Solubility Product

Although some chemical compounds are classified as insoluble, most "insoluble" compounds are actually very slightly soluble. The solubility process reaches a dynamic equilibrium described by an equilibrium constant. This equilibrium constant is called the solubility product constant, and is given the symbol Ksp. The solubility product can be calculated given a compound's solubility. The solubility constant can be used to calculate whether a precipitate will form given the concentrations of ionic species. Compounds having a common ion but different solubility constants can be separated by fractional precipitation.

The solubility product constant Ksp is just one more version of the standard equilibrium constant expression. Here, it's applied to the equilibrium between a solid and the corresponding ions in solution. For example, if there is an equilibrium:

A2B(s) < = > 2A+(aq) + B2-(aq)
the corresponding solubility product constant expression will be:

Ksp = [A+]2[B-]

(Remember that solids do not enter into an equilibrium constant expression.)

To compute the Ksp from given concentrations of ions in solution, use the same method that you have used to compute K: write the balanced equation and expression for Ksp, use stoichiometry to compute the concentration on each ion, then plug in.

Calculate Whether a Precipitate Will Form

The reaction quotient, Q, can be used to determine whether a precipitate will form with a given concentration of ions. (The reaction quotient is also called the ion product when it is calculated using concentrations of species involved in solubility equilibria.) One first calculates Q, then compares it with Ksp.
If Q < Ksp, no precipitate will form.
If Q = Ksp, a precipitate will form.
If Q > Ksp, a precipitate will form.

Note that precipitation may not happen immediately if Q is equal to or greater than Ksp. A solution could be supersaturated for some time until precipitation occurs.

Fractional Precipitation

Fractional precipitation is a technique that separates ions from solution based on their different solubilities. For example, consider an aqueous solution that contains initially Ba2+ and Sr2+ ions. Suppose that an aqueous solution of potassium chromate (K2CrO4) is added slowly. Two precipitates can form: BaCrO4 with Ksp of 1.2 x 10-9 M, and SrCrO4 with Ksp of 3.5 x 10-5 M. (Since all potassium salts are soluble, the potassium ion of the aqueous solution of K2CrO4 is a spectator ion and does not need to be considered in the solubility equilibrium.) Because Ksp for SrCrO4 is smaller than for BaCrO4, SrCrO4 will precipitate first. If K2CrO4 is added to the solution so that the ion product for BaCrO4 is not exceeded, then only SrCrO4 will precipitate. The resulting solution can be filtered and the Sr2+ isolated from the original solution as the chromate salt.

pH and Solubility

If one of the ions appearing in a solubility equilibrium is the conjugate base of a weak acid, then pH can affect the concentrations of ions in the solubility equilibrium. The magnitude of this effect can be estimated from the solubility product.

Example: Consider two slightly soluble salts, calcium carbonate and calcium sulfate. Which of these would have its solubility more affected by the addition of HCl, a strong acid? Would the solubility of the salt increase or decrease?.

A strong acid affects the solubility of a salt of a weak acid by providing hydronium ion that reacts with the anion of the salt in the solution. As the anion is removed by this reaction, more salt dissolves to replenish the anion concentration. You need to ask, which salt has the anion corresponding to the weaker acid?

Calcium carbonate gives the solubility equilibrium

CaCO3(s) <=> Ca2+(aq) + CO32-(aq)

When a strong acid is added, the hydrogen ion reacts with carbonate ion, because it is conjugate to a weak acid (HCO3-).
H3O+(aq) + CO32-(aq) <=> H2O(l) + HCO3-(aq)

As carbonate ion is removed, calcium carbonate dissolves. Moreover, the hydrogen carbonate ion itself is removed in further reaction.

H3O+(aq) + HCO3-(aq) -->H2O(l) + H2CO3(aq) --> 2H2O(l) + CO2(g)

Bubbles of carbon dioxide gas appear as more calcium carbonate dissolves.

For calcium sulfate, the corresponding equilibria are

CaSO4(s) <=> Ca2+(aq) + SO42-(aq)
H3O+(aq) + SO42-(aq) <=> H2O(l) + HSO4-(aq)

Again, the anion of the insoluble salt is removed by reaction with hydronium ion. However, HSO4- is a much stronger acid than HCO3-, as you can see by comparing acid-ionization constants. (Ka for HCO-3 = 4.8 x 10-11 << Ka for HSO4- = 1.1 x 10-2) Thus, calcium carbonate is much more soluble in acidic solution, whereas the solubility of calcium sulfate is only slightly affected.

The effect of pH on solubility can be used to separate metal ions by sulfide precipitation. Hydrogen sulfide is a stronger acid than water, it ionizes in water as a diprotic acid:
H2S(aq) + H2O(l) <=> H3O+(aq) + HS-(aq); Ka1 = 8.9 x 10-8
HS-(aq) + H2O(l) <=> H3O+(aq) + S2-(aq); Ksp = 1.2 x 10-13
The ionization forms sulfide ion, S2-, which can combine with the metal ions to precipitate the sulfides:

Zn2+(aq) + S2-(aq) <=> ZnS(s); Ksp = 1.1 x 10-21
Pb2+(aq) + S2-(aq) <=> PbS(s); Ksp = 2.5 x 10-27

By adjusting the pH of the solution, one can adjust the sulfide-ion concentration in order to precipitate the least soluble metal sulfide while maintaining the other metal ion in solution. The solubility product constant of lead(II) sulfide is much smaller than that of zinc sulfide, so it is the least soluble sulfide. When a solution that is 0.10 M in each metal ion and 0.30 M in hydronium ion is saturated with hydrogen sulfide, lead(II) sulfide precipitates, but zinc ion remains in solution. You can now filter off the precipitate of lead(II) sulfide, leaving a solution containing the zinc ion.

