CHEM1011 · Fundamentals Of Chemistry 1a
Acids, Bases and Titrations
This chapter is the quantitative acid–base toolkit CHEM1011 builds its hardest exam questions on: telling strong (fully dissociating) from weak (equilibrium-governed, via Ka) acids, working the pH = −log[H3O+] scale with pH + pOH = 14, and reading a titration curve region by region. Because the closed-book exam gives you the constants on a datasheet, marks are won on method — knowing which formula belongs to which point on the curve — making this one of the highest-yield topics to drill before the final.
What this chapter covers
- 011. Strong vs weak: full dissociation vs equilibrium, single vs double arrow
- 022. Ka and pKa: Ka = [H₃O⁺][A⁻]/[HA], pKa = −log Ka, pKb = 14 − pKa
- 033. The pH scale: pH = −log[H₃O⁺], pOH, pH + pOH = 14 at 25 °C
- 044. Buffers: weak acid + conjugate base, why they resist pH change
- 055. Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA])
- 066. Titration types and the equivalence-point pH rule (SA/SB = 7, WA/SB > 7, SA/WB < 7)
- 077. Equivalence point vs end point, and the half-equivalence shortcut pH = pKa
- 088. Indicator choice: matching the transition range to the steep jump
Buffer pH by Henderson–Hasselbalch (+ half-equivalence check)
- +1Identify the system: a weak acid (HA = CH3COOH) plus its conjugate base (A− = CH3COO−) is a buffer, so use Henderson–Hasselbalch: pH = pKa + log([A−]/[HA]).
- +1Substitute the concentrations (same volume, so the concentration ratio = the mole ratio): pH = 4.76 + log(0.40 / 0.25) = 4.76 + log(1.60).
- +1Evaluate the log term: log(1.60) = +0.20, so pH = 4.76 + 0.20 = 4.96. More base than acid (ratio > 1) gives pH above pKa — a useful sanity check.
- +1(b) At the half-equivalence point exactly half the acid has been neutralised, so [A−] = [HA], the log term is 0, and pH = pKa = 4.76.
Key terms
- Strong acid / strong base
- An acid or base that dissociates fully in water, so [H₃O⁺] (or [OH⁻]) equals the formal concentration — no equilibrium constant needed. Written with a single arrow, e.g. HCl + H₂O → H₃O⁺ + Cl⁻.
- Weak acid
- An acid that only partially ionises, settling at an equilibrium described by Kₐ = [H₃O⁺][A⁻]/[HA]. Written with a double arrow (⇌); a smaller Kₐ (larger pKₐ) means a weaker acid.
- pH and pOH
- pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻]; at 25 °C they obey pH + pOH = 14. Neutral water has [H₃O⁺] = 10⁻⁷ M, i.e. pH = 7. Each whole pH unit is a tenfold change in [H₃O⁺].
- Buffer
- A mixture of a weak acid and its conjugate base that resists pH change: added acid is neutralised by the base form and added base by the acid form, so the ratio — and hence the pH — barely moves. Its pH is given by Henderson–Hasselbalch.
- Equivalence point vs end point
- The equivalence point is the stoichiometric truth, where moles of titrant added = moles of analyte (n_base = n_acid). The end point is what you observe, the indicator's colour change; a good indicator makes the two coincide.
- Half-equivalence point
- The point in a weak-acid titration where exactly half the acid has been neutralised, so [HA] = [A⁻] and the Henderson–Hasselbalch log term is zero, giving pH = pKₐ.
Acids, Bases and Titrations FAQ
When do I use Henderson–Hasselbalch instead of an ICE table?
Use Henderson–Hasselbalch (pH = pKₐ + log([A⁻]/[HA])) whenever the flask holds appreciable amounts of BOTH a weak acid and its conjugate base — a prepared buffer, or anywhere on the buffer plateau of a weak-acid titration before equivalence. Use a full ICE table on Kₐ for the initial pH of a pure weak acid, and ICE on K_b of the conjugate base at the equivalence point.
Why isn't the equivalence point always at pH 7?
Only a strong-acid/strong-base titration gives equivalence at pH 7, because its salt is neutral. For a weak acid + strong base the flask at equivalence holds only the conjugate base A⁻, which is weakly basic, so the pH is above 7. For a strong acid + weak base it is below 7 because BH⁺ is weakly acidic.
How do I choose the right indicator for a titration?
Pick an indicator whose transition range straddles the equivalence pH on the steep part of the curve. Strong/strong (eq pH 7): bromothymol blue (about 6.0–7.6) works. Weak acid/strong base (eq pH > 7): use phenolphthalein (about 8.3–10.0). Strong acid/weak base (eq pH < 7): use methyl orange (about 3.1–4.4).
What's on the exam for this topic?
Expect pH calculations from [H₃O⁺] or [OH⁻] (route strong bases through pOH first), a buffer pH by Henderson–Hasselbalch, identifying the equivalence pH for each titration type, the half-equivalence shortcut pH = pKₐ, and an n = C × V titration with the vinegar standardisation as the flagship lab example. The datasheet supplies constants, so you are graded on method, not memorising numbers.
How do I get [H₃O⁺] back from a pH?
Invert the log: [H₃O⁺] = 10⁻ᵖᴴ. So pH 4.96 gives [H₃O⁺] = 10⁻⁴˙⁹⁶ ≈ 1.1 × 10⁻⁵ M. For pOH the mirror relation is [OH⁻] = 10⁻ᵖᴼᴴ, and at 25 °C pH + pOH = 14 lets you swap between the two scales.
Exam move
Treat the titration curve as a map and learn which formula lives in which region, because that routing is exactly what the exam tests. Drill the four moves until automatic: a strong acid/base point is just leftover EXCESS (pH = −log of what remains); a buffer point is Henderson–Hasselbalch; half-equivalence is the freebie pH = pKₐ; and a weak-acid equivalence point sits ABOVE 7. Memorise the equivalence-pH rule (SA/SB = 7, WA/SB > 7, SA/WB < 7) and pair each with its indicator, then practise n = C × V with all volumes converted to litres — the single most common slip. Since the datasheet hands you Kₐ, pKₐ and constants, spend your revision on the decision tree, not on rote numbers.
