JC2 Chemistry Tutor Hougang | Weak Acid pH and Ka

Why can a more dilute weak acid have a higher percentage ionisation but still a higher pH? This question reveals whether a JC2 H2 Chemistry student understands Ka and the logarithm rather than memorising one square-root formula. The concentration of hydronium in solution, the fraction of molecules ionised and the acid dissociation constant are related quantities, but they are not interchangeable. A student must also choose the correct model for a weak base, salt or buffer instead of forcing one formula onto every solution.

JC2 Chemistry Tutor Hougang students preparing for Singapore-Cambridge H2 Chemistry 9476 need reliable skills in pH, pOH, Ka, Kb, pKa, Kw, weak-acid approximations and conjugate acid–base systems. The 2027 SEAB syllabus assesses these equilibrium calculations with stated limitations on the algebra required. This article uses original numerical examples and a practical system for recognising which chemical species controls the pH at each stage.

I teach the Chemistry before the calculator. Name the dominant species, write the appropriate equilibrium, identify a valid assumption and only then calculate. Our eduKateSG three-student small-group classes are held at 8 Fourth Avenue near Sixth Avenue MRT, not in Hougang. Families should check current arrangements and travel before considering tuition.

The official 9476 acid–base calculation scope

The 2027 SEAB H2 Chemistry syllabus 9476 covers the theories of acids and bases and, in its aqueous-solution chemistry, Ka, Kb, pKa, pKb, Kw, pH, pOH, buffer behaviour and relevant equilibria. It expects students to handle suitable weak-acid/base and buffer calculations, with mathematical limitations described in the syllabus. A learner should understand which approximations are permitted instead of treating every unfamiliar equilibrium as a university-level algebra problem.

These topics connect to titration curves, salt hydrolysis and solubility equilibria. However, a standalone pH task starts with determining what is actually in the beaker. The formula is a consequence of the species and equilibrium, not a substitute for them.

pH measures hydronium activity through a school approximation

At school level, pH is commonly calculated as −log10[H+] when the concentration approximation is appropriate. Hydronium H3O+ provides a more explicit chemical picture of protonated water, while H+ is a standard shorthand in many equations. For [H+]≈1.0×10−3 mol dm−3, pH≈3.00. A tenfold rise in hydronium concentration lowers pH by one unit.

The logarithm is not an ordinary linear scale. Doubling [H+] does not halve pH. When checking calculations, first estimate the order of magnitude: [H+] near 10−5 suggests pH near 5, not 0.00001 or 50000. A sensible magnitude check can detect an incorrectly used calculator log key.

Strong acid and weak acid are different concepts

A strong acid dissociates essentially completely in the specified aqueous model, whereas a weak acid establishes a significant equilibrium between HA and its ions. Concentration describes how many moles of acid were dissolved per unit volume. A 0.100 mol dm−3 HCl solution can therefore have a very different pH from 0.100 mol dm−3 ethanoic acid, despite their equal analytical concentration.

Students sometimes call a highly concentrated weak acid ‘strong’ because it tastes or behaves corrosively in a broad everyday sense. In acid–base Chemistry, strength has the specific equilibrium meaning. When a comparison is posed, specify both the intrinsic acid strength and the actual concentration before predicting relative pH.

Worked strong monoprotic acid calculation

For a simple aqueous 0.0100 mol dm−3 HCl sample at 25°C, assume effective complete dissociation under the intended school conditions. Then [H+]≈0.0100 mol dm−3 and pH=−log10(0.0100)=2.00. This is a direct concentration-to-log calculation and provides a useful baseline when learning weak-acid equilibria.

The result does not mean every acid with two hydrogen atoms in its formula automatically provides exactly double [H+] at all concentrations. Polyprotic acids have different successive dissociation behaviours. The chemical identity, stoichiometry and assumptions must be checked before applying a simple multiplication.

Write the weak-acid equilibrium first

For a weak monoprotic acid HA, write HA(aq)+H2O(l) ⇌ H3O+(aq)+A−(aq). The school concentration form is Ka=[H+][A−]/[HA] under the appropriate treatment of water activity. The value of Ka describes the acid dissociation equilibrium at a specified temperature, not the measured pH of every possible solution of that acid.

A weak acid can have one Ka value at a fixed temperature yet give different pH values when prepared at different concentrations. This is why knowing the acid’s name and Ka does not determine pH without an analytical concentration or additional mixture information.

