Buffer solutions — Henderson-Hasselbalch equation: Exercises

Where the one tool comes from (don't skip this)

Before we drill, let's earn the equation instead of trusting it. A weak acid $HA$ sits in a tug-of-war with water: $HA ⇌ H^{+} + A^{-}$ The acid dissociation constant $K_{a}$ measures how far right that tug-of-war settles: $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$ We care about pH, and pH is built from $[H^{+}]$ , so solve for $[H^{+}]$ : $[H^{+}] = K_{a} \cdot \frac{[ HA ]}{[ A ^{-} ]}$ This one line is the whole idea in disguise: the free acidity $[H^{+}]$ is $K_{a}$ scaled by the acid-to-base ratio. More acid than base ⇒ more free $H^{+}$ ⇒ lower pH.

Now, why bother with logarithms at all? Because $[H^{+}]$ spans absurd ranges ( $1 0^{- 1}$ to $1 0^{- 14}$ ), and multiplying is clumsy. The $lo g$ turns "multiply" into "add", so the messy product $K_{a} \cdot (ratio)$ becomes a tidy sum. Take $- lo g_{10}$ of both sides: $- lo g_{10} [H^{+}] = - lo g_{10} (K_{a}) - lo g_{10} (\frac{[ HA ]}{[ A ^{-} ]})$ The left side is pH by definition; $- lo g_{10} (K_{a})$ is $p K_{a}$ ; and flipping the fraction flips the sign of its log ( $lo g (1/ x) = - lo g x$ ). That gives:

Picture the log/antilog as a ratio slider

The figure below is the mental model for every problem on this page. Think of a horizontal slider whose position is the ratio $[A^{-}] / [HA]$ on a $lo g_{10}$ scale, and whose readout is pH. Sliding by one decade (ratio ×10) moves pH by exactly +1; sliding the other way (ratio ÷10) moves pH by −1. The centre notch (ratio = 1) is where pH = $p K_{a}$ .

Figure — Buffer solutions — Henderson-Hasselbalch equation

$lo g_{10}$ is the "where am I on the slider?" question: given a ratio, how many decades from the centre?
The antilog $1 0^{x}$ is the reverse: given how far you slid, what ratio is that? If $lo g_{10} (r) = - 0.25$ then $r = 1 0^{- 0.25} = 0.562$ — a bit left of centre.

Definition The two edges where the equation dies

Look at the far ends of the slider figure (shaded red):

$[A^{-}] / [HA] \to 0$ (base almost gone): $lo g_{10}$ dives to $- \infty$ , so pH crashes. Physically you've run out of base to catch $H^{+}$ — it's now essentially a solution of pure weak acid, and the H-H assumption " $[HA] \approx$ initial" fails because dissociation of $HA$ itself becomes the dominant source of $H^{+}$ .
$[A^{-}] / [HA] \to \infty$ (acid almost gone): $lo g_{10}$ soars to $+ \infty$ , pH shoots up. You've run out of acid to donate $H^{+}$ ; the solution behaves like the salt of a weak acid, and water's own $OH^{-}$ matters.

The lesson: near the edges there is almost no reserve of one partner, so the buffer no longer resists change — a tiny addition swings pH hugely. The safe working band is $0.1 \leq ratio \leq 10$ , i.e. $p K_{a} \pm 1$ .

Two undo-buttons you will use constantly:

$lo g_{10}$ asks "10 to what power gives this number?" So $lo g_{10} (10) = 1$ , $lo g_{10} (1) = 0$ , $lo g_{10} (0.1) = - 1$ .
The antilog $1 0^{x}$ undoes the log: if $lo g_{10} (r) = - 0.25$ , then $r = 1 0^{- 0.25}$ .

Level 1 — Recognition

L1.1 — Name the parts

A solution is made from 0.20 M lactic acid ( $H C_{3} H_{5} O_{3}$ ) and 0.20 M sodium lactate ( $Na C_{3} H_{5} O_{3}$ ). Identify which species is $HA$ , which is $A^{-}$ , and predict (no calculator) how pH compares to $p K_{a}$ .

