Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- H₃O⁺ and OH⁻ react completely with each other and with weak bases or weak acids. Treat these reactions as one-way.
- Step 1: convert every volume and molarity to moles. Step 2: let the strong reagent react (limiting reactant). Step 3: look at what is left.
- A buffer is a weak acid with its conjugate base (or a weak base with its conjugate acid) in comparable amounts. Partly neutralising a weak acid or weak base makes one.
- A weak acid and a weak base do not react completely. They reach an equilibrium.
Which calculation?
| What is left after reaction | Method | pH |
|---|---|---|
| Excess strong acid | [H₃O⁺] = excess mol ÷ total volume | below 7 |
| Excess strong base | [OH⁻] = excess mol ÷ total volume, then pOH → pH | above 7 |
| Only water and spectator ions (strong + strong, equal moles) | — | 7.00 at 25 °C |
| HA and A⁻ (or B and HB⁺) | pH = pK_a + log(n base form ÷ n acid form) | near pK_a |
| Only A⁻ (weak acid + strong base, equal moles) | K_b = K_w ÷ K_a; [OH⁻] ≈ √(K_b × c) | slightly above 7 |
| Only HB⁺ (weak base + strong acid, equal moles) | K_a = K_w ÷ K_b; [H₃O⁺] ≈ √(K_a × c) | slightly below 7 |
| Weak acid HA + weak base B | K = K_a(HA) ÷ K_a(HB⁺) | products favoured if HA is the stronger acid |
Assumptions
- 25 °C, so K_w = 1.0 × 10⁻¹⁴.
- Volumes add when solutions are mixed.
- Spectator ions such as Na⁺ and Cl⁻ do not affect pH.
- The weak species’ own reaction with water does not change the buffer’s mole ratio noticeably.
Mistakes to avoid
- Subtracting concentrations instead of moles.
- Dividing excess moles by one solution’s volume, not the total.
- Assuming pH 7 whenever equal moles react.
- Using H–H when one member of the pair has run out.
- Writing the ratio upside down: base form goes on top.
- Using an ICE table for HA + OH⁻; it goes to completion.
Quick self-check
- 10.0 mL of 0.10 M HCl + 10.0 mL of 0.10 M NaOH. pH? (7.00)
- 4.0 × 10⁻³ mol of a weak acid (pK_a 4.20) + 1.0 × 10⁻³ mol of NaOH. pH? (buffer: 4.20 + log(1.0/3.0) = 3.72)
- 5.0 × 10⁻³ mol of a weak acid + 6.0 × 10⁻³ mol of NaOH, total volume 100. mL. pH? (excess OH⁻ 0.010 M; pH 12.00)
Next: practice questions.