Year 12 Chemistry Module 6, Acid/Base Reactions ⏱ ~45 min Lesson 13 of 19 IQ2

Buffer Solutions, Henderson-Hasselbalch and Buffer Capacity

In 1908, Lawrence Henderson at Harvard discovered the equation that bears his name by analysing blood pH data, he found that at normal blood pH of 7.40, the ratio [HCO₃⁻]/[H₂CO₃] is almost exactly 20:1. Karl Hasselbalch rearranged Henderson's formula into the logarithmic form in 1916. Together, they explained how a 0.3 pH unit deviation from 7.40 causes clinical crisis, and why maintaining that ratio is something your body does continuously, every second.

🩸

Today's hook, In 1908, Lawrence Henderson at Harvard analysed blood and found the [HCO₃⁻]/[H₂CO₃] ratio is 20:1 at pH 7.40. Just 0.3 pH units outside this range causes clinical crisis. How does a solution resist pH change when acid or base is added, and how do you calculate the exact pH of a buffer?

0/5QUESTS

You’re here

Get oriented

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First, Two Beakers, One Mystery

A biochemistry student adds 0.001 mol of HCl to two beakers, each containing 100 mL of solution. Beaker A contains pure water (pH 7.00). Beaker B contains a mixture of 0.100 mol/L acetic acid and 0.100 mol/L sodium acetate (pH 4.74). After adding the HCl: Beaker A drops from pH 7.00 to pH 2.00, a change of 5 pH units. Beaker B drops from pH 4.74 to pH 4.66, a change of only 0.08 pH units. Same amount of acid, same volume, dramatically different pH responses.

Before reading on: What is present in Beaker B that is not present in Beaker A, and how does it resist the pH change? What would happen to Beaker B's pH if you added NaOH instead of HCl? What do you think limits a buffer's ability to resist pH changes?

By the end of this lesson you will be able to:

📚 Know

A buffer resists significant pH change when small amounts of acid or base are added
Buffer components: weak acid (HA) and its conjugate base (A⁻)
The Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA])

🔗 Understand

The molecular mechanism: HA reacts with added OH⁻, A⁻ reacts with added H⁺
Why buffer capacity is highest when [A⁻] = [HA] and pH ≈ pKa
Why buffers do not prevent all pH changes, only small ones

✅ Can Do

Calculate buffer pH using the Henderson-Hasselbalch equation
Prepare buffers by partial neutralisation or by mixing known amounts
Identify natural buffer systems and explain their biological importance

Buffer Formulas, Henderson-Hasselbalch & Preparation

Deep dive, Extension. Buffer mechanism, preparation and importance in natural systems are core. The Henderson–Hasselbalch equation, quantitative buffer capacity, and detailed blood and ocean buffer calculations are Extension.

pH = pKa + log([A⁻]/[HA])

pH, pH of the buffer solution pKa = −log₁₀(Ka) of the weak acid component [A⁻], equilibrium concentration of conjugate base (mol/L) [HA], equilibrium concentration of weak acid (mol/L)

When [A⁻] = [HA]: log(1) = 0 → pH = pKa (maximum buffer capacity). Note: mole ratio n(A⁻)/n(HA) = [A⁻]/[HA] because both occupy the same volume, use moles directly after neutralisation calculations.

Buffer capacity, effective range: pH = pKa ± 1

At ratio [A⁻]/[HA] = 10:1 → pH = pKa + 1 (upper limit) At ratio [A⁻]/[HA] = 1:10 → pH = pKa − 1 (lower limit)

Choose a weak acid with pKa within ±1 unit of the target pH. Buffer capacity is maximised when [A⁻] = [HA].

Partial neutralisation (Method 2 buffer preparation):

HA + OH⁻ → A⁻ + H₂O (goes to completion, strong base fully consumed) n(HA)remaining = n(HA)initial − n(OH⁻)added n(A⁻)formed = n(OH⁻)added

Apply Henderson-Hasselbalch to n(A⁻)/n(HA) ratio after neutralisation. Verify: n(OH⁻) < n(HA) or the equivalence point is reached and no buffer exists.

When HH is valid

Henderson-Hasselbalch is a buffer-region approximation, not a universal pH formula. It is only reliable when (1) both the weak acid and its conjugate base are actually present, i.e. between the start and the equivalence point, ideally with the ratio [A⁻]/[HA] within about 0.1–10 (near half-equivalence it is most accurate, where pH = pKa); (2) the concentrations are not extremely dilute, so the "x is negligible" assumption behind it holds; and (3) you use Ka and any tabulated pKa at the same temperature. Do not apply it to a strong acid or base, to a solution of the weak acid alone (no conjugate base yet), or at/after the equivalence point (no buffer remains), use the appropriate ICE or excess-strong-species method there instead.

Blood buffer (Henderson-Hasselbalch form):

pH = pKa(H₂CO₃) + log([HCO₃⁻]/[H₂CO₃]) where pKa = 6.10 Normal ratio: [HCO₃⁻]/[H₂CO₃] ≈ 20:1 → pH = 6.10 + log(20) = 6.10 + 1.30 = 7.40

Buffer mechanism (add H⁺): H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O (CO₂ expelled by lungs). Buffer mechanism (add OH⁻): OH⁻ + H₂CO₃ → HCO₃⁻ + H₂O.

Key Terms, scan these before reading

Buffer solution

A solution that resists changes in pH when small amounts of acid or base are added.

Buffer composition

Contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts.

Henderson-Hasselbalch equation

pH = pKa + log([A⁻]/[HA]); relates buffer pH to the ratio of conjugate base to acid.

Buffer action (acid added)

H⁺ ions are consumed by the conjugate base (A⁻ + H⁺ → HA), preventing large pH drop.

Buffer action (base added)

OH⁻ ions are consumed by the weak acid (HA + OH⁻ → A⁻ + H₂O), preventing large pH rise.

Buffer capacity

The amount of acid or base a buffer can absorb before significant pH change; maximised when [HA] = [A⁻].

📚 Core Content

Cross-lesson links: The Henderson-Hasselbalch equation from L09 Card 5 is fully developed here. The H₂CO₃/HCO₃⁻ blood buffer connects Astrup's altitude data (L08) to Ka values from L12. Buffer preparation calculations use titration stoichiometry from L14. Buffer capacity is tested by titration curves in L15–L16.

What a buffer is

Card 1, What a Buffer Is and What It Is Not

A buffer is one of the most misunderstood concepts in Module 6, students frequently confuse it with a neutralising agent, a diluting agent, or a substance that simply has a particular pH, and none of these misconceptions survives a precise examination of what a buffer actually does.

A buffer is a solution that resists significant changes in pH when small amounts of strong acid or strong base are added, or when the solution is diluted with a small amount of water. The key word is "resists", a buffer does not prevent pH changes entirely, but it dramatically limits them compared to an unbuffered solution. A buffer consists of a weak acid (HA) and its conjugate base (A⁻) present simultaneously in comparable concentrations.

The weak acid component provides resistance to base addition, when OH⁻ is added, HA reacts with it: HA + OH⁻ → A⁻ + H₂O. The conjugate base component provides resistance to acid addition, when H⁺ is added, A⁻ reacts with it: A⁻ + H⁺ → HA. Both reactions consume the added species before it can significantly change [H⁺].

