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Inquiry Question 3: It is all about hydrogen ions

Investigate the application of buffer systems in natural and industrial contexts, including the bicarbonate buffer in blood and the Henderson-Hasselbalch description of buffer pH

A focused answer to the HSC Chemistry Module 6 dot point on buffer applications. Buffer action revisited, the Henderson-Hasselbalch equation, the bicarbonate buffer in blood (HCO3/H2CO3), respiratory and renal compensation, and worked HSC past exam questions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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In plain English

A buffer is like a large, calm crowd standing on both sides of a set of turnstiles, half already through (the conjugate base) and half still waiting (the weak acid). If a handful of extra people (added $H^+$ ) push in from one side, the waiting half simply lets a few more through, one at a time, so the flow barely changes. Blood's bicarbonate buffer is a special version of this: instead of a fixed crowd, there are two gates constantly restocking each side, the lungs topping up or venting the "waiting" gas ( $CO_2$ , which becomes $H_2CO_3$ ) within minutes, and the kidneys topping up or draining the "already through" pool ( $HCO_3^-$ ) over hours to days. The Henderson-Hasselbalch equation is just the maths of the crowd ratio: $pH = pK_a + \log_{10}([A^-]/[HA])$ tells you the pH from how lopsided the two sides are, and because the lungs and kidneys can actively adjust that ratio, blood pH stays locked near 7.40 even when the body is under real physiological stress.

What this dot point is asking

NESA wants you to consolidate buffer chemistry by applying it to natural and industrial contexts, especially the bicarbonate buffer in blood, and to use the Henderson-Hasselbalch equation quantitatively. You should be able to write the buffer equilibria, explain the response to added strong acid or base in equation form, and link the chemistry to physiological situations like exercise, hyperventilation, and acidosis. This builds on conjugate acid-base pairs and the Module 5 page on buffer systems.

The answer

Buffer composition and action (recap)

A buffer is a solution containing significant amounts of both a weak acid and its conjugate base (or a weak base and its conjugate acid). The two species sit in equilibrium:

HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}

Response to added strong acid (extra $H^+$ ). The conjugate base consumes it:

A^-_{(aq)} + H^+_{(aq)} \rightarrow HA_{(aq)}

Response to added strong base (extra $OH^-$ ). The weak acid consumes it:

HA_{(aq)} + OH^-_{(aq)} \rightarrow A^-_{(aq)} + H_2O_{(l)}

In each case, the strong reagent is converted into the corresponding member of the conjugate pair, so $[H^+]$ moves only slightly. The buffer fails when one component is essentially exhausted.

The Henderson-Hasselbalch equation

Take logarithms of the $K_a$ expression:

K_a = \frac{[H^+][A^-]}{[HA]} \quad \Rightarrow \quad pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right)

This is the Henderson-Hasselbalch equation. Three quick consequences:

When $[A^-] = [HA]$ (equimolar buffer), $pH = pK_a$ . Buffer capacity is maximised here.
The useful range of a buffer is roughly $pK_a \pm 1$ (corresponding to a 10:1 ratio of components on either side).
To make a buffer of a target pH, choose a weak acid with $pK_a$ within one unit of the target, then set the ratio.

An owned illustrative buffer-response curve shows why a 20:1 ratio, far from ideal on paper, still keeps blood pH stable in practice: the curve is flattest near $pH = pK_a$ , but it is only truly steep once a component is nearly exhausted.

The bicarbonate buffer in blood

Arterial blood is maintained at pH $7.40 \pm 0.05$ by a coupled buffer-respiratory-renal system. The dominant chemical buffer is the bicarbonate pair:

CO_{2(g)} + H_2O_{(l)} \rightleftharpoons H_2CO_{3(aq)} \rightleftharpoons H^+_{(aq)} + HCO_{3(aq)}^-

For this system $pK_{a1} = 6.10$ at body temperature, with typical concentrations $[HCO_3^-] \approx 24$ mmol/L and $[H_2CO_3] \approx 1.2$ mmol/L (in equilibrium with dissolved $CO_2$ ).

pH = 6.10 + \log_{10}\left(\frac{24}{1.2}\right) = 6.10 + 1.30 = 7.40

The ratio is far from 1:1, so chemically this is a poor buffer in isolation. What makes it physiologically powerful is that both components are continuously regulated:

$[H_2CO_3]$ is controlled by breathing rate (the lungs expel $CO_2$ ).
$[HCO_3^-]$ is controlled by the kidneys (which excrete or retain bicarbonate).

