The Solubility-Product Constant
The equilibria that we have considered thus far have involved gases and acids and bases. Furthermore, they have been homogeneous, that is all the species have been in the same phase. We will now take up equilibria involving another important type of reaction: the dissolution or precipitation of ionic compounds. These reactions are heterogeneous. In our earlier discussion of precipitation reactions, we considered some general rules for predicting the solubility of common salts in water. These rules give us a qualitative sense of whether a compound will have a high or low solubility in water. By considering solubility equilibria we can make quantitative predictions about the amount of a given compound that will dissolve. We can also use these equilibria to analyze the factors that affect solubility.
A saturated solution of a slightly soluble salt in contact with undissolved salt involves an equilibrium like the one below.
CaF2(s) <==> Ca2+(aq) + 2 F-(aq)
In writing the equilibrium constant expression for a heterogeneous equilibria, we ignore the concentrations of pure liquids and solids. So the equilibrium constant expression for the equilibria above is:
Ksp = [Ca2+][F-]2
This equilibrium constant is called a solubility-product constant. Even though [CaF2] is excluded from the equilibrium constant expression, some undissolved CaF2(s) must be present in order for the system to be at equilibrium. In general, the solubility product constant (Ksp) is the equilibrium constant for the equilibrium that exists between a solid ionic solute and its ions in a saturated aqueous solution. The rules for writing the solubility-product expression are the same as those for writing any other equilibrium constant expression: The solubility product is equal to the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation.
Solubility and Ksp
It is important to distinguish carefully between solubility and solubility product. The solubility of a substance is the quantity that dissolves to form a saturated solution. Solubility is often expressed as grams of solute per liter of solution. The molar solubility is the number of moles of the solute that dissolve in forming a liter of saturated solution of the solute. The solubility product constant is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution. The solubility of a substance changes as the concentration of other solutes change. In contrast the solubility product for a given solute is constant at a specific temperature.
In studying solubility equilibria, it is important to be able to interconvert solubility and solubility product. The following examples will illustrate this.
Example 1
The Ksp for CaF2 is 3.9 x 10-11 at 25oC. What is the (a) molar solubility of CaF2 in water? (b) What is the solubility in grams per liter?
First set up an equilibrium table and let x = [Ca2+] (and the molar solubility of CaF2 )
| Concentrations(M) | CaF2(s) | <==> | Ca2+(aq) | 2 F-(aq) | |
| Starting | 0 | 0 | |||
| Change | +x | +2x | |||
| Equilibrium | x | 2x |
Ksp = [Ca2+][F-]2 = (x)(2x)2 = 4x3 = 3.9 x 10-11
(a) x = (3.9 x 10-11 /4)1/3 = 2.1 x 10-4 M
(b) Convert moles per liter of CaF2 to grams per liter.
2.1 x 10-4 mol CaF2 78.1 g CaF2
------------------- x ------------ = 1.6 x 10-2 g CaF2 /L soln
1 L soln 1 mol CaF2
Example 2
Analysis of a saturated solution of silver chromate, Ag2CrO4, indicates that the concentration of silver ion is 1.3 x 10-4 M. What is the Ksp of Ag2CrO4?
The equilibrium equation and the solubility product expression are
Ag2CrO4(s) <==> 2 Ag+(aq) + CrO42- Ksp = [Ag+]2[CrO4]
From the equation we can see that at equilibrium, the concentration of CrO42- is going to be half that of Ag+.
[ CrO42-] = 1/2 [Ag+] = 1/2 (1.3 x 10-4 M) = 6.5 x 10-5
Ksp = [Ag+]2[CrO4] = (1.3 x 10-4)2( 6.5 x 10-5) = 1.1 x 10-12