Acid-Base Reactions and Buffers: Study Guide (Chemistry 8.4)

React first, then think about equilibrium

In Topic 8.3 each solution held one acid or one base. Now you mix an acid with a base. The key fact is that H₃O⁺ and OH⁻ react with each other, and with weak bases and weak acids, essentially to completion. The equilibrium constants are so large that you can treat these reactions as one-way:

• strong acid + strong base: H₃O⁺(aq) + OH⁻(aq) → 2 H₂O(l)
• weak acid + strong base: HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l)
• weak base + strong acid: B(aq) + H₃O⁺(aq) → HB⁺(aq) + H₂O(l)

So every mixing problem has the same shape. First do the stoichiometry in moles, as a limiting-reactant problem. Then look at what is left and choose the right equilibrium calculation.

Two habits make the stoichiometry safe. Work in moles (n = c × V, with V in litres), not in concentrations, because mixing changes the volume. And when you need a concentration at the end, divide by the total volume.

Strong acid with strong base

The reaction H₃O⁺ + OH⁻ → 2 H₂O goes to completion. Whatever is in excess sets the pH. If neither is in excess, only water and spectator ions remain, and the pH is 7.00 at 25 °C.

Worked example 1: excess strong acid

Question. 25.0 mL of 0.200 M HCl is mixed with 30.0 mL of 0.150 M NaOH. Find the pH.

1. n(H₃O⁺) = 0.0250 L × 0.200 M = 5.00 × 10⁻³ mol.
2. n(OH⁻) = 0.0300 L × 0.150 M = 4.50 × 10⁻³ mol. OH⁻ is limiting.
3. Excess H₃O⁺ = 5.00 × 10⁻³ − 4.50 × 10⁻³ = 5.0 × 10⁻⁴ mol.
4. Total volume = 25.0 + 30.0 = 55.0 mL = 0.0550 L, so [H₃O⁺] = 5.0 × 10⁻⁴ ÷ 0.0550 = 9.09 × 10⁻³ M.
5. pH = −log(9.09 × 10⁻³) = 2.04.

Common slip. Dividing by 25.0 mL (the acid volume only) gives pH 1.70. The excess acid is spread through the whole 55.0 mL.

Weak acid with strong base: three possible outcomes

HA + OH⁻ → A⁻ + H₂O goes to completion. What is left depends on which reactant runs out.

• Weak acid in excess. Some HA remains, together with the A⁻ just made. A weak acid and its conjugate base together form a buffer. Its pH follows from the K_a expression rearranged, the Henderson–Hasselbalch equation (studied fully in Topic 8.9):

  pH = pK_a + log([A⁻] / [HA])

  Both species share one volume, so you can put the mole ratio straight in.

• Equal moles. All the HA has become A⁻. A⁻ is a weak base (Topic 8.3), so it reacts slightly with water: A⁻ + H₂O ⇌ HA + OH⁻. The pH is slightly basic. Use K_b = K_w ÷ K_a.

• Strong base in excess. The leftover OH⁻ swamps the tiny amount A⁻ makes. pH comes from the excess OH⁻ moles ÷ total volume.

Worked example 2: one weak acid, three amounts of base

Question. The fictional weak acid HQ has pK_a = 4.80. A 50.0 mL sample of 0.100 M HQ is mixed with 0.100 M NaOH. Find the pH after adding (a) 20.0 mL, (b) 50.0 mL, (c) 60.0 mL of NaOH.

Start: n(HQ) = 0.0500 L × 0.100 M = 5.00 × 10⁻³ mol.

(a) 20.0 mL. n(OH⁻) = 2.00 × 10⁻³ mol, limiting.

	HQ	OH⁻	Q⁻
Before (mol)	5.00 × 10⁻³	2.00 × 10⁻³	0
After (mol)	3.00 × 10⁻³	0	2.00 × 10⁻³

Both HQ and Q⁻ are left: a buffer. pH = 4.80 + log(2.00 / 3.00) = 4.80 − 0.18 = 4.62. Check: there is more acid than conjugate base, so the pH should be a little below pK_a. It is.

(b) 50.0 mL. n(OH⁻) = 5.00 × 10⁻³ mol, exactly equal. Only Q⁻ is left. [Q⁻] = 5.00 × 10⁻³ mol ÷ 0.100 L = 0.0500 M. K_b = 1.0 × 10⁻¹⁴ ÷ 10^(−4.80) = 6.31 × 10⁻¹⁰. [OH⁻] ≈ √(6.31 × 10⁻¹⁰ × 0.0500) = 5.62 × 10⁻⁶ M (0.011% of 0.0500, so the shortcut is fine). pOH = 5.25, so pH = 8.75. Slightly basic, as predicted.

