Education.com
Try
Brainzy
Try
Plus

# Solubility Equilibria for AP Chemistry

(not rated)
By — McGraw-Hill Professional
Updated on Feb 9, 2011

Practice problems for these concepts can be found at:

Many salts are soluble in water, but some are only slightly soluble. These salts, when placed in water, quickly reach their solubility limit and the ions establish an equilibrium system with the undissolved solid. For example, PbSO4, when dissolved in water, establishes the following equilibrium:

The equilibrium constant expression for systems of slightly soluble salts is called the solubility product constant, Ksp. It is the product of the ionic concentrations, each one raised to the power of the coefficient in the balanced chemical equation. It contains no denominator since the concentration of a solid is, by convention, 1 and does not appear in the equilibrium constant expressions. (Some textbooks will say that the concentrations of solids, liquids, and solvents are included in the equilibrium constant.) The Ksp expression for the PbSO4 system would be:

For this particular salt the numerical value of Ksp is 1.6 × 10–8 at 25°C. Note that the Pb2+ and SO42– ions are formed in equal amounts, so the right-hand side of the equation could be represented as [x]2. If the numerical value of the solubility product constant is known, then the concentration of the ions can be determined. And if one of the ion concentrations can be determined, then Ksp can be calculated.

For example, the Ksp of magnesium fluoride in water is 8 × 10–8. How many grams of magnesium fluoride will dissolve in 0.250 L of water?

If a slightly soluble salt solution is at equilibrium and a solution containing one of the ions involved in the equilibrium is added, the solubility of the slightly soluble salt is decreased. For example, let's again consider the PbSO4 equilibrium:

Suppose a solution of Na2SO4 is added to this equilibrium system. The additional sulfate ion will disrupt the equilibrium, by Le Châtelier's principle, and shift it to the left, decreasing the solubility. The same would be true if you tried to dissolve PbSO4 in a solution of Na2SO4 instead of pure water—the solubility would be lower. This application of Le Châtelier's principle to equilibrium systems of slightly soluble salts is called the common-ion effect. Calculations like the ones above involving finding concentrations and Ksp can still be done, but the concentration of the additional common ion will have to be inserted into the solubility product constant expression. Sometimes, if Ksp is very small and the common ion concentration is large, the concentration of the common ion can simply be approximated by the concentration of the ion added.

For example, calculate the silver ion concentration in each of the following solutions:

1. Ag2CrO4(s) + water
2. Ag2CrO4(s) + 1.00 MNa2CrO4
3. Ksp = 1.9 × 10–12

Knowing the value of the solubility product constant can also allow us to predict whether a precipitate will form if two solutions, each containing an ion component of a slightly soluble salt, are mixed. The ion-product, sometimes represented as Q (same form as the solubility product constant), is calculated taking into consideration the mixing of the volumes of the two solutions, and this ion-product is compared to Ksp. If it is greater than Ksp, precipitation will occur until the ion concentrations have been reduced to the solubility level.

If 10.0 mL of a 0.100 M BaCl2 solution is added to 40.0 mL of a 0.0250 M Na2SO4 solution, will BaSO4 precipitate? Ksp for BaSO4 = 1.1 × 10–10

To answer this question, the concentrations of the barium ion and the sulfate ion before precipitation must be used. These may be determined simply from Mdil = Mcon Vcon/Vdil.

For Ba2+: Mdil = (0.100 M)(10.0 mL)/(10.0 + 40.0 mL) = 0.0200 M
For SO42–: Mdil = (0.0250 M)(40.0 mL)/(50.0 mL) = 0.0200 M

Entering these values into the following relation produces:

Q = [Ba2+][SO42–] = (0.0200)(0.0200) = 0.000400

Since Q is greater than Ksp, precipitation will occur.

### Other Equilibria

Other types of equilibria can be treated in much the same way as the ones discussed above. For example, there is an equilibrium constant associated with the formation of complex ions. This equilibrium constant is called the formation constant, Kf. Zn(H2O)42+ reacts with ammonia to form the Zn(NH3)42+ complex ion according to the following equation:

The Kf of Zn(NH3)42+(aq) is 7.8 × 108, indicating that the equilibrium lies to the right.

Practice problems for these concepts can be found at:

### Ask a Question

150 Characters allowed

### Related Questions

#### Q:

See More Questions
Top Worksheet Slideshows