Recall Solution

$HA$ = lactic acid $H C_{3} H_{5} O_{3}$ — it still holds the donatable H.
$A^{-}$ = lactate ion $C_{3} H_{5} O_{3}^{-}$ — sodium lactate is a salt that fully dissociates, releasing the conjugate base.
Ratio $[A^{-}] / [HA] = 0.20/0.20 = 1$ , so $lo g_{10} (1) = 0$ .
Therefore $pH = p K_{a}$ exactly. On the slider figure, we sit dead-centre.

L1.2 — Is it even a buffer?

For each pair, state YES (buffer) or NO, with a one-line reason: (a) $HCl$ + $NaCl$ · (b) $C H_{3} COOH$ + $C H_{3} COONa$ · (c) $N H_{3}$ + $N H_{4} Cl$ · (d) $NaOH$ + $NaCl$ .

Recall Solution

(a) NO — $HCl$ is a strong acid; there is no weak-acid/conjugate-base equilibrium to shift.
(b) YES — weak acid (acetic) + its conjugate base (acetate). Classic buffer.
(c) YES — weak base $N H_{3}$ + its conjugate acid $N H_{4}^{+}$ . Buffer around $p K_{a} (N H_{4}^{+})$ .
(d) NO — $NaOH$ is a strong base and $NaCl$ is a spectator salt; no conjugate pair.

Rule of thumb: a buffer needs a weak acid or base plus a comparable amount of its conjugate partner.

Level 2 — Application

L2.1 — Straight pH calculation

A buffer contains 0.30 M benzoic acid ( $p K_{a} = 4.20$ ) and 0.60 M sodium benzoate. Find the pH.

Recall Solution

$HA$ = benzoic acid, $A^{-}$ = benzoate. $pH = 4.20 + lo g_{10} (\frac{0.60}{0.30}) = 4.20 + lo g_{10} (2)$ $= 4.20 + 0.301 = 4.50$ Ratio 2 is one small step right of centre on the slider ⇒ pH sits above $p K_{a}$ . ✓

L2.2 — Solve for the ratio

You want a buffer at pH 5.00 using acetic acid ( $p K_{a} = 4.76$ ). What ratio $[A^{-}] / [HA]$ do you need?

Recall Solution

$5.00 = 4.76 + lo g_{10} (\frac{[ A ^{-} ]}{[ HA ]})$ Isolate the log: $lo g_{10} (\frac{[ A ^{-} ]}{[ HA ]}) = 5.00 - 4.76 = 0.24$ Antilog (undo the log — this is the "slider readout → ratio" move): $\frac{[ A ^{-} ]}{[ HA ]} = 1 0^{0.24} = 1.74$ pH is above $p K_{a}$ , so we need more base than acid — a ratio bigger than 1 confirms this. ✓

L2.3 — Solve for $p K_{a}$

A buffer measured at pH 7.40 contains $[HP O_{4}^{2 -}] = 0.080$ M and $[H_{2} P O_{4}^{-}] = 0.050$ M. What is the $p K_{a}$ of the relevant dissociation? (See Polyprotic acids.)

Recall Solution

Here $HA = H_{2} P O_{4}^{-}$ (donor) and $A^{-} = HP O_{4}^{2 -}$ (its conjugate base). $7.40 = p K_{a} + lo g_{10} (\frac{0.080}{0.050}) = p K_{a} + lo g_{10} (1.6)$ $lo g_{10} (1.6) = 0.204 \Rightarrow p K_{a} = 7.40 - 0.204 = 7.20$ This matches the known $p K_{a 2}$ of phosphoric acid — the reason phosphate buffers work near physiological pH.

Level 3 — Analysis

L3.1 — Add a strong base

A buffer holds 0.40 mol formic acid and 0.40 mol formate ( $p K_{a} = 3.75$ ) in 1.0 L. Add 0.10 mol $NaOH$ . Find the new pH and the pH change.

Recall Solution

Chemistry first: $O H^{-}$ is a strong base; it converts acid $\to$ base: $O H^{-} + HCOOH \to HCO O^{-} + H_{2} O$ New amounts (0.10 mol reacts, 1 : 1):

$HCOOH$ : $0.40 - 0.10 = 0.30$ mol
$HCO O^{-}$ : $0.40 + 0.10 = 0.50$ mol

Initial pH $= 3.75 + lo g_{10} (1) = 3.75$ . $pH = 3.75 + lo g_{10} (\frac{0.50}{0.30}) = 3.75 + 0.222 = 3.97$ Change $= + 0.22$ pH units. In the stacked-bar figure below, the orange (acid) block shrinks and the teal (base) block grows: the ratio nudges right but the pH barely moves because both reserves are still large. (Volume cancels in the ratio, so moles are fine.)

