Titration: the equivalence point
In a titration you add base until the acid is exactly neutralised. The volume used gives the concentration, and the curve reveals the pKa.
Formulas and rules
- Equivalence: n(base) = n(acid) → c₁V₁ = c₂V₂ (for 1:1)
- Half-equivalence: c(acid) = c(base) → pH = pKa
- Starting pH of a weak acid: ½ · (pKa − log c)
- The equivalence point of a weak acid is basic (pH > 7)
Your turn
Three problems on this topic. Type numbers like 12.5 or 1,250 and elasticities with their sign.
Task
25 mL of acetic acid (c = 0.1 mol/L, pKa = 4.76) are titrated with sodium hydroxide (c = 0.5 mol/L). Work out the starting pH, the volume to the equivalence point and the pH at the half-equivalence point.
Starting pH (2 decimal places)
Volume of NaOH to the equivalence point in mL
pH at the half-equivalence point (2 decimal places)
Numbers like 12.5 or 0.025, with as many decimals as asked. Masses and amounts count within 1%, pH values within ±0.02.
Worked example
20 mL of formic acid (c = 0.05 mol/L, pKa = 3.75) are titrated with sodium hydroxide (c = 0.5 mol/L). Work out the starting pH, the volume to the equivalence point and the pH at the half-equivalence point.
Worked solution
- 1.pH = ½ × (3.75 − lg 0.05) = 2.53
- 2.Equivalence: n(NaOH) = n(acid) → V = c₀ × V₀ / c(NaOH) = 0.05 × 20 / 0.5 = 2 mL
- 3.At half-equivalence half the acid has reacted: c(acid) = c(base) → pH = pKa = 3.75
Typical mistake: Assuming pH 7 at the equivalence point of a weak acid. Only the conjugate base is left there, so the solution is basic.
Common questions
Why is the pH at equivalence not 7?
With a weak acid only its conjugate base is left there, and it is slightly basic.