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Inquiry Question 3: It is all about hydrogen ions

Distinguish between the strength and the concentration of acids and bases, including investigation of the degree of ionisation and the relationship between ionisation, conductivity, and pH

A focused answer to the HSC Chemistry Module 6 dot point on strength vs concentration. The degree of ionisation, Ka and Kb values, conductivity comparison, pH at equal concentration, and worked HSC past exam questions.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-22

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

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What this dot point is asking
The answer
Examples in context
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In plain English

Imagine two doors into a room: one is unlocked and swings wide open every time someone pushes it (that is a strong acid, essentially 100 percent ionised), and the other has a stiff spring that only lets a few people through even when a whole crowd pushes on it (that is a weak acid, only a small percent ionised). Now imagine you can also change how many people are queued outside each door: that is the concentration, completely separate from how "open" the door is. A small queue at the unlocked door can let through the same number of people as a huge queue at the stiff door, which is exactly why a dilute strong acid can have the same pH as a concentrated weak acid. Strength is about the door mechanism (an intrinsic, fixed property of that acid); concentration is about the queue size (how much you dissolved). Mixing the two ideas up, calling a weak acid "just a dilute acid", is the single most common mistake in this dot point.

What this dot point is asking

NESA wants you to clearly distinguish two ideas that students often blur: the strength of an acid (its degree of ionisation, an intrinsic property) and the concentration of an acid (how much is dissolved, an extrinsic property). You should be able to compare pH, conductivity and reactivity of strong and weak acids at equal concentration, and explain everything in terms of the position of the ionisation equilibrium. This builds on properties of acids and bases and is the prerequisite for pH calculations and titration curve shapes.

The answer

Strength vs concentration

Strength describes the degree of ionisation in water. A strong acid ionises essentially 100 percent. A weak acid ionises only a few percent (often less than 1 percent at typical concentrations).
Concentration describes how much acid is dissolved per litre, regardless of how much has ionised.

These properties are independent: you can have a dilute strong acid or a concentrated weak acid.

Strong acids and bases

Strong acids are essentially fully ionised in water. The common examples for HSC:

$HCl$ , $HBr$ , $HI$ (hydrohalic acids except $HF$ , which is weak).
$HNO_3$ (nitric acid).
$H_2SO_4$ (sulfuric acid, fully ionised for the first proton, partially for the second).
$HClO_4$ (perchloric acid).

Strong bases are also fully ionised:

Group 1 hydroxides: $NaOH$ , $KOH$ .
Heavier Group 2 hydroxides: $Ca(OH)_2$ , $Ba(OH)_2$ .

Weak acids and bases

A weak acid only partially ionises, setting up an equilibrium:

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

The acid dissociation constant ( $K_a$ ) measures the position of this equilibrium:

K_a = \frac{[H^+][A^-]}{[HA]}

A small $K_a$ means a weak acid (very little ionised); a large $K_a$ means a stronger acid.

Acid	Formula	$K_a$ at 25 degrees C	Strength
Perchloric	$HClO_4$	very large	strong
Hydrochloric	$HCl$	$\sim 10^6$	strong
Sulfuric (1st)	$H_2SO_4$	very large	strong
Hydrofluoric	$HF$	$6.8 \times 10^{-4}$	weak
Methanoic	$HCOOH$	$1.8 \times 10^{-4}$	weak
Ethanoic	$CH_3COOH$	$1.8 \times 10^{-5}$	weak
Carbonic (1st)	$H_2CO_3$	$4.3 \times 10^{-7}$	weak

For weak bases the equivalent constant is $K_b$ . For ammonia, $K_b = 1.8 \times 10^{-5}$ .

