CHEMISTRY THE CENTRAL SCIENCE

17 ADDITIONAL ASPECTS OF AQUEOUS EQUILIBRIA

17.6 PRECIPITATION AND SEPARATION OF IONS

Equilibrium can be achieved starting with the substances on either side of a chemical equation. For example, the equilibrium that exists between BaSO₄(s), Ba²⁺(aq), and SO₄^2–(aq) (Equation 17.15), can be achieved either by starting with BaSO₄(s) or by starting with solutions containing Ba²⁺ and SO₄^2–. If we mix, say, a BaCl₂ aqueous solution and a Na₂SO₄ aqueous solution, BaSO₄ precipitates if the product of the initial ion concentrations, Q = [Ba²⁺][SO₄^2–], is greater than K_sp.

The use of the reaction quotient Q to determine the direction in which a reaction must proceed to reach equilibrium was discussed in Section 15.6. The possible relationships between Q and K_sp are as follows:

• If Q > K_sp, precipitation occurs, reducing ion concentrations until Q = K_sp.

• If Q = K_sp, equilibrium exists (saturated solution).

• If Q < K_sp, solid dissolves, increasing ion concentrations until Q = K_sp.

SAMPLE EXERCISE 17.15 Predicting Whether a Precipitate Forms

Does a precipitate form when 0.10 L of 8.0 × 10^–3M Pb(NO₃)₂ is added to 0.40 L of 5.0 × 10^–3M Na₂SO₄?

SOLUTION

Analyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined.

Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with K_sp for any potentially insoluble product. The possible metathesis products are PbSO₄ and NaNO₃. Like all sodium salts NaNO₃ is soluble, but PbSO₄ has a K_sp of 6.3 × 10^–7 (Appendix D) and will precipitate if the Pb⁺² and SO₄^2– concentrations are high enough for Q to exceed K_sp.

Solve When the two solutions are mixed, the volume is 0.10 L + 0.40 L = 0.50 L. The number of moles of Pb²⁺ in 0.10 L of 8.0 × 10^–3M Pb(NO₃)₂ is

The concentration of Pb²⁺ in the 0.50-L mixture is therefore

The number of moles of SO₄^2– in 0.40 L of 5.0 × 10^–3M Na₂SO₄ is

Therefore

and

Because Q > K_sp, PbSO₄ precipitates.

PRACTICE EXERCISE

Does a precipitate form when 0.050 L of 2.0 × 10^–2M NaF is mixed with 0.010 L of 1.0 × 10^–2M Ca(NO₃)₂?

Answer: Yes, CaF₂ precipitates because Q = 4.6 × 10^–8 is larger than K_sp = 3.9 × 10^–11

Selective Precipitation of Ions

Ions can be separated from each other based on the solubilities of their salts. Consider a solution containing both Ag⁺ and Cu²⁺. If HCl is added to the solution, AgCl (K_sp = 1.8 × 10^–10) precipitates, while Cu²⁺ remains in solution because CuCl₂ is soluble. Separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or more (but not all) of the ions is called selective precipitation.

SAMPLE EXERCISE 17.16 Calculating Ion Concentrations for Precipitation

A solution contains 1.0 × 10^–2M Ag⁺ and 2.0 × 10^–2M Pb²⁺. When Cl^– is added, both AgCl (K_sp = 1.8 × 10^–10) and PbCl₂ (K_sp = 1.7 × 10^–5) can precipitate. What concentration of Cl^– is necessary to begin the precipitation of each salt? Which salt precipitates first?

SOLUTION

Analyze We are asked to determine the concentration of Cl^– necessary to begin the precipitation from a solution containing Ag⁺ and Pb²⁺ ions, and to predict which metal chloride will begin to precipitate first.

Plan We are given K_sp values for the two precipitates. Using these and the metal ion concentrations, we can calculate what Cl^– concentration is necessary to precipitate each salt. The salt requiring the lower Cl^– ion concentration precipitates first.

Solve For AgCl we have K_sp = [Ag⁺][Cl^–] = 1.8 × 10^–10

Because [Ag⁺] = 1.0 × 10^–2M, the greatest concentration of Cl^– that can be present without causing precipitation of AgCl can be calculated from the K_sp expression:

Any Cl^– in excess of this very small concentration will cause AgCl to precipitate from solution. Proceeding similarly for PbCl₂, we have

Thus, a concentration of Cl^– in excess of 2.9 × 10^–2M causes PbCl₂ to precipitate.

Comparing the Cl^– concentration required to precipitate each salt, we see that as Cl^– is added, AgCl precipitates first because it requires a much smaller concentration of Cl^–. Thus, Ag⁺can be separated from Pb²⁺ by slowly adding Cl^– so that the chloride ion concentration remains between 1.8 × 10^–8M and 2.9 × 10^–2M.

Comment Precipitation of AgCl will keep the Cl^– concentration low until the number of moles of Cl^– added exceeds the number of moles of Ag⁺ in the solution. Once past this point, [Cl^–] rises sharply and PbCl₂ will soon begin to precipitate.

PRACTICE EXERCISE

A solution consists of 0.050 M Mg²⁺ and Cu²⁺. Which ion precipitates first as OH^– is added? What concentration of OH^– is necessary to begin the precipitation of each cation? [K_sp = 1.8 × 10^–11 for Mg(OH)₂, and K_sp = 4.8 × 10^–20 for Cu(OH)₂.]

Answer: Cu(OH)₂ precipitates first, beginning when [OH^–] > 1.5 × 10^–9M. Mg(OH)₂ begins to precipitate when [OH^–] > 1.9 × 10^–5M.

GO FIGURE

What would happen if the pH were raised to 8 first and then H₂S were added?

FIGURE 17.22 Selective precipitation. In this example Cu²⁺ ions are separated from Zn²⁺ ions.

Sulfide ion is often used to separate metal ions because the solubilities of sulfide salts span a wide range and depend greatly on solution pH. For example, Cu²⁺ and Zn²⁺ can be separated by bubbling H₂S gas through an acidified solution containing these two cations. Because CuS (K_sp= 6 × 10^–37) is less soluble than ZnS (K_sp = 2 × 10^–25), CuS precipitates from an acidified solution pH ≈ 1 while ZnS does not (FIGURE 17.22):

The CuS can be separated from the Zn²⁺ solution by filtration. The CuS can then be dissolved by raising the concentration of H⁺ even further, shifting the equilibrium of Equation 17.27 to the left.