Three things a buffer is NOT: (1) a neutral solution, buffers can be acidic or basic depending on their pKa (the acetic acid/acetate buffer has pH 4.74 when [HA] = [A⁻], fully functional, but acidic); (2) a neutralising agent, it absorbs added acid or base into the equilibrium, it does not neutralise them to completion; (3) a substance with unlimited capacity, every buffer has finite capacity determined by the moles of HA and A⁻ present.

Property	Buffer solution	Pure water	Strong acid solution
pH stability	Resists change on small acid/base addition	pH changes dramatically	pH changes dramatically
Components required	Weak acid + conjugate base (comparable concentrations)	H₂O only	One species only
pH determined by	pKa + ratio [A⁻]/[HA] (Henderson-Hasselbalch)	Kw only	Concentration of acid/base
Response to added H⁺	A⁻ + H⁺ → HA (consumes H⁺, minor pH change)	pH drops dramatically	pH drops dramatically
Response to added OH⁻	HA + OH⁻ → A⁻ + H₂O (consumes OH⁻, minor pH change)	pH rises dramatically	pH rises dramatically

Must Do: In any HSC question asking you to explain what a buffer is, your answer must contain four elements: (1) it contains a weak acid AND its conjugate base simultaneously; (2) it resists pH change on addition of small amounts of strong acid or base; (3) the mechanism involves consumption of the added species by either the acid or base component; (4) it has finite capacity. Definitions covering only one or two points earn partial marks.

Common Error: "A buffer has a pH of 7, it keeps solutions neutral." Buffers do not necessarily have a neutral pH. A buffer's pH is determined by the pKa of the weak acid and the ratio [A⁻]/[HA]. The blood buffer maintains pH 7.35–7.45 (close to neutral), but the acetate buffer maintains pH ~4.74, acidic, yet fully functional. Buffers maintain a relatively constant pH at whatever pH they are designed for, not necessarily pH 7.

Buffer = weak acid (HA) + conjugate base (A⁻) in comparable concentrations; resists pH change on small acid/base addition. Add H⁺: A⁻ + H⁺ → HA. Add OH⁻: HA + OH⁻ → A⁻ + H₂O. Buffer is NOT: a neutral solution (pH set by pKa + ratio), a neutralising agent, or unlimited in capacity.

Pause, copy the highlighted definition into your book before moving on.

Which of the following is TRUE about a buffer solution?

How buffers resist pH change

Beyond the syllabus. The buffer mechanism, preparing a buffer and its importance in natural systems are core. Henderson–Hasselbalch calculations, quantitative buffer capacity and the detailed blood/ocean figures are extension — an exam answer needs the equilibrium-shift explanation, not the formula.

Card 2, The Buffer Mechanism at the Molecular Level

We just saw the definition of a buffer, weak acid + conjugate base resisting pH change. That raises a question: Why exactly is the pH change so small, what is the molecular mechanism that prevents H⁺ from accumulating? This card answers it → the ratio [A⁻]/[HA] changes only slightly when both components are present in large amounts; the logarithm in Henderson-Hasselbalch compresses this into a tiny pH shift.

The two equations that describe buffer action are the most important equations in this lesson, understanding each one at the molecular level, rather than just memorising it, is what makes buffer theory accessible under exam pressure.

Consider a buffer made from acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵) and sodium acetate (CH₃COONa). In solution, CH₃COOH (weak acid component) and CH₃COO⁻ (conjugate base component) coexist in equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻.

When a small amount of strong acid (H⁺) is added: H⁺ + CH₃COO⁻ → CH₃COOH. The added H⁺ is consumed by reacting with the abundant CH₃COO⁻, converting it to CH₃COOH. [CH₃COO⁻] decreases slightly and [CH₃COOH] increases slightly, because both were present in large amounts, neither changes significantly in relative terms. The log([A⁻]/[HA]) term in Henderson-Hasselbalch changes only slightly, pH changes by a small amount.

When a small amount of strong base (OH⁻) is added: OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O. The added OH⁻ is consumed by reacting with the abundant CH₃COOH. [CH₃COOH] decreases slightly and [CH₃COO⁻] increases slightly, again, both were present in large amounts, so the ratio changes only modestly, pH changes by a small amount.

Mathematical reason for buffering: pH = pKa + log([A⁻]/[HA]). Adding a small amount of H⁺ or OH⁻ changes the numerator and denominator of the ratio by equal molar amounts, but because both are large, the relative change in the ratio is small. The logarithm compresses this further into a tiny pH change.

Add H⁺ (strong acid)

Species that reacts: CH₃COO⁻ consumes H⁺

Equation: H⁺ + CH₃COO⁻ → CH₃COOH

Effect on [A⁻]/[HA]: [A⁻] ↓; [HA] ↑; ratio decreases

Effect on pH: Decreases slightly

Add OH⁻ (strong base)

Species that reacts: CH₃COOH consumes OH⁻

Equation: OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O

Effect on [A⁻]/[HA]: [HA] ↓; [A⁻] ↑; ratio increases

Effect on pH: Increases slightly

Add water (dilution)

Species that reacts: Both HA and A⁻ diluted equally

Equation: Concentrations decrease proportionally

Effect on [A⁻]/[HA]: Ratio [A⁻]/[HA] unchanged

Effect on pH: Essentially unchanged

Must Do: In a buffer mechanism question, the HSC marking guideline requires three things: (1) identify which component reacts with the added species, A⁻ reacts with added H⁺; HA reacts with added OH⁻; (2) write the equation for the consumption reaction; (3) explain why the pH change is small, the ratio [A⁻]/[HA] changes only slightly because both are present in large amounts. All three elements are required for full marks.

Common Error: "The buffer neutralises the added acid." This implies the acid is completely removed, which is incorrect. The buffer partially consumes the added H⁺ (converting it to HA), but the H⁺ is not neutralised to zero, it shifts the equilibrium slightly, producing a small but non-zero pH change. "Resists" is the correct word, not "prevents" or "neutralises."

Buffer mechanism (add H⁺): A⁻ + H⁺ → HA. Buffer mechanism (add OH⁻): HA + OH⁻ → A⁻ + H₂O. Buffer resists because: both HA and A⁻ are present in large amounts, so the ratio [A⁻]/[HA] changes only slightly; the log term in HH changes by a small amount. Dilution: [A⁻]/[HA] ratio is unchanged → pH essentially unchanged. "Resists", not "prevents" or "neutralises."

Add the highlighted point to your notes before the check below.

In an acetate buffer (CH₃COOH/CH₃COO⁻), which species reacts when a small amount of HCl is added?

Drive It: Resists, Until It Runs Out

+XP for exploring

We just saw that A⁻ mops up added H⁺ and the ratio barely moves. That raises a question: how long can it keep doing that, and what is actually running out? This card answers it → dose the buffer and a beaker of plain water together, and watch the conjugate-base reservoir drain.