(c) 60.0 mL. n(OH⁻) = 6.00 × 10⁻³ mol. Excess OH⁻ = 1.00 × 10⁻³ mol in 110.0 mL. [OH⁻] = 1.00 × 10⁻³ ÷ 0.1100 = 9.09 × 10⁻³ M; pOH = 2.04; pH = 11.96.

Weak base with strong acid: the mirror image

B + H₃O⁺ → HB⁺ + H₂O goes to completion. The three outcomes mirror the weak acid case:

• weak base in excess: a buffer of B and HB⁺, with pH = pK_a(HB⁺) + log([B] / [HB⁺]);
• equal moles: only HB⁺ remains, and HB⁺ + H₂O ⇌ B + H₃O⁺ makes the pH slightly acidic;
• strong acid in excess: pH from the excess H₃O⁺ moles ÷ total volume.

Worked example 3: ammonia and hydrochloric acid

Question. A 40.0 mL sample of 0.250 M NH₃ (K_b = 1.8 × 10⁻⁵) is mixed with 0.250 M HCl. Find the pH after adding (a) 10.0 mL, (b) 40.0 mL of the acid.

Start: n(NH₃) = 0.0400 L × 0.250 M = 1.00 × 10⁻² mol. For NH₄⁺, K_a = K_w ÷ K_b = 5.56 × 10⁻¹⁰ and pK_a = 9.26.

(a) 10.0 mL. n(H₃O⁺) = 2.50 × 10⁻³ mol, limiting. After reaction: 7.50 × 10⁻³ mol NH₃ and 2.50 × 10⁻³ mol NH₄⁺. A buffer. pH = 9.255 + log(7.50 / 2.50) = 9.255 + 0.477 = 9.73 (keep the unrounded pK_a until the end).

(b) 40.0 mL. n(H₃O⁺) = 1.00 × 10⁻² mol, equal to NH₃. Only NH₄⁺ remains. [NH₄⁺] = 1.00 × 10⁻² ÷ 0.0800 L = 0.125 M. [H₃O⁺] ≈ √(5.56 × 10⁻¹⁰ × 0.125) = 8.33 × 10⁻⁶ M, so pH = 5.08.

Interpretation. “Neutralised” does not mean “neutral”. The ammonium ion left behind is a weak acid, so the solution is acidic.

Weak acid with weak base: an equilibrium, not a completed reaction

When neither reactant is strong, the reaction does not go to completion. It reaches equilibrium:

HA(aq) + B(aq) ⇌ A⁻(aq) + HB⁺(aq)

Its constant combines what you know from Topic 8.3: K = K_a(HA) × K_b(B) ÷ K_w, which is the same as K = K_a(HA) ÷ K_a(HB⁺). Read it as a contest between two acids, HA on the left and HB⁺ on the right. The proton ends up mostly on the weaker acid’s conjugate base.

• Acetic acid with ammonia: K = 1.8 × 10⁻⁵ ÷ 5.56 × 10⁻¹⁰ = 3.2 × 10⁴. Acetic acid is the stronger acid, so products are favoured: mostly CH₃COO⁻ and NH₄⁺.
• A fictional acid with pK_a 9.00 and a base whose conjugate acid has pK_a 5.00: K = 10^(−9.00) ÷ 10^(−5.00) = 1.0 × 10⁻⁴. Reactants are favoured; very little proton transfer happens.

The rule: products are favoured when the acid on the left is stronger (smaller pK_a) than the acid on the right.

What makes a buffer, and the boundary of this topic

A buffer contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts. You can make one by mixing the two directly, or, as in Worked examples 2(a) and 3(a), by partly neutralising a weak acid with a strong base or a weak base with a strong acid. Topics 8.8 to 8.10 explain why buffers resist pH change and how much acid or base they can absorb. Calculating the pH change when acid or base is added to an existing buffer is outside the scope of the exam, so it is not covered here.

Common misconceptions

• “Neutralisation always gives pH 7.” True only for strong acid with strong base.
• Subtracting concentrations instead of moles. Mixing changes volumes; always convert to moles first.
• Forgetting the total volume when converting excess moles back to a concentration.
• Running an ICE table on HA + OH⁻. That reaction goes to completion; use a before/after table, then do the equilibrium step.
• Using H–H when one species has run out. A buffer needs both members of the pair. At equal moles, use K_b (or K_a) of what is left.
• Flipping the ratio. It is log(base form ÷ acid form). More base form means pH above pK_a.
• “Any acid plus any base goes to completion.” Two weak species reach an equilibrium whose K may be large or small.

Where this leads

Next, in Topic 8.5, you will follow these same calculations continuously as base is added drop by drop, and see them as a titration curve. Try the practice questions now, then use the revision notes and the checklist to consolidate.