Strong acids and bases
Strong acids ionize completely in solution:
E.g., HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)
0.1 moles 1 L
0.1 M
0.1 M
pH = – log[H3O+] = – log 0.1 = 1.0
→ strong acids have low pH
Strong bases completely ionize in solution:
E.g., Ba(OH)2 (s) + H2O(l) → Ba2+(aq) + 2OH-(aq)
0.1 moles
1L
0.1 M
0.2 M
pOH = – log[OH-] = – log 0.2 = 0.7
pOH = 14.00 – pH
pH
= 14.0 – 0.7 = 13.3
→ strong bases have high pH
Strong acids (in the correct proportions) will
neutralise a basic solution and vice versa:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
(H+ + Cl- + Na+ + OH-) → (Na+ + Cl-)(aq) + H2O
Net ionic equation: H+(aq) + OH-(aq) → H2O(l)
Weak acids and bases
Weak acids and bases ionize partially in solution –
an ionization equilibrium exists
Acid ionization:
Consider a weak acid HA
HA(aq) + H2O(l)
acid 1
base 2
Kc =
H3O+(aq) + A-(aq)
acid 2
base 2
[H3O+][A-]
[HA][H2O]
Ka = Kc[H2O]
[H3O+][A-]
[HA]
Ka is the acid ionization constant
Stronger acids have higher Ka → greater ionization
Ka =
For weak acids: [HA] >> [H+] and [A-]
Polyprotic acids:
Acids that yield more than one H+ ion per molecule
Proton loss occurs in a stepwise manner with each
step having its own equilibrium
E.g.,
Carbonic acid, H2CO3
H2CO3 (aq) + H2O(l)
H3O+(aq) + HCO3-(aq)
Ka = 4.3×10-7
bicarbonate ion
HCO3-(aq) + H2O(l)
H3O+(aq) + CO32-(aq)
Ka = 4.8×10-11 carbonate ion
Note that the conjugate base in the 1st step is the
acid in the 2nd step
Ka1 >> Ka2 >> Ka3
[H3O+] and pH difficult to calculate
Base ionization:
Consider a weak base B:
B(aq) + H2O(l)
base 1 acid 2
HB+(aq) + OH-(aq)
acid 1
base 2
Kc = [HB+][OH-]
[B][H2O]
Kc[H2O] = Kb
Kb = [HB+][OH-]
[B]
Kb is the base ionisation constant
Stronger bases have higher Kb → greater ionization
For weak bases, [B] >> [HB+] and [OH-]
Salt solutions
Salts can dissolve in water to give acidic, neutral, or
basic solutions
Neutral solutions:
Salt of strong base (NaOH) and strong acid (HCl)
NaCl(s) + H2O(l) → Na+(aq) + Cl-(aq)
Neither Na+ nor Cl- react with H2O: solution is
neutral
Basic solutions:
Salt of strong base (NaOH) and weak acid
(CH3COOH)
CH3COONa(s) + H2O(l) → Na+(aq) + CH3COO-(aq)
CH3COO-(aq) + H2O(l)
CH3COOH(aq)+OH-(aq)
Kb can be calculated to give pOH and pH
Acidic solutions:
Salt of a weak base (NH3) and strong acid (HCl)
NH4Cl(s) + H2O(l) → NH4 (aq) + Cl-(aq)
NH4+(aq) + H2O(l)
NH3 (aq) + H3O+(aq)
Ka can be calculated to give pH
Salt solutions
pH of solution depends on the relative strength of
acid base strengths of ions
Kb (anion) > Ka (cation)
Basic
Kb
< Ka
Acidic
Kb
≈ Ka
Neutral
Ka, Kb, and Kw:
Ka and Kb for conjugate acid-base pairs are related
Ka
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
Kb
CH3COO-(aq) + H2O(l)
CH3COOH(aq) + OH-(aq)
Kw
H2O(l) + H2O(l)
H3O+ + OH-
Ka Kb = Kw
Common Ion Effect
Addition of an ion which is already present in an
acid-base equilibrium will affect the pH
Generally for a weak acid HA and a soluble salt of a
the acid (e.g., NaA)
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
Ka = [H3O+][A-]
[HA]
and [H3O+] =
Ka
[HA]
Taking negative logarithms:
–log [H3O+] = – log Ka – log [HA]/[A-]
pH = – log Ka + log [A-]/[HA]
pH = pKa + log [A-]/[HA]
(where – log Ka = pKa)
pH = pKa + log [base]/[acid]
Buffer solutions
A buffer solution resists changes in pH when an
acid or a base are added
Buffers must contain enough acid or base to react
with any added OH- or H3O+
Weak acid-base conjugate pairs are excellent
buffers
E.g., Acetic acid and sodium acetate:
Addition of acid:
CH3COO-(aq)+H2O(l)→CH3COOH(aq)+OH-(aq)
Addition of base:
CH3COOH(aq)+OH-(aq)→CH3COO-(aq)+H2O(l)
For a buffer to be effective there must be
comparable amounts of acid and base present
Effective pH range = pKa ± 1.0
Preparing buffer solutions:
If [base] ≈ [acid] then pH ≈ pKa
To prepare a buffer solution with a specific pH,
choose a weak acid whose pKa ≈ pH