Interactive · Buffer capacity lab

Try this: add 0.010 mol strong acid doses one at a time and count how many the buffer takes before its pH has fallen more than 1 unit — then read what the n(A⁻) bar says at that point. Compare the water beaker after a single dose. Then switch to the ammonium system and watch a buffer work happily at pH 9.25.

Predict then reveal+8 XP

1 · Predict

2 · Reveal

3 · Compare

Before you drive it: after 0.010 mol of HCl goes into the 0.100/0.100 mol acetate buffer, will its pH be unchanged, slightly lower, or much lower?

Confidence 50%

Use the simulator. Starting from 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻ in 1.00 L, add 0.010 mol strong-acid doses one at a time. How many does the buffer absorb before its pH has fallen more than 1 unit, and what does the n(A⁻) bar read after that last absorbed dose?

Use the simulator. Add one strong-acid dose and read the line under the beaker. True or false: the weak acid HA is the component that absorbs the added H⁺.

The Henderson-Hasselbalch equation

Card 3, Henderson-Hasselbalch: Derivation, Use, and Limits

We just saw the molecular mechanism, the ratio [A⁻]/[HA] changes only slightly under buffering action. That raises a question: Is there a single formula that links the ratio [A⁻]/[HA] directly to pH so you can calculate buffer pH without an ICE table? This card answers it → Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]); derived in two steps from the Ka expression.

The Henderson-Hasselbalch equation is not a separate formula to memorise in isolation, it is the Ka expression rearranged to solve for pH directly, and deriving it from Ka in two algebraic steps demonstrates that it is simply the equilibrium expression in a more convenient form.

Derivation: Start with Ka = [H⁺][A⁻]/[HA]. Rearrange: [H⁺] = Ka × [HA]/[A⁻]. Take −log of both sides: −log[H⁺] = −log(Ka) − log([HA]/[A⁻]). pH = pKa − log([HA]/[A⁻]) = pKa + log([A⁻]/[HA]).

Three key cases: (1) When [A⁻] = [HA]: log(1) = 0 → pH = pKa. This is the half-equivalence point, and it is where buffer capacity is maximised. (2) When [A⁻] > [HA]: log > 0 → pH > pKa, more basic than pKa. (3) When [A⁻] < [HA]: log < 0 → pH < pKa, more acidic than pKa.

Buffer capacity: A buffer is effective within approximately one pH unit on either side of pKa, the effective buffering range is approximately pKa ± 1. Outside this range, one component is present in such low relative concentration that it cannot adequately resist pH changes.

[A⁻]/[HA] ratio	log([A⁻]/[HA])	pH relative to pKa	Buffering effectiveness
10:1	+1.00	pH = pKa + 1.00	Weakening, A⁻ mostly consumed if more H⁺ added
2:1	+0.30	pH = pKa + 0.30	Moderate, still effective
1:1	0	pH = pKa	Maximum buffer capacity
1:2	−0.30	pH = pKa − 0.30	Moderate, still effective
1:10	−1.00	pH = pKa − 1.00	Weakening, HA mostly consumed if more OH⁻ added

Must Do: When using Henderson-Hasselbalch after a partial neutralisation, you can use moles directly instead of concentrations: n(A⁻)/n(HA) = [A⁻]/[HA] because volume cancels (both species occupy the same solution). Calculate n(HA)remaining = n(HA)initial − n(OH⁻); n(A⁻)formed = n(OH⁻); then pH = pKa + log(n(A⁻)/n(HA)).

Common Error: Students apply Henderson-Hasselbalch to a pure weak acid solution with no conjugate base ([A⁻] = 0). When [A⁻] = 0, log(0/[HA]) = log(0) = −∞, an impossible result that signals misapplication. Henderson-Hasselbalch requires both HA and A⁻ present in significant amounts. For a pure weak acid solution, use the ICE table method.

Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). Derived from Ka by taking −log of both sides. When [A⁻] = [HA]: pH = pKa (maximum buffer capacity; half-equivalence point). Effective buffering range: pKa ± 1. After partial neutralisation use mole ratio n(A⁻)/n(HA) directly, volume cancels. HH requires both HA and A⁻ present, never apply to pure weak acid solution.

Pause, write the highlighted definition into your book before moving on.

An acetate buffer has pKa = 4.74. At what [A⁻]/[HA] ratio is the buffer capacity maximised, and what pH does this correspond to?

Preparing buffer solutions

Card 4, Preparing Buffer Solutions, Two Methods

We just saw that Henderson-Hasselbalch connects the [A⁻]/[HA] ratio to pH. That raises a question: If you want to make a buffer at a specific target pH in the lab, what are the two practical methods to achieve this? This card answers it → Method 1: direct mixing (calculate ratio from HH, weigh out acid + conjugate salt); Method 2: partial neutralisation (add strong base to weak acid so n(OH⁻) < n(HA)).

A buffer solution is not found ready-made in nature, it is deliberately prepared by combining specific amounts of a weak acid and its conjugate base, and the quantitative control of this ratio is precisely what gives a buffer its target pH.

Method 1, Direct mixing of weak acid and conjugate base salt. Choose a weak acid with a pKa close to the target pH (within ±1 unit). Calculate the required [A⁻]/[HA] ratio using Henderson-Hasselbalch: [A⁻]/[HA] = 10^(pH − pKa). Dissolve the calculated masses of weak acid and its sodium (or potassium) salt in water.

Example: to prepare an acetate buffer at pH 5.00 using CH₃COOH (Ka = 1.8 × 10⁻⁵, pKa = 4.74). Desired ratio = 10^(5.00 − 4.74) = 10^0.26 = 1.82. For a 0.100 mol/L total buffer: [CH₃COO⁻] = 0.0645 mol/L; [CH₃COOH] = 0.0355 mol/L.

Method 2, Partial neutralisation of weak acid with strong base. Add a calculated volume of strong base to a weak acid solution, neutralising the required fraction. Since HA + OH⁻ → A⁻ + H₂O (complete), moles of A⁻ formed = moles of OH⁻ added, and moles of HA remaining = initial moles − moles of OH⁻. Apply Henderson-Hasselbalch to the n(A⁻)/n(HA) ratio. This is the most common method in HSC questions and the method by which buffers form during a titration before the equivalence point.

Method 1 (direct mixing)

Starting materials: Weak acid + conjugate base salt

Key step: Calculate [A⁻]/[HA] from HH; weigh out required amounts

When to use: Precise target pH; both components available

Method 2 (partial neutralisation)

Starting materials: Weak acid + strong base

Key step: n(A⁻) = n(OH⁻) added; n(HA)remaining = n(HA)initial − n(OH⁻)

When to use: During titration before equivalence point; only acid available

Must Do: In partial neutralisation calculations, always verify that the amount of strong base added is LESS than the moles of weak acid, otherwise the equivalence point is reached and the solution is no longer a buffer (no HA remains). If n(OH⁻) = n(HA), all HA is converted to A⁻ and buffer capacity is zero.

Common Error: Students mix equal volumes of equal concentration weak acid and strong base and call the result a buffer, when in fact this is the equivalence point. A buffer requires BOTH HA and A⁻ present. The equivalence point gives a salt solution (e.g. CH₃COONa), not a buffer. For a buffer, the moles of strong base added must be less than the moles of weak acid.

