• Define K_sp, the solubility product.
• Explain solid/solution equilibria using K_sp and Q_sp.
• Calculate molarity of saturated solution from K_sp.
• Calculate K_sp from molarity of saturated solution.

Heterogeneous Equilibria and Precipitation

The equilibrium constants for saturated solution and solid formation (precipitate) are called solubility product, K_sp. For unsaturated and supersaturated solutions, the system is not at equilibrium, and ion products, Q_sp, which have the same expression as K_sp is used.

An oversaturated solution becomes a saturated solution by forming a solid to reduce the dissolved material. The crystals formed are called a precipitate. Often, however, a precipitate is formed when two clear solutions are mixed. For example, when a silver nitrate solution and sodium chloride solution are mixed, silver chloride crystals AgCl(s) (a precipitate) are formed. Na⁺ and NO₃^- are by-stander ions.

Solubility Products, K_sp, and Ion Products Q_sp

Note the expression for the solubility product given above, please. These are special equilibrium constant, because the solid present has a constant tendency of being dissolved. Therefore, their role in K_sp is a constant. They do not appear in the K_sp expression. If the solution is not saturated, no precipitate will form. In this case, the product is called the ion product, Q_sp.

The Table of solubility product is given as Salt, K_sp in the Handbook Section. In this table, the salts are divided into

• Carbonates, salts of CO₃^2-
• Chromates, salts of CrO₄^2-
• Halides, salts of Cl^-, Br^-, and I^-
• Hydroxides, salts of OH^-
• Oxalates, salts of C₂O₄^2-
• Sulfates, salts of SO₄^2-
• Sulfides, salts of S^2-

Q_sp, K_sp and Saturation

Sodium acetate trihydrate, NaCH₃COO^.3H₂O, when heated to 370 K will become a liquid. The sodium acetate is said to be dissolved in its own water of crystallization. The substance stays as a liquid when cooled to room temperature or even below 273 K. As soon as a seed crystal is present, crystallization occur rapidly. In such a process, heat is released, and the liquid feels warm. Thus, the relationship among Q_sp, K_sp and saturation is given below:

Molar Solubilities and Solubility Products

When a salt is dissolved in pure water, solubility products and molar solubilities are related. This is illustrated using calcium carbonate. If x is the concentration of Ca²⁺ (= [CO₃^2-]) in the saturated solution, then

Examples 1

The K_sp for AgCl is 1.8e-10 M². What is the molar solubility of AgCl in pure water?

Solution
Let x be the molar solubility, then

     AgCl  =  Ag+  +  Cl-
              x       x

     x = (1.8e-10 M2)1/2
       = 1.3e-5 M

Discussion

The solubility product, K_sp is a better indicator than the usual solubility specification of g per 100 mL of solvent or moles per unit volume of solvent.

For the AgCl case, when the cation concentration is not the same as the anion concentration ([Ag⁺] =/= [Cl^-]) solubility of AgCl can not be defined in terms of moles per L. In this case, the system can be divided into three zones. The condition [Ag⁺] [Cl^-] = K_sp, is represented by a line which divides the plane into two zones.

When [Ag⁺] [Cl^-] < K_sp, no precipitate will be formed.

When [Ag⁺] [Cl^-] > K_sp, a precipitate will be formed.

When AgCl and NaCl dissolve in a solution, both salts give Cl^- ions. The effect of [Cl^-] on the solubility of AgCl is called the Common ion effect

Example 2

      Ag2CrO4  =  2 Ag+  +  CrO42-
                 2 x        x
      (2 x)2 (x) = Ksp
      x = {(9e-12)/4}(1/3)
        = 1.3e-4 M
      [Ag+] = 2.6e-4 M

Example 3

       Cr(OH)3  =  Cr3+  +  3 OH-
                    x      3 x
Thus,
       x (3 x)3  =  1.2e-15
       x  =  {(1.2e-15)/27}(1/4)
          =  8.2e-5 M

Example 4

    Bi2S3 = 2 Bi(3+) + 3 S2-
           3.6e-15    5.4e-15
    Ksp  =  (3.6e-15)2 (5.4e-15)3
         =  2.0e-72 M5

Review of skills

1. Be able to write the K_sp expression for the ionization of any salts. and calculate K_sp from molar solubility

2. Calculate K_sp from molar solubility and vice versa. Pay attention to the stoichiometry of the salt. You may have to understand what the salt really is in order to know how they ionize in the solution. You have learned the method to calculate molar solubility from K_sp

3. Perform calculations and be able to tell if a precipitate will form. The calculation is an important part of chemistry.

Confidence Building Questions

• Experiment shows that the molar solubility of CuCl is 1.1e-3. What is its solubility product, K_sp?

  Answer 1.2e-6
  Hint...
  The equilibrium equation and concentrations x are shown below

    CuCl  =  Cu+  +  Cl-
             x       x    (x = 1.1e-3)
  

  K_sp = [Cu⁺][Cl^-] = (1.1e-3)² = ? M²
  If CuCl₂ is used, you would have CuCl₂ = Cu²⁺ + 2 Cl^-

• The K_sp for Hg₂I₂ is 4.5e-29 M³. What is its molar solubility in pure water? Remember Hg₂²⁺ is a stable ion in solution.

  Answer 2.2e-10
  Hint...
  Let x be the molar solubility, then [Hg₂²⁺] = x; and [I^-] = 2 x;

      Hg2I2  =  Hg22+  +  2 I-
       x        x        2 x
  

  Molar solubility = [Hg₂²⁺].
  [I^-] = 2 [Hg₂²⁺].
  K_sp = x (2 x)² = 4.5e-29; x = ?

• What is the pH in a saturated solution of Ca(OH)₂?
  K_sp = 5.5e-6 M³ for Ca(OH)₂.

  Answer 12.35
  Hint...
  Let [Ca²⁺] = x, then [OH^-] = 2 x. The equilibrium and concentration are represented below:

     Ca(OH)2  =  Ca2+  +  2 OH-
                  x      2 x
     x = {5.5e-6/4}(1/3)
  

  [OH^-] = 2 x.
  Note the stoichiometry of equilibrium.

• Chemical analysis gave [Pb²⁺] = 0.012 M, and [Br^-] = 0.024 M in a solution. From a table, you find K_sp for PbBr₂ has a value of 4e-5 M³. Is the solution saturated, oversaturated or unsaturated?

  Answer unsaturated
  Hint...
  [0.012] [0.024]² = 6.9e-6 < 4e-5 (K_sp).
  This question deals with the concept of ion product, Q_sp.
  If Q_sp = [Pb²⁺][Br^-]² < K_sp the solution is unsaturated.