Weak-Acid Strong-Base Titration Curves

Introduction

A titration curve is a record of a changing equilibrium. As sodium hydroxide is added to a weak acid, the solution passes through four chemically distinct stages: a weak acid alone, a buffer, a salt solution and finally an excess of strong base. Each stage needs a different calculation. Recognising which stage you are in, and choosing the right equation, is the central skill for predicting and interpreting these curves.

Core explanation

Consider 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74) titrated with 0.100 mol dm⁻³ NaOH. The acid contains 2.50 × 10⁻³ mol, so equivalence occurs at 25.0 cm³ of base. The reaction HA + OH⁻ → A⁻ + H₂O goes essentially to completion because its equilibrium constant, 1/Kb(A⁻), is very large (about 1.8 × 10⁹).

Region 1: before any base. Only the weak acid is present. [H₃O⁺] ≈ √(Ka × c) = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³ mol dm⁻³, so pH = 2.87. This is much higher than the pH of 1.00 for 0.100 mol dm⁻³ HCl.

Region 2: buffer region. Each mole of OH⁻ converts one mole of HA into A⁻. Both species are now present in significant amounts, so the Henderson–Hasselbalch equation applies: pH = pKa + log(n(A⁻)/n(HA)). Moles can be used directly because both species share the same volume. The curve here is shallow, because the mixture resists pH change.

Half-equivalence. When exactly half the acid has reacted, n(A⁻) = n(HA), log 1 = 0 and pH = pKa = 4.74. This gives a direct experimental route to pKa from a titration curve.

Region 3: equivalence. All the HA has become A⁻, now in a total volume of 50.0 cm³. The solution is simply sodium ethanoate at 0.0500 mol dm⁻³, a weak base. Kb = Kw ÷ Ka = 5.6 × 10⁻¹⁰. [OH⁻] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol dm⁻³, pOH = 5.28, pH = 8.72. The equivalence point is basic because the conjugate base hydrolyses.

Region 4: after equivalence. Excess OH⁻ from NaOH swamps the tiny contribution from A⁻. For example, at 30.0 cm³, excess OH⁻ = 0.50 × 10⁻³ mol in 55.0 cm³, giving 9.1 × 10⁻³ mol dm⁻³, pOH = 2.04 and pH = 11.96.

Volume NaOH / cm³ Region pH --- --- --- 0.0 weak acid 2.87 10.0 buffer 4.57 12.5 half-equivalence 4.74 25.0 equivalence 8.72 30.0 excess base 11.96

Step-by-step reasoning

1. Calculate the initial moles of acid and the equivalence volume. 2. For each point, calculate moles of OH⁻ added and do a reaction table (before, change, after). 3. If only HA remains, use the weak-acid expression. 4. If both HA and A⁻ remain, use Henderson–Hasselbalch. 5. If only A⁻ remains, use the weak-base expression with the total volume. 6. If OH⁻ is in excess, use the excess concentration directly.

Visual explanation

Sketch the curve: a start near pH 3, a quick initial rise, a flat buffer plateau centred on pH 4.74 at 12.5 cm³, a steep vertical section from roughly pH 7 to 11 centred on 8.7 at 25.0 cm³, then a gentle levelling towards pH 12 to 13. Mark the half-equivalence point on the plateau and the equivalence point at the steepest part.

Real-world analogy

The titration is like filling a lock on a canal. At first the water level (pH) rises easily; then the lock gates hold it steady while a large volume flows in (the buffer region); once the lock is full, a small extra flow sends the level shooting up to the next reach.

Real-world example

Food chemists determine the acidity of vinegar by titrating it with standard sodium hydroxide using phenolphthalein. Because the main acid is ethanoic acid, the end point is basic, which is why phenolphthalein, changing near pH 8 to 10, is used rather than methyl orange.

Why?

Why is the pH at half-equivalence equal to pKa? At that point the concentrations of HA and A⁻ are equal, so the ratio in Ka = [H₃O⁺][A⁻]/[HA] cancels, leaving Ka = [H₃O⁺]. Taking negative logarithms gives pH = pKa.

Common misconception

"The equivalence point is where the pH is 7." Equivalence is defined by stoichiometry, not by pH. For a weak acid titrated with a strong base, the salt formed is basic, so equivalence lies above pH 7.

Worked example

Question: In the titration above, calculate the pH after 10.0 cm³ of NaOH.

Reasoning: n(OH⁻) = 0.0100 × 0.100 = 1.00 × 10⁻³ mol. n(HA) left = 2.50 × 10⁻³ − 1.00 × 10⁻³ = 1.50 × 10⁻³ mol. n(A⁻) = 1.00 × 10⁻³ mol. pH = 4.74 + log(1.00/1.50) = 4.74 − 0.18 = 4.56 to 4.57.

Answer: pH ≈ 4.57.

Quick check

1. What is the pH at the half-equivalence point when a weak acid of pKa 3.86 is titrated with NaOH? Answer: pH = 3.86, because equal amounts of acid and conjugate base give pH = pKa.

Exam focus

Always show a reaction table in moles before choosing an equation. Remember the total volume at equivalence; forgetting to add the titrant volume is the most common error. Be ready to read pKa from a given curve at half the equivalence volume.

Advanced insight

The Henderson–Hasselbalch equation fails very near the start and very near equivalence, where one partner is tiny and the dissociation or hydrolysis of the other alters the ratio. A single exact equation, derived from charge and mass balances, describes the whole curve and is what curve-plotting software uses. For very weak acids (pKa above about 9) or very dilute solutions, the vertical section becomes too short to locate an end point reliably.

Summary

A weak-acid strong-base titration passes through four regions: weak acid, buffer, salt at equivalence, and excess base. At half-equivalence pH = pKa. The equivalence point is basic because the conjugate base hydrolyses. The correct method depends on which species remain after the stoichiometric reaction, so moles-based reaction tables come first.

Practice questions

1. State the equivalence volume when 20.0 cm³ of 0.150 mol dm⁻³ weak acid is titrated with 0.100 mol dm⁻³ NaOH. Answer: n(HA) = 3.00 × 10⁻³ mol, so 30.0 cm³ of NaOH. 2. Calculate the pH at the start of titrating 0.0500 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵). Answer: [H₃O⁺] = √(9.0 × 10⁻⁷) = 9.5 × 10⁻⁴ mol dm⁻³; pH ≈ 3.02. 3. Explain why the curve is flattest at the half-equivalence point. Answer: Buffer capacity is greatest when [HA] = [A⁻], so added base changes the ratio, and hence the pH, least there. 4. Calculate the pH after 35.0 cm³ of 0.100 mol dm⁻³ NaOH in the titration of 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid. Answer: Excess OH⁻ = 1.00 × 10⁻³ mol in 60.0 cm³ = 0.0167 mol dm⁻³; pOH = 1.78; pH ≈ 12.22.