Weak Acid Strong Base Titration Curve: pH at Every Stage and pKa
Weak acid–strong base titration is a favourite topic in IIT-JAM, GATE, CSIR-NET and CUET-PG physical chemistry. Unlike the strong acid case, the pH at the start is not simply −log C, and the pH at the equivalence point is not 7. This article works through the full curve for acetic acid titrated with NaOH and shows how the same data gives you the Ka of the acid.
The four regions of the curve
Consider a weak acid HA (Ka) titrated with a strong base. The pH is found with a different model in each region.
Before equivalence (buffer): pH = pKa + log([A−] ÷ [HA]) = pKa + log(nbase ÷ (nacid − nbase))
Equivalence: only A−; [OH−] = √(KbCsalt), Kb = Kw ÷ Ka
After equivalence: [OH−] = excess n(OH−) ÷ total volume
Here n means moles, and the ratio in the buffer region can be written with moles because the total volume cancels. The square-root formula at the start assumes the acid is weak enough that ionisation is small compared with C. Kw is taken as 1.0 × 10−14 at 25°C.
Worked example: 25.00 mL of 0.1000 M acetic acid with 0.1000 M NaOH
Take Ka = 1.8 × 10−5, so pKa = 4.74. Initial moles of acid = 0.1000 × 25.00 = 2.500 mmol. The equivalence volume is 25.00 mL.
Point 1: 0.00 mL. [H+] = √(1.8 × 10−5 × 0.1000) = √(1.8 × 10−6) = 1.34 × 10−3 M. pH = 2.87. (Solving the full quadratic gives 2.88; the approximation is good here.)
Point 2: 10.00 mL NaOH. n(OH−) = 1.000 mmol, so A− = 1.000 and HA left = 1.500. pH = 4.74 + log(1.000 ÷ 1.500) = 4.74 − 0.18 = 4.57.
Point 3: 12.50 mL (half-equivalence). A− = 1.250 mmol and HA = 1.250 mmol. The ratio is 1, log 1 = 0, so pH = pKa = 4.74.
Point 4: 24.90 mL. A− = 2.490 and HA = 0.010. pH = 4.74 + log(2.490 ÷ 0.010) = 4.74 + 2.40 = 7.14.
Point 5: 25.00 mL (equivalence). All the acid has become acetate. Total volume = 50.00 mL, so [A−] = 2.500 ÷ 50.00 = 0.0500 M. Kb = 1.0 × 10−14 ÷ 1.8 × 10−5 = 5.56 × 10−10. [OH−] = √(5.56 × 10−10 × 0.0500) = √(2.78 × 10−11) = 5.27 × 10−6 M. pOH = 5.28 and pH = 8.72.
Point 6: 25.10 mL. Excess OH− = 0.100 × 0.1000 = 0.0100 mmol in 50.10 mL, so [OH−] = 2.00 × 10−4 M, pOH = 3.70 and pH = 10.30. (The acetate contributes a negligible extra amount.)
Point 7: 30.00 mL. Excess OH− = 0.500 mmol in 55.00 mL, so [OH−] = 9.09 × 10−3 M. pOH = 2.04, pH = 11.96.
| NaOH added (mL) | Region | pH |
|---|---|---|
| 0.00 | Weak acid only | 2.87 |
| 10.00 | Buffer | 4.57 |
| 12.50 | Half-equivalence | 4.74 (= pKa) |
| 24.90 | Buffer, nearly finished | 7.14 |
| 25.00 | Equivalence | 8.72 |
| 25.10 | Excess base | 10.30 |
| 30.00 | Excess base | 11.96 |
The pH jump near equivalence runs from about 7.1 to 10.3 for a change of just 0.2 mL, but it is centred in the basic region, not at 7. That is why an indicator that changes colour in basic solution, such as phenolphthalein, suits this titration better than one that changes near pH 4.
Getting Ka from the half-equivalence point
Question. A monoprotic weak acid is titrated with NaOH. The equivalence point needs 20.00 mL, and the pH after adding 10.00 mL is measured as 3.80. Find Ka.
Step 1. 10.00 mL is exactly half of 20.00 mL, so this is the half-equivalence point and pH = pKa = 3.80.
Step 2. Ka = 10−3.80 = 1.58 × 10−4.
Answer. Ka is about 1.6 × 10−4. This is a standard method to find Ka experimentally, and it works because the two Henderson–Hasselbalch species are equal there.
Common mistakes
- Assuming equivalence pH = 7. The conjugate base A− hydrolyses, so the solution is basic. For a weak base with a strong acid it would be acidic.
- Using Henderson–Hasselbalch at the very start or exactly at equivalence. At 0 mL there is no A−, and at 25.00 mL there is no HA left, so the log term is undefined.
- Dividing by the wrong volume at equivalence. The salt concentration uses the total volume (acid + base), not the starting volume.
- Using concentrations of reagents instead of moles in the buffer region. Moles are safer, because both species are in the same total volume.
- Applying the formula to a polyprotic acid without handling each stage separately.
Exam relevance
| Feature | Weak acid + strong base |
|---|---|
| Initial pH | Higher than a strong acid of equal C |
| Half-equivalence | pH = pKa |
| Equivalence pH | Greater than 7 (salt hydrolysis) |
| Suitable indicator | Changes colour in the basic range |
| Steepest part of curve | Short and shifted to the basic side |
Check the Ka and the concentrations given in your own question. Different books use slightly different Ka values for acetic acid, which can change the last digit of the pH.
The buffer region before equivalence follows the Henderson–Hasselbalch equation. Check those pH values with the buffer tool.
Open the Buffer (Henderson–Hasselbalch) Calculator →