Weak Acids and Bases
Weak Acids – weak electrolytes that do not ionize completely to form H+ ions
Examples: HF, H2CO3, H2S, H3BO3
Weak Bases – a compound that that reacts incompletely with water to form OHions according to the following equation:
B:(aq) + H2O(l) ↔ OH-(aq) + HB+(aq)
where B: is a base
To act as a Brønsted-Lowry base, a compound must possess an atom with a lone
pair of valence electrons. This lone pair accepts an H+ ion (remember bases are
proton acceptors) from water to form OH-(aq). This is why the base is represented
as B:
According to Le Châtelier’s principle, the hydroxide ions added to water cause a
shift to the left in the autoionization of water, decreasing the hydroxide ion
concentration and producing a pH greater than 7.
H2O(l) ↔ OH-(aq) + H+(aq)
Examples:
ammonia, sodium phosphate
The classic example is ammonia with its lone pair of
electrons:
NH3(g) + H2O(l) ↔ NH4+(aq) + OH-(aq)
base
acid
Determining the Percent Ionization of Weak Acids
1.3%
HC2H3O2(aq) + H2O(l)
↔
C2H3O2-(aq) + H3O+(aq)
p = concentration of acid ionized x 100%
concentration of acid solute
p = percent ionization
p = [H+(aq)] x 100%
[HA(aq)]
or
[H+(aq)] = p x [HA(aq)]
100
[HA(aq)] = conc. of the acid
In 0.1 mol/l Acetic Acid (HC2H3O2(aq))
[H+(aq)] = 1.3 x [0.1 mol/L]
100
+
[H (aq)] = 1.3 x 10-3 mol/L
Since pH = - log [H+(aq)]
pH = -log(1.3 x 10-3 mol/L)
= 3.75
Sample Question:
The pH of a 0.10 mol/L methanoic acid solution is 2.38. Calculate the percent
ionization of methanoic acid.
HCO2H(aq) + H2O(l) ↔ HCO2-(aq) + H3O+(aq)
Or more easily stated:
HCO2H(aq) ↔ HCO2-(aq) + H+(aq)
pH = 2.38,
therefore [H+(aq)] = 10-pH
[H+(aq)] = 10-2.38
[H+(aq)] = 4.2 x 10-3 mol/L
[H+(aq)] = p x [HCO2H(aq)]
[HCO2H(aq)] = concentration of the acid
100
p=
[H+(aq)] x 100%
[HCO2H(aq)]
p = 4.2 x 10-3 mol/L x 100% = 4.2%
0.1 mol/L
Ionization Constants for Weak Acids
Ka = acid ionization constant (equilibrium constant for the ionization of an acid)
HC2H3O2(aq) ↔ H+(aq)
+
C2H3O2-(aq)
Ka = [H+(aq)][C2H3O2-(aq)]
[HC2H3O2(aq)]
see page 554 to calculate Ka from percent ionization
Ionization Constants for Weak Bases
Kb = base ionization constant (equilibrium constant for the ionization of a base)
NH3(g) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Kb = [NH4+(aq)][OH-(aq)]
[NH3(g)][H2O(l)]
Kb = [NH4+(aq)][OH-(aq)]
[NH3(g)]
Relationship between Ka and Kb
KaKb = Kw
Kb = Kw
Ka
(Hess’ Law is used to calculate this on page 559/560)
Ka = Kw
Kb
remember Kw = 1 x 10-14
See page 561 (and appendix C7) for determining Ka given Kb
See page 563 – 568 to determine the pH of an acid given Ka