L3.2 — Dilution twist

Take the original L3.1 buffer (0.40 mol / 0.40 mol in 1.0 L) and dilute it to 4.0 L with pure water. What is the new pH?

Recall Solution

Dilution scales both concentrations by the same factor ( $1/4$ ): $[A^{-}] = 0.10 M, [HA] = 0.10 M$ The ratio is unchanged ( $0.10/0.10 = 1$ ): $pH = 3.75 + lo g_{10} (1) = 3.75$ Because H-H depends on a ratio, not on absolute concentration, dilution leaves pH unchanged to first approximation. What does fall is the buffer capacity — fewer moles to absorb future shocks.

Why "to first approximation"? Two edge effects grow as you dilute further:

Water autoionization. H-H silently ignores the $H^{+}$ (and $OH^{-}$ ) that water itself supplies ( $K_{w} = 1 0^{- 14}$ ). When buffer concentrations drop toward $\sim 1 0^{- 6}$ M, water's own $1 0^{- 7}$ M of $H^{+}$ is no longer negligible, and the true pH drifts toward 7.
Ionic strength. H-H uses concentrations as stand-ins for activities. Diluting changes the ionic atmosphere around each ion, so the effective (active) concentrations shift slightly; at moderate strength this can move a real pH by up to $\sim 0.1$ unit versus the naive prediction.

For ordinary lab buffers (0.01–1 M) these are small; for very dilute or very precise work they are not.

Level 4 — Synthesis

L4.1 — Choose the acid, then the recipe

Design a buffer at pH 4.90 with total concentration $[HA] + [A^{-}] = 0.50$ M. Pick from: formic ( $p K_{a} = 3.75$ ), acetic ( $p K_{a} = 4.76$ ), or ammonium ( $p K_{a} = 9.25$ ). Give the acid and the two concentrations. (Recall the effective range $p K_{a} \pm 1$ ; see Common ion effect for why the salt matters.)

Recall Solution

Choose the acid: target pH must lie within $p K_{a} \pm 1$ .

Formic: range $2.75$ – $4.75$ → 4.90 is (just) outside. ✗
Acetic: range $3.76$ – $5.76$ → 4.90 inside, and close to $p K_{a}$ . ✓ Best.
Ammonium: range $8.25$ – $10.25$ → far off. ✗

Find the ratio with acetic ( $p K_{a} = 4.76$ ): $lo g_{10} (r) = 4.90 - 4.76 = 0.14 \Rightarrow r = 1 0^{0.14} = 1.38$ So $[A^{-}] = 1.38 [HA]$ . Split the total 0.50 M: $[HA] + 1.38 [HA] = 0.50 \Rightarrow [HA] = \frac{0.50}{2.38} = 0.210 M$ $[A^{-}] = 0.50 - 0.210 = 0.290 M$ Recipe: $\approx$ 0.21 M acetic acid + 0.29 M sodium acetate. Quick check: $4.76 + lo g_{10} (0.290/0.210) = 4.76 + 0.14 = 4.90$ . ✓

L4.2 — Buffer from base + strong acid

You only have 0.50 mol $N H_{3}$ in 1.0 L and solid $HCl$ . How many moles of $HCl$ must you add to make a buffer at pH 9.05? ( $p K_{a} (N H_{4}^{+}) = 9.25$ .)

