Practice a College Board-style short-answer question on Acids & Bases. Write your response, then reveal the model answer to see exactly how each part earns its point.
Short Answer Question · Unit 8 · Weak Acid Titration & Buffers
A 25.0 mL sample of 0.100 M acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵) is titrated with 0.100 M NaOH. A student monitors the pH throughout the titration using a pH probe.
A
Calculate the pH of the original 0.100 M acetic acid solution before any NaOH has been added.
✓ Model answer (earns the point)
Set up an ICE table for CH₃COOH ⇌ H⁺ + CH₃COO⁻: Ka = x²/(0.100 − x) ≈ x²/0.100 = 1.8 × 10⁻⁵. Solving, x = [H⁺] = √(1.8 × 10⁻⁶) = 1.3 × 10⁻³ M. pH = −log(1.3 × 10⁻³) ≈ 2.87.
Why it scores: Sets up the correct ICE table/Ka expression for a weak acid AND solves for [H⁺] AND correctly converts to pH.
B
After 12.5 mL of NaOH has been added, the student finds the pH of the solution to be 4.74. Explain why this pH equals the pKa of acetic acid at this specific point in the titration.
✓ Model answer (earns the point)
12.5 mL is exactly half the volume needed to reach the equivalence point (25.0 mL of 0.100 M NaOH would supply the same moles as the original acid). At this half-equivalence point, exactly half of the original acetic acid has been converted to its conjugate base, acetate (CH₃COO⁻), so [CH₃COOH] = [CH₃COO⁻]. By the Henderson-Hasselbalch equation, pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = pKa + log(1) = pKa, since log(1) = 0.
Why it scores: Identifies the point as the half-equivalence point AND states that [acid] = [conjugate base] there AND connects this to the Henderson-Hasselbalch equation reducing to pH = pKa.
C
Predict whether the pH at the equivalence point of this titration will be less than, equal to, or greater than 7, and justify your answer.
✓ Model answer (earns the point)
The pH at the equivalence point will be greater than 7. At the equivalence point, all of the acetic acid has been converted into its conjugate base, acetate (CH₃COO⁻). Acetate is a weak base that reacts with water (hydrolysis): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing excess OH⁻ and making the solution basic.
Why it scores: Correctly predicts pH > 7 AND explains that the solution contains only the conjugate base (acetate) at the equivalence point AND identifies that this conjugate base hydrolyzes water to produce OH⁻.
How to score points on SAQs
Show your work for calculations. Setting up the correct ICE table or Henderson-Hasselbalch expression often earns credit even if the final numerical answer has a small error.
Identify titration landmarks by name. Saying "half-equivalence point" or "equivalence point" explicitly is stronger than just describing volumes.
Always justify pH predictions with hydrolysis or dissociation reasoning. "It will be basic" alone scores less than explaining that the conjugate base reacts with water to produce OH⁻.
Connect every calculated value back to the underlying chemistry. A pH of 4.74 isn't just a number — it equals pKa because of the specific 50:50 ratio of acid to conjugate base at that point.
Keep it tight. 2–4 sentences per part is plenty. Long answers don't score higher; they just waste exam time.