In 2014, the British Dental Association reported that 46% of 15-year-olds in the UK showed visible tooth enamel erosion — and cola was the primary culprit. The reason: cola contains phosphoric acid (Ka1 = 7.5 × 10⁻³), which is 17,000 times stronger than carbonic acid (Ka1 = 4.3 × 10⁻⁷) in sparkling water. Both drinks are below the enamel dissolution threshold of pH 5.5, but knowing the Ka difference explains why cola is so much more damaging.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A dentist tells her patient: "Every time you drink a cola, the carbonic acid and phosphoric acid in it drop your mouth pH to around 3.5 for about 20 minutes. Tooth enamel dissolves below pH 5.5. That's 20 minutes of dissolving per can." The patient asks: "But carbonic acid is in sparkling water too — is that just as bad?" The dentist replies: "Sparkling water has carbonic acid but no phosphoric acid. The pH is around 5.0 — still below the 5.5 threshold, but far less damaging than cola."
Before reading on: What is the difference between carbonic acid (H₂CO₃) and phosphoric acid (H₃PO₄) in terms of how many protons they can donate? Why does cola have a lower pH than sparkling water even though both contain carbonic acid? What does Ka tell you that pH alone does not?
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🔗 Understand
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Larger Ka → greater ionisation → stronger acid. For weak acids: Ka ≪ 1.
Larger Ka → smaller pKa → stronger acid (pKa like golf: lower is better). Smaller Ka → larger pKa → weaker acid.
Applies ONLY to a conjugate pair. Rearrangements: Kb(A⁻) = Kw / Ka(HA) Ka(HA) = Kw / Kb(A⁻)
General rule: Ka1 ≫ Ka2 ≫ Ka3. For pH calculations, use Ka1 only — successive ionisations contribute negligibly to [H⁺].
pH tells you how acidic a specific solution is — Ka tells you how acidic an acid inherently is, regardless of concentration, and this distinction between a solution property and a molecular property is the foundation of every acid strength comparison in this lesson.
Ka is the equilibrium constant for the ionisation of a weak acid in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻]/[HA]. A large Ka means the equilibrium lies far to the right — most acid molecules have ionised — and the acid is relatively strong. A small Ka means most molecules remain intact — the acid is relatively weak.
Because Ka values span many orders of magnitude (from ~10⁻² for strong-ish weak acids to ~10⁻¹⁴ for extremely weak acids), pKa = −log₁₀(Ka) converts these values to a more manageable scale — exactly as pH converts [H⁺]. The inverse relationship: larger Ka → smaller pKa → stronger acid. Smaller Ka → larger pKa → weaker acid. This inverse direction causes one of the most consistent ranking errors in HSC.
Ka is an intrinsic molecular property — it depends only on the identity of the acid and temperature, not on concentration. It is the correct quantity to compare acid strengths, because it eliminates concentration as a confounding variable. pH reflects both Ka and c — two acids with different Ka values can have the same pH if their concentrations are chosen appropriately, which is why pH alone cannot be used to rank acid strengths.
| Acid | Ka | pKa | Relative strength | Degree of ionisation at 0.1 mol/L |
|---|---|---|---|---|
| HClO₄ (strong) | ~10⁷ | ~−7 | Strongest (strong acid) | ~100% |
| H₂SO₄ (1st ionisation) | ~10³ | ~−3 | Strong acid | ~100% |
| HNO₂ | 4.5 × 10⁻⁴ | 3.35 | Moderately weak | ~6.5% |
| HF | 6.8 × 10⁻⁴ | 3.17 | Moderately weak | ~7.6% |
| CH₃COOH | 1.8 × 10⁻⁵ | 4.74 | Weak | ~1.3% |
| H₂CO₃ | 4.3 × 10⁻⁷ | 6.37 | Very weak | ~0.065% |
| HCN | 6.2 × 10⁻¹⁰ | 9.21 | Extremely weak | ~0.00025% |
| HCO₃⁻ | 4.7 × 10⁻¹¹ | 10.33 | Extremely weak | ~negligible |
Ka = [H⁺][A⁻]/[HA] — equilibrium constant for weak acid ionisation; intrinsic molecular property independent of concentration. pKa = −log₁₀(Ka); larger Ka → smaller pKa → stronger acid (golf score: lower pKa = stronger). pH measures solution acidity (depends on Ka AND c); Ka measures intrinsic acid strength. Two acids with different Ka values can produce the same pH at different concentrations — always use Ka to rank acid strength, never pH alone.