This "open" buffer can therefore reset its components in response to disturbances, something a closed buffer in a beaker cannot do.

Acid-base disturbances

Condition	Cause	What happens to pH	Compensation
Respiratory acidosis	Hypoventilation (slow breathing, COPD): $CO_2$ accumulates	falls	Kidneys retain $HCO_3^-$
Respiratory alkalosis	Hyperventilation (panic, altitude): $CO_2$ exhaled too fast	rises	Kidneys excrete $HCO_3^-$
Metabolic acidosis	Excess acid (uncontrolled diabetes, lactic acid build-up)	falls	Lungs increase ventilation to expel $CO_2$
Metabolic alkalosis	Excess base (vomiting, certain antacids)	rises	Lungs reduce ventilation, retain $CO_2$

In each case the body shifts the bicarbonate equilibrium to restore the 20:1 ratio.

Other physiological buffers

Phosphate buffer ( $H_2PO_4^-$ / $HPO_4^{2-}$ ): $pK_a = 7.20$ . Important inside cells where bicarbonate is less effective. Also the basis for laboratory buffers (PBS).
Protein buffers (haemoglobin in particular): histidine side chains have $pK_a$ near 6, so they buffer near physiological pH. Haemoglobin doubles as the oxygen carrier and a major intracellular buffer.

Industrial and laboratory buffers

Buffer	$pK_a$	Useful range	Typical use
Citric acid / citrate	3.13, 4.76, 6.40	2 to 7	Food, soft drinks
Acetate (ethanoate)	4.76	3.7 to 5.7	Enzyme assays, electroplating
Carbonate (HCO3/CO3)	10.33	9.3 to 11.3	Cleaning products
Phosphate (H2PO4/HPO4)	7.20	6.2 to 8.2	Biological PBS
Tris	8.07	7.0 to 9.0	Molecular biology

Worked example

You need 500 mL of a pH 5.00 buffer using ethanoic acid ( $pK_a = 4.74$ ) and sodium ethanoate. Calculate the ratio $[CH_3COO^-]/[CH_3COOH]$ required, and a workable pair of concentrations.

Step 1: Apply Henderson-Hasselbalch.

5.00 = 4.74 + \log_{10}\left(\frac{[A^-]}{[HA]}\right)

\log_{10}\left(\frac{[A^-]}{[HA]}\right) = 0.26 \quad \Rightarrow \quad \frac{[A^-]}{[HA]} = 10^{0.26} = 1.82

Step 2: Choose concentrations. Pick total buffer concentration ~0.20 mol/L for good capacity. With ratio 1.82:1, set $[HA] = 0.071$ mol/L and $[A^-] = 0.129$ mol/L (sum 0.20 mol/L).

Step 3: Check.

pH = 4.74 + \log_{10}(0.129/0.071) = 4.74 + 0.26 = 5.00 \quad \checkmark

Dissolve $0.071 \times 0.500 = 0.0355$ mol $CH_3COOH$ and $0.129 \times 0.500 = 0.0645$ mol $CH_3COONa$ in water and make up to 500 mL.

Common traps

Calling any weak acid a buffer: A weak acid alone is not a buffer. A buffer needs both the weak acid and its conjugate base in comparable amounts.
Forgetting the 20:1 ratio in blood: Most textbook buffers operate near 1:1 ( $pH = pK_a$ ). Blood is held far from 1:1 because the components are open (lungs and kidneys), not closed.
Confusing buffer capacity with buffer pH: Capacity is how much acid or base the buffer can absorb before pH shifts significantly; pH is the operating point. A dilute buffer has low capacity but the same pH as a concentrated one of the same ratio.
Using strong acid in a buffer recipe: Henderson-Hasselbalch assumes the weak acid only partly ionises. A strong acid + strong base mixture is not a buffer.
Misnaming acidosis and alkalosis: Acidosis is pH below 7.35 (more acidic); alkalosis is pH above 7.45 (more basic). Adding "respiratory" or "metabolic" specifies the origin.
Ignoring the sign on log inside Henderson-Hasselbalch: $\log_{10}(ratio < 1)$ is negative, so pH is below $pK_a$ when the acid form dominates.

Examples in context

Example 1. Diabetic ketoacidosis presentation at Westmead ED. A 16-year-old with type-1 diabetes presents at Westmead emergency with rapid breathing and confusion. Arterial blood gases show pH 7.10, $[HCO_3^-]$ at 8 mmol L $^{-1}$ and $pCO_2$ at 18 mmHg. Plugging into Henderson-Hasselbalch with $pK_a = 6.1$ , $pH = 6.1 + \log(8 / (0.03 \times 18)) = 6.1 + 1.17 = 7.27$ , just above the patient's measured value, reflecting partial respiratory compensation. The patient's body is hyperventilating to blow off $CO_2$ and raise the ratio, but renal bicarbonate reserves are exhausted by ketoacid load. Treatment with intravenous saline plus insulin restores the ratio over 8 to 12 hours.

