Weak Acid-Strong Base Titrations

Introduction

A weak acid titrated with a strong base produces a curve unlike a strong-acid titration. The initial pH comes from partial weak-acid dissociation. Added hydroxide converts weak acid into its conjugate base, creating a buffer. At equivalence the conjugate base remains and usually makes the solution basic at 25 °C.

Core explanation

Let HA be a monoprotic weak acid and NaOH the strong-base titrant. The principal neutralization is HA + OH⁻ → A⁻ + H₂O. At zero titrant, determine pH from Ka, acid analytical concentration, and any justified approximation. During pre-equivalence, the added hydroxide consumes HA almost completely. If initial moles of HA are n₀ and hydroxide added is n, then post-reaction HA moles are n₀ − n and A⁻ moles are approximately n. Their ratio can be inserted in pH ≈ pKa + log₁₀(n/(n₀−n)), provided the pair is sufficiently abundant and other contributions are small.

At half-equivalence n = n₀/2. The two post-reaction amounts are equal, so pH ≈ pKa. This relation provides an experimental route to estimating pKa from a measured curve. It is not a statement that the solution is half as acidic, because pH is logarithmic. It applies in the weak-acid/strong-base setting with suitable assumptions, not to all titrations.

At equivalence, essentially all analytical HA has been converted into A⁻. The buffer ratio formula fails because its acid denominator is effectively absent. Instead use A⁻ + H₂O ⇌ HA + OH⁻, with Kb = Kw/Ka under the corresponding standard conditions. If A⁻ hydrolyzes appreciably, the solution is basic relative to neutral water at 25 °C. Calculate the A⁻ analytical concentration from original acid moles divided by total volume, then solve or approximate the base equilibrium. Beyond equivalence, excess strong OH⁻ usually dominates the pH and should be calculated by excess moles over total volume.

The equivalence volume is still stoichiometric: for one proton per HA, n(OH⁻ at equivalence) = initial n(HA). Weakness changes pH behavior, not the mole ratio of neutralization. The endpoint indicator should transition within the steep region near the basic equivalence pH. The jump can be less sharp for very weak or very dilute acids, so instrument choice matters.

Step-by-step reasoning

1. Find initial acid moles and the equivalence volume. 2. Before equivalence, subtract hydroxide moles from HA and add them to A⁻. 3. At half-equivalence, use equal pair amounts and pH near pKa. 4. At equivalence, solve A⁻ hydrolysis; afterward, use excess OH⁻.

Visual explanation

Sketch a rising pH curve with a gradual buffer region. Mark half-equivalence at pH near pKa, then a steeper rise whose stoichiometric center has pH above 7 at 25 °C.

Real-world analogy

Every delivery converts one item from an unprocessed bin to a processed bin. At half the planned deliveries, the bins match; after all deliveries, only processed items remain and their own behavior controls the result.

Real-world example

An acetic-acid sample titrated with standardized sodium hydroxide shows an acetate/acetic-acid buffer region. A pH meter can locate half-equivalence and estimate the acid's pKa from that reading.

Why?

Why is equivalence basic? The remaining A⁻ accepts protons from water, producing OH⁻. Its parent acid was weak, so the conjugate base has nonzero basicity.

Common misconception

“At equivalence, pH must equal pKa.” pH near pKa occurs at half-equivalence; at equivalence the conjugate base hydrolyzes and has a different pH.

Worked example

Titrate 25.0 mL of 0.100 M HA with 0.100 M NaOH; let pKa = 4.76. Equivalence requires 0.00250 mol OH⁻, or 25.0 mL titrant. After 12.5 mL, 0.00125 mol HA and 0.00125 mol A⁻ remain or form. Their ratio is one, giving pH ≈ 4.76. At 25.0 mL, use acetate-like A⁻ hydrolysis in the total 50.0 mL, not the buffer equation.

Quick check

1. What species primarily controls pH at equivalence? Answer: The conjugate base A⁻ through hydrolysis with water.

Exam focus

Use a different model for each region. Check that an alleged half-equivalence point is half the equivalence volume only when titrant concentration is constant and the reaction is one-to-one.

Advanced insight

The Henderson-Hasselbalch estimate near the first additions can be inaccurate because initial weak-acid dissociation already contributes A⁻. Mass and charge balance can correct that approximation.

Summary

Weak-acid titration begins with weak-acid equilibrium, passes through a buffer region, gives pH near pKa at half-equivalence, and reaches a basic conjugate-base solution at equivalence.

Practice questions

1. What is the dominant pre-equivalence reaction with added OH⁻? Answer: HA + OH⁻ → A⁻ + H₂O. 2. Where does pH approximately equal pKa? Answer: At half-equivalence, when HA and A⁻ amounts match. 3. Why does the buffer relation fail at equivalence? Answer: Essentially no analytical HA remains, so its denominator is absent; A⁻ hydrolysis must be considered.