The Effect of Acid–Base Equilibriums on the Solubility of Salts

We begin our discussion by examining the effect of pH on the solubility of a representative salt, M⁺A⁻, where A⁻ is the conjugate base of the weak acid HA. When the salt dissolves in water, the following reaction occurs:

The anion can also react with water in a hydrolysis reaction:

Because of the reaction described in Equation 17.14, the predicted solubility of a sparingly soluble salt that has a basic anion such as S²⁻, PO₄³⁻, or CO₃²⁻ is increased, as described in Section 17.1 "Determining the Solubility of Ionic Compounds". If instead a strong acid is added to the solution, the added H⁺ will react essentially completely with A⁻ to form HA. This reaction decreases [A⁻], which decreases the magnitude of the ion product (Q = [M⁺][A⁻]). According to Le Châtelier’s principle, more MA will dissolve until Q = K_sp. Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH)₂ is relatively insoluble in water:

When acid is added to a saturated solution that contains excess solid Mg(OH)₂, the following reaction occurs, removing OH⁻ from solution:

The overall equation for the reaction of Mg(OH)₂ with acid is thus

As more acid is added to a suspension of Mg(OH)₂, the equilibrium shown in Equation 17.17 is driven to the right, so more Mg(OH)₂ dissolves.

Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF₂ is a sparingly soluble salt:

When strong acid is added to a saturated solution of CaF₂, the following reaction occurs:

Because the forward reaction decreases the fluoride ion concentration, more CaF₂ dissolves to relieve the stress on the system. The net reaction of CaF₂ with strong acid is thus

Example 7 shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt.

Example 7

Lead oxalate (PbC₂O₄), lead iodide (PbI₂), and lead sulfate (PbSO₄) are all rather insoluble, with K_sp values of 4.8 × 10⁻¹⁰, 9.8 × 10⁻⁹, and 2.53 × 10⁻⁸, respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities?

Given: K_sp values for three compounds

Asked for: relative solubilities in acid solution

Strategy:

Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions.

Solution:

The solubility equilibriums for the three salts are as follows:

$$\begin{array}{c}\hfill {\text{PbC}}_2{\text{O}}_4\text{(s)}\rightleftharpoons {\text{Pb}}^{2+}\text{(aq)}+{\text{C}}_2{\text{O}}_4{}^{2-}\text{(aq)}\\ \hfill {\text{PbI}}_2\text{(s)}\rightleftharpoons {\text{Pb}}^{2+}\text{(aq)}+2{\text{I}}^{-}\text{(aq)}\\ \hfill {\text{PbSO}}_4\text{(s)}\rightleftharpoons {\text{Pb}}^{2+}\text{(aq)}+{\text{SO}}_4{}^{2-}\text{(aq)}\end{array}$$

The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI₂ will not greatly affect its solubility; the acid will simply dissociate to form H⁺(aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO₂CCO₂H), which is a weak diprotic acid (pK_a1 = 1.23 and pK_a2 = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions:

$$\begin{array}{c}\hfill {\text{C}}_2{\text{O}}_4{}^{2-}\text{(aq)}+{\text{H}}^{+}\text{(aq)}\to {\text{HO}}_2{\text{CCO}}_2{}^{-}\text{(aq)}\\ \hfill {\text{HO}}_2{\text{CCO}}_2{}^{-}\text{(aq)}+{\text{H}}^{+}\text{(aq)}\to {\text{HO}}_2{\text{CCO}}_2\text{H(aq)}\end{array}$$

These reactions will decrease [C₂O₄²⁻], causing more lead oxalate to dissolve to relieve the stress on the system.The pK_a of HSO₄⁻ (1.99) is similar in magnitude to the pK_a1 of oxalic acid, so adding a strong acid to a saturated solution of PbSO₄ will result in the following reaction:

 

Because HSO₄⁻ has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO₄ will be significantly less than for PbC₂O₄.

Exercise

Which of the following insoluble salts—AgCl, Ag₂CO₃, Ag₃PO₄, and/or AgBr—will be substantially more soluble in 1.0 M HNO₃ than in pure water?

Answer: Ag₂CO₃ and Ag₃PO₄

Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility equilibriums (part (a) in Figure 17.6 "The Chemistry of Cave Formation"). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows:

Figure 17.6 The Chemistry of Cave Formation

(a) This cave in Campanet, Mallorca, Spain, and its associated formations are examples of pH-dependent solubility equilibriums. (b) A cave forms when groundwater containing atmospheric CO₂, forming an acidic solution, dissolves limestone (CaCO₃) in a process that may take tens of thousands of years. As groundwater seeps into a cave, water evaporates from the solution of CaCO₃ in CO₂-rich water, producing a supersaturated solution and a shift in equilibrium that causes precipitation of the CaCO₃. The deposited limestone eventually forms stalactites and stalagmites.

Limestone deposits that form caves consist primarily of CaCO₃ from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO₃ in CO₂-rich water rises toward Earth’s surface or is otherwise heated, CO₂ gas is released as the water warms. CaCO₃ then precipitates from the solution according to the following equation (part (b) in Figure 17.6 "The Chemistry of Cave Formation"):

The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated.

When groundwater-containing atmospheric CO₂ (Equation 17.21 and Equation 17.22) finds its way into microscopic cracks in the limestone deposits, CaCO₃ dissolves in the acidic solution in the reverse direction of Equation 17.24. The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in Equation 17.24 to shift to the right. A circular layer of solid CaCO₃ is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China (Figure 17.7 "Solubility Equilibriums in the Formation of Karst Landscapes").

