Topic 14 · page 18 of 18

Core Practical 11: finding the Ka of a weak acid

Find the equivalence volume by titration, then add exactly half of it to fresh acid and measure the pH. At that point pH = pKa, so Ka = 10 to the power minus pH.

15 min read14.2314.21(ii)

Read first: Buffers on the titration curve

Aim and idea 14.2314.21(ii)

To find the acid dissociation constant, $K_\text{a}$ , of a weak acid, here ethanoic acid, using a pH meter and a titration.

The idea comes from page 16. When exactly half of a weak acid has been neutralised, [HA] = [A⁻], so pH = $\text{p}K_\text{a}$ . You therefore need two things: the volume of NaOH that neutralises all the acid ( $V_\text{eq}$ ), and the pH when half that volume has been added.

Method 14.23

Apparatus: a burette on a stand holds 0.100 mol per cubic decimetre sodium hydroxide and delivers into a beaker of 25.0 cm³ of the weak acid on a magnetic stirrer. A pH probe in the beaker is connected to a pH meter that reads 4.76. — Burette, beaker on a magnetic stirrer, and a pH meter. Schematic, not to scale.

PracticalFinding Ka

Calibrate the pH meter with buffer solutions of known pH (for example pH 4.00 and pH 7.00, which bracket the pH you expect). Rinse the probe with deionised water.
Find the equivalence volume. Pipette 25.0 cm³ of the ethanoic acid into a conical flask, add a few drops of phenolphthalein, and titrate with 0.100 mol dm⁻³ NaOH. Do one rough titration, then repeat until two titres agree within 0.10 cm³. Calculate the mean, $V_\text{eq}$ .
Half-neutralise. Pipette a fresh 25.0 cm³ of the same acid into a beaker. From the burette add exactly $\tfrac{1}{2}V_\text{eq}$ of the NaOH. Stir, and put in the probe.
Measure the pH when the reading is steady, and record it. This is the $\text{p}K_\text{a}$ .
Calculate $K_\text{a}$ = $10^{-\text{pH}}$ .

A variation records the whole curve instead. Place the probe in 25.0 cm³ of acid, add NaOH in 1.0 cm³ portions (0.1 to 0.5 cm³ near equivalence), record the pH after each, plot pH against volume, find $V_\text{eq}$ from the middle of the vertical section and read the pH at $\tfrac{1}{2}V_\text{eq}$ .

Step	Why
Calibrate and rinse the probe	The pH reading is only as good as the calibration, and $K_\text{a}$ depends on it directly
Use the same pipette and the same NaOH for both parts	The ratio of acid to alkali is then exact, so the half-neutralisation is exactly half
Stir before reading	The pH near the probe must match the rest of the solution
Wait for a steady reading	The meter needs a few seconds to settle
Keep the temperature the same	$K_\text{a}$ changes with temperature

Results and processing 14.23

The results below are illustrative, calculated from the equilibrium model, not measured data.

Titration	Volume of NaOH / cm³
1 (rough)	25.60
2	25.05
3	24.95
4	25.00

The concordant titres (2, 3 and 4) have a mean of 25.00 cm³, so $V_\text{eq}$ = 25.00 cm³ and the half-neutralisation volume is 12.50 cm³.

If you recorded the whole curve, you have data like this.

Volume of NaOH / cm³	pH
0	2.88
5.0	4.16
10.0	4.58
12.5	4.76
15.0	4.94
20.0	5.36
24.0	6.14
24.5	6.45
25.0	8.73
25.5	11.00
26.0	11.29
30.0	11.96
40.0	12.36

A graph of pH against volume of sodium hydroxide added, with thirteen data points shown as crosses and a smooth curve through them. The points rise from pH 2.88 at 0 cm³, through 4.76 at 12.5 cm³, to a jump between 24.5 and 25.5 cm³ and a plateau near pH 12.4. A dashed line marks half the equivalence volume, 12.5 cm³, where the pH is 4.76. A second dashed line marks the equivalence volume, 25.0 cm³. — Illustrative results. The pH at half the equivalence volume is the pKa.