Strength versus concentration checkpoints

Strength = degree of ionisation, measured by $K_a$ (or $K_b$ ): an intrinsic, fixed property of the acid or base that does NOT change with dilution.
Concentration = mol/L dissolved: an extrinsic, variable quantity independent of strength.
Strong acid/base: essentially 100 percent ionised, so $[H^+]$ (or $[OH^-]$ ) = the stated concentration.
Weak acid: $[H^+] \approx \sqrt{K_a \cdot c}$ ; percent ionisation $= \dfrac{[H^+]}{c_0} \times 100$ .
Conductivity tracks the TOTAL ion concentration, not the nominal acid/base concentration.
Dilution always increases the PERCENT ionisation of a weak acid or base, even though $K_a$ itself stays constant at a fixed temperature.

Conductivity

Electrical conductivity of a solution depends on the total concentration of ions. At equal acid concentration, a strong acid produces many more ions than a weak acid, so it conducts much better.

This is a definitive HSC experiment: dip a conductivity probe (or a small bulb circuit) into 0.1 mol/L $HCl$ and 0.1 mol/L $CH_3COOH$ . The $HCl$ lights the bulb brightly; the ethanoic acid lights it dimly. Both have the same nominal concentration; only the degree of ionisation differs.

pH at equal concentration

For two solutions of equal concentration:

Strong acid has the lower pH (more ions, lower $[H^+]$ value... wait, higher $[H^+]$ , so lower pH).
Weak acid has a higher pH (less ionised, lower $[H^+]$ ).

For a 0.10 mol/L solution:

$HCl$ : $[H^+] = 0.10$ mol/L, pH = 1.00.
$CH_3COOH$ ( $K_a = 1.8 \times 10^{-5}$ ): $[H^+] \approx 1.34 \times 10^{-3}$ mol/L, pH = 2.87.

Reactivity at equal concentration

A strong acid reacts faster initially because more $H^+$ is present at any instant. However, a weak acid reacts to the same final extent with a stoichiometric excess of, for example, magnesium, because as $H^+$ is consumed the equilibrium shifts right (Le Chatelier) and more weak acid ionises.

Mark schemes reward students who explicitly separate kinetics (initial rate, set by $[H^+]$ ) from stoichiometry (total Mg consumed, set by moles of acid).

Dilution effect on degree of ionisation

When a weak acid is diluted, the percent ionisation increases. Mathematically, if $K_a$ is fixed and $c$ decreases, then $x/c = \sqrt{K_a/c}$ grows. In the limit of infinite dilution every weak acid becomes 100 percent ionised. This is a Le Chatelier consequence: dilution favours the side with more particles.

Worked example

A student is given an unknown 0.10 mol/L acid solution. The pH is measured to be 3.40. Is the acid strong or weak? Justify quantitatively.

Step 1: Calculate $[H^+]$ .

[H^+] = 10^{-3.40} = 3.98 \times 10^{-4} \text{ mol/L}

Step 2: Compare to concentration.

If the acid were strong, $[H^+]$ would equal 0.10 mol/L (pH = 1.00). The measured $[H^+]$ is much smaller, so the acid is weak.

Step 3: Calculate the percent ionisation.

\text{ionisation} = \frac{[H^+]_{\text{eq}}}{c_0} \times 100 = \frac{3.98 \times 10^{-4}}{0.10} \times 100 = 0.40 \text{ percent}

Step 4: Estimate $K_a$ .

K_a \approx \frac{[H^+]^2}{c_0 - [H^+]} \approx \frac{(3.98 \times 10^{-4})^2}{0.10} = 1.58 \times 10^{-6}

This is consistent with a weak acid (small $K_a$ ).

Exam technique for strength versus concentration questions

When a question gives you a pH and an initial concentration, ALWAYS calculate $[H^+]$ first ( $10^{-pH}$ ) and compare it to the stated concentration: if they are (approximately) equal, the acid is strong; if $[H^+]$ is much smaller, the acid is weak. Never answer "strong" or "weak" from the word "concentrated" or "dilute" alone, since concentration and strength are independent variables. In an extended response, explicitly name BOTH quantities you are comparing (e.g. "0.10 mol/L is the concentration; $1.8 \times 10^{-5}$ is $K_a$ , which measures the strength") so the marker can see you have not conflated them. For a calculation, always show the percent-ionisation or $K_a$ working with units, not just the final verdict.

