Weak-Base Strong-Acid Titration Curves

Introduction

Titrating a weak base with a strong acid produces a curve that is, in shape, the mirror image of a weak-acid strong-base titration. The pH falls rather than rises, the buffer plateau lies on the basic side, and the equivalence point is acidic. The same four-region logic applies, but now the key species is the conjugate acid, BH⁺. Recognising this symmetry lets you reuse the methods you already know, simply by viewing every calculation from the protonation side.

Core explanation

Consider 25.0 cm³ of 0.100 mol dm⁻³ aqueous ammonia (Kb = 1.8 × 10⁻⁵, pKb = 4.74) titrated with 0.100 mol dm⁻³ HCl. The ammonia contains 2.50 × 10⁻³ mol, so equivalence is at 25.0 cm³. The conjugate acid NH₄⁺ has pKa = 14.00 − 4.74 = 9.26 (often quoted as 9.25). The titration reaction NH₃ + H₃O⁺ → NH₄⁺ + H₂O is essentially complete, with K = 1/Ka(NH₄⁺), about 1.8 × 10⁹.

Region 1: weak base alone. [OH⁻] ≈ √(Kb × c) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³. pOH = 2.87 and pH = 11.13. This is well below the pH of 13.00 for 0.100 mol dm⁻³ NaOH.

Region 2: buffer region. Each mole of added acid converts one mole of NH₃ into NH₄⁺. With both present, write the Henderson–Hasselbalch equation in terms of the conjugate acid: pH = pKa(BH⁺) + log(n(B)/n(BH⁺)). Note that the base appears on top, just as A⁻ did in the acid case.

Half-equivalence. At 12.5 cm³, n(NH₃) = n(NH₄⁺), so pH = pKa(NH₄⁺) ≈ 9.25. Equivalently, pOH = pKb.

Region 3: equivalence. All ammonia is now NH₄⁺, at 2.50 × 10⁻³ mol in 50.0 cm³, which is 0.0500 mol dm⁻³. Chloride is a spectator. Using Ka = 5.6 × 10⁻¹⁰: [H₃O⁺] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol dm⁻³ and pH = 5.28. The equivalence point is acidic because the conjugate acid donates protons to water.

Region 4: after equivalence. Excess H₃O⁺ from HCl dominates. At 30.0 cm³, excess = 0.50 × 10⁻³ mol in 55.0 cm³ = 9.1 × 10⁻³ mol dm⁻³, so pH = 2.04.

Volume HCl / cm³ Region pH --- --- --- 0.0 weak base 11.13 5.0 buffer 9.85 12.5 half-equivalence 9.25 25.0 equivalence 5.28 30.0 excess acid 2.04

Notice the symmetry with the ethanoic acid titration: 2.87 and 11.13 add to 14.00, as do 8.72 and 5.28. This is no coincidence, because ethanoic acid and ammonia happen to have almost identical Ka and Kb values.

Step-by-step reasoning

1. Find moles of base and the equivalence volume of acid. 2. Convert Kb to Ka of the conjugate acid, since most later steps need it. 3. Complete a moles reaction table for each point. 4. Base only: weak-base square-root expression, then convert pOH to pH. 5. Base and conjugate acid: Henderson–Hasselbalch using pKa(BH⁺). 6. Conjugate acid only: weak-acid expression using the total volume. 7. Excess strong acid: use its concentration directly.

Visual explanation

Sketch a curve starting near pH 11, falling to a flat plateau around pH 9.25, then dropping steeply between roughly pH 7 and 3, centred near 5.3 at 25.0 cm³, before levelling towards pH 1 to 2. It resembles the weak-acid curve turned upside down.

Real-world analogy

Think of an escalator going down instead of up. The journey has the same landings as the upward trip (a starting level, a long flat landing, a sudden drop and a final floor), but it is travelled in reverse order.

Real-world example

The Kjeldahl method for measuring protein nitrogen in food converts nitrogen to ammonia, which is distilled into a known excess of acid or into boric acid and then titrated with standard strong acid. Because the product at equivalence is an ammonium salt, an indicator that changes colour in the acidic range, such as methyl red, is chosen.

Why?

Why is the equivalence point acidic? At equivalence the solution contains only NH₄⁺ and a spectator anion. NH₄⁺ is a weak acid, so it produces H₃O⁺ by reacting with water, pulling the pH below 7.

Common misconception

"At half-equivalence pH equals pKb." It is pOH that equals pKb. The pH equals the pKa of the conjugate acid, which is 14.00 minus pKb at 25 °C. Mixing these up gives a pH on the wrong side of neutral.

Worked example

Question: Calculate the pH after 5.0 cm³ of 0.100 mol dm⁻³ HCl has been added to 25.0 cm³ of 0.100 mol dm⁻³ NH₃. pKa(NH₄⁺) = 9.25.

Reasoning: n(H₃O⁺) = 0.50 × 10⁻³ mol. n(NH₃) left = 2.50 × 10⁻³ − 0.50 × 10⁻³ = 2.00 × 10⁻³ mol; n(NH₄⁺) = 0.50 × 10⁻³ mol. pH = 9.25 + log(2.00/0.50) = 9.25 + 0.60 = 9.85.

Answer: pH ≈ 9.85.

Quick check

1. Which species controls the pH at the equivalence point of an ammonia and hydrochloric acid titration? Answer: The ammonium ion, NH₄⁺, acting as a weak acid in water.

Exam focus

Convert Kb to Ka early and state the conversion. Put the base, not the conjugate acid, in the numerator of the Henderson–Hasselbalch log term. Examiners frequently ask why methyl red or methyl orange suits this titration but phenolphthalein does not.

Advanced insight

For a diprotic base such as carbonate ion, titration with strong acid shows two protonation stages, CO₃²⁻ to HCO₃⁻ and HCO₃⁻ to H₂CO₃, each with its own buffer region and end point. Whether both breaks are visible depends on how far apart the two pKa values are. Loss of CO₂ from the solution near the second end point can shift it, which is why analysts sometimes boil the solution briefly before finishing.

Summary

A weak-base strong-acid titration mirrors the weak-acid case. The pH starts basic, stays nearly constant in a buffer region where pH = pKa(BH⁺) + log(n(B)/n(BH⁺)), equals pKa(BH⁺) at half-equivalence, and is acidic at equivalence because BH⁺ is a weak acid. Beyond equivalence the excess strong acid sets the pH.

Practice questions

1. Calculate pKa of the conjugate acid of a base with Kb = 4.4 × 10⁻⁴. Answer: pKb = 3.36, so pKa = 14.00 − 3.36 = 10.64. 2. Calculate the pH at the start of titrating 0.200 mol dm⁻³ NH₃ (Kb = 1.8 × 10⁻⁵). Answer: [OH⁻] = √(3.6 × 10⁻⁶) = 1.9 × 10⁻³ mol dm⁻³; pOH = 2.72; pH ≈ 11.28. 3. Calculate the pH at equivalence when 20.0 cm³ of 0.100 mol dm⁻³ NH₃ is titrated with 0.100 mol dm⁻³ HCl. Answer: [NH₄⁺] = 0.0500 mol dm⁻³; [H₃O⁺] = 5.3 × 10⁻⁶ mol dm⁻³; pH ≈ 5.28. 4. Explain why the pH at 30.0 cm³ in the example does not depend on Kb. Answer: The excess strong acid supplies far more H₃O⁺ than NH₄⁺ does, so the weak equilibrium is negligible.