Method 1 (direct mixing): target ratio = 10^(pH − pKa); weigh out weak acid and conjugate base salt. Method 2 (partial neutralisation): n(A⁻) = n(OH⁻) added; n(HA)rem = n(HA)init − n(OH⁻); use HH with mole ratio. Critical check: n(OH⁻) must be < n(HA), if equal or greater, buffer is destroyed. Buffer forms in titration between start and equivalence point; half-equivalence point pH = pKa.

Add the highlighted point to your notes before the check below.

A student adds 0.010 mol NaOH to 0.010 mol CH₃COOH in 100 mL water. Is the result a buffer?

Investigation, Preparing a Buffer and Testing What It Resists

Syllabus: conduct a practical investigation to prepare a buffer and demonstrate its properties (ACSCH080)

We just saw how to choose components and ratio for a target pH. That raises a question: does the solution you prepare actually hold its pH, and how would you show that convincingly? This card answers it → by running the same insult on the buffer and on unbuffered water, side by side.

Inquiry question. How effectively does a prepared ethanoate buffer resist pH change on addition of strong acid, strong base and dilution, compared with unbuffered water?

Safety

Teacher supervision and a written risk assessment. Safety glasses and gloves. Use the supplied 1.0 mol L⁻¹ ethanoic acid, not glacial, which is corrosive. 0.1 mol L⁻¹ HCl and NaOH are corrosive to eyes, rinse any splash for 15 minutes and report it. Rinse the pH electrode with distilled water between every reading and return it to its storage solution afterwards.

Equipment. Calibrated pH meter, 1.0 mol L⁻¹ ethanoic acid, 1.0 mol L⁻¹ sodium ethanoate, 0.1 mol L⁻¹ HCl, 0.1 mol L⁻¹ NaOH, distilled water, four 100 mL beakers, measuring cylinders, dropper pipettes, stirring rod, wash bottle.

Calibrate the pH meter against at least two standards, typically pH 4 and pH 7. An uncalibrated meter makes every number below meaningless.

Prepare the buffer. Mix 25.0 mL of 1.0 mol L⁻¹ ethanoic acid with 25.0 mL of 1.0 mol L⁻¹ sodium ethanoate. Equal volumes at equal concentration gives a 1:1 ratio, so the pH should be close to the pKa.

Measure the initial pH and compare it with your Henderson–Hasselbalch prediction before going on.

Split the buffer into two 25 mL portions. Put 25 mL of distilled water into each of two more beakers, these are the unbuffered controls.

Acid challenge. Add 0.1 mol L⁻¹ HCl 1 mL at a time to one buffer beaker, up to 10 mL, stirring and recording the pH after each addition. Repeat identically on one water beaker.

Base challenge. Repeat step 5 with 0.1 mol L⁻¹ NaOH on the second buffer beaker and the second water beaker.

Dilution test. Take a fresh 10 mL of buffer, measure the pH, add 90 mL of distilled water, stir and measure again. A buffer’s pH depends on the ratio of its components, which dilution does not change.

Find the capacity. Keep adding acid beyond 10 mL until the pH falls sharply, and record the volume at which it breaks.

pH after each addition, buffer and water control side by side
Volume added / mL	Buffer + HCl	Water + HCl	Buffer + NaOH	Water + NaOH
0
2
4
6
8
10

Analysis.

Plot pH against volume added, all four series on one set of axes. The comparison is the result; a buffer curve alone demonstrates nothing.
Write the equations showing how the buffer consumes added H⁺ and added OH⁻, and identify which component is used up in each case.
Calculate the pH change per mL for buffer and water over the first 10 mL, and quote the ratio. That is a far stronger claim than “the buffer changed less”.
Explain the dilution result with Henderson–Hasselbalch. Why does a tenfold dilution barely move a buffer’s pH but change a strong acid’s by one unit?
At what volume did the pH break in step 8, and what had run out at that point?

Conclusion. State how much acid and base your buffer absorbed for how much pH change, against water, and the capacity limit you measured. Comment on whether the initial pH matched the pKa prediction and what would explain any difference.

Rinse the electrode between readings: carrying a film of NaOH into the next beaker is a systematic error that always pushes one way, and it looks exactly like a buffer failing.

Common Error: Concluding “the buffer worked” from the buffer series alone. Without the water control on the same axes with the same additions, there is no measure of what resisting means.

Prepare: equal volumes of equal-concentration weak acid and conjugate base gives a 1:1 buffer at pH ≈ pKa. Demonstrate: identical acid and base additions to buffer and to water on one set of axes; dilute, which does not change the ratio and barely changes pH; then exhaust the capacity and record where the pH breaks.

Pause, write the highlighted method into your book.

Natural buffer systems and misconceptions

Card 5, Natural Buffer Systems, Blood, Ocean, and Enzymes

We just saw how buffers are prepared in the laboratory by controlling the HA/A⁻ ratio. That raises a question: Where do natural buffer systems appear, and which ones does NESA explicitly expect you to explain? This card answers it → blood (H₂CO₃/HCO₃⁻ at 20:1, pH 7.40), ocean acidification, and intracellular phosphate buffer.

A patient arrives in hospital with blood pH 7.22, respiratory acidosis from hypoventilation. Her kidneys are retaining HCO₃⁻ and excreting H⁺ in urine to push the ratio [HCO₃⁻]/[H₂CO₃] back toward 20:1. In 1908, Lawrence Henderson watched exactly this process in his blood chemistry data at Harvard, and wrote the equation that quantifies it. Every natural buffer system in this lesson works by the same mechanism: a weak acid and its conjugate base adjusting their ratio to absorb excess H⁺ or OH⁻.

The blood buffer system. Normal blood pH is 7.35–7.45. The primary buffer is the carbonate system: H₂CO₃(aq)/HCO₃⁻(aq), pKa = 6.10. At pH 7.40: 7.40 = 6.10 + log([HCO₃⁻]/[H₂CO₃]) → [HCO₃⁻]/[H₂CO₃] = 20:1. Blood operates far from equal concentrations of the two components, yet it is maintained by continuously replenishing H₂CO₃ from dissolved CO₂: CO₂(aq) + H₂O ⇌ H₂CO₃. Breathing rate regulates CO₂ concentration, faster breathing expels more CO₂, reducing [H₂CO₃] and raising pH (respiratory alkalosis). Slower breathing accumulates CO₂, lowering pH (respiratory acidosis).

When metabolic acids add H⁺ to blood: H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O. The H₂CO₃ formed decomposes to CO₂, which is expelled by the lungs, effectively removing the added H⁺ as gaseous CO₂. This makes blood buffering an "open" system that regenerates capacity far beyond what a closed solution buffer could provide.

Ocean buffering and ocean acidification. Ocean pH is maintained at ~8.1 by the carbonate system: CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. Increased atmospheric CO₂ from fossil fuel combustion dissolves in ocean water, pushing the equilibrium right and increasing [H⁺], lowering ocean pH. Since the industrial revolution, average ocean pH has fallen from ~8.2 to ~8.1, a shift of 0.1 units corresponding to a 26% increase in [H⁺]. Organisms that build calcium carbonate shells are vulnerable: CaCO₃(s) + H⁺ → Ca²⁺(aq) + HCO₃⁻(aq). Increased [H⁺] accelerates shell dissolution and increases shell formation energy cost, threatening marine ecosystems.

