Acid-Base

1. CM4106 Chemical Equilibria & Thermodynamics
Lesson 2
Acid-Base Equilibria
A Chemistry Education Blog by Mr Tan
http://chemistry-mr-tan-yong-yao.blogspot.sg/

2. Fundamentals:
1. Identify acids/ bases
Acid Base
1. proton H+ donor 1. proton H+ acceptor
2. electron acceptor 2. electron donor
(vacant orbital) (lone pair)
2. Identify conjugate acids/ bases
– H+
Acid Conjugate Base
+ H+
+ H+
Base Conjugate Acid
– H+

3. Fundamentals:
3. Identify stronger acids/ bases
(A) In any acid-base reaction, the equilibrium will favor the
reaction where the stronger acid reacts with the stronger
base. i.e. (In any acid-base reaction, the equilibrium will
favor the reaction that moves the proton to the stronger
base.)
(B) The stronger an acid, the weaker its conjugate base.
The stronger a base, the weaker its conjugate acid.
HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq)
CH3CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + CH3CO2- (aq)

5. Fundamentals:
Important relations
p = -log10
At 25ºC:
pH + pOH = pKw = 14
pKa + pKb = 14
[H+][OH–] = 10-14
Ka x Kb = Kw = 10-14

6. (I) Calculations
pH, pKa, [H+], Ka pOH, pKb, [OH–], Kb
Step 1: Determine what is present in the solution.
(A) Acid : Strong Acid vs Weak Acid
Monoprotic Acid / Diprotic Acid / Triprotic Acid
(B) Base: Strong Base vs Weak Base
Take note of concentration of acid / bases
(For dilute solutions, we need to take into consideration of
[H+] / [OH-] from auto-ionization from water)
Step 2: Use the appropriate equations for the respective species.

7. pH of Acid/ Base
Ka = x2
Strong Acid Weak Acid ([HA] – x)
Strong acids dissociate HA(aq) ⇌ H+(aq) + A-(aq)
completely into ions in
aqueous solution.
I [HA] 0 0
C -x +x +x
[H+] = [HA] E [HA] - x +x +x
Kb = x2
Strong Base Weak Base ([B] – x)
Strong bases dissociate B + H2O ⇌ BH+ + OH-
completely into ions in
I [B] - 0 0
aqueous solution.
C -x - +x +x
[OH-] = [B] E [B] - x - x x

8. pH of Acid/ Base
1. Determine if acid/ base is strong or weak (more
common)
2. For weak acids, if asked to determine Ka, pH or [H+], you
can save time by using the formula:
Assumption: x is negligible
[H+] = Ka × c [OH–] = Kb × c
Example:
Calculate the pH of 0.50M HF solution at 25ºC where the Ka = 7.1 x 10-4
[H+][F-] x2
Ka = = = 7.1 x 10-4
[HF] (0.50 – x)
Assumption: For weak acids, x must be very small  0.50 – x ≈ 0.5
x2
= 7.1 x 10-4 x = [H+] = 0.0188 M Assumption is valid;
0.50
x < 5% of [HF]initial
pH = 1.73 (to 2 d.p.)

9. Calculate the pH of the following solutions at 298K
0.10 mol dm−3 CH3COOH (pKa = 4.75) Weak acid solution
(monoprotic acid)
CH3COOH(aq) + H2O (l) ⇌ CH3COO(aq) + H3O+(aq)
Initial (M) 0.10 - 0.00 0.00
Change (M) -x - +x +x
Eqm (M) 0.10 - x - +x +x
[CH3COO][H3O+] x2
Ka = = = 10 4.75 = 1.79 x 10-5
[CH3COOH] (0.10 – x)
Assumption: For weak acids, x must be very small  0.10 – x ≈ 0.10
x2
= 1.79 x 10-5 Assumption is justified,
0.10
x < 5% of [CH3COOH]initial
x = [H3O+] = 1.338 x 10-3 M
Concentration: 2 s.f.
pH = 2.87 ( 2 d.p.) pH: 2 d.p.

10. Calculate the pH of the following solutions at 298K
0.30 mol dm3 ethylamine, CH3CH2NH2 (pKb= 3.27) Weak base solution
(monoprotic base)
CH3CH2NH2 (aq) + H2O (l) ⇌ CH3CH2NH3+(aq) + OH-(aq)
Initial (M) 0.30 - 0.00 0.00
Change (M) -x - +x +x
Eqm (M) 0.30 - x - +x +x
[CH3CH2NH3+][OH-] x2
Kb = = = 10 3.27 = 5.37 x 104
[CH3CH2NH2] (0.30 – x)
Assumption: For weak bases, x must be very small  0.30 – x ≈ 0.30
x2
= 5.37 x 10-4 Assumption is justified,
0.30
x < 5% of [CH3CH2NH2]initial
x = [OH] = 0.01269 M
pH = 14.00 – 1.896
pOH = - log10
[OH] pH = 12.10 (2 d.p.)
= - log10[0.01269 ] Concentration: 2 s.f.
= 1.896 pH: 2 d.p.

11. pH of salt solutions
Salt solutions can be (i) neutral (ii) weak acids or (iii) weak bases
Weak Acid Salt
CH3COONa (aq) → CH3COO- (aq) + Na+ (aq)
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
conjugate base pH > 7
Weak Basic Salt
NH4Cl (aq) → NH4+ (aq) + Cl- (aq)
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
conjugate acid pH < 7