Acidity of Solutions of Weak Bases

James Richard Fromm

An aqueous solution of a weak base is treated by the same methods as used for aqueous solutions of weak acids. It is more convenient to use the base ionization constant Kb rather than the acid ionization constant Ka, calculate pOH, and then calculate pH from pOH. The value of Kb is simply Kw/Ka, where Ka is the ionization constant for the conjugate acid of the weak base whose value of base ionization constant Kb is desired.

Example. Let us calculate the pH of a solution of methylamine whose total concentration is 0.01 molar. The value of the molar ionization constant of methylammonium ion, the conjugate acid of methylamine, is 2.39 x 10-11, so using 1.0 x 10-14 as the ion product of water gives a value of the molar ionization constant Kb for methylamine of 4.18 x 10-4. Again, as in the case of aqueous solutions of weak acids, both an algebraic approach and a reasoning approach may be used. In either case it is necessary to begin with the ionization constant of the weak base:

Kb = [OH-][CH3NH3+]/[CH3NH2] = 4.18 x 10-4

Since all of the methylamine originally present must remain, either as methylamine or as methylammonium ion, [CH3NH2] = 0.01 - [CH3NH3+] and [CH3NH3+] = [OH-]. Substitution of these equations into each other gives

[OH-]2/(0.01 - [OH-]) = 4.18 x 10-4

[OH-]2 + 4.18 x 10-4[OH-] - 4.18 x 10-6 = 0

Applying the quadratic formula and discarding the physically meaningless negative concentration,

[OH-] = 1.85 x 10-3, pOH = -(0.27 - 3) = 2.73

pH + pOH = pKw = 14, pH = 14 - 2.73 = 11.27

Alternatively, the same type of less mathematical reasoning can be used as was used for weak acids. The base is a weak base, and therefore must be only slightly ionized, so [CH3NH2] is approximately equal to 0.01 mol/liter, and:

[OH-][CH3NH3+]/[CH3NH2] = 4.18 x 10-4

[OH-][CH3NH3+]/0.01 = 4.18 x 10-4

[OH-][CH3NH3+] = 4.18 x 10-6, [OH-]2 = 4.18 x 10-6

[OH-] = 2.05 x 10-3, pOH = - (0.31 - 3) = 2.69,

pH = 14 - pOH = 11.31, in excellent agreement with the preceding calculation.

Let us now consider what would happen if the base were even weaker than the methylamine used in the above example. A base ionization constant of 4.18 x 10-5 would give a pH of 10.81 assuming no ionization, while the more accurate quadratic method would give a pH of 10.80 because the no-ionization approximation is closer to being correct. There is a less significant difference between the two values for the weaker base. A stronger base would have to be used before the difference between the no ionization approximation and the more accurate quadratic method would become significant.

Example. If methylamine had a base ionization constant whose value were two orders of magnitude less, 4.18 x 10-6, and a total concentration of 0.01 mol/liter, let us calculate what the pH of the resulting solution would be. The equations needed in this example are the ionization constant, dissociation extent, and ionization of water:

Kb = [OH-][CH3NH3+]/[CH3NH2]

[CH3NH2] + [CH3NH3+] = 0.01

[OH-] = [CH3NH3+] + [H3O+]; Kw = [OH-][H3O+]

In this case the approximation that the ionization of the base proceeds only to a slight extent is very good, so [CH3NH2] is approximately 0.01:

4.18 x 10-6 = [OH-][CH3NH3+]/0.01

4.18 x 10-8 = [OH-][CH3NH3+]

[OH-] = [CH3NH3+] + Kw/[OH-]

[OH-]2 = [CH3NH3+][OH-] + Kw

[OH-]2 = (4.18 x 10-8/[OH-])[OH-] + Kw

[OH-]2 = 4.18 x 10-8 + 1 x 10-14

For all practical purposes, since the ionization of water is still negligible,

[OH-]2 = 4.18 x 10-8, [OH-] = 2.04 x 10-4, pH = 10.31

As the strength of the acid or base increases, however, the assumption that essentially all of it is present in the unionized form begins to become less realistic as the following example shows.

Example. If methylamine had a base ionization constant whose value were two orders of magnitude greater, 4.18 x 10-2, and a total concentration of 0.01 mol/dm3, let us calculate what the pH of the resulting solution would be. Again, we begin with Kb = [OH-][CH3NH3+]/[CH3NH2] and [CH3NH2] + [CH3NH3+] = 0.01. The ionization of water is again neglected so that [CH3NH3+] equals [OH-], and:

4.18 x 10-2 = [OH-][CH3NH3+]/(0.01 - [CH3NH3+])

4.18 x 10-4 - 4.18 x 10-2[CH3NH3+] = [OH-][CH3NH3+]

[OH-]2 + 4.18 x 10-2[OH-] - 4.18 x 10-4 = 0

[OH-] = 0.0083; pOH = 2.08, pH = 11.92

Neglect of the ionization of methylamine would give

4.18 x 10-4 = [OH-]2, [OH-] = 0.0204, pOH = 1.69, pH = 12.31

The difference would be now be more significant. The answers are still of the same order of magnitude.

On the basis of the examples considered above, we may conclude that in solutions which contain a weak acid or weak base neglect of the ionization of water and the ionization of the weak acid or base is reasonable almost all of the time.

Strong and Weak Bases Together

In a mixture of strong and weak bases, the strong base continues to be completely ionized. The OH- ion resulting from this, according to Le Chatelier's Principle, represses the ionization of the weak base so that, given reasonably high concentrations of the strong base, the strong base alone effectively determines the pH of the solution.

In both of the approaches used above, and with both weak acids and weak bases, the ionization of water has been neglected. Only if the concentration and strength of the weak acid or base being considered are great enough that the solution pH is not affected significantly by water ionization can this be done without significant error. However, consideration of water ionization would give a cubic equation and increased mathematical complexity. Only when the product of the ionization constant and the species concentration is approaching the value of 1 x 10-14 can the error due to neglect of water ionization become significant.

Copyright 1997 James R. Fromm