Common traps

Confusing strong with concentrated: "Strong" refers to degree of ionisation; "concentrated" refers to mol/L. Both terms describe the acid, but they describe different things. Concentrated and dilute apply equally to strong and weak.
Stating that a weak acid is "less reactive" without qualification: It is less reactive initially per mole of dissolved acid. Given time and excess reactant, the total chemistry can be the same.
Forgetting that diluting a weak acid increases percent ionisation: Dilution does not turn a weak acid into a strong one, but it does increase the fraction ionised.
Ignoring the second ionisation of $H_2SO_4$: For dilute solutions HSC typically treats both protons as fully ionised. For concentrated solutions the second $K_a$ matters. State your assumption.
Reading too much from low conductivity: Pure water has very low conductivity but is not "weakly acidic" (its $[H^+] = 10^{-7}$ comes from auto-ionisation). Always relate conductivity to total ion concentration, not to acidity alone.

Examples in context

Example 1. Cleaning toilets: HCl-based versus citric-acid-based products. Domestic toilet cleaners sold across NSW supermarkets split into two camps: hydrochloric acid (Harpic Power Plus, around 10 percent w/v) and citric acid (Earth Choice, around 5 percent w/v). At equimolar concentration HCl ionises fully and gives pH near 0.8, while citric acid ( $K_a = 7.4 \times 10^{-4}$ ) ionises only 4 percent and gives pH near 2.1. Conductivity meters reflect the same difference, with HCl solutions conducting more than 20 times better. The HCl product attacks limescale 30 times faster but also corrodes chrome, while the citric product is gentler and biodegradable. Strength versus concentration explains every consumer trade-off.

Example 2. NSW HSC depth study comparing acetic and hydrochloric acid. A typical Stage 6 lab task gives students 0.10 mol L $^{-1}$ acetic acid and 0.10 mol L $^{-1}$ HCl. Measurements at 25 degrees C consistently give pH 1.0 for HCl and pH 2.87 for acetic acid, conductivity 39 vs 1.3 mS cm $^{-1}$ , and reaction with magnesium ribbon producing visible hydrogen in 12 seconds for HCl but 90 seconds for acetic acid. Students compute the percent ionisation of acetic acid as $10^{-2.87} / 0.10 = 1.3$ percent, neatly demonstrating that strength and concentration are independent variables. The depth study is the canonical practical NESA examines.

Try this

Q1. Distinguish between the strength and concentration of an acid using the terms ionisation and moles per litre. [3 marks]

Cue. Strength: degree of ionisation (intrinsic property, measured by $K_a$ ); concentration: moles per litre (how much is present).

Q2. A 0.20 mol L $^{-1}$ solution of a weak acid HA has pH 3.40. Calculate the percent ionisation and the $K_a$ of HA. [3 marks]

Cue. $[H^+] = 10^{-3.40} = 4.0 \times 10^{-4}$ mol L $^{-1}$ ; percent ionisation = 0.20 percent; $K_a = (4.0 \times 10^{-4})^2 / 0.20 = 8.0 \times 10^{-7}$ .

Q3. Two solutions are tested with a conductivity probe: 0.10 mol L $^{-1}$ HCl reads 39 mS cm $^{-1}$ and 0.10 mol L $^{-1}$ $CH_3COOH$ reads 1.3 mS cm $^{-1}$ . (a) Account for the difference. (b) Predict the conductivity of 0.010 mol L $^{-1}$ HCl. (c) State whether reacting either acid with excess magnesium gives the same volume of $H_2$ at completion. [2+1+2 marks]

Cue. (a) HCl fully ionised, more ions; acetic acid only 1.3 percent ionised. (b) Roughly 4 mS cm $^{-1}$ (ten-fold dilution drops conductivity roughly proportionally). (c) Same volume eventually because both deliver the same total moles of $H^+$ .

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.