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We just saw that Ka is the intrinsic measure of acid strength — independent of concentration. That raises a question: Once you have a Ka ranking for a list of acids, what does it tell you about the strength of each acid's conjugate base? This card answers it → Ka × Kb = Kw, so acid and conjugate base strengths are perfectly inverted: the weakest acid has the strongest conjugate base.
Once acid strengths are ranked by Ka, the strength of every conjugate base is immediately determined — because the inverse relationship Ka × Kb = Kw means that the weakest acid has the strongest conjugate base, and vice versa, without any additional data.
The conjugate pair relationship Ka(HA) × Kb(A⁻) = Kw = 1.0 × 10⁻¹⁴ has a profound consequence: the strength ordering of acids and the strength ordering of their conjugate bases are perfectly inverted. The strongest acid has the weakest conjugate base (e.g. HCl, very strong → Cl⁻, Kb ≈ 10⁻²¹). The weakest acid has the strongest conjugate base (e.g. HCN, Ka = 6.2 × 10⁻¹⁰ → CN⁻, Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵ — a moderately strong weak base).
This inversion predicts the direction of acid-base reactions: a reaction proceeds in the direction that produces the weaker acid and weaker base on the product side. For example, HF + CN⁻ ⇌ HCN + F⁻: Ka(HF) = 6.8 × 10⁻⁴, Ka(HCN) = 6.2 × 10⁻¹⁰. HF is a much stronger acid than HCN → equilibrium lies far to the right. The weaker acid (HCN) and weaker base (F⁻) are on the product side — consistent with the general rule.
Ka(HA) × Kb(A⁻) = Kw = 1.0 × 10⁻¹⁴ — applies ONLY to a conjugate acid-base pair (A⁻ = HA − H⁺). Acid and conjugate base strengths are inversely related: strongest acid → weakest conjugate base. Kb = Kw/Ka; Ka = Kw/Kb. Reactions favour the side with the weaker acid and weaker base as products. Applying Ka × Kb = Kw to non-conjugate pairs (e.g. CH₃COOH and NH₃) is the most common error in this topic.
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We just saw that Ka × Kb = Kw links acid and conjugate base strengths within a conjugate pair. That raises a question: What happens when an acid has more than one ionisable proton — does each H⁺ leave with equal ease? This card answers it → each successive Ka is ~10⁵ smaller due to increasing electrostatic attraction; pH calculations use Ka1 only.
A polyprotic acid can donate more than one proton — but each successive proton is dramatically harder to remove than the one before, and understanding why at the molecular level is what distinguishes a superficial from a deep understanding of acid strength.
A polyprotic acid has more than one ionisable proton, each removed in a separate step with its own Ka. For phosphoric acid (H₃PO₄, triprotic): Ka1 = 7.5 × 10⁻³ (removing H⁺ from the neutral molecule); Ka2 = 6.2 × 10⁻⁸ (removing H⁺ from H₂PO₄⁻); Ka3 = 4.8 × 10⁻¹³ (removing H⁺ from HPO₄²⁻). Each successive Ka is approximately 10⁵ smaller than the previous.
Electrostatic explanation: Removing a proton (H⁺, positive) from an already-negative ion requires overcoming greater electrostatic attraction between the proton and the negatively charged species. H₂PO₄⁻ (charge −1) holds its proton more tightly than H₃PO₄ (charge 0). HPO₄²⁻ (charge −2) holds it even more tightly. Each additional negative charge on the conjugate base increases the electrostatic opposition to further proton donation.
Practical consequence: In pH calculations for polyprotic acid solutions, only Ka1 is needed. For H₃PO₄, Ka2/Ka1 ≈ 10⁻⁵ — the second ionisation contributes less than 0.001% of the H⁺ from the first. This simplifies all polyprotic acid pH calculations to a single ICE table using Ka1 only.
Polyprotic acids: Ka1 ≫ Ka2 ≫ Ka3 — each successive ionisation is ~10⁴–10⁵× weaker. Electrostatic reason: removing H⁺ from a negatively charged species requires more energy. pH calculations use Ka1 only (Ka2 contributes <0.001% to total [H⁺]). Each ionisation step is a separate equilibrium with its own Ka expression — never combine into a single Ka.