Enzyme function. Most enzymes have an optimal pH range within which their active site geometry and ionisation state support catalysis. Outside this range, the ionisation of acidic and basic amino acid side chains changes, altering binding affinity and catalytic rate. For example, pepsin (stomach protease) is optimally active at pH 1.5–2.5; trypsin (intestinal protease) at pH 7.5–8.5. Intracellular phosphate buffers (H₂PO₄⁻/HPO₄²⁻, pKa = 7.21) maintain cytoplasmic pH within the range where metabolic enzymes function optimally.

Blood (carbonate)

Components: H₂CO₃ / HCO₃⁻

pKa: 6.10

Normal pH range: 7.35–7.45

Buffer equation (add H⁺): H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O

Blood (phosphate)

Components: H₂PO₄⁻ / HPO₄²⁻

pKa: 7.21

Normal pH range: 7.35–7.45

Buffer equation (add H⁺): H⁺ + HPO₄²⁻ → H₂PO₄⁻

Ocean (carbonate)

Components: CO₂/H₂CO₃ / HCO₃⁻ / CO₃²⁻

pKa: Multiple

Normal pH range: ~8.1

Buffer equation (add H⁺): H⁺ + CO₃²⁻ → HCO₃⁻

Intracellular (phosphate)

Components: H₂PO₄⁻ / HPO₄²⁻

pKa: 7.21

Normal pH range: pH 6.8–7.4

Buffer equation (add H⁺): H⁺ + HPO₄²⁻ → H₂PO₄⁻

Must Do: For HSC extended response questions on the blood buffer, always include: (1) both buffer equations, H⁺ consumed by HCO₃⁻; OH⁻ consumed by H₂CO₃; (2) the connection to CO₂/breathing rate and Le Chatelier's principle; (3) the normal blood pH range (7.35–7.45) and what happens outside this range (acidosis < 7.35; alkalosis > 7.45). All three components are required for full marks.

Common Error: "The blood buffer maintains pH 7." The correct value is 7.35–7.45, which is slightly basic. Also wrong: "the blood buffer works because H₂CO₃ neutralises added base." More precisely, H₂CO₃ donates H⁺ to OH⁻ (Brønsted-Lowry acid), producing HCO₃⁻ and H₂O, not a simple neutralisation.

Insight: Diabetic ketoacidosis (DKA) causes metabolic acidosis, ketone bodies accumulate in blood, consuming HCO₃⁻ faster than it can be replenished. Blood HCO₃⁻ falls from normal ~24 mmol/L to <10 mmol/L in severe DKA. The body responds with Kussmaul breathing, deep, rapid respirations to expel CO₂, lowering [H₂CO₃] and partially compensating. Without treatment, the buffer fails entirely.

Blood buffer: H₂CO₃/HCO₃⁻ (pKa 6.10); normal pH 7.35–7.45; [HCO₃⁻]/[H₂CO₃] ≈ 20:1 at pH 7.40. Add H⁺: H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O (expelled by lungs, open system). Add OH⁻: OH⁻ + H₂CO₃ → HCO₃⁻ + H₂O. Ocean acidification: increased CO₂ → increased H₂CO₃ → lower [HCO₃⁻]/[H₂CO₃] ratio → lower pH → CaCO₃ shells dissolve.

Pause, copy the highlighted definition into your book before moving on.

Normal blood [HCO₃⁻]/[H₂CO₃] ≈ 20:1 and pKa = 6.10. Using Henderson-Hasselbalch, what is the normal blood pH?

⚠ Common Misconceptions, Buffers

✗

"A buffer prevents pH changes." A buffer resists, not prevents. There is always a small pH change, the buffer minimises it. "Prevents" earns no credit in an explanation; "resists" is the required term.

✗

"A buffer must have pH 7." Buffer pH is determined by pKa + log([A⁻]/[HA]). The acetate buffer operates at pH ~4.74, acidic, not neutral. Blood buffer operates at 7.35–7.45, basic, not exactly neutral.

✗

"The HA component reacts with added H⁺." Wrong, the conjugate base A⁻ reacts with added H⁺ (A⁻ + H⁺ → HA). The weak acid HA reacts with added OH⁻ (HA + OH⁻ → A⁻ + H₂O). Many students reverse these roles.

✗

"Equal volumes of weak acid and strong base makes a buffer." If concentrations are equal, this is the equivalence point, all HA is consumed, no buffer exists. Only if the strong base is added in less than stoichiometric amount does a buffer form.

✗

"Use pH = −log(Ka) to find buffer pH." This is wrong for all solutions. For a buffer, use Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). pH = pKa only when [A⁻] = [HA], a special case, not the general formula.

Worked examples and activities

🔬 Worked Examples

Worked Example 1, Henderson-Hasselbalch: Buffer pH, Required Ratio, and pH After Adding Acid

Apply Band 4

(a) Calculate the pH of a buffer prepared by mixing 0.150 mol/L CH₃COOH and 0.200 mol/L CH₃COONa. Ka(CH₃COOH) = 1.8 × 10⁻⁵. (b) Calculate the ratio [CH₃COO⁻]/[CH₃COOH] required to prepare an acetate buffer at pH 5.20. (c) The buffer in (a) is prepared using 500 mL of solution. Calculate the new pH after adding 5.00 mL of 1.00 mol/L HCl. Assume volume change is negligible.

GIVEN: [CH₃COOH] = 0.150 mol/L; [CH₃COO⁻] = 0.200 mol/L; Ka = 1.8 × 10⁻⁵; V = 500 mL = 0.500 L; V(HCl) = 5.00 mL; c(HCl) = 1.00 mol/L

METHOD and ANSWER (a), Buffer pH:

pKa = −log(1.8 × 10⁻⁵) = 5 − log(1.8) = 5 − 0.255 = 4.74

pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.200/0.150) = 4.74 + log(1.333) = 4.74 + 0.125 = 4.87

METHOD and ANSWER (b), Required ratio for pH 5.20:

pH = pKa + log([A⁻]/[HA]) → log([A⁻]/[HA]) = pH − pKa = 5.20 − 4.74 = 0.46

[A⁻]/[HA] = 10^0.46 = 2.88 approximately 3 parts acetate to 1 part acetic acid

METHOD and ANSWER (c), pH after adding HCl:

n(CH₃COOH) = 0.150 × 0.500 = 0.0750 mol

n(CH₃COO⁻) = 0.200 × 0.500 = 0.100 mol

n(HCl) = 1.00 × 0.00500 = 5.00 × 10⁻³ mol H⁺ added

H⁺ reacts with CH₃COO⁻: CH₃COO⁻ + H⁺ → CH₃COOH

n(CH₃COO⁻) after = 0.100 − 5.00 × 10⁻³ = 0.0950 mol

n(CH₃COOH) after = 0.0750 + 5.00 × 10⁻³ = 0.0800 mol

pH = 4.74 + log(0.0950/0.0800) = 4.74 + log(1.188) = 4.74 + 0.075 = 4.81

Change in pH = 4.81 − 4.87 = −0.06 pH units. Compare: same HCl in pure water → pH 7.00 → pH 2.00, a change of 5.00 pH units. The buffer reduced the pH change by a factor of ~83.

