Chemistry · General chemistry II · Worked example
Plot a weak acid titration curve
25.0 mL of 0.200 M acetic acid (Ka = 1.8 × 10⁻⁵) is titrated with 0.200 M NaOH. Find the pH at the start, after 5.00 mL, at the half-equivalence point, at the equivalence point and after 30.0 mL.
Find the equivalence volume
The acid is 5.00 × 10⁻³ mol, so it takes 5.00 × 10⁻³ mol of NaOH: 25.0 mL of 0.200 M base.
Before any base
This is the weak acid alone. As in the weak-acid example, [H₃O⁺] = 1.9 × 10⁻³ M and the pH is 2.72.
In the buffer region
After 5.00 mL, 1.00 × 10⁻³ mol OH⁻ has turned that much acid into acetate, leaving 4.00 × 10⁻³ mol acid beside 1.00 × 10⁻³ mol acetate.
Half-equivalence
At 12.5 mL half the acid is neutralized, so [A⁻] = [HA] and pH = pKa = 4.74. Reading this point from a measured curve is one way to find pKa.
Equivalence
At 25.0 mL all the acid has become acetate: 5.00 × 10⁻³ mol in 50.0 mL, or 0.100 M. That is the weak-base problem: Kb = 5.6 × 10⁻¹⁰, [OH⁻] = 7.48 × 10⁻⁶ M and pH 8.87.
Past equivalence
At 30.0 mL, 1.00 × 10⁻³ mol OH⁻ is left over in 55.0 mL: [OH⁻] = 0.0182 M, pOH 1.74 and pH 12.26. The excess strong base now sets the pH.
Trace the curve
Plotted against the volume of base, the points trace an S-shaped curve: a slow rise through the buffer region, a steep jump around 25.0 mL and a level tail.
| NaOH added (mL) | 0 | 5.00 | 12.5 | 25.0 | 30.0 |
|---|---|---|---|---|---|
| pH | 2.72 | 4.14 | 4.74 | 8.87 | 12.26 |
Result
pH 2.72 at the start, 4.14 after 5.00 mL, 4.74 at half-equivalence (12.5 mL), 8.87 at equivalence (25.0 mL) and 12.26 after 30.0 mL.
Your turn
What is the pH after 15.0 mL of base has been added?
Show the answer and explanation
pH = 4.92.
15.0 mL of 0.200 M NaOH is 3.00 × 10⁻³ mol, leaving 2.00 × 10⁻³ mol acid beside 3.00 × 10⁻³ mol acetate. pH = 4.74 + log(3.00/2.00) = 4.92.
Keep exploring
In Buffers & titration curves, switch the analyte to a strong acid of the same concentration. The curve starts lower, at pH 0.70, has no buffer region, and passes pH 7.00 at equivalence.
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