VII. Precipitation Equilibria
VII-1. Introduction and Background VII-2. Sample Problem VII-3. List of Problems VII-4. Simple Problems VII-5. Advanced Problems

Introduction

A precipitate is a solid compound that forms from ions in solution. Precipitation occurs when the concentration of ions in solution exceed the solubility limit for that particular compound. A common example that is used to analyze for lead in solution is the formation of solid lead chromate, PbCrO4.

The figure to the right represents a precipitate of solid PbCrO4 in equilibrium with Pb2+ and CrO42- ions in solution. At equilibrium the concentrations of ions in solution are constant. Pb2+ and CrO42- ions continue to form solid PbCrO4, and solid PbCrO4 continues to dissolve. Because the rate of precipitation and dissolution are the same, there is no change in the concentrations of the ions in solution.


Solubility Product Constant, Ksp

When a precipitate is present, equilibrium concentrations of the cations and anions remain in solution. The equilibrium concentrations depend on the equilibrium constant, which is called the solubility product constant, Ksp. (Ksp values are usually tabulated for pure water at 25oC.)

Precipitation equilibria are always written in the direction of the solid dissolving. For PbCrO4 example the equilibrium reaction is:

PbCrO4 (s) <--> Pb2+(aq) + CrO42-(aq)

For this reaction, the equilibrium constant expression is:

Ksp = [Pb2+][CrO42-]

Use the usual rules for equilibrium constant expressions: the products go over the reactants, and the concentration of the pure solid is not included.

The Ksp value for PbCrO4 is 1.8x10-14. This very small Ksp indicates that the equilibrium concentrations remaining in solution are very low. PbCrO4 is therefore a good precipitate to use to get all of the lead out of a solution for analysis.


Dissolving Precipitates

How can we dissolve precipitates?
Remember that the precipitate is in equilibrium with ions in solution and think about LeChatelier's principle. If you remove ions from solution, the system will be out of equilibrium, and the precipitate will dissolve to replace the ions that are removed.

Examples:

Adding strong acid:

      CaCO3(s)  <-->  Ca2+(aq) + CO32-(aq)
CO32-(aq) + 2H+  <-->  H2CO3(aq)
--------------------------------------
CaCO3(s) + 2H+  <-->  H2CO3(aq) + Ca2+(aq)
Adding NH3 or strong base to complex metals:
          Zn(OH)2(s)  <-->  Zn2+(aq) + 2OH-(aq)
  Zn2+(aq) + 4NH3(aq)  <-->  Zn(NH3)42+(aq)
------------------------------------------------
Zn(OH)2(s) + 4NH3(aq)  <-->  Zn(NH3)42+(aq) + 2OH-(aq)


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