2020 HSC5 marksTwo solutions are prepared: 0.10 mol/L hydrochloric acid and 0.10 mol/L ethanoic acid. Compare the two solutions in terms of pH, electrical conductivity, and reactivity with magnesium metal. Justify each comparison using the concept of degree of ionisation.

Show worked answer →

A 5 mark answer needs the three comparisons, with justification linked to the degree of ionisation each time.

Degree of ionisation. Hydrochloric acid is a strong acid and ionises completely: $HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)}$ . Ethanoic acid is a weak acid and only partly ionises: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$ , with $K_a = 1.8 \times 10^{-5}$ .

At equal concentration (0.10 mol/L), the strong acid has $[H^+] = 0.10$ mol/L while the weak acid has $[H^+] \approx \sqrt{K_a \cdot c} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3}$ mol/L.

pH: Hydrochloric acid has pH = 1.00. Ethanoic acid has pH = 2.87. The strong acid has a lower pH because more of it has ionised at the same concentration.
Electrical conductivity: Conductivity depends on the total ion concentration. Hydrochloric acid has 0.20 mol/L of total ions (0.10 each of $H^+$ and $Cl^-$ ). Ethanoic acid has only about $2.7 \times 10^{-3}$ mol/L of ions. The hydrochloric acid conducts noticeably better (lights a bulb brightly; the ethanoic acid lights it dimly).
Reaction with Mg: Both produce hydrogen gas, but the strong acid reacts much faster initially because $[H^+]$ is higher. As the ethanoic acid reacts, the equilibrium shifts right (Le Chatelier responds to consumed $H^+$ ), so the weak acid eventually consumes the same amount of Mg per mole of acid, but more slowly.

Markers reward (1) explicit reference to degree of ionisation as the underlying cause, (2) a numerical or qualitative comparison for each of pH, conductivity and reaction rate, (3) the insight that initial rate and total amount reacted are different things (kinetics vs stoichiometry).

2018 HSC2 marksExplain why dilute hydrochloric acid and concentrated ethanoic acid can have the same pH despite one being a strong acid and the other a weak acid.

Show worked answer →

pH depends on $[H^+]$ , not directly on the acid concentration. Strong acids fully ionise, so $[H^+]$ equals the acid concentration. Weak acids only partly ionise, so $[H^+]$ is a small fraction of the acid concentration.

A dilute (low concentration) strong acid can have the same $[H^+]$ as a concentrated weak acid. For example, $10^{-3}$ mol/L $HCl$ (pH 3) has the same pH as roughly 0.05 mol/L $CH_3COOH$ ( $K_a = 1.8 \times 10^{-5}$ ), because both give $[H^+] \approx 10^{-3}$ mol/L.

Markers reward (1) the distinction between strength (degree of ionisation) and concentration (mol/L), (2) a numerical or worked illustration showing they can give the same pH.

Practice questions

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

foundation2 marksState whether each pair of terms describes the STRENGTH or the CONCENTRATION of an acid: (a) 'dilute' and 'concentrated'; (b) 'strong' and 'weak'. Explain your answer for (b) in one sentence.

Show worked solution →

(a) Dilute and concentrated describe concentration (mol/L dissolved).

(b) Strong and weak describe strength, the degree of ionisation: a strong acid or base ionises essentially completely in water, while a weak one only partially ionises, reaching an equilibrium with undissociated molecules present.

Marking criteria: 1 mark for correctly assigning (a) to concentration, 1 mark for correctly assigning (b) to strength with a one-sentence definition referencing degree of ionisation.

foundation3 marksClassify each acid as strong or weak and justify using its approximate

K_a

: (a) nitric acid,

HNO_3

; (b) hydrofluoric acid,

HF

K_a = 6.8 \times 10^{-4}

; (c) carbonic acid (first ionisation),

H_2CO_3

K_a = 4.3 \times 10^{-7}

Show worked solution →

(a) Nitric acid: Strong: $HNO_3$ ionises essentially completely in water (no meaningful $K_a$ value is quoted because the equilibrium lies fully to the right).
(b) Hydrofluoric acid: Weak: $K_a = 6.8 \times 10^{-4}$ is small (much less than 1), showing only a small fraction of $HF$ molecules ionise at equilibrium, despite $HF$ being a hydrohalic acid.
(c) Carbonic acid: Weak: $K_a = 4.3 \times 10^{-7}$ is even smaller than that of $HF$ , showing carbonic acid ionises to a lesser extent again.

