2.6.14 · D5Equilibrium
Question bank — Buffer solutions — Henderson-Hasselbalch equation
This is a companion to Buffer solutions — Henderson-Hasselbalch equation. Prerequisites worth having open: Weak acids and bases, Common ion effect, pH and pOH.
True or false — justify
Adding water to a buffer changes its pH significantly
False — dilution scales and by the same factor, so the ratio in is unchanged; pH stays essentially constant (until dilution is extreme and the approximation fails).
A buffer at pH = pKₐ has equal moles of acid and conjugate base
True — when the ratio is 1 and , so pH pKₐ exactly.
A solution of pure weak acid (no conjugate base added) is a buffer
False — a buffer needs appreciable, comparable amounts of both partners; pure HA has almost no , so it cannot catch added OH⁻ effectively.
Mixing a strong acid with excess weak base can produce a buffer
True — the strong acid partially converts the weak base into its conjugate acid, leaving both forms present; this is exactly how buffers are made by partial neutralization.
Two buffers with the same pKₐ but different total concentrations have the same pH
True (if their ratios match) — pH depends on the ratio , not the absolute concentrations; but the more concentrated one has higher buffer capacity.
Doubling both and raises the buffer capacity but leaves pH unchanged
True — pH is set by the ratio (unchanged), while capacity depends on how many moles of each reserve exist to absorb added acid/base.
A buffer works equally well against added acid and added base only when pH = pKₐ
True — equal reserves of (fights added OH⁻) and (fights added H⁺) exist only at the 1:1 point.
The Henderson-Hasselbalch equation gives the exact pH of any buffer
False — it is an approximation assuming equilibrium concentrations equal the analytical (mixed) amounts and that H⁺ from water is negligible; both fail for very dilute or very weak/strong cases.
Spot the error
"For an ammonia buffer I use pH = pKₐ + log([NH₄⁺]/[NH₃])."
Wrong: is the acid and is the base, so it must be — base on top, always.
"pKₐ = 4.76, so Kₐ = 4.76 × 10⁻¹."
Wrong: pKₐ , so — you must take the antilog, not read the digits off directly.
"Since I added acid, the pH must rise."
Backwards: adding H⁺ lowers pH; the buffer only makes the drop small by converting some into .
"[A⁻]/[HA] = 100 is fine, the equation still applies, so it's still a good buffer."
The equation still computes a pH, but a 100:1 ratio is outside the pKₐ ± 1 range — buffer capacity is poor because the acid reserve is nearly exhausted.
"pOH = pKₐ + log([base]/[acid]) for a basic buffer."
Error: Henderson-Hasselbalch is written in terms of pKₐ and pH; if you use the base form you need , a different constant and a flipped ratio.
"I added 0.02 mol strong base, so both forms just shift by 0.02 mol in the same direction."
Error: added OH⁻ converts HA into A⁻, so decreases by 0.02 and increases by 0.02 — opposite directions, conserving total.
"A 1:1 buffer sits at pH 7."
Only if the pKₐ happens to be 7; a 1:1 acetate buffer sits at pH 4.76, and a 1:1 ammonium buffer at pH 9.25.
Why questions
Why does the ratio set the pH rather than the individual concentrations?
Because contains the concentrations only as a ratio, so any common factor (like dilution) cancels out.
Why is buffer capacity greatest exactly at pH = pKₐ?
There the two reserves are equal and largest-balanced, so a given amount of added acid or base shifts the ratio (and pH) the least in either direction.
Why must a buffer contain a weak acid, not a strong one?
A strong acid dissociates completely, leaving essentially no undissociated to donate H⁺ against added base — there is no reserve pair, hence no buffering (Common ion effect needs an equilibrium that can shift).
Why does adding a tiny bit of strong acid barely move the pH, while adding it to pure water swings it hugely?
In the buffer the added H⁺ is consumed by (only the ratio nudges slightly); in water there is nothing to absorb it, so jumps directly.
Why do we take a logarithm at all in the derivation?
pH is defined as ; taking of both sides turns the multiplicative into an additive — easy to read as "start at pKₐ, then shift."
Why does flipping the fraction change the sign in front of the log?
Because , so writing base-over-acid instead of acid-over-acid converts the into a , giving the standard form.
Why is the effective range pKₐ ± 1 and not pKₐ ± 3?
At pKₐ ± 1 the ratio is 10:1 or 1:10; push to ±2 (100:1) and one component is so scarce it is nearly used up, so it can no longer neutralize incoming acid or base (Le Chatelier's principle can no longer shift much).
Edge cases
What happens to the Henderson-Hasselbalch prediction when the buffer is diluted a millionfold?
It breaks down — as concentrations approach from water autoionization, the "equilibrium ≈ analytical" assumption fails and the pH drifts toward 7.
If you add enough strong acid to convert all the conjugate base to acid, is it still a buffer?
No — once the ratio and log blow up; you now have just a weak acid solution past its buffering point (the equivalence-region behaviour of a titration curve).
Can a solution of only NaCl and water act as a buffer?
No — and are the ions of a strong base and strong acid, neither has a usable conjugate equilibrium, so there is nothing to catch added H⁺ or OH⁻.
What is the pH ratio requirement at the extreme edge of the useful range?
At pH = pKₐ + 1 the ratio ; at pH = pKₐ − 1 it is — these are the practical cutoffs for "still a functional buffer."
For a diprotic acid, why can a buffer form near either pKₐ value?
Each dissociation step has its own conjugate pair (e.g. and ), so buffering is possible around each pKₐ separately (Polyprotic acids).
Does blood being buffered near pH 7.4 by a system whose pKₐ is 6.1 violate the pKₐ ± 1 rule?
It stretches it, but works because the body continuously removes CO₂ to keep the ratio high — an open system, unlike a sealed beaker (Biological bufers).
Recall Quick self-check
Which form goes on top of the log, acid or base? ::: The conjugate base always sits on top. pH relative to pKₐ when base exceeds acid? ::: pH is above pKₐ (positive log term). Where is buffer capacity maximal? ::: At pH pKₐ, where the two reserves are equal.