Acid-Base Titration Curves: Worked Examples With Equivalence Points

A titration curve plots pH (y-axis) against volume of titrant added (x-axis). The shape reveals the strength of the acid and base, the equivalence point (where moles of acid = moles of base), the half-equivalence point (where pH = pKa for weak-acid titrations), and the buffer region (a flat segment where pH changes slowly). Four canonical curve shapes cover most introductory acid-base chemistry: strong acid + strong base, weak acid + strong base, strong acid + weak base, and polyprotic acid + strong base. Each shape has a characteristic equivalence-point pH that determines the choice of indicator.

Both species ionize completely. The pH calculation is straightforward at every stage. Worked Example. Titrate 25.0 mL of 0.10 M HCl with 0.10 M NaOH. Initial: 25.0 mL of 0.10 M HCl. Moles H+ = 25.0 × 0.10 = 2.50 mmol. pH = −log(0.10) = 1.0. At 12.5 mL NaOH (half to equivalence): mmol NaOH added = 1.25; remaining mmol H+ = 1.25; total volume = 37.5 mL. [H+] = 1.25/37.5 = 0.0333 M. pH = 1.48. At 25.0 mL NaOH (equivalence): all H+ neutralized. Solution is just NaCl in water. pH = 7.00. At 35.0 mL NaOH: excess NaOH = 1.00 mmol; total volume = 60.0 mL. [OH−] = 1.00/60.0 = 0.0167 M. pOH = 1.78. pH = 12.22. The curve has a steep rise at the equivalence point, jumping from acidic to basic over a small volume of titrant. Equivalence pH = 7.00. Choose indicator with color change near pH 7 — phenolphthalein (pH 8.3-10.0) or bromothymol blue (pH 6.0-7.6) both work because the steep rise crosses both ranges quickly.

The weak acid partially ionizes; pH calculations require equilibrium analysis at most stages. Worked Example. Titrate 25.0 mL of 0.10 M acetic acid (Ka = 1.8 × 10^−5) with 0.10 M NaOH. Initial: weak acid alone. [H+] from Ka equilibrium. Approximation: x² / 0.10 = 1.8 × 10^−5; x = 1.34 × 10^−3 M. pH = 2.87. Before equivalence (e.g., 5.0 mL NaOH added): mmol acetic acid initially = 2.50; mmol NaOH added = 0.50. NaOH consumes acetic acid forming acetate: mmol HA remaining = 2.00; mmol A− formed = 0.50. Buffer region. Use Henderson-Hasselbalch: pH = pKa + log([A−]/[HA]) = 4.74 + log(0.50/2.00) = 4.74 + log(0.25) = 4.74 − 0.60 = 4.14. Half-equivalence point (12.5 mL NaOH): mmol HA = 1.25; mmol A− = 1.25. Henderson-Hasselbalch: pH = pKa + log(1.25/1.25) = pKa + 0 = pKa = 4.74. THIS IS THE KEY OBSERVATION: at half-equivalence, pH = pKa. This is how we determine pKa experimentally from a titration curve. At equivalence (25.0 mL NaOH): all acetic acid converted to acetate. mmol acetate = 2.50; total volume = 50.0 mL; [A−] = 0.050 M. Acetate is the conjugate base of a weak acid; it hydrolyzes: A− + H2O ⇌ HA + OH−. Kb = Kw/Ka = 10^−14 / 1.8 × 10^−5 = 5.56 × 10^−10. x² / 0.050 = 5.56 × 10^−10; x = [OH−] = 5.27 × 10^−6. pOH = 5.28. pH = 8.72. KEY: equivalence point pH > 7 for weak-acid + strong-base titration. Choose phenolphthalein (color change pH 8.3-10.0) — fits the basic equivalence point. Past equivalence: excess strong base; same as type 1 calculation. The complete curve has FOUR distinguishing features: (1) Initial pH about 2.87 (above 1.0 because weak acid only partially ionized); (2) Buffer region from before equivalence to half-equivalence point; (3) Half-equivalence point where pH = pKa; (4) Equivalence pH > 7 due to conjugate base hydrolysis.

Mirror of type 2. Worked Example. Titrate 25.0 mL of 0.10 M ammonia (Kb = 1.8 × 10^−5) with 0.10 M HCl. Initial: weak base. From Kb equilibrium: x² / 0.10 = 1.8 × 10^−5; x = [OH−] = 1.34 × 10^−3. pOH = 2.87. pH = 11.13. Before equivalence (buffer region): NH3 + HCl → NH4Cl + (excess NH3). Use Henderson-Hasselbalch (for the conjugate acid form): pOH = pKb + log([BH+]/[B]); equivalent to pH = pKa + log([B]/[BH+]) where pKa = 14 − pKb = 9.26. Half-equivalence point: pH = pKa = 9.26 (or pOH = pKb = 4.74). Equivalence point: all ammonia converted to NH4+. NH4+ is conjugate acid of weak base; hydrolyzes acid-side. Ka = Kw/Kb = 5.56 × 10^−10. [NH4+] = 0.050 M. x² / 0.050 = 5.56 × 10^−10; x = [H+] = 5.27 × 10^−6. pH = 5.28. KEY: equivalence point pH < 7 for strong-acid + weak-base titration. Choose methyl red (pH 4.4-6.2) or methyl orange (pH 3.1-4.4) — fits the acidic equivalence point.

Polyprotic acids (H3PO4, H2CO3, H3C6H5O7 citric, H2SO4) have multiple ionization steps. Each step has its own Ka and produces its own equivalence point. For phosphoric acid H3PO4: Ka1 = 7.5 × 10^−3 (pKa1 = 2.12) — first proton released Ka2 = 6.2 × 10^−8 (pKa2 = 7.21) — second proton Ka3 = 4.2 × 10^−13 (pKa3 = 12.38) — third proton The Ka values typically differ by factors of 10^4-10^6. This means the steps are well-separated on the titration curve — one proton is released completely before the next begins ionizing. Titration curve for diprotic H2A with Ka1 >> Ka2: - Initial pH: from Ka1 equilibrium - First half-equivalence: pH = pKa1 - First equivalence: half-way through; pH ≈ (pKa1 + pKa2)/2 - Second half-equivalence: pH = pKa2 - Second equivalence: pH controlled by hydrolysis of A^2− - Past second equivalence: pH from excess strong base The curve shows two stepped equivalence points (small pH jumps in the middle, since polyprotic acid is buffered between equivalence points). Physiological example: phosphate buffer in blood operates around pKa2 = 7.21, very close to physiological pH 7.4. The HPO4^2−/H2PO4^− buffer is one of the body's primary blood buffer systems.

Titration curve problems are heavily tested on ACS Gen Chem, AP Chemistry, MCAT Chem/Phys, and intermediate chemistry courses. The challenge is identifying which type of titration is occurring (which determines the equivalence-point pH) and computing pH at each stage. Snap a photo of any titration problem and ChemistryIQ identifies the type, sets up the appropriate equilibrium, computes pH at each stage (initial, buffer region, half-equivalence, equivalence, past equivalence), and visualizes the complete titration curve with the key features labeled. This content is for educational purposes only and does not constitute chemistry advice.