Chapter 17 Solubility and Complexation Equilibriums

Example 1

Calcium oxalate monohydrate [Ca(O₂CCO₂)·H₂O, also written as CaC₂O₄·H₂O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca₃(PO₄)₂]. Its solubility in water at 25°C is 7.36 × 10⁻⁴ g/100 mL. Calculate its K_sp.

Given: solubility in g/100 mL

Asked for: K _sp

Strategy:

A Write the balanced dissolution equilibrium and the corresponding solubility product expression.

B Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate K_sp.

Solution:

A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox²⁻) are as follows:

$$\text{Ca}({\text{O}}_2{\text{CCO}}_2)\text{⋅}{\text{H}}_2\text{O(s)}\rightleftharpoons {\text{Ca}}^{2+}\text{(aq)}+{}^{-}\text{O}{}_2{\text{CCO}}_2{}^{-}\text{(aq)}+{\text{H}}_2\text{O(l) }{K}_{\text{sp}}=[{\text{Ca}}^{2+}][{\text{ox}}^{2-}]$$

Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.

B Next we need to determine [Ca²⁺] and [ox²⁻] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:

$$\frac{7.36\times {10}^{-4}\cancel{\text{g}}}{146.1\cancel{\text{g}}\text{/mol}}=5.04\times {10}^{-6}\text{ mol Ca}({\text{O}}_2{\text{CCO}}_2)\text{⋅}{\text{H}}_2\text{O}$$

The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:

$$\left(\frac{5.04\times {10}^{-6}\text{ mol Ca}({\text{O}}_2{\text{CCO}}_2\text{⋅}){\text{H}}_2\text{O}}{100\cancel{\text{mL}}}\right)\left(\frac{1000\cancel{\text{mL}}}{1.00\text{ L}}\right)=5.04\times {10}^{-5}\text{ mol/L}=5.04\times {10}^{-5}\text{ M}$$

Because of the stoichiometry of the reaction, the concentration of Ca²⁺ and ox²⁻ ions are both 5.04 × 10⁻⁵ M. Inserting these values into the solubility product expression,

$$K_{\text{sp}}=[{\text{Ca}}^{2+}][{\text{ox}}^{2-}]=(5.04\times {10}^{-5})(5.04\times {10}^{-5})=2.54\times {10}^{-9}$$

In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.

Exercise

One crystalline form of calcium carbonate (CaCO₃) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its K_sp.

Answer: 4.5 × 10⁻⁹

Example 2

We saw that the K_sp for Ca₃(PO₄)₂ is 2.07 × 10⁻³³ at 25°C. Calculate the aqueous solubility of Ca₃(PO₄)₂ in terms of the following:

1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C

Given: K _sp

Asked for: molar concentration and mass of salt that dissolves in 100 mL of water

Strategy:

A Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.

B Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.

Solution:

1. A The dissolution equilibrium for Ca₃(PO₄)₂ (Equation 17.1) is shown in the following table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca₃(PO₄)₂ that dissolves, 3 mol of Ca²⁺ and 2 mol of PO₄³⁻ ions are produced in solution. If we let x equal the solubility of Ca₃(PO₄)₂ in moles per liter, then the change in [Ca²⁺] will be +3x, and the change in [PO₄³⁻] will be +2x. We can insert these values into the table.

   ${\text{Ca}}_3{({\text{PO}}_4)}_2\text{(s)}\rightleftharpoons 3{\text{Ca}}^{2+}\text{(aq)}+2{\text{PO}}_4{}^{3-}\text{(aq)}$
   	Ca3(PO4)2	[Ca2+]	[PO43−]
   initial	pure solid	0	0
   change	—	+3x	+2x
   final	pure solid	3x	2x

   Although the amount of solid Ca₃(PO₄)₂ changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):

   $$\begin{array}{cc}\hfill K_{\text{sp}}={[{\text{Ca}}^{2+}]}^3{[{\text{PO}}_4{}^{3-}]}^2& ={(3x)}^3{(2x)}^2\hfill \\ \hfill 2.07\times {10}^{-33}& =108x^5\hfill \\ \hfill 1.92\times {10}^{-35}& =x^5\hfill \\ \hfill 1.14\times {10}^{-7}\text{ M}& =x\hfill \end{array}$$

   This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO₄³⁻] = 2.28 × 10⁻⁷ and [Ca²⁺] = 3.42 × 10⁻⁷.

2. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:

   $$\left(\frac{1.14\times {10}^{-7}\cancel{\text{mol}}}{1\cancel{\text{L}}}\right)100\cancel{\text{mL}}\left(\frac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\right)\left(\frac{310.18{\text{g Ca}}_3{({\text{PO}}_4)}_2}{1\cancel{\text{mol}}}\right)=3.54\times {10}^{-6}{\text{ g Ca}}_3{({\text{PO}}_4)}_2$$

Exercise

The solubility product of silver carbonate (Ag₂CO₃) is 8.46 × 10⁻¹² at 25°C. Calculate the following:

1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature

Answer:

1. 1.28 × 10⁻⁴ M
2. 3.54 mg

Example 3

We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10⁻¹⁰ at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na₂SO₄ is added to 100 mL of 3.2 × 10⁻⁴ M BaCl₂? Recall that NaCl is highly soluble in water.

Given: K_sp and volumes and concentrations of reactants

Asked for: whether precipitate will form

Strategy:

A Write the balanced equilibrium equation for the precipitation reaction and the expression for K_sp.

B Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).

C Compare the values of Q and K_sp to decide whether a precipitate will form.

Solution:

A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO₄ because NaCl is highly soluble. The equation for the precipitation of BaSO₄ is as follows:

$${\text{BaSO}}_4\text{(s)}\rightleftharpoons {\text{Ba}}^{2+}\text{(aq)}+{\text{SO}}_4{}^{2-}\text{(aq)}$$

The solubility product expression is as follows:

$$K_{\text{sp}}=[{\text{Ba}}^{2+}][{\text{SO}}_4{}^{2-}]=1.08\times {10}^{-10}$$

B To solve this problem, we must first calculate the ion product—Q = [Ba²⁺][SO₄²⁻]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba²⁺ when the solutions are mixed is the total number of moles of Ba²⁺ in the original 100 mL of BaCl₂ solution divided by the final volume (100 mL + 10.0 mL = 110 mL):

$${\text{moles Ba}}^{2+}=100\cancel{\text{mL}}\left(\frac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\right)\left(\frac{3.2\times {10}^{-4}\text{ mol}}{1\cancel{\text{L}}}\right)=3.2\times {10}^{-5}{\text{ mol Ba}}^{2+}$$ $$[{\text{Ba}}^{2+}]=\left(\frac{3.2\times {10}^{-5}{\text{ mol Ba}}^{2+}}{110\cancel{\text{mL}}}\right)\left(\frac{1000\cancel{\text{mL}}}{1\text{ L}}\right)=2.9\times {10}^{-4}{\text{ M Ba}}^{2+}$$

Similarly, the concentration of SO₄²⁻ after mixing is the total number of moles of SO₄²⁻ in the original 10.0 mL of Na₂SO₄ solution divided by the final volume (110 mL):

$${\text{moles SO}}_4{}^{2-}=10.0\cancel{\text{mL}}\left(\frac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\right)\left(\frac{0.0020\text{ mol}}{1\cancel{\text{L}}}\right)=2.0\times {10}^{-5}{\text{ mol SO}}_4{}^{2-}$$ $$[{\text{SO}}_4{}^{2-}]=\left(\frac{2.0\times {10}^{-5}{\text{ mol SO}}_4{}^{2-}}{110\cancel{\text{mL}}}\right)\left(\frac{1000\cancel{\text{mL}}}{1\text{ L}}\right)=1.8\times {10}^{-4}{\text{ M SO}}_4{}^{2-}$$

We can now calculate Q:

$$Q=[{\text{Ba}}^{2+}][{\text{SO}}_4{}^{2-}]=(2.9\times {10}^{-4})(1.8\times {10}^{-4})=5.2\times {10}^{-8}$$

C We now compare Q with the K_sp. If Q > K_sp, then BaSO₄ will precipitate, but if Q < K_sp, it will not. Because Q > K_sp, we predict that BaSO₄ will precipitate when the two solutions are mixed. In fact, BaSO₄ will continue to precipitate until the system reaches equilibrium, which occurs when [Ba²⁺][SO₄²⁻] = K_sp = 1.08 × 10⁻¹⁰.

