What is the pH at the equivalence point for the titration between sodium hydroxide and acetic acid?

Question

BlackTom AI · Accepted Answer

Acetic acid is a weak acid (Ka = 1.8×10^-5) and sodium hydroxide is a strong base, so at the equivalence point the solution contains the conjugate base acetate (A–) in water, which hydrolyzes to produce OH– and makes the solution basic.

1) Relevant knowledge points and constants:
- Ka(acetic acid) ≈ 1.8 × 10^-5.
- Kw = 1.0 × 10^-14, so Kb for acetate = Kw/Ka.
- At equivalence all original acetic acid has been converted to acetate; the pH is set by hydrolysis: A– + H2O ⇌ HA + OH–.
- Approximation: x («) [A–], so Kb ≈ x^2/[A–] and [OH–] = x = √(Kb·[A–]).

2) Compute Kb:
Kb = Kw/Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10.

3) Choose a typical concentration to compute [A–]. For a common titration with equal concentrations of acid and base (for example 0.10 M CH3COOH titrated by 0.10 M NaOH), at equivalence volumes double the original volume so [A–]final = 0.10/2 = 0.050 M. (The pH depends weakly on concentration; this is a standard example.)

4) Solve hydrolysis for [OH–]:
[OH–] = √(Kb · [A–]) = √(5.56 × 10^-10 × 0.050)
= √(2.78 × 10^-11) ≈ 5.27 × 10^-6 M.

5) Convert to pH:
pOH = −log(5.27 × 10^-6) ≈ 5.28
pH = 14.00 − 5.28 ≈ 8.72 ≈ 8.7–9.0.

6) Why this is correct and why other answers (e.g., pH = 7 or very high pH) are wrong:
- pH = 7 would be expected for a strong acid–strong base titration; here the conjugate base hydrolyzes, producing OH– and giving a basic pH.
- Very high pH values (≥11–14) are incorrect because the conjugate base is only a weak base (Kb ~10^-10), so [OH–] is small.
- The computed value (~8.7) is close to 9, so the correct answer (~9) is confirmed.

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