CHEM1011 · Chemistry 1a
Acids, Bases and Buffers
Week 7 applies equilibrium to protons: Brønsted–Lowry acids and bases and conjugate pairs, pH/pOH and the constants Ka, Kb and Kw, and the weak-acid and buffer calculations that dominate assessment. The Henderson–Hasselbalch equation, buffer design and the interpretation of titration and speciation curves (flagged Mastery) make this one of the most calculation-heavy — and heavily examined — topics in the course.
What this chapter covers
- 01Brønsted–Lowry acids (proton donors) and bases (proton acceptors); conjugate pairs differ by one H⁺
- 02Water autoionisation Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C; pH = −log[H⁺], pOH = −log[OH⁻], pH + pOH = 14
- 03Strong vs weak: strong acids/bases dissociate completely; weak ones partially (Ka, Kb)
- 04Conjugate relations Ka·Kb = Kw and pKa + pKb = 14; lower pKa = stronger acid
- 05Weak-acid pH via ICE and the small-x form: pH ≈ ½(pKa − log c₀)
- 06Buffers: a weak acid and its conjugate base resist pH change; Henderson–Hasselbalch pH = pKa + log([A⁻]/[HA])
- 07Buffer design: choose an acid with pKa near the target pH; buffer capacity greatest at pKa, range ≈ pKa ± 1
- 08Titration and speciation curves; salt hydrolysis and the Lewis acid/base definition (Mastery)
Buffer pH from the Henderson–Hasselbalch equation
- +1A weak acid with its conjugate base is a buffer, so use Henderson–Hasselbalch: pH = pKa + log₁₀([A⁻]/[HA]).
- +1Substitute the concentrations: [A⁻]/[HA] = 0.30/0.20 = 1.5.
- +1Evaluate the log term: log₁₀(1.5) = 0.18.
- +1pH = 4.76 + 0.18 = 4.94. The pH sits just above pKa because there is more conjugate base than acid ([A⁻] > [HA]).
Key terms
- Brønsted–Lowry acid/base
- An acid is a proton (H⁺) donor and a base is a proton acceptor; a conjugate acid–base pair differs by exactly one H⁺ (HA/A⁻, B/BH⁺).
- Ka and pKa
- For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA] measures acid strength; pKa = −log Ka, and a lower pKa means a stronger acid. Conjugate pairs satisfy Ka·Kb = Kw and pKa + pKb = 14 at 25 °C.
- pH and pOH
- pH = −log₁₀[H⁺] and pOH = −log₁₀[OH⁻]; since Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C, pH + pOH = 14. Neutral is 7, acidic below, basic above.
- Buffer
- A solution of a weak acid and its conjugate base (or weak base and conjugate acid) in comparable amounts that resists pH change, because added strong acid is consumed by A⁻ and added strong base by HA.
- Henderson–Hasselbalch equation
- pH = pKa + log₁₀([A⁻]/[HA]); the buffer pH is set by the base-to-acid ratio, so equal amounts give pH = pKa.
- Buffer capacity and range
- Buffer capacity — the amount of acid or base a buffer can absorb with little pH change — is greatest when [A⁻] = [HA] (pH = pKa); the useful buffer range is about pKa ± 1 (Mastery).
Acids, Bases and Buffers FAQ
How do I calculate the pH of a weak acid versus a strong acid?
For a strong acid, dissociation is complete, so [H⁺] equals the acid concentration and pH = −log[H⁺] directly. For a weak acid, only a fraction ionises: set up the ICE table for HA ⇌ H⁺ + A⁻, giving Ka = x²/(c₀ − x), and with the small-x approximation x = √(Ka·c₀), so pH ≈ ½(pKa − log c₀). Always check the ionisation is under about 5% before trusting the approximation.
What makes a solution a buffer, and how do I get its pH?
A buffer contains a weak acid and its conjugate base (or the base/conjugate-acid version) in comparable amounts, so it can neutralise added acid or base with only a small pH change. Its pH comes from the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]) — a function of the base-to-acid ratio. Equal amounts give pH = pKa, which is also where the buffer works best.
How do I design a buffer for a target pH?
Choose the weak acid whose pKa is closest to the target pH, because capacity is greatest when [A⁻] ≈ [HA] and the useful range is only about pKa ± 1. Then rearrange Henderson–Hasselbalch to get the required [A⁻]/[HA] ratio, and convert that ratio into the volumes or masses you need. There are several equivalent recipes (weak acid + its salt, weak acid + some strong base, or conjugate-base salt + some strong acid) that all reach the same pH.
Can Sia help me with pH, Ka and buffer calculations?
Yes. Sia can run a weak-acid pH from the ICE table, apply Henderson–Hasselbalch to a buffer, help you pick an acid and solve for the mixing ratio in a buffer-design problem, and check a buffer-capacity calculation after adding strong base. It explains the method and checks your working; it does not do graded assessment, and UNSW academic-integrity rules apply.
Exam move
This is a calculation-dense topic, so build a small set of reflexes. Fix the p-scale relations (pH = −log[H⁺], pH + pOH = 14, Ka·Kb = Kw, pKa + pKb = 14) and the strong-versus-weak split (complete vs partial ionisation). Drill the weak-acid pH routine — ICE, Ka = x²/(c₀ − x), small-x to pH ≈ ½(pKa − log c₀), then the under-5% check. Make Henderson–Hasselbalch automatic and internalise its sanity checks (equal amounts → pH = pKa; more base → pH above pKa). For buffer design, practise the full chain: pick the acid with pKa nearest the target, solve the ratio, convert to volumes/masses, then verify capacity by recomputing the ratio after a strong-acid or strong-base addition. For the Mastery layer, be able to annotate a titration curve (buffer region, half-equivalence pH = pKa, equivalence point) and read a speciation curve crossing at pKa. When a log or ICE step slips, ask Sia to redo it step by step.
Working through Acids, Bases and Buffers in CHEM1011? Sia is AskSia’s AI Chemistry tutor — ask any CHEM1011 Acids, Bases and Buffers question and get a clear, step-by-step explanation grounded in how CHEM1011 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
