SCH 4U- Acids-Bases Equilibria - Weak Acids/Bases & Equilibrium (Ka & Kb)
Weak Acids/Bases
Strong acids and bases ionize in aqueous solutions completely.
100%
HClO3 + H2O →
ClO3-
100%
+
KOH → K+ + OH-
+ H3O
Weak acids/bases do NOT dissociate 100% in aqueous solutions.
<100%
HC2H3O2 + H2O ⇄
C2H3O2-
<100%
NH3 + H2O ⇄ NH4+ + OH-
+
+ H3O
Weak acids and bases do not dissociate completely because they are somewhat stable molecules, and often
have some covalent character to them. HF is a special case; it is so strongly ionic that fluoride ion is capable of
‘taking’ a proton from water molecules.
Table 1 – Some Weak Acids and Bases
Weak Acids
Weak Bases
Acetic acid – HC2H3O2
Ammonia – NH3
Methanoic acid - HCOOH
Methylamine – CH3NH2
Hydrofluoric acid – HF
Dimethylamine – CH3CH3NH
Citric acid – H3C6H5O7
Trimethylamine – CH3CH3CH3N
Nitrous acid – HNO2
Hydrazine – N2H4
Benzoic acid – C6H5COOH
Hydroxylamine – NH2OH
The percentage to which a weak acid/base ionizes is calculated as the percentage of [H+] or [OH-] that appears in
solution.
For Acids:
[H + ]
%i =
× 100%
[HA]
For Bases:
[OH − ]
%i =
× 100%
[B]
Example:
In a 0.10 M solution of acetic acid, the pH is measured to be 2.89. What is the % ionization?
Acid-Base Equilibrium Constant, Ka and Kb
How do we determine the H+/OH- concentration of a weak acid/base if we do not know the % ionization? No
problem! The dissociation of a weak acid/base in solution is an example of a dynamic equilibrium! Therefore, it
is an equilibrium problem 
Consider acetic acid:
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
As some acetic acid molecules are donating protons from water, acetate ions are accepting protons from
hydronium ions. Therefore, the forward and reverse are occurring simultaneously.
The equilibrium expression for this reaction is called the Ka since it is an acid.
+
[C2 H3 O−
2 ][H3 O ]
𝐾𝑎 =
[HC2 H3 O2 ]
If we know the Ka of acetic acid, we can determine the pH of a given concentration of this weak acid.
Example problem
If the Ka of acetic acid is 1.8  10-5, what is the pH of a 0.20 M solution of acetic acid?
The solution is solved in three steps:
1. Set up an ICE table
2. Substitute the expressions for concentrations in the equilibrium constant expression
3. Solve for x.
Consider ammonia:
NH3(aq) + H2O(l) ⇄ NH4+(aq) + OH-(aq)
As some ammonia molecules are accepting protons from water, ammonium ions are donating protons to
hydroxide ions. Therefore, the forward and reverse are occurring simultaneously.
The equilibrium expression for this reaction is called the Kb since it is a base.
[NH4+ ][OH − ]
𝐾𝑏 =
[NH3 ]
If we know the Kb of ammonia, we can determine the pH of a given concentration of this weak base.
Example problem
If the Kb of ammonia is 1.8  10-5, what is the pH of a 0.20 M solution of ammonia?
The solution is solved in three steps:
4. Set up an ICE table
5. Substitute the expressions for concentrations in the equilibrium constant expression
6. Solve for x.
The Relationship of Ka and Kb
One interesting consequence of all this is the relationship between the K of a weak acid or base, and the K of its
conjugate…
Consider acetic acid, HC2H3O2:
HC2H3O2 + H2O ⇄ C2H3O2- + H3O+
The conjugate base of acetic acid, the acetate ion, is a weak base.
When you multiply the K of a weak acid or base with the K of its conjugate, it always equals Kw!
𝐾𝑊 = 𝐾𝑎 ∙ 𝐾𝑏
Example problem:
The Ka of acetic acid is 1.8  10-5. What is its conjugate base, and what is the value of its Kb?
Often times in chemistry literature, the Ka or Kb of a weak acid or base will be communicated as the pKa or pKb.
In this case, the ‘p’ works exactly as it does in pH and pOH. That is:
𝑝𝐾𝑎 = −𝑙𝑜𝑔𝐾𝑎
𝑝𝐾𝑏 = −𝑙𝑜𝑔𝐾𝑏
This also yields the same relationship as pH and pOH:
14 = 𝑝𝐾𝑎 + 𝑝𝐾𝑏
Example problems:
a. What is the pKa of acetic acid? (Ka = 1.8x10-5)
b. What is the Kb of methylamine? (pKb = 3.36)
c. What is the pKb of the conjugate base of hydrofluoric acid? (pKa = 3.15)
Weak Acid & Base Problems
1. At 25oC, a 0.500 M solution of ammonia (NH3) is 2.50% dissociated. Calculate the pH.
2. Calculate the percent dissociation of a 0.20 mol/L solution of the weak acid HNO2 if the pH of the solution is
5.10.
3. Calculate the percent dissociation of a 3.1 x 10-4 mol/L solution of the weak acid, HA, if the pH of the
solution is 6.10.
4. Calculate the Ka of a general weak acid, HA, if a 0.500 M solution has a pH of 3.300.
5. A 0.75 M solution of weak base has a pH of 11.00. Find the Kb.
6. A solution having a pH of 9.00 contains 0.20 M of fluoride ion (F-) which is a weak base. Find the Kb of
fluoride ion.
7. What is the equilibrium concentration of a hypochlorous acid (HClO) solution with Ka = 3.0 x 10-8 if the pH is
4.10?
8. Acetate ion, CH3COO-, has a concentration of 2.0 M. Its Kb is 7.7 x 10-5. What is the pH?
9. Lactic acid, HC3H5O3(aq), is a weak acid that gives yogurt its sour taste. Calculate the pH of a 0.0010 M
solution of lactic acid. The Ka for lactic acid is 1.4  10-4.
10. Ascorbic acid, HC6H7O6(aq), is a weak organic acid, also known as vitamin C. A student prepares a 0.20 M
aqueous solution of ascorbic acid, and measures its pH as 2.40. Calculate the Ka of ascorbic acid.
Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
pH= 12.097
4.0  10-3 % (note that Ka here is extremely small: 6.1 x 10-12)
0.26 %
Ka = 5.02 x 10-7
Kb = 1.3 x 10-6
Kb = 5.0 x 10-10
[HClO] = 0.21 M
pH = 12.09
pH = 3.51
Ka = 7.9  10-5