Acetic Acid Titration: What Should the pH Be at Half-Equivalence?
During a unit on acids and bases, a teacher asks students to predict the pH at three points in the titration of 50.0 mL of 0.100 M acetic acid (Ka = 1.8 × 10−5) with 0.100 M NaOH: after 0 mL, 25.0 mL, and 50.0 mL of base. One student predicts that the pH at 25.0 mL will be 7.00 because “half of the acid has been neutralized, so the solution is neutral.” Another student predicts that the pH at 50.0 mL will also be 7.00 because “all acid and base have reacted.” The teacher wants to use the predictions to diagnose whether students are applying equilibrium and hydrolysis ideas or only recalling the strong acid–strong base pattern. The class has already calculated that the initial pH is about 2.87, but the students disagree about the equivalence point.
1. Calculate the pH after 25.0 mL of NaOH has been added and evaluate the first student’s pH = 7.00 prediction.
Read the worked answer
At 25.0 mL, moles OH− added = 0.0250 L × 0.100 M = 2.50 × 10−3 mol. Initial HA = 0.0500 L × 0.100 M = 5.00 × 10−3 mol. After reaction, HA remaining = 5.00 × 10−3 − 2.50 × 10−3 = 2.50 × 10−3 mol, and A− formed = 2.50 × 10−3 mol. This is the half-equivalence point, so [HA] = [A−]. Henderson–Hasselbalch gives pH = pKa + log([A−]/[HA]) = −log(1.8 × 10−5) + log(1) = 4.74. The student’s pH 7.00 is wrong because the solution is a buffer of comparable weak acid and conjugate base, not a neutral salt solution. At half-equivalence the pH is approximately pKa; neutrality would require [H3O+] = [OH−], which is not imposed by equal HA and A− concentrations.
2. Calculate the pH after 50.0 mL of NaOH has been added. Then recommend a suitable indicator and justify the choice using the equivalence-point pH.
Read the worked answer
At 50.0 mL, moles OH− added = 0.0500 L × 0.100 M = 5.00 × 10−3 mol, exactly equal to initial HA. The acid is converted to acetate: 5.00 × 10−3 mol A− in a total volume of 100.0 mL, so [A−] = 0.0500 M. Acetate is a weak base: Kb = Kw/Ka = 1.0 × 10−14 / 1.8 × 10−5 = 5.6 × 10−10. Let x = [OH−]; x2/(0.0500 − x) ≈ x2/0.0500 = 5.6 × 10−10, so x = 5.3 × 10−6 M, pOH = 5.28, pH = 8.72. The equivalence point is basic, not neutral, because the conjugate base hydrolyzes. Phenolphthalein (color change roughly pH 8.2–10.0) is suitable; bromothymol blue (6.0–7.6) would change too early and mislead students about the equivalence point.
3. After 10.0 mL of NaOH has been added, calculate the pH and explain why it is not simply the pH of the diluted acetic acid.
Read the worked answer
After 10.0 mL, OH− added = 0.0100 L × 0.100 M = 1.00 × 10−3 mol. This converts 1.00 × 10−3 mol HA to A−. Remaining HA = 5.00 × 10−3 − 1.00 × 10−3 = 4.00 × 10−3 mol; A− = 1.00 × 10−3 mol. The total volume is 60.0 mL, but the ratio is what enters the buffer equation: pH = 4.74 + log(1.00 × 10−3 / 4.00 × 10−3) = 4.74 + log(0.250) = 4.74 − 0.602 = 4.14. Dilution alone would only slightly raise the pH of acetic acid (from about 2.87 toward 2.9) because it lowers [H3O+] modestly. The large change to pH 4.14 comes from converting a strong base into the conjugate base A−, which shifts the acid equilibrium and creates a buffer. Students who ignore the reaction and calculate only dilution miss the chemistry.