Why does adding acid to a buffer change moles of both forms, not just one?

Every mole of strong acid added reacts completely with one mole of conjugate base, converting it to the weak acid form. So conjugate base moles go down by exactly the moles of acid added, and weak acid moles go up by the same amount — both forms change, by the same quantity, in opposite directions.

Do I use concentrations or moles for the strong-acid-addition problem?

Moles (or mmol) are safer, since the added strong acid or base is a fixed amount regardless of the buffer's volume. If the total volume doesn't change, the concentration ratio and the mole ratio give the same Henderson-Hasselbalch result anyway, since the volume cancels out of the ratio — but tracking moles avoids mistakes if volume does change.

How much strong acid or base can a buffer absorb before it stops working?

Roughly until one form runs out. In practice, a buffer is considered to have good capacity as long as the resulting pH stays within about pKa ± 1 of its pKa — beyond that, small further additions cause much larger pH swings, the same way an unbuffered solution would behave.

The 3 MCAT Buffer Question Types, Explained

The 3 types

Buffer selection — "which of these buffers works best at pH X?"
Acid/base ratio — "what fraction of the buffer is in each form at pH X?"
Strong acid/base addition — "you add this much HCl or NaOH — what's the new pH?"

All three use the same equation — pH = pKa + log([A⁻]/[HA]) — but they ask for different things, and type 3 trips up far more students than the other two because it looks like a two-step problem when it's actually three.

Type 1: Buffer selection

You're given a target pH and a list of candidate buffers with their pKa values. The rule: pick the buffer whose pKa is closest to the target, and confirm it's within about pKa ± 1 — outside that range, a buffer has too little of one form left to resist a pH change.

Worked example: best buffer for pH 9.0

Buffer	pKa	\|pH − pKa\|	In range?
Acetate	4.76	4.24	No
Bicarbonate	6.35	2.65	No
Phosphate	7.21	1.79	No
HEPES	7.48	1.52	No
Tris	8.06	0.94	Yes
Ammonium	9.25	0.25	Yes

Two candidates fall within range, but ammonium's pKa (9.25) is closer to 9.0 than Tris's (8.06) — ammonium is the better choice.

Type 2: Acid/base ratio

Given a buffer's pKa and a target pH, find what fraction is conjugate base versus weak acid: ratio [A⁻]/[HA] = 10^(pH − pKa).

Worked example: ammonium buffer at pH 9.0

Ratio [A⁻]/[HA] = 10^(9.0 − 9.25) = 10^(−0.25) = 0.562 Fraction A⁻ = 0.562/1.562 = 0.360 (36.0%) Fraction HA = 1/1.562 = 0.640 (64.0%)

Ratio below 1 makes sense: the target pH (9.0) is below the pKa (9.25), so the weak acid form (NH₄⁺) dominates.

Type 3: strong acid/base addition — the one that trips people up

This is the type that looks straightforward but actually has three steps: figure out the starting moles of each form, apply the stoichiometry of the addition, then plug the new mole amounts into Henderson-Hasselbalch. Skipping straight to the equation with the original ratio is the single most common mistake on this question type.

The key fact: strong acid reacts completely with the conjugate base (A⁻ + H⁺ → HA), and strong base reacts completely with the weak acid (HA + OH⁻ → A⁻ + H₂O). Every mole added shifts one mole from one form to the other.

Worked example: adding strong acid

500 mL of 0.10 M acetate buffer (pKa 4.76), currently at pH 4.76 — meaning the two forms start equal. Add 10 mmol of HCl. Find the new pH.

Step 1 — starting moles: Total buffer = 0.10 M × 500 mL = 50 mmol At pH = pKa, the two forms are equal: A⁻ = 25 mmol, HA = 25 mmol Step 2 — apply the stoichiometry of adding 10 mmol HCl: A⁻ + H⁺ → HA, so 10 mmol of A⁻ converts to HA New A⁻ = 25 − 10 = 15 mmol New HA = 25 + 10 = 35 mmol Step 3 — Henderson-Hasselbalch with the new amounts: pH = pKa + log(A⁻/HA) = 4.76 + log(15/35) = 4.76 + log(0.4286) pH = 4.76 + (−0.368) = 4.39

Notice the pH dropped by only 0.37 units even though a meaningful amount of strong acid was added — that's the buffer doing its job. Without any buffer present, that much HCl in 500 mL would drop the pH far more sharply.

Worked example: adding strong base (the mirror case)

Same starting buffer (25 mmol A⁻, 25 mmol HA, pKa 4.76). This time add 8 mmol of NaOH instead.

HA + OH⁻ → A⁻ + H₂O, so 8 mmol of HA converts to A⁻ New A⁻ = 25 + 8 = 33 mmol New HA = 25 − 8 = 17 mmol pH = 4.76 + log(33/17) = 4.76 + log(1.941) = 4.76 + 0.288 = 5.05

Same mechanism, opposite direction: adding base raises the ratio (more conjugate base, less weak acid), which raises the pH.

Check any ratio instantly in the buffer calculator →

Common mistakes

Using concentrations from before the reaction with moles from after it. Do the stoichiometry in moles (or mmol) first, then convert to concentration or plug straight into the pH ratio — don't mix the two.

Forgetting both forms change. Added acid doesn't just create more HA — it also destroys an equal amount of A⁻. Track both.

Applying Henderson-Hasselbalch before doing the stoichiometry. The equation needs the post-reaction mole amounts. Using the original ratio skips the step that makes this a "buffer" question instead of a plain ratio lookup.

Practice problem

1. You have 400 mL of a 0.20 M phosphate buffer (pKa 7.21), starting with equal amounts of both forms. You add 15 mmol of NaOH. What is the new pH?

Show answer

Total buffer = 0.20 M × 400 mL = 80 mmol Starting equal: A⁻ = 40 mmol, HA = 40 mmol Adding 15 mmol NaOH converts HA → A⁻: New A⁻ = 40 + 15 = 55 mmol New HA = 40 − 15 = 25 mmol pH = 7.21 + log(55/25) = 7.21 + log(2.2) = 7.21 + 0.342 = 7.55

FAQ

Why does adding acid change moles of both forms, not just one?
Every mole of strong acid reacts completely with one mole of conjugate base, converting it to weak acid. Conjugate base goes down by exactly the moles added; weak acid goes up by the same amount.

Should I use concentrations or moles for the addition problem?
Moles are safer, since the added strong acid or base is a fixed amount regardless of volume. If total volume doesn't change, concentration and mole ratios give the same result anyway — but tracking moles avoids mistakes if volume changes too.

How much acid or base can a buffer absorb before it stops working well?
Roughly until the resulting pH drifts outside pKa ± 1 of the buffer's pKa. Past that point, further additions cause much larger pH swings, similar to an unbuffered solution.