Diluting a weak acid with water usually raises its pH even though a larger fraction of the acid molecules becomes ionised. This sounds contradictory until we separate two quantities: the percentage of molecules that dissociate and the concentration of H₃O⁺ ions present. When a weak acid is diluted, its degree of ionisation can increase, but the total amount of acid per unit volume decreases enough that [H₃O⁺] ordinarily falls. The pH therefore rises.
This is a powerful JC Chemistry diagnostic for anyone looking for an H2 Chemistry tutor in Hougang, weak-acid Ka calculations, dilution and pH worked examples, ethanoic acid dissociation questions, degree of ionisation explanations or 9476 A-Level Chemistry equilibrium tutorials. A student may know that dilution shifts a weak-acid equilibrium toward ionisation, yet still predict the wrong pH change. The problem is not a missing formula; it is that two different kinds of measurement have been confused.
I like to start by asking, ‘More of which thing?’ More ionised molecules as a fraction of the original acid? More H₃O⁺ ions per litre? A stronger acid? Those claims are not equivalent. Once the student names the quantity, the equilibrium and the mathematics begin to agree with each other.
Hougang JC Chemistry learning library · Official SEAB H2 Chemistry 9476 syllabus
The reaction behind a weak acid in water
Let HA represent a generic monoprotic weak acid. Its equilibrium in water is:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq).
The familiar acid dissociation constant is:
Ka = [H₃O⁺][A⁻]/[HA].
For the school-level expression, the activity of pure liquid water is incorporated into the constant and dilute-solution concentrations are used as approximations. Ka is characteristic of the acid at the specified temperature. The value of Ka does not become larger merely because the acid solution is diluted.
The acid molecules exist in a dynamic equilibrium between undissociated HA and the ions H₃O⁺ and A⁻. A ‘weak acid’ is weak because its ionisation is incomplete under typical conditions, not because each individual acid molecule has only a small amount of acidity once it donates a proton.
In ethanoic acid, HA is CH₃COOH and A⁻ is CH₃COO⁻. The chemical reaction is:
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq).
The word ‘weak’ refers to its equilibrium behaviour, not to whether its solution is concentrated, safe to handle or harmless. A concentrated weak acid can still be hazardous and cause harm.
The key distinction: fraction ionised versus concentration of H₃O⁺
Suppose we have 100 acid molecules and, in a simplified picture, one has donated a proton. The degree of ionisation is 1%. If after dilution the equilibrium adjusts so that four out of those 100 molecules donate a proton, the degree of ionisation becomes 4%.
But now imagine we have increased the solution volume tenfold. The concentration of all acid-derived species per unit volume is much lower. Four ionised molecules dispersed in ten times the volume need not give a higher H₃O⁺ concentration than one ionised molecule in the smaller starting volume.
The distinction is the difference between a proportion and an amount per unit volume. pH depends on hydronium activity, commonly approximated by [H₃O⁺] in school calculations. It does not directly depend on the percentage of acid molecules that are ionised.
I ask students to write that distinction in their own words before calculating. It prevents a surprisingly persistent error.
Why dilution favours a greater degree of ionisation
For HA ⇌ H₃O⁺ + A⁻, the products contain two independently dissolved ionic species while the reactant side has one dissolved acid species in the simplified stoichiometric representation. Dilution changes the concentration terms and the equilibrium re-establishes itself at the same Ka.
Under the familiar weak-acid model, the ratio of ionised to unionised acid generally increases as the analytical concentration falls. This is reflected mathematically by an increasing degree of ionisation.
It is tempting to say ‘more ions form, so the acid must become more acidic’. That second clause does not follow. More as a percentage does not mean more per cubic decimetre. The volume change must be accounted for.
The most reliable way to see this is to calculate an actual example and compare both quantities side by side.
Worked example 1: ethanoic acid at 0.100 mol dm⁻³
Let Ka(CH₃COOH) = 1.8 × 10⁻⁵ at approximately 25 °C for this illustrative calculation. Consider a solution with initial analytical concentration C = 0.100 mol dm⁻³.
Assume the usual monoprotic weak-acid model and initially neglect the very small hydronium contribution from water. Let x be the equilibrium concentration of H₃O⁺ formed by the weak acid.
The equilibrium concentrations are:
[HA] = 0.100 − x.
[H₃O⁺] ≈ [A⁻] = x.
The dissociation expression becomes:
1.8 × 10⁻⁵ = x²/(0.100 − x).