Exercise

The solubility product of calcium fluoride (CaF₂) is 3.45 × 10⁻¹¹. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10⁻⁵ M solution of Ca(NO₃)₂, will CaF₂ precipitate?

Answer: yes (Q = 4.7 × 10⁻¹¹ > K_sp)

Example 4

Calculate the solubility of calcium phosphate [Ca₃(PO₄)₂] in 0.20 M CaCl₂.

Given: concentration of CaCl₂ solution

Asked for: solubility of Ca₃(PO₄)₂ in CaCl₂ solution

Strategy:

A Write the balanced equilibrium equation for the dissolution of Ca₃(PO₄)₂. Tabulate the concentrations of all species produced in solution.

B Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca₃(PO₄)₂.

Solution:

A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca₃(PO₄)₂ in moles per liter, then the change in [Ca²⁺] is once again +3x, and the change in [PO₄³⁻] is +2x. We can insert these values into the table.

B The K_sp expression is as follows:

$$K_{\text{sp}}={[{\text{Ca}}^{2+}]}^3{[{\text{PO}}_4{}^{3-}]}^2={(0.20+3x)}^3{(2x)}^2=2.07\times {10}^{-33}$$

Because Ca₃(PO₄)₂ is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the K_sp expression as follows:

$$\begin{array}{ccc}\hfill K_{\text{sp}}& ={(0.20)}^3{(2x)}^2\hfill & =2.07\times {10}^{-33}\hfill \\ \hfill & x^2& =6.5\times {10}^{-32}\hfill \\ \hfill & x& =2.5\times {10}^{-16}\text{ M}\hfill \end{array}$$

This value is the solubility of Ca₃(PO₄)₂ in 0.20 M CaCl₂ at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Châtelier’s principle. With one exception, this example is identical to Example 2—here the initial [Ca²⁺] was 0.20 M rather than 0.

Exercise

Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10⁻¹² at 25°C.

Answer: 2.9 × 10⁻⁶ M (versus 1.3 × 10⁻⁴ M in pure water)

Example 5

If 12.5 g of Cu(NO₃)₂·6H₂O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu²⁺(aq)?

Given: mass of Cu²⁺ salt and volume and concentration of ammonia solution

Asked for: equilibrium concentration of Cu²⁺(aq)

Strategy:

A Calculate the initial concentration of Cu²⁺ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation 17.7 to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.

B Substitute the final concentrations into the expression for the formation constant (Equation 17.8) to calculate the equilibrium concentration of Cu²⁺(aq).

Solution:

Adding an ionic compound that contains Cu²⁺ to an aqueous ammonia solution will result in the formation of [Cu(NH₃)₄]²⁺(aq), as shown in Equation 17.7. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.

A The initial concentration of Cu²⁺ from the amount of added copper nitrate prior to any reaction is as follows:

$$12.5\cancel{\text{g}}\text{ Cu}{\left({\text{NO}}_3\right)}_2\text{•}6{\text{H}}_2\text{O}\left(\frac{1\text{ mol}}{295.65\cancel{\text{g}}}\right)\left(\frac{1}{500\cancel{\text{mL}}}\right)\left(\frac{1000\cancel{\text{mL}}}{1\text{ L}}\right)=0.0846\text{ M}$$

Because the stoichiometry of the reaction is four NH₃ to one Cu²⁺, the amount of NH₃ required to react completely with the Cu²⁺ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 10¹³), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH₃)₄]²⁺ in the first step is complete and allow some of it to dissociate into Cu²⁺ and NH₃ until equilibrium has been reached. If we define x as the amount of Cu²⁺ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH₃)₄]²⁺ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.

B Substituting the final concentrations into the expression for the formation constant (Equation 17.8) and assuming that x << 0.0846, which allows us to remove x from the sum and difference,

$$\begin{array}{cc}\hfill K_{\text{f}}& =\frac{\left[{[\text{Cu}{({\text{NH}}_3)}_4]}^{2+}\right]}{[{\text{Cu}}^{2+}]{[{\text{NH}}_3]}^4}=\frac{0.0846-x}{x{(0.66+4x)}^4}\approx \frac{0.0846}{x{(0.66)}^4}=2.1\times {10}^{13}\hfill \\ \hfill x& =2.1\times {10}^{-14}\hfill \end{array}$$

The value of x indicates that our assumption was justified. The equilibrium concentration of Cu²⁺(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10⁻¹⁴ M.

