pH Calculations for Strong and Weak Acids
pH Calculations for Strong and Weak Acids
Defining pH
pH is a logarithmic scale that measures the concentration of hydrogen ions in solution:
pH = −log₁₀[H⁺]
Conversely: [H⁺] = 10⁻ᵖᴴ
A low pH means a high [H⁺] (acidic). A high pH means a low [H⁺] (alkaline). pH 7 is neutral at 25 °C.
The Ionic Product of Water (Kw)
Water undergoes a very slight self-ionisation:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
The equilibrium expression for this is:
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ (at 25 °C)
This relationship holds in all aqueous solutions (acidic, neutral, or alkaline).
In pure water at 25 °C: [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, so pH = 7.
At higher temperatures, Kw increases (the forward reaction is endothermic), so [H⁺] increases and the pH of pure water falls below 7 — but the water is still neutral because [H⁺] = [OH⁻].
pH of Strong Acids
Strong acids dissociate completely in water. Every molecule ionises:
HCl(aq) → H⁺(aq) + Cl⁻(aq)
H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)
For a monoprotic strong acid (HCl, HNO₃):
[H⁺] = concentration of the acid
Worked Example: Calculate the pH of 0.050 mol dm⁻³ HCl.
[H⁺] = 0.050 mol dm⁻³
pH = −log₁₀(0.050) = 1.30
For a diprotic strong acid (H₂SO₄ — assuming both dissociations are complete):
[H⁺] = 2 × concentration of the acid
Worked Example: Calculate the pH of 0.020 mol dm⁻³ H₂SO₄.
[H⁺] = 2 × 0.020 = 0.040 mol dm⁻³
pH = −log₁₀(0.040) = 1.40
pH of Strong Bases
For strong bases like NaOH (fully dissociates):
[OH⁻] = concentration of NaOH
Then use Kw to find [H⁺]:
[H⁺] = Kw / [OH⁻]
Worked Example: Calculate the pH of 0.10 mol dm⁻³ NaOH at 25 °C.
[OH⁻] = 0.10 mol dm⁻³
[H⁺] = (1.00 × 10⁻¹⁴) / 0.10 = 1.00 × 10⁻¹³ mol dm⁻³
pH = −log₁₀(1.00 × 10⁻¹³) = 13.00
Weak Acids and Ka
Weak acids only partially dissociate in water. An equilibrium is established:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
The acid dissociation constant (Ka) is:
Ka = [H⁺][A⁻] / [HA]
A larger Ka means a stronger (more dissociated) weak acid.
pKa = −log₁₀(Ka) and Ka = 10⁻ᵖᴷᵃ
A smaller pKa means a stronger acid.
Calculating pH of a Weak Acid
Two simplifying assumptions are made:
1. [H⁺] = [A⁻] — because the only significant source of both is the dissociation of HA (the contribution from water self-ionisation is negligible)
2. [HA] at equilibrium ≈ initial concentration — because the acid is weak, very little HA dissociates
With these assumptions:
Ka = [H⁺]² / [HA]
Rearranging: [H⁺] = √(Ka × [HA])
Then: pH = −log₁₀[H⁺]
Worked Example
Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³).
[H⁺] = √(1.74 × 10⁻⁵ × 0.100)
[H⁺] = √(1.74 × 10⁻⁶)
[H⁺] = 1.32 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.32 × 10⁻³) = 2.88
Check the assumption: % dissociation = (1.32 × 10⁻³ / 0.100) × 100 = 1.32%. This is small (well below 5%), so the assumption that [HA] ≈ initial concentration is valid.
Calculating Ka from pH
Given the pH and initial concentration, work backwards:
1. [H⁺] = 10⁻ᵖᴴ
2. [A⁻] = [H⁺] (assumption 1)
3. [HA] ≈ initial concentration (assumption 2)
4. Ka = [H⁺]² / [HA]
Worked Example: A 0.200 mol dm⁻³ solution of a weak acid has pH 3.50. Find Ka.
[H⁺] = 10⁻³·⁵⁰ = 3.16 × 10⁻⁴ mol dm⁻³
Ka = (3.16 × 10⁻⁴)² / 0.200 = 9.99 × 10⁻⁸ / 0.200 = 5.00 × 10⁻⁷ mol dm⁻³
Dilution of Strong vs Weak Acids
If you dilute a strong acid by a factor of 10:
- [H⁺] decreases by a factor of 10
- pH increases by exactly 1
If you dilute a weak acid by a factor of 10:
- The equilibrium shifts right (Le Chatelier), partially restoring [H⁺]
- pH increases by less than 1 (approximately 0.5)
Exam Tips
- Always state your assumptions when calculating pH of a weak acid
- Be careful with sig figs — pH is usually given to 2 decimal places
- Remember that Kw changes with temperature — if asked for pH at a non-standard temperature, use the Kw value given
- For diprotic acids like H₂SO₄, the second dissociation is sometimes treated as incomplete at A-Level; follow the question's guidance
- The logarithmic relationship means a pH change of 1 = a tenfold change in [H⁺]