Example 2. Aquaculture water buffering at the Port Stephens prawn farm. Black Tiger prawn larvae from NSW DPI hatcheries require pond water held at pH 8.0 to 8.3. Operators buffer ponds with a sodium carbonate / sodium bicarbonate system, the $CO_3^{2-}$ / $HCO_3^-$ pair with $pK_a$ near 10.3. The Henderson-Hasselbalch ratio at pH 8.2 is therefore $10^{8.2 - 10.3} = 10^{-2.1} = 0.0079$ , so $[HCO_3^-]$ overwhelms $[CO_3^{2-}]$ . Adding 50 g of $Na_2CO_3$ per 1000 L raises buffering capacity without spiking pH. Without the buffer, algal photosynthesis during the day would push pond pH above 9.0 and dissolve larval shells.

Try this

Q1. State the Henderson-Hasselbalch equation and explain what it predicts for a buffer with equal concentrations of weak acid and conjugate base. [3 marks]

Cue. $pH = pK_a + \log([A^-] / [HA])$ ; when concentrations are equal the log is 0 and $pH = pK_a$ .

Q2. Calculate the volume of 0.10 mol L $^{-1}$ HCl that can be added to 250 mL of a phosphate buffer (containing 0.050 mol $HPO_4^{2-}$ and 0.050 mol $H_2PO_4^-$ ) before the pH drops by more than 0.1 unit. [3 marks]

Cue. Use Henderson-Hasselbalch: the ratio change $\log(0.045/0.055) = -0.087$ , so adding about 0.005 mol HCl corresponds to 50 mL of 0.10 mol L $^{-1}$ HCl.

Q3. Describe the bicarbonate buffer system and its compensation mechanisms. (a) Write the buffer equilibrium. (b) Explain respiratory compensation. (c) Explain renal compensation. [1+2+2 marks]

Cue. (a) $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$ . (b) Breathing rate adjusts $CO_2$ and hence $H_2CO_3$ within minutes. (c) Kidneys reabsorb or excrete $HCO_3^-$ over hours to days.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC5 marksThe bicarbonate buffer in human blood is described by the equilibrium CO2(g) + H2O(l) <-> H2CO3(aq) <-> H+(aq) + HCO3-(aq). Normal arterial blood has pH 7.40, [HCO3-] = 24 mmol/L, and dissolved [H2CO3] = 1.2 mmol/L. (a) Verify the pH using the Henderson-Hasselbalch equation, given pKa1 of carbonic acid = 6.10 at body temperature. (b) Explain what happens to the equilibrium when CO2 is exhaled more rapidly (hyperventilation), and how this affects blood pH.

Show worked answer →

A 5 mark answer needs the Henderson-Hasselbalch verification, the Le Chatelier shift on hyperventilation, and the pH consequence with direction.

(a) pH from Henderson-Hasselbalch.

pH = pK_a + \log_{10}\left(\frac{[HCO_3^-]}{[H_2CO_3]}\right) = 6.10 + \log_{10}\left(\frac{24}{1.2}\right) = 6.10 + \log_{10}(20) = 6.10 + 1.30 = 7.40

The calculated pH matches the measured arterial pH.

(b) Hyperventilation. Exhaling $CO_2$ faster removes the leftmost species in the equilibrium chain. By Le Chatelier, the equilibrium shifts left to replace $CO_2$ , consuming $H_2CO_3$ and (via the second step) consuming $H^+$ from $HCO_3^-$ . The ratio $[HCO_3^-]/[H_2CO_3]$ rises, so pH rises.

Quantitatively, if $[H_2CO_3]$ falls to 0.8 mmol/L while $[HCO_3^-]$ stays near 24 mmol/L momentarily, $pH = 6.10 + \log_{10}(30) = 7.58$ . This is respiratory alkalosis, a condition seen in panic attacks and at high altitude.

Markers reward (1) substitution into Henderson-Hasselbalch with correct logarithm, (2) the Le Chatelier shift on exhaling, (3) the direction of pH change with naming of alkalosis.

2018 HSC3 marksExplain how a buffer made from 0.10 mol/L ethanoic acid and 0.10 mol/L sodium ethanoate resists a small addition of strong acid. Include a balanced equation.