ANSWERS: (a) pH = 4.87. (b) [A⁻]/[HA] = 2.88. (c) pH after HCl = 4.81; ΔpH = −0.06 units (vs −5.00 units in pure water).

Worked Example 2, Buffer Preparation by Partial Neutralisation and Capacity

Analyse Band 5

A student prepares a phosphate buffer by adding 30.0 mL of 0.200 mol/L NaOH to 50.0 mL of 0.200 mol/L NaH₂PO₄. Ka₂(H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻) = 6.2 × 10⁻⁸. (a) Identify the weak acid and conjugate base. (b) Calculate the pH of the resulting buffer. (c) Calculate the maximum moles of NaOH that can be added before the buffer fails. (d) Explain why a phosphate buffer is appropriate for intracellular biological applications.

GIVEN: V(NaOH) = 30.0 mL = 0.0300 L; c(NaOH) = 0.200 mol/L; V(NaH₂PO₄) = 50.0 mL = 0.0500 L; c(NaH₂PO₄) = 0.200 mol/L; Ka₂ = 6.2 × 10⁻⁸

(a) Buffer components:

NaH₂PO₄ provides H₂PO₄⁻, the weak acid (Ka₂ = 6.2 × 10⁻⁸; ionisation: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻)

When NaOH added: H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O → HPO₄²⁻ is the conjugate base

Buffer pair: H₂PO₄⁻ (weak acid) / HPO₄²⁻ (conjugate base)

(b) Buffer pH:

n(H₂PO₄⁻) initial = 0.200 × 0.0500 = 0.0100 mol

n(OH⁻) = 0.200 × 0.0300 = 6.00 × 10⁻³ mol

H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O (complete)

n(H₂PO₄⁻) remaining = 0.0100 − 6.00 × 10⁻³ = 4.00 × 10⁻³ mol

n(HPO₄²⁻) formed = 6.00 × 10⁻³ mol

pKa₂ = −log(6.2 × 10⁻⁸) = 8 − log(6.2) = 8 − 0.792 = 7.21

pH = 7.21 + log(6.00 × 10⁻³ / 4.00 × 10⁻³) = 7.21 + log(1.50) = 7.21 + 0.176 = 7.39 ✓

(c) Maximum NaOH capacity:

The buffer fails when all H₂PO₄⁻ is consumed. Remaining n(H₂PO₄⁻) = 4.00 × 10⁻³ mol.

Maximum additional NaOH = 4.00 × 10⁻³ mol

Maximum HCl that can be absorbed = n(HPO₄²⁻) = 6.00 × 10⁻³ mol

(d) Why phosphate buffer suits biological use:

pKa₂(H₂PO₄⁻) = 7.21 is within 0.2 pH units of physiological pH (7.35–7.45). Effective buffering range = pKa ± 1 = 6.21–8.21, which comfortably spans physiological pH. At pH 7.39, [HPO₄²⁻]/[H₂PO₄⁻] = 1.50, both components present in comparable concentrations, providing capacity to resist both acid and base loads. Additionally, phosphate is a naturally occurring biological molecule, non-toxic at physiological concentrations, and participates directly in metabolic pathways (ATP/ADP cycle).

ANSWERS: (a) Weak acid = H₂PO₄⁻; conjugate base = HPO₄²⁻. (b) pH = 7.39. (c) Maximum NaOH = 4.00 × 10⁻³ mol; maximum HCl = 6.00 × 10⁻³ mol. (d) pKa₂ = 7.21 close to physiological pH; effective range 6.21–8.21; non-toxic and naturally present.

Worked Example 3, Extended Response: Blood Buffer and Diabetic Ketoacidosis (8 marks)

Evaluate Band 6 (8 marks)

A patient presents with diabetic ketoacidosis (DKA). Blood tests show: pH = 7.10; [HCO₃⁻] = 6.0 mmol/L (normal: 22–26 mmol/L); [H₂CO₃] = 0.60 mmol/L (reduced due to hyperventilation). pKa(H₂CO₃) = 6.10. (a) Verify the patient's blood pH using Henderson-Hasselbalch. (b) Explain why [HCO₃⁻] has fallen from normal (24 mmol/L) to 6.0 mmol/L. (c) Explain the physiological significance of the reduced [H₂CO₃] in terms of respiratory compensation. (d) Calculate blood pH without respiratory compensation ([H₂CO₃] = 1.20 mmol/L; [HCO₃⁻] = 6.0 mmol/L). Explain why compensation is insufficient to fully normalise pH.

GIVEN: pH = 7.10 (measured); [HCO₃⁻] = 6.0 mmol/L; [H₂CO₃] = 0.60 mmol/L (compensated); [H₂CO₃] = 1.20 mmol/L (uncompensated); pKa = 6.10

(a) Verify blood pH:

pH = pKa + log([HCO₃⁻]/[H₂CO₃]) = 6.10 + log(6.0/0.60) = 6.10 + log(10) = 6.10 + 1.00 = 7.10 ✓

(b) Why [HCO₃⁻] has fallen:

In DKA, ketone bodies dissociate in blood, releasing H⁺. The released H⁺ reacts with bicarbonate via the buffer reaction:

H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O

Each mole of metabolic H⁺ produced consumes one mole of HCO₃⁻, producing CO₂ (exhaled). As ketone body production exceeds the rate of HCO₃⁻ replenishment by the kidneys, [HCO₃⁻] falls progressively, from 24 mmol/L to 6.0 mmol/L in this patient. This depletion of the conjugate base component destroys buffer capacity.

(c) Respiratory compensation:

The patient's reduced [H₂CO₃] (0.60 vs normal 1.20 mmol/L) indicates respiratory compensation, the body is breathing faster and deeper (Kussmaul breathing) to expel CO₂. Expelling CO₂ removes the acid component from the buffer equilibrium: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻. Reducing [CO₂(aq)] shifts the equilibrium to the left (Le Chatelier's principle), decreasing [H⁺] and raising pH toward normal.

(d) Uncompensated pH and why compensation is insufficient:

Without respiratory compensation: pH = 6.10 + log(6.0/1.20) = 6.10 + log(5.0) = 6.10 + 0.699 = 6.80

Without respiratory compensation, blood pH would be 6.80, severely life-threatening. Respiratory compensation raised pH from 6.80 to 7.10 (+0.30 units) by halving [H₂CO₃].

Why compensation is insufficient: [HCO₃⁻] has fallen from 24 to 6.0 mmol/L, even if [H₂CO₃] were reduced to near zero, the ratio [HCO₃⁻]/[H₂CO₃] cannot reach the required ~20:1 because the numerator ([HCO₃⁻] = 6.0) is the limiting factor. The kidneys replenish HCO₃⁻ slowly (over hours to days) while respiration acts over minutes. Without insulin to stop ketone body production and IV fluids to restore [HCO₃⁻], the buffer system continues to fail.

