AP Chemistry Titration Curve pH Calculations

Three points on the same titration curve, three different calculation methods.

A titration curve isn't one calculation, it's three, depending on which species actually dominate the solution at that point. Here's the same weak acid/strong base titration worked through all three regions: before equivalence, at equivalence, and after it.

The Titration Setup

25.0 mL of 0.100 M acetic acid (CH₃COOH, K_a = 1.8 × 10⁻⁵, pK_a = 4.74) is titrated with 0.100 M NaOH. Since both solutions have the same concentration, the equivalence point is reached at exactly 25.0 mL of NaOH added.

Worked Example 1: Before Equivalence

What is the pH after 10.0 mL of NaOH has been added?

This is a stoichiometry problem first, then a Henderson-Hasselbalch problem:
mol HA (initial) = (0.0250 L)(0.100 M) = 2.50 × 10⁻³ mol
mol OH⁻ added = (0.0100 L)(0.100 M) = 1.00 × 10⁻³ mol
mol HA remaining = 2.50 × 10⁻³ − 1.00 × 10⁻³ = 1.50 × 10⁻³ mol
mol A⁻ formed = 1.00 × 10⁻³ mol

pH = pK_a + log(mol A⁻/mol HA) = 4.74 + log(1.00/1.50) = 4.74 − 0.18 = 4.56

Worked Example 2: At Equivalence

What is the pH at the equivalence point (25.0 mL of NaOH added)?

All of the original acid has become acetate ion, A⁻. Total volume = 25.0 + 25.0 = 50.0 mL:
[A⁻] = (2.50 × 10⁻³ mol) / (0.0500 L) = 0.0500 M

Acetate hydrolyzes water, so this needs an ICE table with K_b = K_w/K_a = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰:
K_b = x²/0.0500 → x = [OH⁻] = 5.27 × 10⁻⁶ M
pOH = 5.28, so pH = 8.72

This confirms the equivalence point is basic, not neutral, exactly as expected for a weak acid titrated with a strong base.

Worked Example 3: After Equivalence

What is the pH after 30.0 mL of NaOH has been added, 5.0 mL past the equivalence point?

Every bit of the original acid is already consumed, so this is now a direct excess-strong-base calculation, not an equilibrium one:
mol NaOH (total added) = (0.0300 L)(0.100 M) = 3.00 × 10⁻³ mol
mol OH⁻ in excess = 3.00 × 10⁻³ − 2.50 × 10⁻³ = 5.00 × 10⁻⁴ mol
Total volume = 25.0 + 30.0 = 55.0 mL

[OH⁻] = (5.00 × 10⁻⁴ mol) / (0.0550 L) = 9.09 × 10⁻³ M
pOH = 2.04, so pH = 11.96

Across all three points, pH rises from 4.56 to 8.72 to 11.96, consistent with the shape of a real titration curve.

Common Titration Calculation Mistakes

Using Henderson-Hasselbalch past the equivalence point. Once all the original acid is consumed, there's no more HA/A⁻ buffer, the calculation switches to direct excess-base concentration instead.
Forgetting the equivalence point isn't neutral for a weak acid/strong base titration. The leftover conjugate base hydrolyzes water and pushes pH above 7.
Mixing up total volume at each stage. The denominator changes every time more titrant is added, don't reuse an earlier total volume.
Attempting full quantitative calculations on a polyprotic acid's titration curve. That's explicitly excluded; everything on this page assumes a monoprotic acid.

This same stoichiometry-then-equilibrium approach applies across the rest of AP Chemistry too -- for every other free tool and guide on this site, start from the AP Chem Score Calculator.

Frequently Asked Questions

What determines the pH calculation method at each point on a titration curve?

Which species are actually present in solution at that volume of titrant added. Before equivalence, a weak acid/conjugate base buffer is present, so Henderson-Hasselbalch applies. At equivalence, only the conjugate base remains, which hydrolyzes water, requiring a Kb/ICE table calculation. After equivalence, excess strong base dominates, and pH comes directly from its concentration.

Why is the equivalence point basic, not neutral, for a weak acid/strong base titration?

Because the only species left in solution at that point is the weak acid's conjugate base, which is itself a weak base. It reacts with water (hydrolyzes) to produce a small amount of OH-, pushing the pH above 7.

Do I need to do full quantitative pH calculations for a polyprotic acid titration curve?

No. Computing the exact concentration of every species at each point along a polyprotic acid's titration curve is explicitly excluded. Full quantitative work for monoprotic (single-proton) titrations, like every example on this page, remains in scope.

Why does pH = pKa at the half-equivalence point?

At half-equivalence, exactly half the original weak acid has been converted to its conjugate base, so [HA] = [A-]. In the Henderson-Hasselbalch equation, that ratio equals 1, log(1) = 0, so pH = pKa exactly.

What changes in the calculation once you're past the equivalence point?

The weak acid and its conjugate base are no longer the deciding factor, every bit of the original acid has already reacted. pH is now set entirely by how much excess strong base has been added beyond the equivalence point, a direct concentration calculation, not an equilibrium one.

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