The ICE table for a weak acid

Suppose the initial analytical concentration of HA is C, with negligible initially added A− and hydronium from other sources under the approximation. Let x mol dm−3 dissociate. The equilibrium concentrations are [HA]=C−x, [H+]≈x and [A−]≈x. Therefore Ka=x²/(C−x). This equation follows from chemical amounts and equilibrium stoichiometry.

If x is small relative to C, the familiar approximation C−x≈C gives x≈√(Ka C). The square-root relation is not an independent law for every acid-base system. It follows from the stated assumptions and should be checked for self-consistency, particularly as C becomes small or Ka becomes relatively large.

Worked ethanoic acid pH at 0.100 M

Take an illustrative ethanoic-acid Ka=1.8×10−5 at 25°C and C=0.100 mol dm−3. The approximation gives [H+]≈√(1.8×10−5×0.100)=√(1.8×10−6)=1.34×10−3 mol dm−3. Taking negative base-ten logarithm gives pH≈2.87. This is higher than the pH≈1.00 of a 0.100 M simple strong monoprotic acid under the comparable model.

The calculation is only useful if a student can explain why ethanoic acid produces fewer hydronium ions than a strong acid at equal analytical concentration. The un-ionised weak acid remains abundant. Ka and the equilibrium balance—not merely the word organic—determine the difference.

Check the weak-acid approximation quantitatively

After estimating x≈1.34×10−3 for 0.100 M ethanoic acid, the fraction dissociated is x/C≈0.0134, or about 1.34%. That is small relative to the starting concentration, so replacing C−x by C is a reasonable school approximation for the data. Checking the resulting ratio is better than declaring every weak acid has negligible dissociation.

Students sometimes calculate a valid square root but fail to test whether the assumption holds. If the predicted fractional change is large, the same approximate method may not be justified. The current H2 syllabus’s intended algebraic scope should be followed; identifying an invalid approximation is a useful conceptual answer even when a complicated exact calculation is not required.

Dilution raises percentage ionisation but usually raises pH

Using the same weak acid at lower analytical concentration, the fraction ionised can increase as the dissociation equilibrium adjusts. Yet the absolute [H+] can still decrease, so the solution becomes less acidic and pH rises. This is not a contradiction: percentage of molecules ionised and number of hydronium ions per litre are different quantities.

A powerful diagnostic question asks the student to decide which of two statements can both be true: ‘the weak acid is more ionised in percentage terms’ and ‘its pH is higher after dilution’. Both can be true, and the distinction helps prevent a common MCQ trap.

An example where the simple approximation becomes weak

If C=0.00100 M while Ka remains 1.8×10−5, the simple square-root estimate gives x≈1.34×10−4 M. That is about 13.4% of C, so the assumption x is negligible relative to C is no longer particularly accurate. Do not present the approximate pH as though it were an exact high-precision result.

The useful examination skill is recognising the limitation and referring back to the full equilibrium relation or the method specified by the question and syllabus. Avoid silently treating a significant dissociation as one per cent just because the acid is labelled weak.

pKa is a logarithmic description of Ka

pKa=−log10 Ka. For Ka=1.8×10−5, pKa≈4.74. At a fixed temperature, a smaller pKa corresponds to a larger Ka and hence a stronger acid among appropriately comparable systems. The pKa is not automatically the pH of a bottle containing that weak acid. It equals pH in the familiar buffer half-equivalence condition under relevant assumptions, but that is a specific relationship.

Students who swap pKa and pH in every question may obtain a plausible number for the wrong quantity. Read which variable is requested and which conditions are present. An acid’s pKa is a chemical equilibrium characteristic; a sample’s pH describes its current solution environment.

The conjugate base has its own equilibrium

The conjugate base A− can accept a proton from water: A−+H2O ⇌ HA+OH−. Its base dissociation constant Kb is related to the corresponding weak-acid Ka through Ka Kb=Kw at the same temperature under the usual conjugate-pair conventions. Therefore the strength of a weak acid and the proton-accepting tendency of its conjugate base are connected.

Do not use the Ka of HA directly as Kb for A−. The two quantities describe opposite proton-transfer equilibria and have different numerical values. When the question involves sodium ethanoate instead of ethanoic acid, start by recognising the conjugate base that is actually present.

What Kw means

Water undergoes a small self-ionisation equilibrium. At about 25°C in the usual school concentration treatment, Kw≈1.0×10−14, with [H+][OH−]=Kw. For pure neutral water at that temperature, the two concentrations are equal and each is approximately 1.0×10−7 mol dm−3, giving pH≈7.00.