ANSWERS: (a) pH = 6.10 + log(6.0/0.60) = 7.10 ✓. (b) Ketone body H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O; [HCO₃⁻] falls 24 → 6.0 mmol/L. (c) Kussmaul breathing expels CO₂ → [H₂CO₃] ↓ → ratio ↑ → pH rises (Le Chatelier). (d) Uncompensated pH = 6.80; compensation raised pH to 7.10; insufficient because [HCO₃⁻] = 6.0 mmol/L is too depleted, renal HCO₃⁻ replenishment and insulin required.

Key Definitions & Rules

Buffer: a solution containing a weak acid (HA) and its conjugate base (A⁻) that resists pH change when small amounts of strong acid or base are added
Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). When [A⁻] = [HA] → pH = pKa (maximum buffer capacity)
Buffer mechanism, add H⁺: A⁻ + H⁺ → HA (conjugate base consumes H⁺)
Buffer mechanism, add OH⁻: HA + OH⁻ → A⁻ + H₂O (weak acid consumes OH⁻)
Effective buffer range: pKa ± 1. Choose weak acid with pKa within 1 unit of target pH
Blood buffer: H₂CO₃/HCO₃⁻ (pKa = 6.10); normal pH 7.35–7.45; [HCO₃⁻]/[H₂CO₃] ≈ 20:1; CO₂ expelled by lungs regenerates capacity
Partial neutralisation: n(A⁻) = n(OH⁻) added; n(HA)rem = n(HA)init − n(OH⁻); then use Henderson-Hasselbalch with mole ratio

✏️ Activities

Analyse: Buffer pH and Ratio Problems

Use the Henderson-Hasselbalch equation to solve each scenario. Show all working.

Scenario 1: A formate buffer is prepared by mixing 0.120 mol/L HCOOH (Ka = 1.77 × 10⁻⁴) and 0.080 mol/L HCOONa. Calculate the pH of this buffer and determine what volume of 1.00 mol/L NaOH must be added to 250 mL of this buffer to raise its pH by exactly 0.20 units. Assume volume change is negligible.

Scenario 2: A student needs to prepare a buffer at pH 7.20 using the phosphate pair H₂PO₄⁻/HPO₄²⁻ (Ka₂ = 6.2 × 10⁻⁸). She has 100 mL of 0.500 mol/L NaH₂PO₄. What volume of 0.500 mol/L NaOH must be added to achieve the desired pH? Why is this buffer suitable for simulating intracellular conditions?

Connect: Blood Buffer and Ocean Acidification

Answer each question in full sentences, referencing the Henderson-Hasselbalch equation where relevant.

Part 1: A patient with severe pneumonia cannot expel CO₂ efficiently. Explain, using the Henderson-Hasselbalch equation and Le Chatelier's principle, how this will affect blood pH. Name the clinical condition and state whether it is acidosis or alkalosis.

Part 2: Atmospheric CO₂ has increased from 280 ppm (pre-industrial) to 420 ppm (current), and ocean pH has fallen from 8.2 to 8.1. (a) Write the equilibrium showing how CO₂ dissolves in ocean water and produces H⁺. (b) Use the Henderson-Hasselbalch equation to explain qualitatively why increasing [H₂CO₃] lowers ocean pH. (c) Explain why organisms with CaCO₃ shells are particularly vulnerable.

🔬 Predict, Then Reveal +8 XP

A buffer solution contains CH₃COOH and CH₃COO⁻Na⁺. A small amount of HCl is added. Predict: (a) which component reacts with the added H⁺ and (b) why the pH barely changes.

Your predictionExpert answerCompare

Complete the Learn phase to unlock Practice.

✓

Practice questions

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

Short Answer Questions

Understand Band 3–4 (4 marks)

Q6. A student claims: "Any solution that contains both an acid and a base is a buffer." Using a specific counterexample, explain why this claim is incorrect. Then write the two molecular equations that describe how a CH₃COOH/CH₃COO⁻ buffer resists pH change when (i) HCl is added and (ii) NaOH is added. (4 marks)

Apply Band 4–5 (5 marks)

Q7. A student adds 20.0 mL of 0.300 mol/L NaOH to 50.0 mL of 0.300 mol/L lactic acid (CH₃CH(OH)COOH, Ka = 1.38 × 10⁻⁴). (a) Show that a buffer forms and identify its components. (b) Calculate the pH of the resulting buffer using the Henderson-Hasselbalch equation. (c) Calculate the pH if an additional 2.00 mL of 0.300 mol/L NaOH is added to the buffer in (b). (5 marks)

Evaluate high-mark (7 marks)

Q8. Blood pH is maintained at 7.35–7.45 primarily by the H₂CO₃/HCO₃⁻ buffer system (pKa = 6.10). (a) Calculate the normal [HCO₃⁻]/[H₂CO₃] ratio at blood pH 7.40. (b) During intense exercise, lactic acid enters the bloodstream, lowering blood pH to 7.30. Using Henderson-Hasselbalch, calculate the new [HCO₃⁻]/[H₂CO₃] ratio and identify whether this represents acidosis or alkalosis. (c) Explain at the molecular level why blood pH at 7.30 is classified as acidosis even though the solution is still basic, and describe two physiological mechanisms the body uses to restore pH to the normal range. (7 marks)

Show All Answers

MC Q1: A pKa = −log(6.5 × 10⁻⁵) = 4.19. pH = 4.19 + log(0.120/0.080) = 4.37. Adding 0.010 mol H⁺: reacts with C₆H₅COO⁻ (conjugate base). n(A⁻) after = 0.110; n(HA) after = 0.090. New pH = 4.19 + log(0.110/0.090) = 4.28. ΔpH = −0.09.

MC Q2: D[H₂CO₃] is halved → new [H₂CO₃] = 0.5× original. New ratio = 20/0.5 = 40:1. pH = 6.10 + log(40) = 6.10 + 1.60 = 7.70. Since 7.70 > 7.45, this is respiratory alkalosis. Option A incorrectly uses 20/2 (as if HCO₃⁻ halved instead of H₂CO₃).

MC Q3: C In NH₃/NH₄Cl, the weak acid is NH₄⁺ (proton donor: NH₄⁺ ⇌ H⁺ + NH₃). Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.56 × 10⁻¹⁰; pKa = 9.26. Equal concentrations → pH = pKa = 9.26. Option A identifies NH₃ as the weak acid, wrong; NH₃ is the conjugate base.

MC Q4: B Buffer fails when all CH₃COOH is consumed. n(CH₃COOH) = 0.050 mol. Each mole of NaOH consumes one mole of HA. Buffer fails at exactly 0.050 mol NaOH.

MC Q5: B Henderson-Hasselbalch: pH = pKa + log([HCO₃⁻]/[H₂CO₃]). Increased CO₂ → increased [H₂CO₃] (denominator). [HCO₃⁻]/[H₂CO₃] ratio decreases → log term decreases → pH falls. This is ocean acidification.

Q6, Sample Answer:

Counterexample (1 mark): Mixing equal moles of HCl and NaOH gives NaCl(aq). Contains H⁺ and OH⁻, yet NaCl has no buffering ability, no weak acid/conjugate base pair is present.

Correct definition (1 mark): A buffer requires a weak acid (HA) and its conjugate base (A⁻) in comparable concentrations.

