1. Starter question.
Consider propanoic acid (CH3CH2COOH) reacting with water…..
1. Write a symbol equation for the equilibrium that is established.
2. Identify the conjugate base and acid on the right hand side of the equation.
3. Write a Kc expression for the equilibrium.

2. Starter answers CH3CH2COOH (aq) + H20(l) ↔ H3O+(aq) + CH3CH2COO-(aq)
Conjugate base….CH3CH2COO-
Conjugate acid H3O+
Ka = [H+(aq)][CH3CH2COO-(aq)]
[CH3CH2COOH (aq)]

3. Calculating the pH of a weak acid.

4. Calculating the pH of a weak acid. An example.
What is the pH of 0.1moldm-3 ethanoic acid, which has a Ka of….1.74X10-5moldm-3?)
You need to…
1. Write out the equilibrium expression (Ka)
2. convert hydrogen concentration [H+] to pH

5. Step 1 the Ka expression. Equilibrium equation.
CH3COOH (aq) ↔ H+ (aq) + CH3COO-(aq)
So the general Ka expression will be……
Ka = [CH3COO-(aq)][H+(aq)]
[CH3COOH(aq)]
But we need to consider when full equilibrium has been reached and take account of our initial concentration of acid.

6. Step 2 the equilibrium position.
CH3COOH ↔ H CH3COO-
CH3COOH H+
CH3COO-
Start Conc.
0.100
At equilibrium.
[CH3COO-]
[H+]
[CH3COO-]
CH3COOH
For every ethanoic acid molecule that dissociates an ethanoate ion and a H+ also forms. H+
CH3COO-

7. Writing the new Ka expression at equilibrium.
Initial concentration minus the dissociation concentration of the ions.
Ka = [CH3COO-(aq)]eq[H+ (aq)]eq
0.100 – [CH3COO-(aq)]eq
As [CH3COO-] = [H+] at equilibrium we can write…
Ka = [H+]2
0.100 – [H+]
Because the disassociation of the weak acid is so small we assume – [H+] = 0.100

8. Plugging the numbers in..
Substituting in the values gives….
1.74 X 10-5 = [H+]2
0.100
[H+]2= X 1.74 X 10-5
[H+] = √(1.74 X 10-6)
[H+]= X 10-3moldm-3
= pH = (using –log)