Marking criteria: 1 mark per acid for the correct strong/weak classification with a justification that references the size of $K_a$ (or, for the strong acid, the absence of a meaningful equilibrium constant).

core6 marksA 0.25 mol L

^{-1}

solution of hydrofluoric acid (

K_a = 6.8 \times 10^{-4}

) is prepared. Calculate (a) the equilibrium

[H^+]

, (b) the pH to 2 decimal places, and (c) the percent ionisation to 3 significant figures.

Show worked solution →

Step 1: set up the equilibrium expression.

HF_{(aq)} \rightleftharpoons H^+_{(aq)} + F^-_{(aq)} \qquad K_a = \frac{[H^+][F^-]}{[HF]}

Let $[H^+] = [F^-] = x$ and $[HF]_{eq} = 0.25 - x$ .

Step 2: substitute and solve the quadratic (the 5 percent approximation is checked below and fails, so the full quadratic is used).

6.8 \times 10^{-4} = \frac{x^2}{0.25 - x}

x^2 + 6.8 \times 10^{-4}x - 1.70 \times 10^{-4} = 0

Using the quadratic formula $x = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$ , $b = 6.8 \times 10^{-4}$ , $c = -1.70 \times 10^{-4}$ :

x = \frac{-6.8 \times 10^{-4} + \sqrt{(6.8 \times 10^{-4})^2 + 4(1.70 \times 10^{-4})}}{2}

x = \frac{-6.8 \times 10^{-4} + \sqrt{4.62 \times 10^{-7} + 6.80 \times 10^{-4}}}{2} = \frac{-6.8 \times 10^{-4} + 0.02610}{2} = 0.01271\ \text{mol L}^{-1}

(a) Equilibrium $[H^+]$ .

[H^+] = 1.27 \times 10^{-2}\ \text{mol L}^{-1} \ (3 \text{ s.f.})

(b) pH.

pH = -\log_{10}[H^+] = -\log_{10}(0.01271) = 1.90 \ (2 \text{ d.p.})

(c) Percent ionisation.

\text{percent ionisation} = \frac{[H^+]_{eq}}{c_0} \times 100 = \frac{0.01271}{0.25} \times 100 = 5.08 \text{ percent} \ (3 \text{ s.f.})

Marking criteria: 1 mark for the correct equilibrium expression and ICE-style substitution, 1 mark for recognising the approximation is invalid and solving the quadratic (or an equivalent iterative method) correctly, 1 mark for the correct $[H^+]$ to 3 s.f. with units, 1 mark for the correct pH to 2 d.p., 1 mark for the correct percent ionisation to 3 s.f., 1 mark for showing every step with units. Note that using the simpler $[H^+] \approx \sqrt{K_a c}$ approximation here gives $[H^+] \approx 0.01304$ mol L $^{-1}$ , about 2.6 percent high, because $HF$ 's percent ionisation (about 5 percent) is right at the edge of where the approximation is considered valid; the full quadratic is the more defensible method for full marks.

core5 marksThe bar chart below is owned illustrative ExamExplained data comparing the measured conductivity of 0.10 mol L

^{-1}

HCl (39 mS cm

^{-1}

) and 0.10 mol L

^{-1}

CH_3COOH

(1.3 mS cm

^{-1}

) at 25 degrees C. (a) Explain the large difference in conductivity despite equal nominal concentration. (b) Predict, with reasoning, whether 0.10 mol L

^{-1}

NaOH

would have a conductivity closer to the HCl bar or the

CH_3COOH

bar.