(i) HCl added (1 mark): H⁺ + CH₃COO⁻ → CH₃COOH

(ii) NaOH added (1 mark): OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O

Q7, Sample Answer:

(a) Buffer formation (1 mark): n(HA) = 0.300 × 0.0500 = 0.0150 mol. n(NaOH) = 0.300 × 0.0200 = 6.00 × 10⁻³ mol. Since n(NaOH) < n(HA), buffer forms. Components: lactic acid (HA) and lactate ion (A⁻).

(b) Buffer pH (2 marks): n(A⁻) formed = 6.00 × 10⁻³ mol; n(HA) remaining = 9.00 × 10⁻³ mol. pKa = −log(1.38 × 10⁻⁴) = 3.86. pH = 3.86 + log(6.00 × 10⁻³/9.00 × 10⁻³) = 3.86 + log(0.667) = 3.86 − 0.176 = 3.68.

(c) Additional NaOH (2 marks): n(NaOH) extra = 0.300 × 0.00200 = 6.00 × 10⁻⁴ mol. n(HA) after = 9.00 × 10⁻³ − 6.00 × 10⁻⁴ = 8.40 × 10⁻³ mol. n(A⁻) after = 6.00 × 10⁻³ + 6.00 × 10⁻⁴ = 6.60 × 10⁻³ mol. pH = 3.86 + log(6.60/8.40 × 10⁻³) = 3.86 + log(0.786) = 3.86 − 0.105 = 3.76. ΔpH = +0.08 units.

Q8, Sample Answer:

(a) Normal ratio (1 mark): 7.40 = 6.10 + log([HCO₃⁻]/[H₂CO₃]) → log = 1.30 → ratio = 10^1.30 = 20:1.

(b) New ratio at pH 7.30 and classification (2 marks): 7.30 = 6.10 + log(ratio) → log = 1.20 → ratio = 10^1.20 = 15.8:1. pH 7.30 < 7.35 → metabolic acidosis.

(c) Explanation and mechanisms (4 marks): Blood at pH 7.30 is still basic ([H⁺] < [OH⁻]), but acidosis is defined clinically relative to the normal range 7.35–7.45. At pH 7.30, increased [H⁺] destabilises enzyme active sites, disrupts protein ionisation, and impairs oxygen delivery by haemoglobin. Mechanism 1, respiratory compensation: brain detects pH fall → increased breathing rate → more CO₂ expelled → [H₂CO₃] decreases → [HCO₃⁻]/[H₂CO₃] ratio increases → pH rises (Le Chatelier). Mechanism 2, renal compensation: kidneys increase HCO₃⁻ reabsorption and H⁺ excretion in urine over hours to days → [HCO₃⁻] replenished → ratio returns toward 20:1 → pH normalises to 7.35–7.45.

Activity A, Answers:

Scenario 1: pKa(HCOOH) = −log(1.77 × 10⁻⁴) = 3.75. pH = 3.75 + log(0.080/0.120) = 3.75 − 0.176 = 3.57. To raise pH by 0.20 to 3.77: required ratio = 10^(3.77−3.75) = 10^0.02 = 1.047. In 250 mL: n(HA) = 0.0300 mol; n(A⁻) = 0.0200 mol. After x mol NaOH: (0.0200+x)/(0.0300−x) = 1.047 → x = 5.57 × 10⁻³ mol. Volume = 5.57 mL of 1.00 mol/L NaOH.

Scenario 2: pKa₂ = 7.21. log([HPO₄²⁻]/[H₂PO₄⁻]) = 7.20 − 7.21 = −0.01 → ratio = 0.977. n(NaH₂PO₄) = 0.0500 mol. x/(0.0500−x) = 0.977 → x = 0.0247 mol. Volume = 0.0247/0.500 = 49.5 mL NaOH. Biological suitability: pKa₂ = 7.21 within 0.01 units of target; effective range 6.21–8.21; non-toxic and native to cells.

Now You Know, Henderson's 1908 Discovery

Return to your Think First. Henderson's 1908 Harvard analysis gave: pH 7.40 = pKa(H₂CO₃) + log([HCO₃⁻]/[H₂CO₃]) → 20:1 ratio. Check your Beaker B answers:

What is in Beaker B that resists pH change: Beaker B contains CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base) simultaneously. When HCl (H⁺) is added, CH₃COO⁻ + H⁺ → CH₃COOH, the conjugate base consumes the added H⁺ before it can significantly change [H⁺]. The ratio [A⁻]/[HA] changes slightly, and Henderson-Hasselbalch shows pH changes by only ~0.08 units vs 5 units in pure water.
What happens if NaOH is added to Beaker B: The weak acid component reacts: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The OH⁻ is consumed, [A⁻]/[HA] ratio increases slightly, and pH rises by only a small amount, the buffer resists base addition as effectively as acid addition.
What limits buffer capacity: Finite moles of HA and A⁻. If enough H⁺ is added to exhaust all A⁻, the ratio [A⁻]/[HA] → 0 and the buffer fails, pH then drops dramatically. Buffer capacity = the moles of acid or base the buffer can absorb before failing.

✓

Retrieve and reflect

Check what actually stuck

🏆 Review

Define a buffer in one sentence using the required HSC terms.

A buffer is a solution containing a weak acid (HA) and its conjugate base (A⁻) in comparable concentrations that resists significant pH change when small amounts of strong acid or base are added, due to the finite capacity of each component to consume the added species.

Write the two equations showing how an acetate buffer (CH₃COOH/CH₃COO⁻) responds to (a) added H⁺ and (b) added OH⁻.

(a) H⁺ + CH₃COO⁻ → CH₃COOH (conjugate base consumes added H⁺). (b) OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O (weak acid consumes added OH⁻).

What is the Henderson-Hasselbalch equation and what condition gives maximum buffer capacity?

pH = pKa + log([A⁻]/[HA]). Maximum buffer capacity when [A⁻] = [HA], giving log(1) = 0 → pH = pKa. At this point (half-equivalence point) the buffer has equal amounts of both components and maximum resistance to both acid and base addition.

Calculate the normal blood pH using the H₂CO₃/HCO₃⁻ buffer (pKa = 6.10, normal ratio [HCO₃⁻]/[H₂CO₃] = 20:1).

pH = 6.10 + log(20) = 6.10 + 1.30 = 7.40 ✓. This confirms the Henderson-Hasselbalch equation predicts physiological blood pH from the 20:1 buffer ratio. If the ratio falls (more H⁺ consuming HCO₃⁻), pH falls below 7.35 = acidosis.

Explain ocean acidification using the Henderson-Hasselbalch equation.

Increased atmospheric CO₂ → increased dissolved CO₂ → increased [H₂CO₃] (denominator of HH). pH = pKa + log([HCO₃⁻]/[H₂CO₃]) → larger denominator → smaller ratio → smaller log → lower pH. Ocean pH has fallen from 8.2 to 8.1 since industrialisation (26% increase in [H⁺]), threatening CaCO₃ shell-forming organisms: CaCO₃ + H⁺ → Ca²⁺ + HCO₃⁻.

I've completed this lesson

← Lesson 12 Lesson 14 →