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We just saw that polyprotic acids each have separate Ka values, with Ka1 dominating pH. That raises a question: Where does Ka reasoning appear in real-world contexts that NESA explicitly expects students to discuss? This card answers it → dental erosion: cola (H₃PO₄, Ka1 = 7.5 × 10⁻³) vs sparkling water (H₂CO₃, Ka1 = 4.3 × 10⁻⁷); enamel dissolves below pH 5.5.
The Ka values of carbonic acid and phosphoric acid directly determine at what pH tooth enamel dissolves, why cola is more damaging than sparkling water, and why the dentist's 20-minute warning is quantitatively justified — making this the most directly applicable Ka calculation in everyday chemistry.
Tooth enamel is composed principally of hydroxyapatite, Ca₅(PO₄)₃OH. It dissolves in acid: Ca₅(PO₄)₃OH + 7H⁺ → 5Ca²⁺ + 3H₂PO₄⁻ + H₂O. The critical pH below which enamel dissolution occurs is approximately 5.5 — the pH at which the solution becomes undersaturated with respect to hydroxyapatite. Above pH 5.5, saliva re-deposits hydroxyapatite and remineralises enamel. Below pH 5.5, net dissolution occurs.
Sparkling water contains dissolved CO₂ → H₂CO₃, Ka1 = 4.3 × 10⁻⁷. A realistic carbonation level of ~0.001 mol/L gives pH ≈ 4.68 — below 5.5, so sparkling water does erode enamel, but moderately. Cola contains both H₂CO₃ AND H₃PO₄ (Ka1 = 7.5 × 10⁻³ — a much larger Ka) plus citric and malic acids. The stronger Ka of H₃PO₄ compared to H₂CO₃ (Ka ratio ≈ 10⁴) means H₃PO₄ contributes far more H⁺ at equivalent concentrations. Cola pH typically reaches 3.0–3.5 — well below 5.5, with far more aggressive enamel dissolution.
Enamel dissolution threshold: pH 5.5 (below this, hydroxyapatite dissolves; above this, saliva remineralises). Sparkling water: H₂CO₃ only (Ka1 = 4.3 × 10⁻⁷) → pH ~4.7–5.0; moderate risk. Cola: H₂CO₃ + H₃PO₄ (Ka1 = 7.5 × 10⁻³, ~17,000× larger) + citric acid → pH 3.0–3.5; far greater risk. Compare acids at same concentration: larger Ka → greater ionisation → lower pH.
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We just saw how Ka explains why cola is 17,000× more acidically aggressive than sparkling water at equivalent carbonic acid concentrations. That raises a question: Beyond dental erosion, where else does industry use Ka-to-pH reasoning in real analytical practice? This card answers it → food (wine tartaric acid), pharmaceutical (aspirin titration), environmental (pH probes for water quality and acid rain).
In 2014, the British Dental Association surveyed 15-year-olds and found 46% had enamel erosion. The culprit: cola drinks at pH 3.0–3.5 containing H₃PO₄ (Ka1 = 7.5 × 10⁻³) — 17,000 times stronger than H₂CO₃ (Ka1 = 4.3 × 10⁻⁷) in sparkling water. Ka is the number that makes this comparison quantitative. Across food, pharmaceutical, and environmental chemistry, the same Ka-to-pH reasoning controls quality decisions made every day.
Food and beverage industry: Wine acidity is measured by titrating total acidity (primarily tartaric acid, Ka1 = 1.0 × 10⁻³, and malic acid) against NaOH. The Ka of tartaric acid determines how the acid-base curve will look during titration (equivalence point pH, buffer region shape). Dairy acidity in cheese and yoghurt production is measured by titration of lactic acid (Ka = 1.4 × 10⁻⁴) — the Ka determines the equivalence point pH and which indicator is appropriate.
Pharmaceutical industry: Aspirin tablets are quality controlled by dissolving a known mass and titrating with NaOH. The Ka of aspirin (acetylsalicylic acid, Ka = 3.0 × 10⁻⁴) determines how sharp the endpoint will be and which indicator gives an accurate result. pH probes measure drug solution acidity to verify purity — deviations from expected pH indicate hydrolysis or contamination.
