James Richard Fromm
In a solution of a weak acid, which by definition is not completely ionized, there is always a significant amount of the acid present in the protonated form. We will examine the situation of an aqueous solution of a weak acid by considering a typical example in detail.
Example. For the weak acid HF, hydrofluoric acid, the ionization equilibrium is
HF(aq) + H2O H3O+(aq) + F-(aq)
for which the acid ionization constant Ka = 6.94 x 10-4 = [H3O+][F-]/[HF]. We will calculate the pH of an aqueous solution which was made up as 0.1 molar HF.
If the solution has a total concentration of 0.1 molar HF, part of this HF is present as fluoride ion and part of it remains as HF. From the principle of conservation of matter, or in this case simply the statement that all of the fluoride has to be somewhere, [HF] + [F-] = 0.1; [HF] = 0.1 - [F-]. Substitution into the ionization constant gives:
6.94 x 10-4 = [H3O+][F-]/(0.1 - [F-])
6.94 x 10-5 - 6.94 x 10-4[F-] = [H3O+][F-]
For every fluoride ion produced by ionization of HF (the only way it can be produced in this solution), one hydrogen ion is also produced according to the overall balanced ionization equation. If this is the only source of protons, and it is if the ionization of water is negligible, then [H3O+] = [F-] and
6.94 x 10-5 - 6.94 x 10-4[H3O+] = [H3O+][H3O+]
[H3O+]2 + 6.94 x 10-4[H3O+] - 6.94 x 10-5 = 0
This is in the standard quadratic form ax2 + bx + c = 0, which is solved by the quadratic formula:
[H3O+] = (-6.94 x 10-4 +/- (the square root of)(48.16 x 10-8 + 27.76 x 10-5))/2
[H3O+] = (-6.94 x 10-4 +/- (the square root of)(2.776 x 10-4))/2
[H3O+] = (-1.74 x 10-2 or + 1.60 x 10-2)/2
Negative concentrations have no physical meaning, so [H3O+] = 8.0 x 10-3 and thus
log[H3O+] = -3 + 0.90 = - 2.10 and pH = 2.10. This solution is so acidic that we can neglect any contribution from ionization of the water.
Alternatively, a less mathematical reasoning approach may be used to consider the same example. This is a solution of a weak acid at reasonably high concentrations. If the acid is so weak that very little of it is in the unprotonated form, then as a good approximation [HF] = 0.1. Then substitution into the ionization constant gives [H3O+][F-]/0.1 = 6.94 x 10-4, or [H3O+][F-] = 6.94 x 10-5.
Again, for every fluoride ion produced there is also produced one hydrogen ion and so if ionization of HF is the only significant source of H3O+, one concentration can be substituted for the other:
[H3O+][H3O+] = 6.94 x 10-5
[H3O+] = 8.33 x 10-3, log[H3O+] = - 3 + 0.92, pH = 2.08
A very similar answer is obtained either way. The reasoning procedure is easier and nearly always works to within an accuracy of about half a pH unit.
When a solution contains both strong acids and weak acids, the acidity of the solution is affected only slightly by the weaker acids. Strong acids always ionize completely, but the H3O+ they produce does, by Le Chatelier's Principle, act to repress the ionization of the weak acid. For reasonable concentration ratios such solutions behave as solutions of strong acid alone. When more than one acid is present in a solution, the dominant contribution to the acidity is that of the stronger acid alone.
Example. Let us calculate the pH of an aqueous solution which is 0.01 molar in acetic acid and also 0.01 molar in ammonium ion. We begin with the ionization constant of acetic acid and the ionization constant of ammonium ion:
Ka(CH3COOH) = 1.75 x 10-5 = [H3O+][CH3COO-]/0.01
Ka(NH4+ = 5.60 x 10-10 = [H3O+][NH3]/0.01
1.75 x 10-7 = [H3O+][CH3COO-]; 5.60 x 10-12 = [H3O+][NH3]
Neglecting contributions to the concentration of H3O+ from ammonium ion and water,
[H3O+]2 = 1.75 x 10-7 = 17.5 x 10-8;[H3O+] = 4.18 x 10-4, pH = 3.38
At this pH,
[NH3] = 5.60 x 10-12/4.18 x 10-4 = 1.34 x 10-8
[OH-] = 1.01 x 10-14/4.18 x 10-4 = 2.14 x 10-11
The molar concentration of hydrogen ion which would arise from formation of ammonia (1.3 x 10-8 mol/L) and from formation of hydroxide ion (2.14 x 10-11 mol/L) is negligible compared to the concentration of hydrogen ion due to ionization of acetic acid (4.18 x 10-4 mol/L).