Chapter 15 Equilibria of Other Reaction Classes
15.2 Predicting Precipitation
Learning Objectives
By the end of this section, you will be able to:
- Predict if and at concentration precipitate will form
- Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentration
Predicting Precipitation
The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:
It is important to realize that this equilibrium is established in any aqueous solution containing Ca2+ and CO32– ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate (Na2CO3) and calcium nitrate (Ca(NO3)2). Because both Ca2+ and CO32- are present in the solution, precipitation of CaCO3 can occur if concentrations are high enough.
If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, Qsp, that exceeds the solubility product, Ksp, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (Qsp = Ksp). The comparison of Qsp to Ksp to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:
- Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)
- Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)
This predictive strategy and related calculations are demonstrated in the next few example exercises.
Example 15.7 – Precipitation of Mg(OH)2
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH– ion:
Solution
Calculation of the reaction quotient under these conditions is shown here:
[latex]Q_{sp} = [Mg^{2+}] \, [OH^-]^2 = (0.0537) \, (0.0010)^2 = 5.4 \times 10^{-8}[/latex]
Because Qsp is greater than Ksp (Qsp = 5.4 × 10–8 is larger than Ksp = 8.9 × 10–12), the reverse reaction will proceed. Mg(OH)2 will precipitate until the dissolved ion concentrations have been sufficiently lowered, so that Qsp = Ksp.
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[latex]Q_{sp} = [Ca^{2+}] \, [HPO4^{2-}] = (0.0001) \, (0.001) = 1 \times 10^{-7}[/latex]
Because Qsp < Ksp, the solution is unsaturated and precipitation will not occur.
Example 15.8 – Precipitation of AgCl
Does silver chloride precipitate when equal volumes of a 2.0 × 10–4 M solution of AgNO3 and a 2.0 × 10–4 M solution of NaCl are mixed?
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. Because the volume doubles when equal volumes of AgNO3 and NaCl solutions are mixed, each concentration is reduced to half its initial value
The reaction quotient, Qsp, is greater than Ksp for AgCl, so a supersaturated solution is formed:
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The total solution volume is 100 mL. The new concentrations of K+ and ClO4– are
[latex]\left[K^+\right]=\frac{\left(20\ mL\right)\left(0.050\ M\right)}{100\ mL}=0.010\ M[/latex]
[latex]\left[ClO_4^-\right]=\frac{\left(80\ mL\right)\left(0.50\ M\right)}{100\ mL}=0.40\ M[/latex]
The Qsp is
[latex]Q_{sp}=\left[K^+\right]\left[ClO_4^-\right]=\left(0.010\right)\left(0.40\right)=4.0\times10^{-3}[/latex]
which is less than Ksp = 1.05 × 10–2. The solution is unsaturated; no precipitate will form.
Example 15.9 – Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, C2O42−, for this purpose (Figure 15.5). At sufficiently high concentrations, the calcium and oxalate ions form solid CaC2O4·H2O (calcium oxalate monohydrate). The concentration of Ca2+ in a sample of blood serum is 2.2 × 10–3 M.
What concentration of C2O42− ion must be established before CaC2O4·H2O begins to precipitate?
Solution
The equilibrium expression is:
For this reaction:
(see Appendix J)
Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:
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The dissolution reaction of interest is
Ag2CrO4 (s) ⇌ 2 Ag+(aq) + CrO42−(aq)
and the Ksp is
Ksp = [Ag+]2[CrO42-] = 9.0 x 10-12
Substitute the provided chromate concentration into the solubility product expression and solve for silver concentration:
Example 15.10 – Concentrations Following Precipitation
Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 × 10–6 M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be decreased by adding a base to precipitate Mn(OH)2.
What pH is required to keep [Mn2+] equal to 1.8 × 10–6 M?
Solution
The dissolution of Mn(OH)2 is described by the equation:
or
so
[latex]\text {pOH} = - \log {[OH^-]} = - \log {(3 \times 10^{-4})} = 3.5[/latex]
[latex]\text {pH} = 14.00- \text {pOH} = 14.00 - 3.5 = 10.5[/latex]
(final result rounded to one significant digit, limited by the certainty of the Ksp)
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The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 × 10–2 M.
Calculate the pH at which [Mg2+] is decreased to 1.0 × 10–5 M
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The dissolution of Mg(OH)2 is described by the equation:
or
so
[latex]\text {pOH} = - \log {[OH^-]} = - \log {(9.4 \times 10^{-4})} = 3.03[/latex]
[latex]\text {pH} = 14.00- \text {pOH} = 14.00 - 3.03 = 10.97[/latex]
In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.
Example 15.11 – Precipitation of Silver Halides
A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
Solution
The two equilibria involved are:
AgBr(s) ⇌ Ag+(aq) + Br−(aq) Ksp = 5.0 × 10−13
If the solution contained about equal concentrations of Cl– and Br–, then the silver salt with the smaller Ksp (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag+] precipitates first.
AgBr precipitates when Qsp equals Ksp for AgBr
AgBr begins to precipitate when [Ag+] is 5.0 × 10–9 M.
For AgCl: AgCl precipitates when Qsp equals Ksp for AgCl (1.6 × 10–10). When [Cl–] = 0.10 M:
AgCl begins to precipitate when [Ag+] is 1.6 × 10–9 M.
AgCl begins to precipitate at a lower [Ag+] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a Ksp greater than that of silver bromide.
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The two equilibria are:
AgBr(s) ⇌ Ag+(aq) + Br−(aq) Ksp = 5.0 × 10−13
Because [Cl–] = [Br–], the compound with the smaller Ksp (AgBr) will precipitate first.
The [Ag+] when precipitation of AgBr begins is
Chemistry in Everyday Life
The Role of Precipitation in Wastewater Treatment
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure 15.6). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions (PO43−) are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.
One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)2. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3OH, which then precipitates out of the solution:
Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.
View this site for more information on how phosphorus is removed from wastewater.
Common Ion Effect
Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le ChÂtelier’s principle. Consider the dissolution of silver iodide:
This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag+ and I–. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.
This effect may also be explained in terms of mass action as represented in the solubility product expression:
The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion’s concentration must be balanced by a proportional decrease in the other.
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Example 15.12 – Common Ion Effect on Solubility
What is the effect on the amount of solid Mg(OH)2 and the concentrations of Mg2+ and OH– when each of the following are added to a saturated solution of Mg(OH)2?
(a) MgCl2
(b) KOH
(c) NaNO3
(d) Mg(OH)2
Solution
The solubility equilibrium is
(a) Adding a common ion, Mg2+, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.
(b) Adding a common ion, OH–, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.
(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.
(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.
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Example 15.13 – Common Ion Effect
Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 × 10–28.
Solution
This calculation can be performed using the ICE approach:
| CdS(s) | ⇌ | Cd2+(aq) | + | S2−(aq) | |
| I (M) | — | 0.010 | 0 | ||
| C (M) | (−x) | +x | +x | ||
| E (M) | — | 0.010 + x | x |
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This calculation can be performed using the ICE approach:
| Al(OH)3(s) | ⇌ | Al3+(aq) | + | 3 OH– (aq) | |
| I (M) | — | 0.015 | 0 | ||
| C (M) | (−x) | +x | +3x | ||
| E (M) | — | 0.015 + x | 3x |