Observation	Explanation
The start is at pH 2.88, not 1	ethanoic acid is weak, so it is only partly dissociated
The pH rises slowly up to about 24 cm³	HA and A⁻ are both present: a buffer
A sharp rise at 25.0 cm³	equivalence: all the HA has been used
pH 4.76 at 12.5 cm³	[HA] = [A⁻], so pH = $\text{p}K_\text{a}$

K_\text{a} = 10^{-4.76} = 1.7 \times 10^{-5}\ \text{mol dm}^{-3}

There is a cross-check from the start of the curve. A pH of 2.88 for 0.100 mol dm⁻³ acid gives, by the method of page 6, $K_\text{a}$ = ( $10^{-2.88}$ )² ÷ 0.100 = 1.7 × 10⁻⁵. The two methods agree.

Evaluation 14.23

Source of error	Effect	Improvement
Meter badly calibrated	every pH shifts by the same amount, so $K_\text{a}$ is wrong	calibrate with two buffers that bracket the pH
Equivalence volume read from a curve	$V_\text{eq}$ is less exact, so half of it is too	take readings every 0.1 cm³ near the jump, or use concordant titrations
NaOH has absorbed CO₂ from the air	its true concentration is lower than stated	use the same NaOH for both parts, as the method does
Temperature different from the data-book value	$K_\text{a}$ changes with temperature	record the temperature and compare with a value at that temperature
Assumes [HA] = [A⁻] exactly at the half point	a small error, because some of the HA also dissociates	negligible here, as [H⁺] is far smaller than [HA] and [A⁻]

Two points about the size of the errors.

A pH uncertainty of ±0.01 gives a $K_\text{a}$ uncertainty of about 2.3%, because $K_\text{a}$ = $10^{-\text{pH}}$ . On page 6 it was doubled to about 5% because [H⁺] was squared. Here there is no squaring.
The curve is almost flat at 12.5 cm³ (about 0.07 pH units per cm³), so a burette error of ±0.05 cm³ changes the pH by less than 0.004. The result is not sensitive to small errors in the volume.

Check yourself

Explain why the pH meter is calibrated with buffers at pH 4.00 and pH 7.00 for this experiment.
The mean equivalence volume is 21.40 cm³. What volume of NaOH is added to a fresh 25.0 cm³ of acid in the second part?
The pH at half-neutralisation is 4.62. Calculate $K_\text{a}$ and $\text{p}K_\text{a}$ .
Explain why the pH at half-neutralisation equals the $\text{p}K_\text{a}$ .
A student's result is $K_\text{a}$ = 1.55 × 10⁻⁵ mol dm⁻³ and the data-book value is 1.74 × 10⁻⁵. Calculate the percentage difference, and the error in the pH reading that would explain it.
The pH reading is uncertain by ±0.02. Calculate the percentage uncertainty in $K_\text{a}$ .

Answers cover these until you have tried the questions

The expected pH of the half-neutralised acid is about 4.8. Buffers at pH 4.00 and pH 7.00 bracket that value, so the meter reads accurately across the range used.
Half of 21.40 = 10.70 cm³ of NaOH, added to 25.0 cm³ of the same acid.
$\text{p}K_\text{a}$ = 4.62. $K_\text{a}$ = $10^{-4.62}$ = 2.40 × 10⁻⁵, so 2.4 × 10⁻⁵ mol dm⁻³.
At half-neutralisation half the acid has been converted to its conjugate base, so [HA] = [A⁻]. In $K_\text{a}$ = [H⁺][A⁻] ÷ [HA] these cancel, so $K_\text{a}$ = [H⁺], and pH = $\text{p}K_\text{a}$ .
Percentage difference = (1.74 − 1.55) ÷ 1.74 × 100 = 10.9% low. $\text{p}K_\text{a}$ for 1.55 × 10⁻⁵ = 4.81, against 4.76, so a pH reading 0.05 units too high would explain it (for example from imperfect calibration).
$K_\text{a}$ = $10^{-\text{pH}}$ , so the factor is $10^{0.02}$ = 1.047, an uncertainty of about 4.7% in $K_\text{a}$ .