HSO₄^– can be assumed to go to completion) and Kw. A mass balance can be written for total sulfate in the system (let Co = initial [H₂SO₄]:

\[\ce{[Ca^{2+}] + Co = [HSO4^-] + [SO4^{2-}]}\tag{339}\]

The key is to recognize that the two sources of sulfate. Sulfate and bisulfate from H₂SO₄ dissociation = Co; sulfate arising from CaSO₄·2H₂O dissolution = [Ca⁺²]. Finally a charge balance equation can be written:

\[\ce{[H^+] + 2[Ca^{2+}] = [OH^-] + [HSO4^-] + 2[SO4^{2-}]}\tag{340}\]

This can be solved by elimination of variables and suitable substitutions.

5. Metal Oxide/Hydroxide-Metal Salt Buffering.

This type of buffer was mentioned briefly in the section on buffers. Consider an important hydrometallurgical example. ZnO is a sparingly soluble salt, and as an oxide also a weak base.

\[\ce{ZnO + H2O = Zn^{2+} + 2OH^- \qquad K_{sp} = 3.7 \times 10^{-17}}\tag{341}\]

\[\ce{2OH^- + 2H^+ = 2H2O \qquad \frac{1}{K_w^2} = 1 \times 10^{28}}\tag{342}\]

\[\ce{ZnO + 2H^+ = Zn^{2+} + H2O \qquad K’ = \frac{K_{sp}}{K_w^2}}\tag{343}\]

Reversing the reaction,

\[\ce{Zn^{2+} + H2O = ZnO + 2H^+ \qquad K = \frac{K_w^2}{K_{sp}}}\tag{344}\]

This is an acid dissociation reaction, except that 2H⁺ are involved rather than just one. Strictly speaking, ZnO and Zn⁺² are not conjugate acid-base pairs, but they do differ by two protons and a water. Consider what will happen if we add strong acid to a mixture of ZnSO₄ (a soluble salt) and ZnO_(s). The ZnO will react as per reaction (343) and be converted into Zn⁺², which is a weak acid; reaction (344). Hence the pH will drop, but only a little. Similarly if we add some strong base, Zn⁺² will react according to the reverse of reaction (341) to form ZnO, a weak base. The pH will drop only a little. Hence this system is a buffer. It’s pH can be calculated as follows:

\[\ce{K = \frac{[H^+]^2}{[Zn^{2+}]} = \frac{K_w^2}{K_{sp}}}\tag{345}\]

\[\ce{\log[H^+]^2 – \log[Zn^{2+}] = \log K_w^2 – \log K_{sp}}\tag{346}\]

\[\ce{2\log[H^+] = \log[Zn^{2+}] + \log K_w^2 – \log K_{sp}}\tag{347}\]

\[\ce{-2\log[H^+] = -\log[Zn^{2+}] – \log K_w^2 + \log K_{sp}}\tag{348}\]

\[\ce{pH = \tfrac{1}{2}\log\!\left(\frac{K_{sp}}{K_w^2}\right) – \tfrac{1}{2}\log[Zn^{2+}]}\tag{349}\]

Note the similar form of the last equation to that for a buffer involving a soluble acid and it soluble conjugate base. The activity of the base (ZnO) is one, since it is a pure solid. For a solution of 0.5 M ZnSO₄ in contact with solid ZnO the pH will be:

\[\ce{pH = 0.5\log\!\left(\frac{3.7 \times 10^{-17}}{1 \times 10^{-28}}\right) – 0.5\log(0.5) = 5.93}\tag{350}\]

This is actually a very important aspect of the zinc production industry. ZnS from an ore is concentrated, then roasted in air to form ZnO. The ZnO is leached in sulfuric acid. The leaching process involves a zinc sulfate solution plus sulfuric acid contacted with ZnO. The solution pH is maintained at a pH close to that calculated above. This is beneficial for subsequent processing steps since strong acid would interfere. The process is called the “neutral leach,” because it occurs at a relatively high pH (although certainly not pH 7 per se).

In general metal oxides may exhibit this kind of buffering when in contact with a soluble salt of the metal ion. The same applies for metal hydroxides, and since Ksp has the identical form to that for metal oxides, the equation is identical:

\[\ce{M(OH)2(s) = M^{2+} + 2OH^- \qquad K_{sp} = [M^{2+}][OH^-]^2}\tag{351}\]

In general for a divalent metal ion and either its insoluble oxide or hydroxide,

\[\ce{pH = \tfrac{1}{2}\log\!\left(\frac{K_{sp}}{K_w^2}\right) – \tfrac{1}{2}\log[M^{2+}]}\tag{352}\]

Of course, this is not restricted to divalent metal oxides/hydroxides. A similar expression can be derived for Ag₂O/Ag+ buffers (although Ag₂SO₄ is not very soluble, and such a system would be of little practical value since silver grades in ores are low). Trivalent metal oxides tend to be too insoluble to establish equilibrium rapidly enough for such buffering to be manifest in a reasonable time, although it is theoretically possible.

There are numerous possible oxo/hydroxo complexes and compounds and these two in principle may get involved in buffering. For example, MO_(s)/M(O)OH^–_(aq), where the oxide-hydroxide anion is suitably stable. Likewise M(OH)_2(s)/M(O)OH^–_(aq), and other similar combinations for other valence states. In fact, metal ions may have a large number of weak acids, for example,

\[\ce{M^{2+} + H2O = MOH^+ + H^+}\tag{353}\]

\[\ce{MOH^+ = MO(s) + H^+ \quad\text{or}\quad MOH^+ + H2O = M(OH)2(s) + H^+}\tag{354}\]

\[\ce{MO + H2O = M(O)OH^- + H^+ \quad\text{or}\quad M(OH)2 = M(O)OH^- + H^+}\tag{355}\]

etc. Not all such species have significant stability for all metal ions, but a number may for any given metal ion and this may lead to a number of buffer points.