Since x is small compared with 0.100, use the standard approximation 0.100 − x ≈ 0.100:
x ≈ √(Ka × C) = √(1.8 × 10⁻⁵ × 0.100) ≈ 1.34 × 10⁻³ mol dm⁻³.
The estimated pH is:
pH ≈ −log₁₀(1.34 × 10⁻³) ≈ 2.87–2.88.
The degree of ionisation is approximately:
α ≈ x/C = 0.00134/0.100 ≈ 0.0134 = 1.34%.
Because x/C is only about 1.3%, the simple approximation is reasonable for the requested precision. An exact quadratic solution gives a slightly different value but does not alter the key conclusion.
Worked example 2: dilute the same acid tenfold
Now reduce the analytical concentration from 0.100 to 0.0100 mol dm⁻³ by dilution with water at the same temperature. Ka remains 1.8 × 10⁻⁵.
Using the same approximate model:
[H₃O⁺] ≈ √(1.8 × 10⁻⁵ × 0.0100) = √(1.8 × 10⁻⁷) ≈ 4.24 × 10⁻⁴ mol dm⁻³.
The estimated pH is:
pH ≈ −log₁₀(4.24 × 10⁻⁴) ≈ 3.37.
The degree of ionisation becomes approximately:
α ≈ 0.000424/0.0100 = 0.0424 = 4.24%.
Here is the result to underline: the degree of ionisation has risen from roughly 1.34% to roughly 4.24%, while the hydronium concentration has dropped from about 1.34 × 10⁻³ to about 4.24 × 10⁻⁴ mol dm⁻³. Since [H₃O⁺] has decreased, the pH has increased.
At the more dilute concentration, the approximation is near the commonly used 5% ionisation rule. An exact calculation would give slightly different values. It is sensible to check whether the approximation is adequate when a question requires high precision.
Use a side-by-side table to see both changes
| Quantity | Before dilution | After tenfold dilution |
|---|---|---|
| Analytical [CH₃COOH] at preparation | 0.100 mol dm⁻³ | 0.0100 mol dm⁻³ |
| Ka at fixed temperature | 1.8 × 10⁻⁵ | 1.8 × 10⁻⁵ |
| Approximate equilibrium [H₃O⁺] | 1.34 × 10⁻³ mol dm⁻³ | 4.24 × 10⁻⁴ mol dm⁻³ |
| Approximate pH | 2.87–2.88 | 3.37 |
| Approximate degree of ionisation | 1.34% | 4.24% |
| Direction of change | Starting state | Higher percentage ionised, lower [H₃O⁺], higher pH |
The exact numbers depend on the stipulated Ka, concentration and model. The key trend is the important lesson: ionisation fraction and acid-ion concentration answer different questions.
A more accurate check with the quadratic equation
The weak-acid approximation starts with Ka = x²/(C − x). If the degree of ionisation is not negligible, multiplying and rearranging gives:
x² + Ka x − Ka C = 0.
The physically meaningful positive solution is:
x = [−Ka + √(Ka² + 4KaC)]/2.
For C = 0.100 mol dm⁻³ and Ka = 1.8 × 10⁻⁵, this gives x ≈ 1.33 × 10⁻³ mol dm⁻³.
For C = 0.0100 mol dm⁻³, it gives x ≈ 4.15 × 10⁻⁴ mol dm⁻³. The corresponding pH is about 3.38 and the degree of ionisation about 4.15%.
The exact version makes the same prediction. It also demonstrates why a chemist should check when an approximation is valid rather than trust the square-root formula in every setting.
A useful shortcut: why tenfold dilution increases weak-acid pH by about 0.5
Where the weak-acid approximation is valid, [H₃O⁺] ≈ √(KaC). At constant temperature, Ka is constant, so [H₃O⁺] is proportional to √C.
If we dilute the acid tenfold, C is divided by 10. Hydronium concentration is then divided by approximately √10, about 3.16, rather than by exactly 10.
Taking negative logarithms means that pH rises by approximately:
log₁₀(√10) = 0.5 pH unit.
That is why the illustrative change from about pH 2.88 to pH 3.38 is roughly half a pH unit. This is a useful pattern, not an exact universal law. It depends on the weak-acid approximation, consistent temperature and neglect of water’s autoionisation. At much lower concentrations or larger fractions of dissociation, the simple 0.5 rule becomes less accurate.
How a strong acid behaves differently
For a fully dissociated monoprotic strong acid in the familiar concentration range, the acid-derived hydronium concentration is approximately equal to the analytical acid concentration. Diluting it tenfold therefore reduces the hydronium concentration about tenfold and raises its pH by about 1.0 unit.