Figure 17.7 Solubility Equilibriums in the Formation of Karst Landscapes

Landscapes such as the steep limestone pinnacles of the Stone Forest in Yunnan Province, China, are formed from the same process that produces caves and their associated formations.

Acidic, Basic, and Amphoteric Oxides and Hydroxides

One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. Basic oxidesAn oxide that reacts with water to produce a basic solution or dissolves readily in aqueous acid. and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. Acidic oxidesAn oxide that reacts with water to produce an acidic solution or dissolves in aqueous base. or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. As shown in Figure 17.8 "Classification of the Oxides of the Main Group Elements According to Their Acidic or Basic Character", there is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water:

Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed!

The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete Mn⁺ cations and O²⁻ anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: E^δ+–O^δ−. The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO₃, which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO₄²⁻):

As shown in Figure 17.8 "Classification of the Oxides of the Main Group Elements According to Their Acidic or Basic Character", there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called amphoteric oxidesAn oxide that can dissolve in acid to produce water and dissolve in base to produce a soluble complex. (from the Greek ampho, meaning “both,” as in amphiprotic, which was defined in Chapter 16 "Aqueous Acid–Base Equilibriums", Section 16.1 "The Autoionization of Water"). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in Figure 17.9 "Chromium(III) Hydroxide [Cr(OH)", for example, mixing the amphoteric oxide Cr(OH)₃ (also written as Cr₂O₃·3H₂O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH)₃ to dissolve to give a bright violet solution of Cr³⁺(aq), which contains the [Cr(H₂O)₆]³⁺ ion, whereas adding strong base gives a green solution of the [Cr(OH)₄]⁻ ion. The chemical equations for the reactions are as follows:

Figure 17.8 Classification of the Oxides of the Main Group Elements According to Their Acidic or Basic Character

There is a gradual transition from basic oxides to acidic oxides from the lower left to the upper right in the periodic table. Oxides of metallic elements are generally basic oxides, which either react with water to form a basic solution or dissolve in aqueous acid. In contrast, oxides of nonmetallic elements are acidic oxides, which either react with water to form an acidic solution or are soluble in aqueous base. Oxides of intermediate character, called amphoteric oxides, are located along a diagonal line between the two extremes. Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex ion. (Radioactive elements are not classified.)

Figure 17.9 Chromium(III) Hydroxide [Cr(OH)₃ or Cr₂O₃·3H₂O] Is an Example of an Amphoteric Oxide

All three beakers originally contained a suspension of brownish purple Cr(OH)₃(s) (center). When concentrated acid (6 M H₂SO₄) was added to the beaker on the left, Cr(OH)₃ dissolved to produce violet [Cr(H₂O)₆]³⁺ ions and water. The addition of concentrated base (6 M NaOH) to the beaker on the right caused Cr(OH)₃ to dissolve, producing green [Cr(OH)₄]⁻ions.

Selective Precipitation Using pH

Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate.

Suppose, for example, we have a solution that contains 1.0 mM Zn²⁺ and 1.0 mM Cd²⁺ and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilibriums can be written as follows:

Because the S²⁻ ion is quite basic and reacts extensively with water to give HS⁻ and OH⁻, the solubility equilibriums are more accurately written as $\text{MS(s)}\rightleftharpoons {\text{M}}^{2+}\text{(aq)}+{\text{HS}}^{-}(\text{aq)}+{\text{OH}}^{-}$ rather than $\text{MS(s)}\rightleftharpoons {\text{M}}^{2+}\text{(aq)}+{\text{S}}^{2-}\text{(aq)}\text{.}$ Here we use the simpler form involving S²⁻, which is justified because we take the reaction of S²⁻ with water into account later in the solution, arriving at the same answer using either equilibrium equation.

The sulfide concentrations needed to cause ZnS and CdS to precipitate are as follows:

Thus sulfide concentrations between 1.6 × 10⁻²¹ M and 8.0 × 10⁻²⁴ M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H₂S contains 0.10 M H₂S at 20°C. The pK_a1 for H₂S is 6.97, and pK_a2 corresponding to the formation of [S²⁻] is 12.90. The equations for these reactions are as follows:

We can show that the concentration of S²⁻ is 1.3 × 10⁻¹³ by comparing K_a1 and K_a2 and recognizing that the contribution to [H⁺] from the dissociation of HS⁻ is negligible compared with [H⁺] from the dissociation of H₂S. Thus substituting 0.10 M in the equation for K_a1 for the concentration of H₂S, which is essentially constant regardless of the pH, gives the following:

Substituting this value for [H⁺] and [HS⁻] into the equation for K_a2,

Although [S²⁻] in an H₂S solution is very low (1.3 × 10⁻¹³ M), bubbling H₂S through the solution until it is saturated would precipitate both metal ions because the concentration of S²⁻ would then be much greater than 1.6 × 10⁻²¹ M. Thus we must adjust [S²⁻] to stay within the desired range. The most direct way to do this is to adjust [H⁺] by adding acid to the H₂S solution (recall Le Châtelier's principle), thereby driving the equilibrium in Equation 17.34 to the left. The overall equation for the dissociation of H₂S is as follows:

Now we can use the equilibrium constant K for the overall reaction, which is the product of K_a1 and K_a2, and the concentration of H₂S in a saturated solution to calculate the H⁺ concentration needed to produce [S²⁻] of 1.6 × 10⁻²¹ M:

Thus adding a strong acid such as HCl to make the solution 0.94 M in H⁺ will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H₂S.