Kw changes with temperature. Neutrality is defined by equal hydronium and hydroxide concentrations, not by a pH number fixed at seven under every thermal condition. If a data-based question supplies a different Kw, calculate the neutral pH from the given equilibrium rather than reuse the room-temperature result.

The pH and pOH relationship depends on temperature

At 25°C, pH+pOH≈14.00 under the conventional concentration treatment. More generally, pH+pOH=pKw, where pKw=−log10 Kw at the specified temperature. If an examination supplies a new Kw at another temperature, use the corresponding pKw; the fixed fourteen shortcut may no longer be appropriate.

A student should identify whether they are calculating hydronium or hydroxide concentration and whether the requested final answer is pH or pOH. This simple audit prevents returning a correct pOH numerical value with the wrong label.

Worked ammonia weak-base pH

Take ammonia with illustrative Kb=1.8×10−5 at 25°C and C=0.100 M. For NH3+H2O ⇌ NH4+ + OH−, the approximation gives [OH−]≈√(Kb C)=1.34×10−3 M. Thus pOH≈2.87 and pH≈11.13 using pKw≈14.00. The calculation is similar in structure to a weak acid, but the observed pH lies on the basic side.

One frequent error is to call 2.87 the pH after computing an OH− concentration. It is the pOH in this setting. Track which species was calculated and perform the appropriate conversion. The similarity of the formulae makes that distinction particularly important.

A conjugate-acid constant from Kb

If NH3 has Kb≈1.8×10−5 at 25°C, its conjugate acid NH4+ has Ka≈Kw/Kb=(1.0×10−14)/(1.8×10−5)≈5.56×10−10. This small Ka describes why ammonium can acidify water weakly under the appropriate conditions without acting like a fully dissociated strong acid.

The calculation also explains why NH4Cl solution can be acidic while sodium chloride solution is approximately neutral in the simple model. The dissolved ion undergoing proton transfer matters more than whether the formula contains the word chloride.

Salt hydrolysis requires identifying the active ion

In ammonium chloride, NH4+ can donate H+ to water, producing NH3 and H3O+ in an equilibrium description. In sodium ethanoate, CH3COO− can accept H+ from water, producing ethanoic acid and OH−. The Na+ and Cl− spectator ions are not the drivers of those particular weak-acid/base hydrolysis effects.

Students sometimes assume any salt made from acid and alkali must have pH 7. That is true for selected strong acid–strong base salts under ordinary conditions, not a universal rule. Identify whether the remaining ion is a conjugate weak acid or base.

A worked sodium ethanoate equivalence-type calculation

Suppose 0.0500 M sodium ethanoate is the principal dissolved species at 25°C. Using Kb≈5.56×10−10, the weak-base approximation gives [OH−]≈√(5.56×10−10×0.0500)≈5.27×10−6 M. This corresponds to pOH≈5.28 and pH≈8.72. The alkaline result is due to ethanoate hydrolysis, not necessarily leftover NaOH.

This calculation is useful in analysing weak-acid/strong-base titration equivalence. But the analytical concentration must first be determined from the moles of salt produced and total solution volume. Copying 8.72 into another equivalence problem without checking conditions is not a valid method.

How a buffer controls pH

A suitable buffer contains appreciable amounts of a weak acid and its conjugate base, or the analogous weak base and conjugate acid. Added small amounts of strong acid can be consumed by the conjugate base, while added strong base can be consumed by the weak acid. Buffer capacity is finite; the solution cannot resist unlimited reagent addition.

The Henderson–Hasselbalch form pH=pKa+log([A−]/[HA]) is useful for an appropriate weak-acid buffer under its assumptions. The ratio must describe the species actually remaining, not amounts from before a strong reagent reacted with the buffer.

Worked buffer ratio without unnecessary dilution

Take an illustrative mixture containing 0.0200 mol ethanoic acid and 0.0100 mol ethanoate in a common solution volume, at 25°C. If pKa≈4.74 and the buffer assumptions apply, the ratio of conjugate base to acid is 0.5, so pH≈4.74+log(0.5)≈4.44. The common volume cancels in the ratio, so the mole ratio is sufficient here.

Students sometimes divide each component by different volumes because the reagents were originally in separate bottles. After they have been mixed, both species occupy the same total solution volume. The appropriate ratio can be simplified only after the chemistry and mixing have been accounted for.