For a weak acid in the simple approximation, tenfold dilution raises pH by about 0.5 unit because the degree of ionisation increases as the solution is diluted.
This contrast is useful only when the strong-acid assumptions, weak-acid assumptions and concentration ranges are valid. At extreme dilution, water’s own ions become significant, and neither classroom shortcut should be used mechanically.
Also remember that acid strength refers to ionisation equilibrium, whereas acid concentration refers to the amount dissolved per unit volume. A dilute strong acid and a concentrated weak acid can have very different pH values depending on the numbers.
Does a weak acid become stronger when diluted?
No. Its Ka at a specified temperature does not increase just because more water has been added. A larger fraction of molecules becomes ionised, but that is a change in the equilibrium composition at the new concentration.
I sometimes hear a student say, ‘It is stronger because more of it dissociates.’ This confuses the fraction dissociated at the current concentration with the intrinsic acid-strength parameter. A more accurate sentence is: the degree of ionisation increases upon dilution while Ka remains constant at the given temperature.
If temperature changes, the equilibrium constant can change. If activity effects are significant, concentration-based approximations need care. But in the standard fixed-temperature aqueous question, dilution itself is not a reason to write a new Ka.
Can the total amount of ionised acid increase even if [H₃O⁺] falls?
Yes, and this is the part that makes the apparent paradox most interesting.
Suppose we take 100.0 cm³ of the original 0.100 mol dm⁻³ ethanoic acid. It contains an analytical total of 0.0100 mol of ethanoic acid species. With approximately 1.34% ionisation, the initial amount of ionised acid is roughly 0.000134 mol.
Now dilute that same amount of acid to a total volume of 1.00 dm³. Its analytical concentration becomes 0.0100 mol dm⁻³. At approximately 4.15% ionisation, about 0.000415 mol of acid is ionised.
So the absolute number of ionised acid molecules in this particular fixed sample can increase while the hydronium concentration falls, because those ions occupy a much larger volume. The approximate calculations assume no loss of acid and use the same temperature.
That is the conceptual centre of the entire topic: number, fraction and concentration are three different measurements.
Why extreme dilution needs an additional caution
If a student carries the weak-acid square-root approximation down to fantastically low concentrations, they may predict a pH that no longer makes physical sense for an ordinary pure weak-acid solution in water. At those concentrations, the contribution from water autoionisation can become comparable to or larger than the acid-derived hydronium concentration.
At 25 °C, pure neutral water has approximately pH 7. A sufficiently diluted solution of an ordinary nonreacting acid approaches the neutral-water condition from the acidic side, rather than turning alkaline merely because the acid is diluted.
Real laboratory water may contain dissolved carbon dioxide, impurities or buffering species, so measured pH near neutrality requires careful apparatus and sample handling. In an A-Level question, check the given conditions and whether the approximation is justified instead of extending it indefinitely.
This limitation is not usually the first thing a JC1 student needs to memorise; it is a useful reminder that all simplified models have a domain where they work.
How this differs from diluting a buffer solution
A buffer contains appreciable concentrations of a conjugate acid–base pair, commonly HA and A⁻. In the Henderson–Hasselbalch approximation, pH depends on pKa plus the logarithm of the ratio [A⁻]/[HA].
If an ideal buffer is diluted without changing that ratio, its pH may change relatively little while its capacity to resist added acid or base falls. That is different from diluting a solution containing only the weak acid as its primary solute.
Students sometimes borrow the buffer dilution rule to answer a weak-acid-alone question. They should first identify what species were actually placed in the beaker. Was a significant concentration of conjugate base added? If not, it is not automatically a conventional buffer.
What the question wording can reveal
| Question says | Quantity being discussed | What to check |
|---|---|---|
| Degree of dissociation increases | Fraction of total acid molecules ionised | Calculate α = x/C |
| pH increases | Hydronium activity or its concentration approximation decreases | Evaluate −log[H₃O⁺] |
| Ka is unchanged | Acid equilibrium constant at fixed temperature | Do not confuse dilution with temperature change |
| Weak acid becomes more ionised | Equilibrium composition shifts in relative terms | Ask whether the claim is a fraction or concentration |
| Total moles ionised increase | Absolute amount of dissociated species in the vessel | Include the vessel volume |
| Very dilute acid approaches neutral pH | Water autoionisation becomes important | Check the model’s limits |
I encourage students to underline ‘percentage’, ‘concentration’, ‘moles’ or ‘pH’ before solving. Often the calculation error happens because they answer a different quantity from the one actually requested.
