CHEM3120 · Environmental and Analytical Chemistry
Acid-Base Speciation & Distribution Diagrams
Lecture 18 covers acid-base equilibria in environmental systems for CHEM3120: weak-acid and weak-base pH from Ka and Kb, the Henderson-Hasselbalch equation, and reading species-distribution (alpha) diagrams where the fraction of HA and A- crosses over at pH = pKa. These are core quantitative Part A short-answer items and the foundation for the carbonate and metal-ion speciation that follows.
What this chapter covers
- 01Weak-acid pH via Ka = x²/(C - x) ≈ x²/C when dissociation is small
- 02Weak-base / salt hydrolysis via Kb = Kw/Ka
- 03The Henderson-Hasselbalch equation pH = pKa + log([A-]/[HA])
- 04Species-distribution (alpha) diagrams: fraction of HA vs A- against pH
- 05Crossover at pH = pKa (each species 50%); one species dominates about 2 pH units either side
- 06When to add water autoionisation at very low acid concentration ([H+] = √(Ka[HA] + Kw))
- 07Applications: benzoic acid (pKa ~4.2), chlorophenols
pH of a weak acid, and its speciation
- +1Set up the weak-acid equilibrium with x = [H+] = [A-]: Ka = x²/(C - x) ≈ x²/C for small dissociation. So x = √(Ka·C) = √(6.3×10^-5 × 0.10) = √(6.3×10^-6).
- +1Evaluate: x = 2.51×10^-3 mol/L. Check the approximation: x/C = 2.5%, safely under 5%, so the ≈ is valid. pH = -log(2.51×10^-3) = 2.60.
- +1Speciation: at pH 2.60, which is well below pKa 4.20, the protonated form (benzoic acid, HA) dominates. Buffered to pH 6.2 (about pKa + 2) the equilibrium shifts to the conjugate base: [A-]/[HA] = 10^(pH - pKa) = 10^(6.2 - 4.20) = 10^2 = 100, so benzoate now dominates roughly 100:1.
Key terms
- Weak-acid dissociation (Ka)
- Ka = [H+][A-]/[HA]; for a weak acid of concentration C, [H+] ≈ √(Ka·C) when dissociation is small (x/C < 5%).
- Kb from Ka
- For a conjugate acid-base pair, Ka·Kb = Kw = 1.0×10^-14, so a salt's base strength is Kb = Kw/Ka — used for hydrolysis pH calculations.
- Henderson-Hasselbalch equation
- pH = pKa + log([A-]/[HA]); rearranges to [A-]/[HA] = 10^(pH - pKa), giving the species ratio directly from pH and pKa.
- Distribution (alpha) diagram
- A plot of the fraction of each acid-base species versus pH; HA and A- cross over (each 50%) at pH = pKa, and one form dominates ~2 pH units either side.
- pKa
- -log Ka; the pH at which an acid is half dissociated. Lower pKa means a stronger acid and a lower crossover pH on the distribution diagram.
- Water autoionisation correction
- At very low acid concentration the water equilibrium is no longer negligible, so [H+] = √(Ka[HA] + Kw) rather than √(Ka[HA]) alone.
Acid-Base Speciation & Distribution Diagrams FAQ
When can I use [H+] = √(Ka·C) for a weak acid?
When dissociation is small — the rule of thumb is that x = [H+] should be under about 5% of the initial concentration C. Then Ka = x²/(C - x) simplifies to x²/C and gives [H+] = √(Ka·C). Always check x/C afterwards; if it exceeds 5%, or the acid is very dilute, solve the full quadratic (supplied on the exam formula sheet) or add the water term [H+] = √(Ka[HA] + Kw).
What does a species-distribution (alpha) diagram tell me?
It shows the fraction of each acid-base species as a function of pH. The single most useful feature is the crossover: HA and A- are each 50% exactly at pH = pKa. Move about two pH units below pKa and the protonated form dominates (~100:1); two units above and the conjugate base dominates. Reading off which species prevails at a given pH is a common short-answer task.
How does Henderson-Hasselbalch connect to the distribution diagram?
They are the same relationship in two forms. Henderson-Hasselbalch, pH = pKa + log([A-]/[HA]), rearranges to [A-]/[HA] = 10^(pH - pKa) — which is precisely the ratio the distribution diagram plots. Set pH = pKa and the ratio is 1 (the crossover); each pH unit away multiplies the ratio by ten. Knowing this lets you compute a species ratio without drawing the curve.
Can AI help me with acid-base speciation in CHEM3120?
Yes. Sia can take you through a weak-acid or hydrolysis pH calculation step by step, check whether your small-x approximation is valid, and explain how the Henderson-Hasselbalch ratio maps onto a distribution diagram. It explains the method and checks your reasoning; it does not do graded work for you, and University of Sydney academic-integrity rules apply.
Exam move
Acid-base speciation is the engine for the carbonate and metal-ion chapters, so master it now. Drill the weak-acid pH routine (x ≈ √(Ka·C), then verify x/C < 5% and fall back to the quadratic if not) and the hydrolysis route via Kb = Kw/Ka. Above all, internalise the distribution diagram: crossover at pH = pKa, and [A-]/[HA] = 10^(pH - pKa) for any pH. Practise stating which species dominates and by roughly what ratio, since that is a fast, recurring Part A mark. Because these methods reappear throughout Block 3, keep them warm rather than isolating them. The exam is closed-book with a formula sheet, so rehearse speed. Confirm the exam date and permitted materials on Canvas and the exam timetable.
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