Show worked answer →

A buffer contains a weak acid ( $CH_3COOH$ ) and its conjugate base ( $CH_3COO^-$ ) in comparable amounts.

When a small amount of strong acid ( $H^+$ ) is added, the added $H^+$ is consumed by the conjugate base:

CH_3COO^-_{(aq)} + H^+_{(aq)} \rightarrow CH_3COOH_{(aq)}

The strong acid is converted to a weak acid, which only partly re-ionises. The added $H^+$ is therefore not all "free" in solution; most of it has been mopped up. The ratio $[CH_3COO^-]/[CH_3COOH]$ shifts slightly toward the acid side, so pH falls only marginally.

Markers reward (1) the correct weak-acid plus conjugate-base composition, (2) the reaction consuming added $H^+$ , (3) explaining that the strong acid is replaced by a much weaker one, so pH changes little.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksA 250 mL buffer is prepared from 3.00 g of ethanoic acid (

CH_3COOH

M = 60.05\ \text{g mol}^{-1}

K_a = 1.8 \times 10^{-5}

) and 4.10 g of sodium ethanoate (

CH_3COONa

M = 82.03\ \text{g mol}^{-1}

). Calculate the pH of the buffer, to 2 decimal places.

Show worked solution →

Step 1: find pKa from Ka.

pK_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74

Step 2: moles of each component.

n(CH_3COOH) = \frac{3.00\ \text{g}}{60.05\ \text{g mol}^{-1}} = 0.04996\ \text{mol}

n(CH_3COO^-) = \frac{4.10\ \text{g}}{82.03\ \text{g mol}^{-1}} = 0.04998\ \text{mol}

Step 3: ratio of conjugate base to acid. Both are in the same 250 mL volume, so volume cancels and the mole ratio equals the concentration ratio:

\frac{[A^-]}{[HA]} = \frac{0.04998}{0.04996} = 1.000

Step 4: apply Henderson-Hasselbalch.

pH = pK_a + \log_{10}(1.000) = 4.74 + 0 = 4.74

Marking criteria: 1 mark for correctly finding pKa from Ka, 1 mark for correct moles of both components, 1 mark for the correct pH to 2 decimal places (4.74), recognising the near-1:1 ratio makes pH essentially equal to pKa.

foundation3 marksExplain, using an equation, how the bicarbonate buffer resists a small addition of metabolic acid (extra

H^+

) in the blood.

Show worked solution →

The conjugate base of the bicarbonate pair, $HCO_3^-$ , consumes the added $H^+$ :

HCO_{3(aq)}^- + H^+_{(aq)} \rightarrow H_2CO_{3(aq)}

The strong acid's $H^+$ is converted into the weak acid $H_2CO_3$ , which only partly re-ionises, so free $[H^+]$ rises only slightly and pH falls only marginally rather than sharply.

Marking criteria: 1 mark for identifying $HCO_3^-$ as the component that reacts, 1 mark for a correctly balanced equation, 1 mark for explaining why converting strong acid to a weak acid limits the pH change.

core4 marksThe buffer-response curve below plots blood pH against the ratio [HCO3-]/[H2CO3] on a logarithmic axis, with pKa1 = 6.10. Using the curve, (a) estimate the pH at ratio 1:1, and (b) explain why the curve is much steeper near ratio 100:1 than near ratio 5:1.

Show worked solution →

(a) Reading the curve at ratio 1:1. The curve passes through pH 6.10 at the marked ratio 1:1 point, matching $pH = pK_{a1}$ exactly (since $\log_{10}(1) = 0$ ).

(b) Why the curve steepens near ratio 100:1. Near ratio 100:1, the weak acid component ( $H_2CO_3$ ) is nearly used up relative to the conjugate base, so the buffer has little capacity left to absorb further base; a small further shift in the ratio produces a much larger change in $\log_{10}(\text{ratio})$ and hence in pH. Near ratio 5:1, both components remain present in comparable amounts, so the buffer still has plenty of capacity and the curve stays shallow.

Marking criteria: 1 mark for reading pH = 6.10 (= pKa1) at ratio 1:1, 1 mark for linking this to $\log_{10}(1) = 0$ , 1 mark for identifying that one component is nearly exhausted near ratio 100:1, 1 mark for explaining this in terms of buffer capacity/the logarithmic relationship rather than just asserting "it gets steeper".

core5 marksA patient in diabetic ketoacidosis has arterial [HCO3-] = 10.0 mmol/L and pCO2 = 24 mmHg, with dissolved [H2CO3] related to pCO2 by [H2CO3] = 0.030 x pCO2 (mmol/L, mmHg). Using pKa1 = 6.10, calculate the predicted arterial pH to 2 decimal places, and state whether this represents acidosis, alkalosis, or normal pH.