Four changed questions to test transfer
- At fixed temperature, a weak acid is diluted tenfold. Does Ka change? Does the degree of ionisation generally increase or decrease?
- A weak acid has C = 0.0400 mol dm⁻³ and Ka = 1.0 × 10⁻⁵. Estimate [H₃O⁺] using the small-dissociation approximation.
- A strong monoprotic acid at 0.0100 mol dm⁻³ is diluted tenfold. What is its approximate pH change under the ideal complete-ionisation assumption at 25 °C?
- A weak-acid solution contains a greater fraction of dissociated molecules after dilution. Can its pH still become higher? Explain the difference between the two quantities.
Answers
1. Ka stays constant; the degree of ionisation generally increases. 2. √(1.0 × 10⁻⁵ × 0.0400) = √(4.0 × 10⁻⁷) ≈ 6.32 × 10⁻⁴ mol dm⁻³, giving pH ≈ 3.20 if the model applies. 3. Approximately +1.0 pH unit, from about pH 2 to pH 3. 4. Yes. The percentage ionised can rise while hydronium concentration per unit volume falls.
If a student gets Question 1 right but Question 4 wrong, they may have memorised the weak-acid dilution rule without understanding concentration. That is an opportunity for a short targeted explanation, not a reason to restart the entire acids-and-bases syllabus.
A six-session study plan for JC1 and JC2
- Write the weak-acid dissociation equilibrium and label HA, H₃O⁺ and A⁻.
- Calculate Ka from given equilibrium concentrations, then explain what each bracket represents.
- Solve one weak-acid pH problem using Ka and C with units displayed.
- Dilute the original solution tenfold and compare pH, [H₃O⁺] and percentage ionisation.
- Solve a changed problem that asks for total moles ionised after making the solution up to a larger volume.
- Contrast the result with a strong acid and then with a prepared conjugate-pair buffer.
A good JC2 revision session then mixes these with titration, salt hydrolysis and data-based reasoning so the chapter title no longer tells the student which formula to apply.
How parents in Hougang can evaluate tuition help
Bring a marked Ka or pH problem and ask the tutor to identify the first incorrect step. Did the student mistake acid strength for concentration, calculate a degree of ionisation as a concentration, apply a square-root approximation outside its valid range, or misuse the logarithm?
The right next practice question depends on that diagnosis. A three-student class can be effective when every learner explains their own independent numerical result; one-to-one tuition should provide the same direct feedback. Check current H2 syllabus expertise, actual lesson location, schedule and travel needs rather than assuming that a Hougang-targeted article proves a specific class is located in Hougang.
Why scientists need this distinction beyond the examination
In analytical chemistry, environmental testing, pharmaceuticals, food science and biochemical systems, the fraction of a substance present in a chemical form can be very different from that form’s concentration. Chemists must distinguish intrinsic equilibrium behaviour from sample concentration and measure the right quantity for a given decision.
Learning that habit now prepares students for more demanding university reasoning, whatever science or engineering course they later consider. Entry requirements differ by programme, and no tuition approach can promise admission, but the discipline of separating percentage, amount and concentration remains useful.
Frequently asked questions
Why does pH rise when a weak acid is diluted?
The hydronium concentration usually falls, even though the percentage of acid molecules ionised increases.
Does dilution increase Ka?
No. Ka is a property of the specified equilibrium at a given temperature, not a direct function of how much water is added.
Why is a weak acid more dissociated when diluted?
The equilibrium adjusts at the new lower analytical concentration, typically favouring a higher fraction of ionised molecules.
Does a tenfold dilution always raise weak-acid pH by 0.5?
Only approximately when the small-dissociation weak-acid model is valid. Use the exact equilibrium and water autoionisation if the assumptions fail.
Can dilution turn an ordinary acid solution alkaline?
Not merely by adding pure neutral water under the ideal acid-only model. At extreme dilution, pH approaches that of neutral water rather than crossing into alkalinity for that reason alone.
Is a weak acid solution the same as a buffer?
No. A useful buffer ordinarily contains substantial amounts of a weak acid and its conjugate base, or the analogous weak base/conjugate acid pair.
Continue the H2 Chemistry route
Why NH₄Cl is acidic and sodium ethanoate is alkaline · Why diluting a buffer affects capacity more than pH · Why titration equivalence-point pH may not be 7 · Hougang Chemistry learning library · eduKate Singapore tuition enquiries.
All numerical results use specified illustrative equilibrium constants and textbook dilute-solution assumptions. Students should use the temperature, data and approximation requirements provided in their examination questions.