Add strong acid to a buffer: react first, logarithm second

Now add 0.00500 mol strong acid to that illustrative buffer without exhausting its ethanoate, and assume relevant volume effects cancel in the ratio. Ethanoate is consumed to form ethanoic acid. Its moles fall from 0.0100 to 0.00500, while ethanoic acid rises from 0.0200 to 0.0250. Then pH≈4.74+log(0.005/0.025)=4.74+log(0.2)≈4.05.

A common wrong move uses the original 0.0100/0.0200 ratio after strong acid was added. The resulting calculation has ignored the neutralisation step. The mole ledger is the reason the buffer equation remains meaningful rather than a formula to insert stock numbers into.

What happens after conjugate base is exhausted

If the added strong acid exceeds the available conjugate base in the same simple buffer, the usual buffer ratio calculation is no longer appropriate because the weak-acid/conjugate-base pair is no longer present in the required proportions. Excess strong acid may then dominate hydronium concentration. The solution has crossed a chemical boundary, not merely entered a different numeric range.

Students should determine whether the conjugate base remains positive in the post-reaction mole ledger before applying Henderson–Hasselbalch. This check prevents undefined logarithms and false claims that a buffer can neutralise unlimited strong acid.

Dilution and buffer capacity are different questions

An ideal weak-acid/conjugate-base buffer ratio can remain similar after equal dilution of both components, leaving pH nearly unchanged under suitable assumptions. But the actual amount of strong acid or base the diluted solution can absorb per unit volume without a large pH change is reduced. Thus pH and buffer capacity are not identical quantities.

Students who see the same initial pH may incorrectly assume two buffers contain equal amounts of acid/base material. An exam can provide two solutions with similar ratio but very different total concentrations and ask which resists a fixed added amount more effectively. A mole ledger reveals the distinction.

A five-step equation-selection decision tree

First identify all dissolved chemical species. Second decide whether there is excess strong acid or strong base. Third, if not, ask whether a weak acid/base stands alone, a conjugate salt is present, or a genuine buffer pair remains. Fourth write the corresponding equilibrium and identify which concentration is sought. Fifth check the mathematical approximation and report the correct pH or pOH.

This method applies across weak-acid, salt, buffer and titration problems without forcing one universal formula. It also identifies when supplied information is insufficient for a unique pH. Clear classification at the beginning often saves more time than faster arithmetic.

First decide whether the substance is an acid or its conjugate base

Ethanoic acid, CH3COOH, is not the same dissolved species as sodium ethanoate, CH3COONa. In the former sample, proton donation by a weak acid is the main simplified equilibrium; in the latter, ethanoate ions can accept protons from water and generate OH−. The analytical concentration alone does not tell you whether a pH or pOH calculation is the correct starting point.

Write the dissociation or dissolution reaction before using Ka or Kb. For sodium ethanoate, sodium ions are usually spectators for this acid–base purpose, and the relevant base is CH3COO−. A student who plugs Ka directly into a weak-acid square root for the salt has selected the wrong chemical model.

A pH calculation must match the reference temperature

At 25°C, pKw is about 14, allowing the familiar pH+pOH=14 conversion. At another temperature, Kw changes and the neutral pH need not be 7 even though neutral water retains equal hydronium and hydroxide concentrations. If a question gives a new Kw, compute pKw from that value before converting pOH to pH.

The key distinction is that neutrality is a comparison of concentrations in the relevant model, not a fixed numerical pH across all temperatures. A student who rigidly writes 14 in every calculation may be answering the 25°C version of a problem that explicitly changed temperature.

The half-equivalence identity is conditional

For a monoprotic weak acid titrated with strong base, the ordinary buffer approximation at half the stoichiometric titrant volume gives equal weak-acid and conjugate-base moles. The Henderson–Hasselbalch relation then gives pH≈pKa. This does not say that pH equals pKa at half the vertical height of the plotted titration curve.

The equivalence volume is an x-axis quantity based on moles, so half-equivalence means half the required titrant amount. A student should determine that volume from the balanced equation, not pick a visually central point on the graph. This logical step is more reliable than memorising one specific pH value.

When a diprotic acid needs two equilibria

An acid capable of donating more than one proton can have successive dissociation constants. The first and second ionisation steps need not be equally extensive. For sulfuric acid in water, the first dissociation is commonly treated as strong, while the second is not always quantitatively complete at the concentration of interest. A student should not automatically double the analytical concentration to find exact [H+] in every problem without considering the second equilibrium.

This is an example of why stoichiometric potential and equilibrium behaviour are not the same thing. If a data-based item provides Ka for a second proton step, use the information rather than ignore it. The correct method is dictated by the acid and conditions.

