Solubility of Ionic Salts in Water: The Common Ion Effect

James Richard Fromm


The concentrations of ions in solution are affected by all equilibria and all species present in the solution. The simplest and most significant such effect is called the common ion effect. The common ion effect is observed whenever an ion in solution is common to two different salts which serve as its sources. Addition of the second salt adds the common ion, which is a product of the dissolution of the first. The effect of adding the product ion will be to decrease the solubility of the first salt. The data required to calculate the magnitude of this effect in aqueous solutions are the molar solubility constants.

In an example in a previous section, the solubility of AgCl in water was shown to be 1.33 x 10-5 mol/dm3. Now if a quantity of the much more soluble salt NaCl is added, chloride ion is common to both salts and the chloride ion from NaCl will repress the ionization of AgCl according to the Principle of Le Chatelier. If the concentration of Na+ is 1.00 x 10-3 mol/litre then the concentration of Cl- from NaCl is 1.00 x 10-3 mol/litre also, and that is one hundred times as great as the concentration of chloride which would be produced by dissolution of solid AgCl. Ignoring the negligible contribution of solid AgCl to the actual concentration of chloride, we can say that [Cl-] = 1.0 x 10-3 = [Na+]. Since the actual concentration of chloride is what affects all equilibria present,

Ksp = 1.76 x 10-10 = 1.00 x 10-3[Ag+], [Ag+] = 1.76 x 10-7

The addition of sodium chloride has reduced the molar solubility of silver chloride, which is still equal to [Ag+] but no longer equal to [Cl-], by two orders of magnitude. A common ion effect is observed whenever any constituent ion is added to a saturated solution of a slightly soluble salt. The molar solubility of silver chloride would be reduced by adding a source of aqueous silver ion, such as the soluble salt AgNO3.


Example. The molar solubility product of Ag2CrO4 is 1.12 x 10-12. Let us compute the solubility of silver chromate in (a) pure water, and (b) a solution of 0.1 molar sodium chromate.

a. The molar solubility product Ksp = [Ag+]2[CrO42-] for Ag2CrO4(s) rarrow.gif (63 bytes) 2Ag+ + CrO42- and the molar solubility, or the molar concentration of silver chromate going into solution, is equal to [CrO42-] and is also equal to [Ag+]/2. In terms of the concentration of chromate ion, Ksp = (2[CrO42-])2[CrO42-] = 4[CrO42-]3[CrO42-] = (the cube root of)(Ksp/4) = (the cube root of)0.279 x 10-12 = 6.53 x 10-5

b. The molar solubility product is the same in this case as it was before, but the molar solubility is not. The molar solubility is not equal to [CrO42-] since chromate ion comes from both the slightly soluble silver chromate and the very soluble sodium chromate. In this case twice the molar solubility is equal to [Ag+]. It is important to note the additional source of chromate, which to a very good approximation will establish the actual molar concentration of chromate as 0.1. Then

Ksp = 1.12 x 10-12 = [Ag+]2[CrO42-]. The molar concentration of silver ion is then (the square root of)(1.12 x 10-12/0.1) which is 3.34 x 10-6 and the molar solubility of silver chromate in 0.1 molar sodium chromate is 1.67 x 10-6 mol/litre.


Solubility equilibria are present whenever a solid substance is in contact with a solution, and as with all equilibrium constants they reflect the actual concentrations in solution. Thus if these concentrations are altered, as by addition of reagents, removal of material, addition of water, or by other chemical reactions or equilibria, solid material may come out of solution, or precipitate from solution, or on the other hand solid material may dissolve, or go into solution.

A table of molar solubility products is useful in understanding these examples.


Example. Let us calculate the equilibrium concentrations of Ag+(aq) and Cl-(aq) when 50 mL of 0.01 molar silver nitrate solution are added to 100 mL of 0.02 molar sodium chloride solution.

The initial conditions are given and the equilibrium concentrations are required, so the actual equilibrium conditions must first be established. These, except for chemical reaction, are: volume, 150 mL; Ag+, 0.5 mmol; NO3-, 0.5 mmol; Na+, 2.0 mmol; and Cl-, 2.0 mmol. The nitrate ion and sodium ion have nothing to react with, so they will remain unchanged. The silver ion and chloride ion can react with each other, forming insoluble AgCl, to the extent of about 0.5 mmol. This would reduce the amount of silver ion to a very low value and the amount of chloride ion to 1.5 mmol. The approximate equilibrium molar concentrations are then about 0.0033, nitrate; 0.0133, sodium ion; and 0.010, chloride ion. The very low concentration of silver ion can be calculated from the molar solubility product:

Ksp = 1.76 x 10-10 = [Ag+][Cl-],[Ag+] = 1.76 x 10-8


Addition of an ion which can precipitate more than one possible slightly soluble salt will precipitate more than one slightly soluble salt if enough of the precipitating ion is added. The slightly soluble salts will precipitate in the order least soluble to most soluble. For example, addition of silver ion to an aqueous solution which is equimolar in chloride ion and bromide ion will precipitate AgBr first and will only then precipitate AgCl.


Example. Let us calculate the concentration of silver ion required to begin precipitation of AgBr from a solution originally 0.01 molar in both bromide ion and chloride ion. Then calculate (a) the concentration of silver ion required to begin precipitation of AgCl, and (b) the concentration of bromide ion remaining when precipitation of AgCl begins.

The concentration of silver ion required to begin precipitation of the AgBr is 5.35 x 10-13/0.01 = 5.35 x 10-11 mol/litre; that required to begin the precipitation of AgCl is 1.76 x 10-10/0.01 = 1.76 x 10-8 mol/litre. The concentration of bromide ion remaining when AgCl begins to precipitate is 5.35 x 10-13/1.76 x 10-8 = 3.0 x 10-5. Since the original concentration of bromide was 0.01 molar, there remains only 0.3% of it. About 99.7% of the original amount or concentration of bromide ion precipitated as AgBr before any of the chloride precipitated as AgCl.


Copyright 1997 James R. Fromm