8.5 Acid-Base Titrations

Polyprotic Acid Titration

Titrate diprotic and triprotic acids against NaOH. A synced speciation chart below the curve tracks the mole fraction of every species as volume increases.

Two EPsSpeciationAmphoteric6 Acids
Topic 8.5

Acid–Base Titrations

Explain results from the titration of a mono- or polyprotic acid or base solution, in relation to the properties of the solution and its components.

An acid–base reaction carried out under controlled conditions is a titration; a titration curve plots pH against volume of titrant added.

At the equivalence point, the moles of titrant added equal the moles of analyte originally present. This relationship gives the analyte's concentration, and it holds for both strong and weak systems.

The half-equivalence point is the most useful landmark on a weak-acid curve. Halfway to equivalence, exactly half the acid has been converted, so [HA] = [A⁻]. Since pH = pKa + log([A⁻]/[HA]) and log(1) = 0:

pH = pKa at the half-equivalence point

Reading the pH there gives you the pKa straight off the graph — a standard AP task.

pH at the equivalence point is determined by the major species present:

  • Strong acid + strong base → only spectator ions and water → pH = 7.
  • Weak acid + strong base → the conjugate base A⁻ remains and undergoes proton transfer with water → pH > 7.
  • Weak base + strong acid → the conjugate acid HB⁺ remains → pH < 7.

Polyprotic acids: the curve shows one steep jump per acidic proton, so counting jumps counts the protons. The major species can be identified at any point, and each pKa is read at the corresponding half-equivalence plateau.

The buffer region is the relatively flat stretch surrounding the half-equivalence point, where both HA and A⁻ are present in comparable amounts.

Indicator choice (see 8.7): pick an indicator whose pKa is close to the pH at the equivalence point.

Key points

  • pH = pKa at the half-equivalence point — read the pKa directly off the graph.
  • Equivalence pH: 7 for strong/strong, above 7 for weak acid/strong base, below 7 for weak base/strong acid.
  • Number of steep jumps = number of acidic protons.
  • Weak and strong acids of the same concentration and volume need the same titrant volume to reach equivalence.

Equations

  • not on the sheetFollows from Henderson–Hasselbalch when [A⁻] = [HA].
  • not on the sheetScaled by the stoichiometric ratio.

Common mistakes

  • Assuming the equivalence point is always pH 7. Only for strong/strong.
  • Confusing the equivalence point with the endpoint.
  • Thinking a weak acid needs less titrant. Equal moles of acid need equal moles of base regardless of strength.
  • Reading pKa at the equivalence point. It is at the half-equivalence point.
  • Misreading a polyprotic curve. The second equivalence point is at twice the volume of the first.

Worked example

25.00 mL of a weak monoprotic acid is titrated with 0.100 M NaOH. The equivalence point occurs at 32.60 mL, and the pH at 16.30 mL is 4.85. Determine (a) the concentration of the acid, (b) its Ka, and (c) whether the equivalence point pH is above or below 7, with justification.

(a) Concentration of the acid.
At equivalence, n(NaOH) = n(HA):
n = (0.100 mol/L)(0.03260 L) = 3.260 × 10⁻³ mol
[HA] = 3.260 × 10⁻³ mol ÷ 0.02500 L = 0.1304 M

(b) Ka.
16.30 mL is exactly half of 32.60 mL, so this is the half-equivalence point, where [HA] = [A⁻] and therefore pH = pKa.
pKa = 4.85
Ka = 10−4.85 = 1.4 × 10⁻⁵

(c) Equivalence-point pH. Above 7.

At the equivalence point all the weak acid has been converted to its conjugate base A⁻, and the only other species present are Na⁺ (a spectator) and water. A⁻ is the conjugate base of a weak acid, so it is itself a weak base and undergoes proton transfer with water:
A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq)
This generates hydroxide ions, making the solution basic.

Full notes for topic 8.5 →