Exercise

The ferrocyanide ion {[Fe(CN)₆]⁴⁻} is very stable, with a K_f of 1 × 10³⁵. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K₄[Fe(CN)₆].

Answer: 2 × 10⁻⁶ M

Example 6

Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl₂⁻). Calculate the solubility of AgCl in each situation:

1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl₂⁻ is the only Ag⁺ complex that forms in significant concentrations

At 25°C, K_sp = 1.77 × 10⁻¹⁰ for AgCl and K_f = 1.1 × 10⁵ for AgCl₂⁻.

Given: K_sp of AgCl, K_f of AgCl₂⁻, and KCl concentration

Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions

Strategy:

A Write the solubility product expression for AgCl and calculate the concentration of Ag⁺ and Cl⁻ in water.

B Calculate the concentration of Ag⁺ in the KCl solution.

C Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl₂⁻ complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.

D Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.

Solution:

1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag⁺] = [Cl⁻] = x M. Substituting this value into the solubility product expression,

   $$\begin{array}{ccc}\hfill K_{\text{sp}}=& [{\text{Ag}}^{+}][{\text{Cl}}^{-}]& =(x)(x)=x^2=1.77\times {10}^{-10}\hfill \\ \hfill & x& =1.33\times {10}^{-5}\hfill \end{array}$$

   Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10⁻⁵ M.

2. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag⁺] = x  M and [Cl⁻] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,

   $$K_{\text{sp}}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=(x)(1.0+x)\approx x(1.0)=1.77\times {10}^{-10}=x$$

   If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.

3. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to K_sp and K_f gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of K_sp and K_f:

   $$\begin{array}{ccc}\hfill \text{AgCl(s)}\rightleftharpoons \cancel{{\text{Ag}}^{+}\text{(aq)}}+\cancel{{\text{Cl}}^{-}\text{(aq)}}& K_{\text{sp}}& =1.77\times {10}^{-10}\hfill \\ \hfill \cancel{{\text{Ag}}^{+}\text{(aq)}}+\cancel{2}{\text{Cl}}^{-}\rightleftharpoons {[{\text{AgCl}}_2]}^{-}& K_{\text{f}}& =1.1\times {10}^5\hfill \\ \hfill \text{AgCl(s)}\text{ + }{\text{Cl}}^{-}\rightleftharpoons {[{\text{AgCl}}_{{}^2}]}^{-}& K& =K_{\text{sp}}{K}_{\text{f}}=1.9\times {10}^{-5}\hfill \end{array}$$

   D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl₂⁻] = x and [Cl⁻] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,

   $$K=\frac{[{\text{AgCl}}_2{}^{-}]}{[{\text{Cl}}^{-}]}=\frac{x}{1.0-x}\approx 1.9\times {10}^{-5}=x$$

   That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10⁻⁵ M solution of the AgCl₂⁻ complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 10⁵ times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)

Exercise

Calculate the solubility of mercury(II) iodide (HgI₂) in each situation:

1. pure water
2. a 3.0 M solution of NaI, assuming [HgI₄]²⁻ is the only Hg-containing species present in significant amounts

K_sp = 2.9 × 10⁻²⁹ for HgI₂ and K_f = 6.8 × 10²⁹ for [HgI₄]²⁻.

Answer:

1. 1.9 × 10⁻¹⁰ M
2. 1.4 M

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

Example 8

Aluminum hydroxide, written as either Al(OH)₃ or Al₂O₃·3H₂O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base.

Given: amphoteric compound

Asked for: dissolution reactions in acid and base

Strategy:

Using Equation 17.28 and Equation 17.29 as a guide, write the dissolution reactions in acid and base solutions.

Solution:

1. An acid donates protons to hydroxide to give water and the hydrated metal ion, so aluminum hydroxide, which contains three OH⁻ ions per Al, needs three H⁺ ions:

   Al(OH)₃(s) + 3H⁺(aq) → Al³⁺(aq) + 3H₂O(l)

   In aqueous solution, Al³⁺ forms the complex ion [Al(H₂O)₆]³⁺.

2. In basic solution, OH⁻ is added to the compound to produce a soluble and stable poly(hydroxo) complex:

   Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq)

Exercise

Copper(II) hydroxide, written as either Cu(OH)₂ or CuO·H₂O, is amphoteric. Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base.