Show worked solution →

Step 1: find [H2CO3] from pCO2.

[H_2CO_3] = 0.030 \times 24 = 0.72\ \text{mmol/L}

Step 2: apply Henderson-Hasselbalch.

pH = pK_{a1} + \log_{10}\left(\frac{[HCO_3^-]}{[H_2CO_3]}\right) = 6.10 + \log_{10}\left(\frac{10.0}{0.72}\right)

= 6.10 + \log_{10}(13.89) = 6.10 + 1.14 = 7.24

Step 3: classify. Normal arterial pH is $7.40 \pm 0.05$ (i.e. 7.35 to 7.45). A calculated pH of 7.24 is below 7.35, so this is acidosis. Since [HCO3-] is low (bicarbonate has been consumed buffering the excess ketoacids), the primary disturbance is metabolic acidosis, with partial respiratory compensation shown by the low pCO2 (hyperventilation blowing off CO2 to raise the ratio back towards normal).

Marking criteria: 1 mark for correctly converting pCO2 to [H2CO3], 1 mark for correct substitution into Henderson-Hasselbalch, 1 mark for the correct pH to 2 decimal places (7.24), 1 mark for correctly classifying this as acidosis against the 7.35 to 7.45 reference range, 1 mark for identifying the metabolic origin with respiratory compensation.

exam6 marksAssess the effectiveness of the bicarbonate buffer system, compared with a simple laboratory acetate buffer, at maintaining a stable pH in the human body during vigorous exercise.

Show worked solution →

This is a 6-mark ASSESS: markers reward a judgement supported by contrasted evidence, not just a description of each buffer.

Band 6 PLAN.

Thesis: the bicarbonate buffer is far more effective than a closed acetate buffer for the body because it is an OPEN system, continuously regulated by two independent organs, even though its chemical ratio (20:1) is less favourable on paper than an ideal 1:1 buffer.
Chemistry of the challenge: vigorous exercise produces lactic acid and increased CO2 from cellular respiration, both adding H+ (directly, and via the CO2/H2CO3 equilibrium) faster than a fixed-composition buffer could absorb.
Acetate buffer (closed): a beaker of ethanoic acid/ethanoate can only absorb H+ until one component is consumed; once used up, it cannot regenerate and pH will fall sharply, per the steep region of a Henderson-Hasselbalch buffer-response curve.
Bicarbonate buffer (open): breathing rate increases within seconds to expel excess CO2 (shifting the equilibrium left, consuming H+, restoring the ratio), while the kidneys make a slower, sustained adjustment to HCO3- over hours. This means the "used" component of the buffer is constantly resupplied or removed, unlike a sealed system.
Judgement: despite a chemically "worse" 20:1 ratio (further from the ideal 1:1 buffer optimum), the bicarbonate system is more effective in the body specifically because physiological regulation, not the intrinsic chemistry, gives it renewable buffer capacity that a closed acetate buffer cannot match.

Model paragraph (excerpt). Although a 1:1 acetate buffer sits exactly at its pKa and would, in an isolated beaker, resist pH change more efficiently per mole than the 20:1 bicarbonate system, this comparison breaks down under a sustained acid load such as vigorous exercise. A fixed acetate buffer has a finite capacity: once the conjugate base is consumed neutralising incoming H+, no more remains and pH falls steeply, following the sharply increasing region of a Henderson-Hasselbalch buffer-response curve. The bicarbonate buffer avoids this because it is open at both ends of its equilibrium: as lactic acid and CO2 raise [H+], the equilibrium $CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$ shifts right, but increased ventilation immediately vents the CO2 being generated, preventing $H_2CO_3$ from accumulating, while the kidneys make a slower correction to [HCO3-] over the following hours. The buffer's "capacity" is therefore continuously renewed rather than fixed, which is why blood pH typically changes by only hundredths of a unit during intense exercise despite a large acid load.

Marker's note: top-band answers (1) explicitly contrast OPEN versus CLOSED buffer behaviour rather than just describing each system, (2) use the correct equilibrium and both compensation mechanisms with their characteristic timescales, (3) acknowledge the apparent paradox that a 20:1 ratio is chemically further from ideal than 1:1 yet physiologically more effective, and (4) end with an explicit judgement, not a neutral summary.