Quantitative Analysis

Qualitative analysis involves the determination of the identity of the substances present in a mixture. One scheme for the qualitative analysis of a solution of metal ions is based on the relative solubilities of the metal sulfides. The figure below shows a flowchart illustrating how the metal ions in an aqueous solution are first separated into five analytical groups.

Complex Ions

In addition to forming precipitates, some metal ions in solution can react with ligands to form soluble complex ions. The metal ions, ligands, and complex ions are in dynamic equilibrium described by a formation constant. The solubility of normally insoluble metal salts can be increased by the formation of complex ions.

A complex ion is an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. A complex is a compound containing complex ions. A ligand is a Lewis base that bonds to a metal ion to form a complex ion. For example, Ag(NH3)2+ is a complex ion formed from the Ag+ ion and two NH3 molecules. The NH3 molecules are the ligands.

The overall equation for the formation of the complex ion Ag(NH3)2+is given by:

Ag+(aq) + 2NH3(aq) <=> Ag(NH3)2+(aq)

The formation constant, or stability constant, Kf, of a complex ion is the equilibrium constant for the formation of the complex ion from the aqueous metal ion and the ligands.

The dissociation constant (Kd) for a complex ion is the reciprocal, or inverse, value of Kf. The equation for the dissociation of Ag(NH3)2+ is Ag(NH3)2+(aq) <=> Ag+ (aq) + 2NH3(aq) .

 Kd = 1 = [Ag+][NH3]2 Kf [Ag(NH3)2+]
and
 Kf = [Ag(NH3)2+] [Ag+][NH3]2

The solubility of normally insoluble metal salts can be increased by the formation of complex ions.

Example1: What is the concentration of Ag+(aq) ion in 0.010 M AgNO3 that has also 1.00 M NH3? Kf for Ag(NH3)2+ ion is 1.7 x 107.

The formation constant for Ag(NH3)2+ is large, so silver exists primarily as this ion. This suggests that you do this problem in two parts: 1) the stoichiometry calculation, in which you assume that Ag+(aq) reacts completely to form Ag(NH3)2+(aq); 2) the equilibrium calculation, in which this complex ion dissociates to give a small amount of Ag+(aq).
1) Stoichiometry calculation: In 1 L of solution, you initially have 0.010 mol Ag+(aq) from AgNO3. This reacts to give 0.010 mol Ag(NH3)2+, leaving (1.00 - 2 x 0.010) mol NH3, which equals 0.98 mol NH3 to two significant figures. You now look at the equilibrium for the dissociation of Ag(NH3)2+.
2) Equilibrium calculation:

 Concentration (M) Ag(NH3)2+(aq) <=> Ag+(aq) + 2NH3(aq) Starting 0.010 0 0.98 Change - x + x + 2x Equilibrium 0.010 - x x 0.98 + 2x

The dissociation constant is

 [Ag+] [NH3]2 = Kd = 1 1 = ~ 5.9 x 10-8 [Ag(NH3)2+] Kf 1.7 x 10 7

Substituting into this equation gives:

 x(0.98)2 x(0.98)2 = 5.9 x 10-8 ~ (0.010 - x) 0.010

and x ~ 5.9x10-8 x 0.010/(0.98)2 = 6.1 x 10-10

Example2: (a) Will silver chloride precipitate from a solution that is 0.010 M AgNO3 and 0.010 M NaCl?

(a) To determine whether a precipitate should form, you calculate the ion product and compare it with Ksp for AgCl (1.8 x 10-10).

Qi = [Ag+]i [Cl-]i= (0.010) (0.010) = 1.0 x 10-4

This is greater than Ksp = 1.8 x 10-10, so a precipitate should form.

(b) Wold that precipitate form if we also had in the solution 1.00 M NH3? You first need to calculate the concentration of Ag+(aq) in a solution containing 1.00 M NH3. We did this in Example 1 and found that [Ag+] equals 6.1 x 10-10. Hence,

Qi = [Ag+]i [Cl-]i = (6.1 x 10-10)(0.010) = 6.1 x 10-12

Because the ion product is smaller than Ksp = 1.8 x 10-10, no precipitate should form.

Amphoteric hydroxides can react with hydroxide ions to form soluble complex ions, as shown by the acid-base reactions of Zn(OH)2.

Zinc hydroxide reacts with a strong acid, as you would expect of a base, and the metal hydroxide dissolves.

Zn(OH)2(s) + 2H3O+(aq) <=> Zn2+(aq) + 4H2O(l)

With a base, however, Zn(OH)2 reacts to form the complex ion Zn(OH)42-.

Zn(OH)2(s) + 2OH-(aq) <=> Zn(OH)42-(aq)