Show worked solution →

(a) Explaining the difference. Conductivity depends on the TOTAL concentration of mobile ions, not on the nominal (stated) concentration of acid dissolved. $HCl$ is a strong acid and ionises essentially completely, so 0.10 mol L $^{-1}$ $HCl$ produces about 0.10 mol L $^{-1}$ each of $H^+$ and $Cl^-$ (0.20 mol L $^{-1}$ total ions). $CH_3COOH$ is a weak acid and only about 1.3 percent ionises at this concentration, so it produces only about $1.3 \times 10^{-3}$ mol L $^{-1}$ of ions, roughly 150 times fewer mobile charge carriers, giving far lower conductivity even though the acid concentration dissolved is identical.

(b) Prediction for $NaOH$ . Closer to the HCl bar. $NaOH$ is a strong base and, like $HCl$ , ionises essentially completely: $NaOH_{(aq)} \rightarrow Na^+_{(aq)} + OH^-_{(aq)}$ . A 0.10 mol L $^{-1}$ solution would therefore also produce about 0.20 mol L $^{-1}$ of total ions, giving a conductivity of a similar order of magnitude to the HCl bar (the exact value differs slightly because $Na^+$ and $OH^-$ have different ionic mobilities to $H^+$ and $Cl^-$ , but the qualitative comparison, strong electrolyte versus weak electrolyte, is what matters).

Marking criteria: (a) 1 mark for stating conductivity depends on total ion concentration, 1 mark for correctly quantifying/comparing the ion concentrations from each acid's degree of ionisation, 1 mark for linking this explicitly to the observed bar heights. (b) 1 mark for the correct prediction (closer to HCl), 1 mark for justifying via $NaOH$ being a strong, fully-ionising base rather than just asserting the answer.

exam7 marksThe graph below is owned illustrative ExamExplained data showing the percent ionisation of ethanoic acid (

K_a = 1.8 \times 10^{-5}

) as its concentration is decreased from 1.0 mol L

^{-1}

to 0.0001 mol L

^{-1}

on a logarithmic concentration axis. (a) Describe the trend shown. (b) Explain the trend using Le Chatelier's principle and the equilibrium expression for

K_a

. (c) Justify why the curve does NOT contradict the fact that

K_a

is constant at a fixed temperature.

Show worked solution →

(a) Description: As the concentration of ethanoic acid decreases (moving right along the log scale), the percent ionisation steadily increases, from about 0.42 percent at 1.0 mol L $^{-1}$ up to about 34.4 percent at 0.0001 mol L $^{-1}$ ; the curve rises more steeply at the more dilute end.
(b) Le Chatelier explanation: The ionisation equilibrium is $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$ , with one particle on the reactant side and two particles on the product side. Diluting the solution (adding water) reduces the concentration of every species; by Le Chatelier's principle, the equilibrium shifts to the side with MORE dissolved particles to partially counteract the change, which is the ionised (right-hand) side. This shift means a larger FRACTION of the original acid ends up ionised at equilibrium, even though the absolute equilibrium concentration of $H^+$ falls in dilute solution.
(c) Why $K_a$ stays constant: $K_a = \dfrac{[H^+][A^-]}{[HA]}$ is a ratio of equilibrium concentrations, and its numeric value depends only on temperature, not on concentration. As the solution is diluted, $[H^+]$ , $[A^-]$ and $[HA]$ all fall, but they fall in a way that keeps the RATIO $\dfrac{[H^+][A^-]}{[HA]}$ fixed at $1.8 \times 10^{-5}$ ; percent ionisation is a different quantity ( $[H^+]/c_0$ ), and it is entirely consistent for that ratio to increase with dilution while $K_a$ itself remains unchanged, because percent ionisation compares $[H^+]$ to the INITIAL concentration $c_0$ , not to the other equilibrium concentrations.

