Acetic Acid Titration: Half-Equivalence and Indicator Choice
A 25.0 mL sample of 0.100 mol/L acetic acid (CH3COOH, Ka = 1.8 × 10^-5 at 25 °C) is titrated with 0.100 mol/L NaOH from a burette. The student records pH after each addition and wants to stop at the equivalence point, but the burette readings are only reliable to ±0.05 mL. Before the lab, the student must choose an indicator from the following: methyl orange (colour change pH 3.1–4.4), bromothymol blue (6.0–7.6), and phenolphthalein (8.2–10.0). The student also plans to report the pH at half-equivalence and at equivalence, and to explain why the half-equivalence pH is a useful check on the Ka value. The laboratory is at 25 °C, and activity coefficients are assumed to be 1. Carbonate contamination of the NaOH is ruled out.
1. Calculate the pH at half-equivalence and explain why this value is a check on Ka rather than on the acid concentration.
Read the worked answer
The initial amount of acetic acid is
[n_{\text{HA}}=(0.100)(0.0250)=2.50\times10^{-3}\ \text{mol}.]
Half-equivalence requires (1.25\times10^{-3}\ \text{mol}) of NaOH, which is 12.5 mL of 0.100 mol/L NaOH. At that point,
[n_{\text{HA}}=1.25\times10^{-3}\ \text{mol},\qquad n_{\text{A}^-}=1.25\times10^{-3}\ \text{mol}.]
Since the two amounts are equal and they are in the same total volume,
[\text{pH}=\text{p}K_a+\log\frac{[\text{A}^-]}{[\text{HA}]} =\text{p}K_a+\log1=\text{p}K_a.]
Thus
[\text{pH}=-\log(1.8\times10^{-5})\approx4.74.]
The result depends on the ratio ([\text{A}^-]/[\text{HA}]), not on the absolute concentration, because at half-equivalence that ratio is 1 regardless of the starting concentration. So it checks (K_a), not the molarity. The approximation neglects water autoionization, which is reasonable here because the buffer concentrations are much larger than (10^{-7}) M.
2. Calculate the pH at the equivalence point. Is the solution acidic, neutral, or basic? Justify your answer chemically.
Read the worked answer
Equivalence requires 25.0 mL of NaOH, so the total volume is 50.0 mL. All acetic acid has been converted to acetate:
[[\text{CH}_3\text{COO}^-]=\frac{2.50\times10^{-3}}{0.0500}=0.0500\ \text{mol/L}.]
Acetate is the conjugate base of a weak acid, so it hydrolyses:
[\text{CH}_3\text{COO}^-+\text{H}_2\text{O}\rightleftharpoons\text{CH}_3\text{COOH}+\text{OH}^-.]
[K_b=\frac{K_w}{K_a}=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}=5.56\times10^{-10}.]
Then
[[\text{OH}^-]\approx\sqrt{K_bC} =\sqrt{(5.56\times10^{-10})(0.0500)} =5.27\times10^{-6}\ \text{M}.]
So
[\text{pOH}=-\log(5.27\times10^{-6})=5.28,\qquad \text{pH}=14.00-5.28=8.72.]
The solution is basic, not neutral, because the conjugate base produces hydroxide. The approximation is valid because (C/K_b) is large and the hydroxide concentration greatly exceeds the autoionization contribution.
3. Choose the most suitable indicator and judge whether methyl orange would give a sharp endpoint for this titration.
Read the worked answer
The equivalence pH is about 8.7, so the indicator should change colour in the basic region. Phenolphthalein, with a range of 8.2–10.0, brackets the equivalence pH and is the best choice. Bromothymol blue changes too early, around pH 6–7.6, and methyl orange changes at pH 3.1–4.4, far before equivalence. At pH 4, the Henderson–Hasselbalch ratio is
[\frac{[\text{A}^-]}{[\text{HA}]}=10^{\text{pH}-\text{p}K_a}=10^{4.0-4.74}\approx0.18,]
so only about 15% of the acid has been neutralized. Methyl orange would therefore signal an endpoint long before the stoichiometric point, and the titration would give a falsely low volume. Phenolphthalein gives a sharp change because the pH rises steeply near equivalence and passes through its colour-change interval. The ±0.05 mL burette uncertainty is still best handled by adding near the endpoint dropwise and reading the bottom of the meniscus consistently.