Recall Solution

Adding $HCl$ converts base to acid: $N H_{3} + H^{+} \to N H_{4}^{+}$ . Let $x$ = mol $HCl$ added. Then $N H_{4}^{+} = x$ , $N H_{3} = 0.50 - x$ . Target: $9.05 = 9.25 + lo g_{10} (\frac{0.50 - x}{x})$ $lo g_{10} (\frac{0.50 - x}{x}) = - 0.20 \Rightarrow \frac{0.50 - x}{x} = 1 0^{- 0.20} = 0.631$ $0.50 - x = 0.631 x \Rightarrow 0.50 = 1.631 x \Rightarrow x = 0.307 mol$ Check: $N H_{3} = 0.193$ , $N H_{4}^{+} = 0.307$ ; $9.25 + lo g_{10} (0.193/0.307) = 9.25 - 0.20 = 9.05$ . ✓ pH below $p K_{a}$ ⇒ acid form must dominate, and indeed $0.307 > 0.193$ . ✓

Level 5 — Mastery

L5.1 — Sequential shocks with a capacity check

Start with 0.60 mol acetic acid + 0.40 mol acetate in 1.0 L ( $p K_{a} = 4.76$ ). (a) Find the starting pH. (b) Add 0.15 mol $NaOH$ ; find the new pH. (c) Then add 0.35 mol $HCl$ to the result of (b); find the pH. (d) Comment on whether the buffer is still "in range" after (c).

Recall Solution

(a) Start: $pH = 4.76 + lo g_{10} (\frac{0.40}{0.60}) = 4.76 + (- 0.176) = 4.58$

(b) Add 0.15 mol $O H^{-}$ (acid → base):

Acid: $0.60 - 0.15 = 0.45$ mol
Base: $0.40 + 0.15 = 0.55$ mol $pH = 4.76 + lo g_{10} (\frac{0.55}{0.45}) = 4.76 + 0.087 = 4.85$

(c) Add 0.35 mol $H^{+}$ (base → acid), starting from acid = 0.45, base = 0.55:

Base: $0.55 - 0.35 = 0.20$ mol
Acid: $0.45 + 0.35 = 0.80$ mol $pH = 4.76 + lo g_{10} (\frac{0.20}{0.80}) = 4.76 + (- 0.602) = 4.16$

(d) In range? Effective range is $p K_{a} \pm 1 = 3.76$ to $5.76$ . Final pH $4.16$ is inside, but the ratio is now $0.20/0.80 = 0.25$ — on the slider we've drifted well left of centre, approaching the lower edge ( $0.1$ ). The buffer still works but its remaining capacity toward more acid is getting thin (only 0.20 mol base left to absorb H).

The figure tracks the pH across all three states against the shaded safe band.

L5.2 — Compare against unbuffered water

For the buffer after step (c) above (pH 4.16), and for 1.0 L of pure water (pH 7.00), add 0.05 mol $HCl$ to each. Find both new pH values and compare the shocks.

Recall Solution

Buffer (acid = 0.80, base = 0.20; convert 0.05 mol base → acid):

Base: $0.20 - 0.05 = 0.15$ ; Acid: $0.80 + 0.05 = 0.85$ $pH = 4.76 + lo g_{10} (\frac{0.15}{0.85}) = 4.76 + (- 0.753) = 4.01$ Change $= 4.16 \to 4.01 = - 0.15$ pH units.

Pure water: $0.05$ mol $HCl$ in $1.0$ L $\Rightarrow [H^{+}] = 0.05$ M: $pH = - lo g_{10} (0.05) = 1.30$ Change $= 7.00 \to 1.30 = - 5.70$ pH units.

Comparison: the same acid dose moves water by 5.7 units but the (already partly spent) buffer by only 0.15 units — a ~38× smaller swing. This is Le Chatelier's principle at work: the added H is consumed by shifting $A^{-} \to HA$ , so free $[H^{+}]$ barely rises.

Recall Self-test cloze

A buffer resists pH change because it holds a weak acid and its conjugate base in comparable amounts. When the ratio $[A^{-}] / [HA] = 1$ , then $pH$ equals == $p K_{a}$ . The effective buffer range is $p K_{a} \pm 1$ . Dilution changes buffer capacity but not (to first order) the pH. H-H is really the equation $K_{a}$ == rewritten with logarithms.

Undo the log — say it aloud:

lo g_{10} (r) = - 0.25

, then

r

equals?

1 0^{- 0.25} = 0.562

Why does pH stay fixed on dilution?

H-H depends on a ratio, and dilution scales both terms equally.

Where is buffer capacity maximal?

pH = p K_{a}

, where

[A^{-}] = [HA]

What breaks H-H at the ratio extremes?

near ratio → 0 or → ∞ one partner runs out, plus water autoionization dominates.

Related: Titration curves · Biological bufers · pH and pOH