Marking criteria: (a) 1 mark for correctly describing the increasing trend with at least one numeric reference point. (b) 1 mark for correctly writing/using the ionisation equilibrium, 1 mark for identifying that dilution favours the side with more particles, 1 mark for linking this shift explicitly to increased percent ionisation. (c) 1 mark for stating $K_a$ depends only on temperature, 1 mark for distinguishing $K_a$ (an equilibrium ratio) from percent ionisation (a ratio to initial concentration) rather than treating them as the same quantity.

exam8 marksAssess the claim that 'a strong acid is always more dangerous or reactive than a weak acid' by comparing 0.10 mol L

^{-1}

hydrochloric acid with glacial (pure, highly concentrated) ethanoic acid, in terms of pH, total available

H^+

, and real-world hazard.

Show worked solution →

This is an 8-mark ASSESS: markers reward a judgement backed by a worked, two-sided comparison, not a one-line answer.

Band 6 PLAN.

Thesis: the claim is FALSE as a blanket statement; strength (degree of ionisation) and concentration jointly determine hazard, so a dilute strong acid can be less dangerous than a concentrated weak acid, even though the strong acid ionises more completely.
Numerical comparison 1 (dilute strong acid): 0.10 mol L $^{-1}$ $HCl$ is essentially 100 percent ionised, giving $[H^+] = 0.10$ mol L $^{-1}$ and pH = 1.00.
Numerical comparison 2 (concentrated weak acid): glacial ethanoic acid is close to pure liquid ethanoic acid, roughly 17.4 mol L $^{-1}$ before dilution. Even at only about 0.4 percent ionisation (extrapolating the dilution trend to a very concentrated weak acid), $[H^+] \approx 0.004 \times 17.4 \approx 0.07$ mol L $^{-1}$ , giving a pH in a similar low range to the dilute strong acid, and a much larger TOTAL moles of $HA$ available to react further as $H^+$ is consumed (Le Chatelier keeps replenishing $H^+$ from the huge undissociated reserve).
Real-world hazard evidence: glacial acetic acid is corrosive and causes severe skin and eye burns, and is classified as a hazardous corrosive substance in SDS documentation, comparable in practical danger to dilute mineral acids despite being "weak"; meanwhile very dilute strong acid (e.g. 0.001 mol L $^{-1}$ HCl, pH 3) is roughly as mild as vinegar and poses little hazard.
Judgement: the claim conflates strength with danger. Real hazard depends on the ACTUAL $[H^+]$ delivered (a function of BOTH strength and concentration) and on the total reactive reserve (moles of acid present), not on the strong/weak label alone; a rigorous assessment must therefore reject the blanket claim and instead evaluate strength and concentration together, case by case.

Model paragraph (excerpt). The claim that a strong acid is always more dangerous than a weak acid fails to withstand scrutiny once concentration is held fixed as an independent variable. A dilute 0.10 mol L $^{-1}$ solution of hydrochloric acid, though 100 percent ionised, delivers only 0.10 mol L $^{-1}$ of $H^+$ and a correspondingly modest total reactive capacity, whereas glacial ethanoic acid, though ionising only a small percentage, is present at such a high total concentration (approximately 17.4 mol L $^{-1}$ undiluted) that its equilibrium continually replenishes $H^+$ as it reacts, driven by Le Chatelier's principle, giving it comparable corrosive hazard in practice; safety data sheets for glacial acetic acid confirm this, classifying it as severely corrosive to skin and eyes. The claim is therefore an oversimplification: hazard is a function of the delivered $[H^+]$ and the reactive reserve of acid present, which requires strength AND concentration to be considered jointly, not strength in isolation.

Marker's note: top-band answers (1) use actual numbers for both solutions rather than vague comparison, (2) explicitly invoke Le Chatelier to explain why a weak acid's reserve keeps supplying $H^+$ during reaction, (3) bring in a real-world hazard reference (SDS classification or an equivalent) rather than asserting danger without evidence, and (4) end with an explicit judgement that rejects the blanket claim, rather than a neutral "it depends" with no resolution.