11.2 Acid-Base Titrations & Buffer Systems

Titration Curves and Buffer Action

Quick Answer: A titration curve plots pH vs. volume of titrant added. Three landmarks matter on the PA-CAT: the half-equivalence point (pH = pKa for a weak acid), the equivalence point (moles acid = moles base), and the buffer region (±1 pH unit around pKa) where pH resists change.

The PA-CAT Bulletin of Information, rev. 20240815 weights Chemistry at 16% (≈38 items) and Bloom's Application at 54%, so you should expect to compute pH, interpret curves, and choose buffer ratios—not just define terms.

Curve Shapes

For a weak acid + strong base curve, four regions appear:

1. Initial pH — from the weak acid Ka: [H+] = √(Ka·CHA).
2. Buffer region — after partial neutralization, HA/A⁻ coexist; pH ≈ pKa + log([A⁻]/[HA]).
3. Half-equivalence — exactly half the acid neutralized, [A⁻] = [HA], so pH = pKa.
4. Equivalence and beyond — all HA converted to A⁻; pH set by A⁻ hydrolysis (Kb = Kw/Ka), then by excess strong base.

Worked Titration: 25.0 mL of 0.100 M Acetic Acid titrated with 0.100 M NaOH

Ka(CH3COOH) = 1.8 × 10⁻⁵, so pKa = 4.74.

Step 1 — Initial pH (0 mL NaOH): [H+] = √(Ka·CHA) = √(1.8×10⁻⁵ × 0.100) = √(1.8×10⁻⁶) = 1.34×10⁻³ M → pH = 2.87.

Step 2 — After 10.0 mL NaOH (buffer region): moles HA remaining = (0.0250)(0.100) − (0.0100)(0.100) = 0.00150 mol; moles A⁻ formed = 0.00100 mol. pH = 4.74 + log(0.00100/0.00150) = 4.74 + log(0.667) = 4.74 − 0.176 = 4.56.

Step 3 — Half-equivalence (12.5 mL NaOH): [HA] = [A⁻], so pH = pKa = 4.74. This is the cleanest way to read pKa off a curve.

Step 4 — Equivalence (25.0 mL NaOH): All HA → A⁻. Total volume = 50.0 mL; [A⁻] = 0.00250 mol / 0.0500 L = 0.0500 M. Kb = Kw/Ka = 1.0×10⁻¹⁴/1.8×10⁻⁵ = 5.6×10⁻¹⁰. [OH⁻] = √(Kb·[A⁻]) = √(5.6×10⁻¹⁰ × 0.0500) = √(2.8×10⁻¹¹) = 5.3×10⁻⁶ M. pOH = 5.28, so pH = 8.72 (basic, as expected for a weak acid/strong base equivalence).

Step 5 — 30.0 mL NaOH (excess base): Excess OH⁻ = (0.00500 L)(0.100 M) = 5.0×10⁻⁴ mol in 0.0550 L → [OH⁻] = 9.1×10⁻³ M → pOH = 2.04 → pH = 11.96.

Henderson–Hasselbalch and Buffer Design

pH = pKa + log([A⁻]/[HA]) — the equation is valid when both forms are present in appreciable amount (ratio 0.1–10). To prepare a pH 5.0 acetate buffer with pKa 4.74, solve 5.0 = 4.74 + log([A⁻]/[HA]) → log(ratio) = 0.26 → ratio = 10^0.26 ≈ 1.82. So mix acetate and acetic acid in a 1.82 : 1 molar ratio.

Buffer capacity = moles of acid or base needed to shift pH by 1 unit. It is highest at pH = pKa (where the ratio is 1) and scales with total concentration. A 0.10 M acetate buffer at pH 4.74 has roughly 10× the capacity of a 0.010 M buffer at the same pH.

Physiological Relevance

Blood pH is held near 7.40 by the bicarbonate buffer (pKa = 6.35). Because pH − pKa = 1.05, the buffer ratio is ~11 : 1 (HCO3⁻ : H2CO3). Although this is outside the ideal 0.1–10 window, the open-system exhalation of CO2 makes the bicarbonate buffer effective in vivo—a point PA-CAT items occasionally raise in clinical framing.

Titration Curves and Indicator Choice

A strong-acid–strong-base titration (e.g., HCl + NaOH) has a pH curve that starts low, rises gradually, then jumps almost vertically through pH 7 at the equivalence point. A weak acid–strong base titration (e.g., acetic acid + NaOH) has an equivalence point above pH 7 because the conjugate base hydrolyzes, and a buffer region near pKa where pH changes slowly — the flat shoulder that lets half-equivalence double as a pKa measurement.

An indicator changes color over about two pH units centered on its pKin, so it must bracket the equivalence-point pH:

Worked example: 25.0 mL of 0.100 M acetic acid titrated with 0.100 M NaOH reaches equivalence at 25.0 mL and pH ≈ 8.7, so phenolphthalein is appropriate; methyl orange would change color far before equivalence and produce a large systematic error. The PA-CAT tests whether you can match indicator to titration chemistry and compute titrant volume from M1V1 = M2V2 at equivalence.