A diluted strong acid still needs a mole ledger

Suppose 10.0 cm3 of 0.100 M HCl is diluted to 100.0 cm3 total under the ordinary volume-additivity assumption. The HCl amount is 0.00100 mol, and its final concentration is 0.0100 M. In the simple strong-acid model, pH≈2.00. A student who uses the original 0.100 M after the dilution will report pH≈1.00 and miss the changed concentration.

Notice that acid strength has not changed during dilution. The analytical concentration changed, and hence the resulting hydronium concentration and pH. Calling the diluted sample a weak acid is a conceptual error, even though its pH is now higher.

The meaning of percentage ionisation

Percentage ionisation describes the fraction of analytical weak-acid molecules that have dissociated in an appropriate equilibrium model. If a 0.100 M weak acid produces 0.00134 M hydronium from its own dissociation, the approximate fraction is 0.00134/0.100=0.0134, or 1.34%. This is not the same number as pH≈2.87.

Students sometimes report pH as a percentage because they have memorised both results from one worksheet. Ask which quantity is dimensionless ratio, which is concentration and which is logarithmic scale. The units and definitions can separate them immediately.

When Ka is large relative to C

The weak-acid small-x approximation assumes the hydrogen-ion concentration generated is small compared with the analytical acid concentration. If Ka is large compared with C, the acid may dissociate appreciably and the approximation becomes poor. This does not make the equilibrium expression invalid; it means the simplified denominator C−x≈C is no longer an accurate representation.

Under the prescribed H2 scope, questions are designed for appropriate methods without unnecessary quadratic-equation solving. The student should still know why the small-x assumption is being made and check whether the calculated fraction supports it. A carefully qualified estimate is more honest than a highly precise result from invalid algebra.

A weak base uses hydroxide as the primary unknown

For NH3(aq)+H2O⇌NH4++OH−, Kb describes hydroxide production through a weak-base equilibrium. The usual small-ionisation calculation estimates [OH−], from which pOH and then pH can be found at the stated temperature. It is not correct to call an OH− concentration the value of [H+] simply because both appear in water.

After calculating [OH−], multiply it by the appropriate [H+] in Kw=[H+][OH−] or use pH+pOH=pKw. Naming the first chemical species correctly prevents a final answer on the wrong side of neutrality.

Two solutions with the same pH can differ greatly in amount

A concentrated weak acid and a dilute strong acid can sometimes have similar hydronium concentrations and therefore similar pH, while containing very different total amounts of acid available for subsequent neutralisation. The pH gives a current acidity measure; it does not uniquely determine how many moles of weak-acid molecules are still present.

A titration question can therefore require much more strong base to neutralise one sample than another despite similar starting pH. The mole ledger controls that quantity. This is a useful bridge between weak-acid equilibria and titration interpretation.

A buffer’s effectiveness depends on total amounts

Two buffer solutions can share the same conjugate-base-to-acid ratio and similar pH while having quite different total concentrations. The more concentrated buffer generally has more acid and conjugate base per unit volume to consume added strong reagent, giving greater buffer capacity under relevant conditions. Equal starting pH is therefore not proof of equal resistance to a specified acid addition.

Ask the student to calculate moles rather than compare only logarithms. If one mixture has 0.010 mol of each buffer partner and another has 0.100 mol of each in the same volume, the second has much more material available for reaction even if the starting pH is similar.

Conjugate-pair identification is worth practising alone

The conjugate base of HCO3− is CO3²−, while its conjugate acid is H2CO3 in the familiar carbonic-acid system. The conjugate acid of NH3 is NH4+, and the conjugate base of CH3COOH is CH3COO−. Each conjugate pair differs by one proton, with charges adjusted accordingly.

Before attempting a numerical Ka or Kb question, ask students to name the relevant pair. A wrong conjugate species leads to a wrong equilibrium even if the algebra afterward is flawless. Short structure-and-charge drills are an efficient repair.

The mole ledger after strong base addition

Start with 0.0200 mol weak monoprotic acid HA and add 0.00500 mol strong base. Under complete stoichiometric neutralisation for that step, 0.0150 mol HA remains and 0.00500 mol A− is formed. The solution may then behave as a weak-acid buffer under the relevant assumptions. The Henderson–Hasselbalch ratio is 0.00500/0.0150, not the original initial ratio of 0/0.0200.

