Acid-Base Mixing Explorer

8.4 Mixing & Buffers

Acid-Base Mixing Explorer

A 2×2 matrix of acid × base types shows the titration curve and a live BCA table at chosen volumes. Read pH, regime, and major species for each mixing case.

BCA + H-H4 Mixing CasesMajor SpeciesRegime pH

Topic 8.4

Acid–Base Reactions and Buffers

Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.

This topic is the decision tree for "what happens when I mix these?" The CED walks through all four combinations.

Strong acid + strong base: they react quantitatively by H⁺(aq) + OH⁻(aq) → H₂O(l). The pH of the resulting solution is determined from the concentration of the excess reagent. If they are equimolar, the result is neutral.

Weak acid + strong base: HA(aq) + OH⁻(aq) ⇌ A⁻(aq) + H₂O(l). Three cases:

Weak acid in excess → a buffer forms; use the Henderson–Hasselbalch equation.
Strong base in excess → pH from the moles of excess hydroxide and the total volume.
Equimolar → only A⁻ remains; the pH is slightly basic, determined by A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq).

Weak base + strong acid: B(aq) + H₃O⁺(aq) ⇌ HB⁺(aq) + H₂O(l). Mirror image:

Weak base in excess → buffer; use Henderson–Hasselbalch.
Strong acid in excess → pH from excess hydronium and total volume.
Equimolar → only HB⁺ remains; pH is slightly acidic from HB⁺(aq) + H₂O(l) ⇌ B(aq) + H₃O⁺(aq).

Weak acid + weak base: they react to an equilibrium state, HA(aq) + B(aq) ⇌ A⁻(aq) + HB⁺(aq). This does not go to completion, and the AP Exam treats it qualitatively.

The universal method: work in moles, not molarity. Convert everything to moles, run the neutralization to completion, identify what is left, then decide which pH calculation applies. Only convert back to concentration at the end, using the combined volume.

Key points

Work in moles through the neutralization; convert to molarity only at the end with the total volume.
Excess strong reagent → pH from the excess. Excess weak reagent + its conjugate → buffer.
Equimolar weak acid + strong base → basic salt solution, pH > 7.
Equimolar weak base + strong acid → acidic salt solution, pH < 7.

Equations

$n = M V$
not on the sheetConvert to moles before neutralizing.
$pH = p K_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]}$
on the exam sheetUse when a buffer results.

Common mistakes

Using initial volumes instead of the combined volume when converting back to molarity.
Assuming an equimolar weak-acid/strong-base mixture is neutral. It is basic.
Averaging pH values. pH is logarithmic; never average it.
Forgetting the strong reagent always wins the neutralization before any equilibrium is considered.

Worked example

50.0 mL of 0.100 M CH₃COOH (K_a = 1.8 × 10⁻⁵) is mixed with 20.0 mL of 0.100 M NaOH. Determine the pH of the resulting solution.

Step 1 — moles before reaction.
n(CH₃COOH) = (0.100)(0.0500) = 5.00 × 10⁻³ mol
n(OH⁻) = (0.100)(0.0200) = 2.00 × 10⁻³ mol

Step 2 — neutralization goes to completion (BCA table, mol):

	CH₃COOH	OH⁻	CH₃COO⁻
B	5.00×10⁻³	2.00×10⁻³	0
C	−2.00×10⁻³	−2.00×10⁻³	+2.00×10⁻³
A	3.00×10⁻³	0	2.00×10⁻³

Step 3 — identify the situation. Both the weak acid and its conjugate base remain in significant amounts: this is a buffer.

Step 4 — Henderson–Hasselbalch. Because both species are in the same solution, the mole ratio equals the concentration ratio and the volume cancels:
pK_a = −log(1.8 × 10⁻⁵) = 4.74
pH = 4.74 + log(2.00 × 10⁻³ / 3.00 × 10⁻³)
pH = 4.74 + log(0.667) = 4.74 − 0.176
pH = 4.56

Below pK_a, as expected — there is more acid than conjugate base.

Full notes for topic 8.4 →