Buffer solutions — Henderson-Hasselbalch equation: Worked examples

This page hunts down every kind of buffer question the parent topic can throw at you. We first draw a map of all the cases, then solve one example per region of the map so no scenario can surprise you in an exam.

Everything rests on one equation you already met:

If any of "pH", "log", "conjugate base", or " $K_{a}$ " feels shaky, revisit pH and pOH, Weak acids and bases, and Acid-base equilibria first.

The scenario matrix

Every buffer problem lands in exactly one of these cells. The last column names the example that covers it.

Cell	What makes it special	What the log term does	Example
A. Base-rich	$[A^{-}] > [HA]$	ratio $> 1$ , log $> 0$ → pH above $p K_{a}$	Ex 1
B. Acid-rich	$[A^{-}] < [HA]$	ratio $< 1$ , log $< 0$ → pH below $p K_{a}$	Ex 2
C. Exact centre	$[A^{-}] = [HA]$	ratio $= 1$ , log $= 0$ → pH $= p K_{a}$	Ex 3
D. Degenerate: pure acid	$[A^{-}] \to 0$	log $\to - \infty$ → equation breaks, not a buffer	Ex 4
E. Base-buffer (flip the roles)	given a weak base + its salt	must pick the acid form correctly	Ex 5
F. Add strong acid	H⁺ eats the base form	ratio shifts down a little	Ex 6
G. Add strong base	OH⁻ eats the acid form	ratio shifts up a little	Ex 7
H. Design for target pH	pH given, ratio unknown	invert with antilog $1 0^{x}$	Ex 8
I. Real-world / edge of range	ratio outside 0.1–10	buffer is weak, warn the reader	Ex 9

Figure — Buffer solutions — Henderson-Hasselbalch equation

The figure above is the "number line" for the whole page: the log term simply slides you left or right of $p K_{a}$ . Keep it in your head for every example.

Base-rich, acid-rich, and dead centre (cells A, B, C)

The degenerate case — when it is NOT a buffer (cell D)

Worked example Example 4 — Cell D (pure acid,

$[A^{-}] \to 0$ ) You dissolve $0.10 M$ acetic acid alone (no acetate). Can you use Henderson–Hasselbalch?

Forecast: what does $lo g$ of a number heading to zero do?

Here $[A^{-}] \to 0$ , so the ratio $\frac{[ A ^{-} ]}{[ HA ]} \to 0$ . Why this step? We must test the limit before blindly plugging in.
$lo g_{10} (0^{+}) \to - \infty$ , so the formula predicts $pH \to - \infty$ — physically impossible. Why? The derivation assumed $[A^{-}]$ came mostly from an added salt, not from the acid's own tiny dissociation. With no salt, that assumption collapses.
Correct route: treat it as a plain weak acid via the equilibrium $K_{a} = \frac{x ^{2}}{0.10 - x} \approx \frac{x ^{2}}{0.10}, x = [H^{+}] .$ With $K_{a} = 1 0^{- 4.76} = 1.74 \times 1 0^{- 5}$ : $x = 1.74 \times 1 0^{- 5} \times 0.10 = 1.32 \times 1 0^{- 3} .$
$pH = - lo g_{10} (1.32 \times 1 0^{- 3}) = 2.88$ .

Verify: A single weak acid should be clearly acidic and its pH must lie below $p K_{a}$ ; $2.88 < 4.76$ . ✓ The lesson: no conjugate base ⇒ no buffer ⇒ don't use the equation. ✓

Flipping the roles — a base buffer (cell E)

Recall Roles cheat-sheet

Which species is $HA$ ? ::: The one that can donate a proton (the charged/protonated form for amine buffers). Which species is $A^{-}$ ? ::: The one that can accept a proton (the neutral amine, or the salt's anion for acid buffers).

The buffer under attack (cells F and G)

When you add strong acid or base, first do the stoichiometry (a one-way reaction that runs to completion), then apply Henderson–Hasselbalch to the leftovers.

Designing a buffer & the edge of range (cells H, I)

Worked example Example 8 — Cell H (target pH, solve for the ratio)

You need a phosphate buffer at $pH = 7.00$ using $H_{2} P O_{4}^{-}$ / $HP O_{4}^{2 -}$ , where the relevant $p K_{a} = 7.21$ . What ratio $[HP O_{4}^{2 -}] / [H_{2} P O_{4}^{-}]$ do you need? (See Polyprotic acids for why phosphate has several $p K_{a}$ values.)

Forecast: pH is below $p K_{a}$ — should you need more acid or more base?

Set up with $HA = H_{2} P O_{4}^{-}$ , $A^{-} = HP O_{4}^{2 -}$ : $7.00 = 7.21 + lo g_{10} (r), r = \frac{[ A ^{-} ]}{[ HA ]} .$
Isolate the log: $lo g_{10} (r) = 7.00 - 7.21 = - 0.21$ . Why this step? Subtract the fixed $p K_{a}$ to expose the unknown ratio.
Undo the log with the antilog $1 0^{x}$ : $r = 1 0^{- 0.21} = 0.617.$ Why? $lo g$ and $1 0^{()}$ are inverse operations — $1 0^{x}$ answers "which ratio has this log?"
So $[HP O_{4}^{2 -}] : [H_{2} P O_{4}^{-}] \approx 0.62 : 1$ , i.e. more acid form.

Verify: pH below $p K_{a}$ ⇒ ratio $< 1$ ⇒ acid-rich; $0.62 < 1$ . ✓ Sanity: $7.21 + lo g_{10} (0.617) = 7.21 - 0.21 = 7.00$ . ✓

Worked example Example 9 — Cell I (real-world, edge of buffer range)

Blood plasma uses the $C O_{2} /HC O_{3}^{-}$ system with $p K_{a} = 6.1$ . Healthy blood is $pH = 7.4$ . What ratio $[HC O_{3}^{-}] / [C O_{2}]$ does the body maintain, and is this a "textbook-safe" buffer? (More at Biological bufers.)

Forecast: pH is a full $1.3$ units above $p K_{a}$ — is the ratio inside the comfy $0.1$ – $10$ window?

$7.4 = 6.1 + lo g_{10} (r)$ so $lo g_{10} (r) = 1.3$ .
$r = 1 0^{1.3} = 20.0$ . Why this step? Antilog recovers the actual concentration ratio.
Interpretation: $[HC O_{3}^{-}] : [C O_{2}] \approx 20 : 1$ — outside the ideal $0.1$ – $10$ range, so on paper it looks like a poor buffer. Why? At ratio $20$ , there is little dissolved $C O_{2}$ reserve to fight added base.
Why the body still wins: it is an open system — the lungs blow off or retain $C O_{2}$ , constantly resupplying the acid form, so the effective capacity is huge.

Verify: Reverse-check: $6.1 + lo g_{10} (20) = 6.1 + 1.301 = 7.40$ . ✓ And $20 > 10$ confirms it sits past the classical range — the "open lungs" caveat is essential. ✓

Recall Full-matrix self-test

Ratio $> 1$ ⇒ pH is where vs $p K_{a}$ ? ::: Above it (log positive). Ratio $= 1$ ⇒ pH? ::: Exactly $p K_{a}$ (log $= 0$ ). Add strong acid to a buffer — which form grows? ::: The acid form $HA$ (base is eaten). Pure weak acid, no salt — use Henderson–Hasselbalch? ::: No, $[A^{-}] \to 0$ breaks it; use $K_{a} = x^{2} / C$ . To hit a target pH you invert the log with…? ::: The antilog $1 0^{x}$ .