This simple example is the bridge between ordinary weak-acid Ka questions and mixed buffer tasks. Chemical conversion happens before equilibrium recalculation, so the student’s first operation should be balancing and subtracting moles.

How pKa changes the buffer’s preferred range

A buffer based on a particular weak acid is often most effective for controlling pH in a region near that acid’s pKa when both conjugate partners are present in appreciable quantities. That does not mean any buffer can maintain every pH or that pH is fixed exactly at pKa regardless of ratio. The logarithmic relationship makes the mixture composition relevant.

Students should match a proposed buffer to the target chemical pH and consider how much reagent the buffer must absorb. In practical applications such as water chemistry, choosing a buffer requires both chemical compatibility and sufficient capacity.

A salt of a weak base can be acidic

Ammonium chloride supplies NH4+ ions and chloride ions when dissolved. The ammonium ion is a conjugate acid and can donate a proton to water, making the solution acidic under appropriate conditions. The chloride ion is a very weak conjugate base of strong HCl and does not account for the acidity in the intended explanation.

This helps students understand why a salt’s name is not a complete pH prediction. Ask which ion reacts with water. The correct species-to-equilibrium link is much more valuable than memorising ‘ammonium chloride is acidic’.

A salt of a weak acid can be basic

Sodium ethanoate supplies CH3COO−, which can accept a proton from water and generate OH−. The sodium ion is generally a spectator in this school acid–base context. The solution can therefore be alkaline even though no NaOH was added as a separate reagent.

The first question for the student is whether the dissolved anion is a conjugate base of a weak acid. If yes, a Kb relation may be appropriate. The calculation then depends on concentration and temperature rather than one fixed pH for every sodium ethanoate solution.

A data-table question about Ka trends

Suppose an unfamiliar table lists three Ka values measured at the same temperature. The largest Ka corresponds to the greatest dissociation tendency among the comparable acids under the model, while the smallest pKa corresponds to the largest Ka. The student need not invent each acid’s chemical structure to rank the data; the equilibrium definitions supply the reasoning.

However, a ranking of intrinsic acid strengths does not by itself establish the ranking of pH if the prepared solution concentrations differ. Ask which variable the data actually compare before reaching a conclusion.

A numerical plausibility check for basic pH

If a calculation for a simple 0.100 M weak base at 25°C produces [OH−] around 0.001 M, its pOH is around 3 and pH around 11, an alkaline result. If the student reports pH 3, they may have stopped at pOH. If a weak base is predicted to make a strongly acidic solution without any other chemistry, reconsider the chosen species or logarithmic conversion.

Quick sense checks do not replace correct equations, but they catch frequent sign and label errors. Use them only within the physical assumptions of the question.

What to do when there is insufficient information

A prompt giving only the identity of a weak acid, with no concentration or relevant equilibrium data, may not determine one numerical pH. Similarly, an unspecified buffer ratio cannot yield one unique logarithmic result. A careful student should identify which quantity is missing rather than invent a concentration from an earlier example.

This is scientific honesty and an examination skill. Many data-based questions supply enough information but conceal it in a table; locate that evidence first. If it is genuinely absent, explain the dependency rather than manufacture precision.

A seven-day weak-acid calculation route

Session one is pH, pOH and logarithmic units. Session two constructs weak-acid equilibria and the small-x relation. Session three checks percentage dissociation and approximation validity. Session four uses weak-base Kb and Kw. Session five calculates conjugate salt pH. Session six applies a mole ledger to buffer additions. Session seven uses unseen mixed questions that require choosing the correct model before calculating.

Spread the work over more days if school commitments require it. The measure of readiness is correct model selection and reasonable pH under changed conditions, not memorising one acid’s answer. A tutor should ask for an independent retest after a delay.

Frequently asked JC2 pH questions

Does dilution make a weak acid less ionised in percentage terms? Often the opposite: percentage ionisation can rise while [H+] falls. Is pKa the same as pH? No, except under specific relationships such as suitable half-equivalence. Is Kw always 10−14? That familiar approximate value is at 25°C. Why can NH4Cl be acidic? Ammonium acts as a weak conjugate acid.

Can sodium ethanoate be basic? Yes, ethanoate can accept protons from water. Is the square-root weak-acid approximation always valid? No; check the resulting fraction dissociated. Does a buffer resist unlimited acid? No. Can pOH be reported as pH directly? Not without conversion under the relevant Kw.

JC2 Chemistry Tutor Hougang | H2 Weak Acid pH, Ka, pKa, Kb and Kw Worked Calculations