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Heterogeneous Equilibria

Skills to develop

Heterogeneous Equilibria and Precipitation

Phase transitions such as sublimation, deposition, melting, solidification, vaporization, and condensation are heterogeneous equilibria, so are the formation of crystals from a saturated solution, because a solid and its solution are separated phases. The picture shown here is the formation of solid from a gaseous solution.

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

Ag+(aq) + Cl-(aq) ® AgCl(s)   (precipitate) Silver chloride is one of the few chloride that has a limited solubility. A precipitate is also formed when sodium carbonate is added to a sample of hard water, Ca2+(aq) + CO32-(aq) ® CaCO3(s)     (precipitate). When a solid dissolves, we have the reverse reaction: AgCl(s) ® Ag+(aq) + Cl-(aq)     (dissolving)
CaCO3(s) ® Ca2+(aq) + CO32-(aq)     (dissolving)

Solubility Products, Ksp, and Ion Products Qsp

Formations of precipitates are chemical equilibria phenomena, and we usually write these heterogeneous equilibrium in the following manner, and call the equilibrium constants solubility products, Ksp:
Equilibrium Expression for Ksp and Qsp
AgCl(s) = Ag+(aq) + Cl-(aq) [Ag+] [Cl-]
CaCO3(s) = Ca2+(aq) + CO32-(aq) [Ca2+] [CO3-]
Li2CO3(s) = 2 Li+(aq) + CO32-(aq) [Li+]2 [CO3-]
The solubility product, Ksp, is a special type of equilibrium constant given to a solution containing sparingly soluble salts. The symbols (aq) indicate that these ions are surrounded by water molecules. These ions are in the solution.

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 Ksp is a constant. They do not appear in the Ksp expression. If the solution is not saturated, no precipitate will form. In this case, the product is called the ion product, Qsp.

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

Qsp, Ksp and Saturation

For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated. These substances have a tendency to form oversaturated solutions. For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids. For oversatureated solutions, Qsp is greater than Ksp. When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring Qsp to approach Ksp.

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 Qsp, Ksp and saturation is given below:

Qsp < Ksp Unsaturated solution
Qsp = Ksp Saturate solution
Qsp > Ksp Oversaturate solution

Molar Solubilities and Solubility Products

Solubility products, Ksp, of salts are indirect indication of their solubilities expressed in mol/L (called molar solubility). However, the solubility products are more useful than molar solubility. The molar solubilities are affected when there are common ions present in the solution. We need to employ the solubility products to estimate the molar solubilities in these cases.

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+ (= [CO32-]) in the saturated solution, then

Ksp = x2 In this case, x is also called the molar solubility of CaCO3. The following examples illustrate the relationship between solubility products, Ksp, and molar solubilities.

Examples 1

The Ksp 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, Ksp 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-] = Ksp, is represented by a line which divides the plane into two zones.

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

When [Ag+] [Cl-] > Ksp, 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

The Ksp for Ag2CrO4 is 9e-12 M3. What is the molar solubility of Ag2CrO4 in pure water?

Solution
Let x be the molar solubility of Ag2CrO4, then

      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
and the molar solubility is 1.3e-4 M.

Discussion
A similar diagram to the one given for AgCl can be drawn, but the shape of the curve representing the Ksp is different.

Example 3

The Ksp for Cr(OH)3 is 1.2e-15 M4. What is the molar solubility of Cr(OH)3 in pure water?

Solution

Let x be the molar solubility of Cr(OH)3, then you have

       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

Very careful experiment indicates that the molar solubility of Bi2S3 is 1.8e-15 M, what value of Ksp does this compound have?

Solution
If the molar solubility of Bi2S3 is 1.8e-15, then

    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

Discussion
You should be able to calculate the Ksp if you know the molar solubility.

Review of skills

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

  2. Calculate Ksp 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 Ksp

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

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