Answer:

Cu(OH)₂(s) + 2H⁺(aq) → Cu²⁺(aq) + 2H₂O(l) Cu(OH)₂(s) + 2OH⁻(aq) → [Cu(OH)₄]²⁻(aq)

Example 9

A solution contains 0.010 M Ca²⁺ and 0.010 M La³⁺. What concentration of HCl is needed to precipitate La₂(C₂O₄)₃·9H₂O but not Ca(C₂O₄)·H₂O if the concentration of oxalic acid is 1.0 M? K_sp values are 2.32 × 10⁻⁹ for Ca(C₂O₄) and 2.5 × 10⁻²⁷ for La₂(C₂O₄)₃; pK_a1 = 1.25 and pK_a2 = 3.81 for oxalic acid.

Given: concentrations of cations, K_sp values, and concentration and pK_a values for oxalic acid

Asked for: concentration of HCl needed for selective precipitation of La₂(C₂O₄)₃

Strategy:

A Write each solubility product expression and calculate the oxalate concentration needed for precipitation to occur. Determine the concentration range needed for selective precipitation of La₂(C₂O₄)₃·9H₂O.

B Add the equations for the first and second dissociations of oxalic acid to get an overall equation for the dissociation of oxalic acid to oxalate. Substitute the [ox²⁻] needed to precipitate La₂(C₂O₄)₃·9H₂O into the overall equation for the dissociation of oxalic acid to calculate the required [H⁺].

Solution:

A Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox²⁻ for oxalate, we write the solubility product expression for calcium oxalate as follows:

$$\begin{array}{cc}\hfill K_{\text{sp}}=[{\text{Ca}}^{2+}][{\text{ox}}^{2-}]& =(0.010)[{\text{ox}}^{2-}]=2.32\times {10}^{-9}\hfill \\ \hfill [{\text{ox}}^{2-}]& =2.32\times {10}^{-7}\text{ M}\hfill \end{array}$$

The expression for lanthanum oxalate is as follows:

$$\begin{array}{cc}\hfill K_{\text{sp}}={[{\text{La}}^{3+}]}^2{[{\text{ox}}^{2-}]}^3& ={(0.010)}^2{[{\text{ox}}^{2-}]}^3=2.5\times {10}^{-27}\hfill \\ \hfill [{\text{ox}}^{2-}]& =2.9\times {10}^{-8}\text{ M}\hfill \end{array}$$

Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between 2.9 × 10⁻⁸ M and 2.32 × 10⁻⁷ M.

B To prevent Ca²⁺ from precipitating as calcium oxalate, we must add enough H⁺ to give a maximum oxalate concentration of 2.32 × 10⁻⁷ M. We can calculate the required [H⁺] by using the overall equation for the dissociation of oxalic acid to oxalate:

$${\text{HO}}_2{\text{CCO}}_2\text{H(aq)}\rightleftharpoons 2{\text{H}}^{+}\text{(aq)}+{\text{C}}_2{\text{O}}_4{}^{2-}\text{(aq)}$$ $$K=K_{\text{a1}}{K}_{\text{a2}}=({10}^{-1.25})({10}^{-3.81})={10}^{-5.06}=8.7\times {10}^{-6}$$

Substituting the desired oxalate concentration into the equilibrium constant expression,

$$\begin{array}{ccc}\hfill 8.7\times {10}^{-6}& =\frac{{[{\text{H}}^{+}]}^2[{\text{ox}}^{2-}]}{[{\text{HO}}_2{\text{CCO}}_2\text{H}]}\hfill & =\frac{{[{\text{H}}^{+}]}^2(2.32\times {10}^{-7})}{1.0}\hfill \\ \hfill & [{\text{H}}^{+}]& =6.1\text{ M}\hfill \end{array}$$

Thus adding enough HCl to give [H⁺] = 6.1 M will cause only La₂(C₂O₄)₃·9H₂O to precipitate from the solution.

Exercise

A solution contains 0.015 M Fe²⁺ and 0.015 M Pb²⁺. What concentration of acid is needed to ensure that Pb²⁺ precipitates as PbS in a saturated solution of H₂S, but Fe²⁺ does not precipitate as FeS? K_sp values are 6.3 × 10⁻¹⁸ for FeS and 8.0 × 10⁻²⁸ for PbS.

Answer: 0.018 M H⁺