Consider the lead(II) ion concentration in this saturated solution of PbCl₂. The balanced reaction is

PbCl₂(s) ? Pb²⁺(aq) + 2Cl^-(aq)

Assuming the concentration of dissolved lead(II) chloride is s, then:

[Pb²⁺] = s

[Cl^-] = 2s

Put these values into the solubility product expression, and do the sum.

So the concentration of lead(II) ions in the solution is 1.62 x 10^-2 M. What happens if you add some sodium chloride to this saturated solution? Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".

Look at the original equilibrium expression again:

PbCl₂(s) ? Pb²⁺(aq) + 2Cl^-(aq)

What would happen to that equilibrium if you added extra chloride ions? According to Le  Châtelier, the position of equilibrium will shift to counter what you have just done. In this case, it would tend to remove the chloride ions by making extra solid lead(II) chloride.

 

Note:  Actually, of course, the concentration of lead(II) ions in the solution is so small to start with, that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride.

The lead(II) chloride will become even less soluble - and, of course, the concentration of lead(II) ions in the solution will decrease. Something similar happens whenever you have a sparingly soluble substance. It will be less soluble in a solution which contains any ion which it has in common. This is the common ion effect.

The common ion effect of H₃O⁺ on the ionization of acetic acid

The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.

Consider the common ion effect of OH^- on the ionization of ammonia

Therefore, adding the common ion of hydroxide, shifts the reaction towards the left to decrease the stress (LeChâtelier's Principle), forming more reactants. This decreases the reaction quotient, for the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, K_b=1.8*10^-5, does not change. The reaction is put out of balance, or equilibrium.

Q_A= [NH⁴⁺][OH^-]/[NH₃]

At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing Q to decrease towards K.

Example 1

A 0.150 M solution of formic acid at 25°C (pK_a = 3.75) has a pH of 2.28 and is 3.5% ionized.

1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?

Worked Problem Here

Subpages (1): Example 1