Environmental monitoring: Water quality is assessed by measuring pH (calibrated glass electrode probes) and titrating for total alkalinity (primarily HCO₃⁻, Ka2 = 4.7 × 10⁻¹¹). Acid rain is characterised by pH below 5.6 (pH of CO₂-saturated pure water). Industrial wastewater must be neutralised to pH 6.5–8.5 before discharge — continuous pH monitoring uses digital probes with data logging, comparing real-time pH against calculated equilibrium values derived from known Ka and concentration.
Industrial acid-base analysis: pH probe (primary method) + titration (quantitative). Ka determines equivalence point pH, buffer region shape, and indicator choice. Applications: food (wine, dairy titration), pharmaceutical (aspirin purity verification), environmental (acid rain threshold pH < 5.6; wastewater must reach pH 6.5–8.5). Glass electrode probe calibrated with pH 4.00 and 7.00 buffers; reads H⁺ concentration via Nernst equation.
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"Larger pKa = stronger acid." The opposite is true. pKa = −log(Ka), so larger pKa means smaller Ka means weaker acid. Use the golf score mnemonic: lower pKa = stronger acid.
"Ka × Kb = Kw applies to any acid-base pair." It applies ONLY to a conjugate pair — Ka(HA) × Kb(A⁻) where A⁻ = HA − H⁺. It does not apply to unrelated acids and bases such as CH₃COOH and NH₃.
"For polyprotic acids, sum up all Ka contributions to get total [H⁺]." Only Ka1 is needed. Ka2 and Ka3 are so much smaller (ratios of ~10⁴–10⁵) that their contributions to [H⁺] are negligible and ignored in all standard HSC calculations.
"pH can be used to rank acid strength." pH depends on both Ka AND concentration. Two acids with different Ka values can have the same pH at different concentrations. Only Ka (or pKa) is an intrinsic measure of acid strength.
Four acids have the following Ka values: Acid P: Ka = 1.3 × 10⁻²; Acid Q: Ka = 8.4 × 10⁻⁵; Acid R: Ka = 2.1 × 10⁻⁸; Acid S: Ka = 6.7 × 10⁻¹².
(a) Rank the four acids from strongest to weakest. (b) Calculate pKa for each acid. (c) Calculate Kb for the conjugate base of each acid. (d) Rank the conjugate bases from strongest to weakest.
GIVEN: Ka(P) = 1.3 × 10⁻², Ka(Q) = 8.4 × 10⁻⁵, Ka(R) = 2.1 × 10⁻⁸, Ka(S) = 6.7 × 10⁻¹² FIND: Ranking, pKa values, Kb values, conjugate base ranking
(a) Ranking by Ka: Largest Ka = strongest acid.
Ranking: P (1.3 × 10⁻²) > Q (8.4 × 10⁻⁵) > R (2.1 × 10⁻⁸) > S (6.7 × 10⁻¹²). P is strongest; S is weakest.
(b) pKa values: pKa = −log(Ka)
P: pKa = −log(1.3 × 10⁻²) = 2 − log(1.3) = 2 − 0.114 = 1.89
Q: pKa = −log(8.4 × 10⁻⁵) = 5 − log(8.4) = 5 − 0.924 = 4.08
R: pKa = −log(2.1 × 10⁻⁸) = 8 − log(2.1) = 8 − 0.322 = 7.68
S: pKa = −log(6.7 × 10⁻¹²) = 12 − log(6.7) = 12 − 0.826 = 11.17
Verify: P (1.89) < Q (4.08) < R (7.68) < S (11.17) — smaller pKa = stronger acid ✓
(c) Kb for conjugate bases: Kb = Kw/Ka
P⁻: Kb = 1.0 × 10⁻¹⁴ / 1.3 × 10⁻² = 7.7 × 10⁻¹³
Q⁻: Kb = 1.0 × 10⁻¹⁴ / 8.4 × 10⁻⁵ = 1.2 × 10⁻¹⁰
R⁻: Kb = 1.0 × 10⁻¹⁴ / 2.1 × 10⁻⁸ = 4.8 × 10⁻⁷
S⁻: Kb = 1.0 × 10⁻¹⁴ / 6.7 × 10⁻¹² = 1.5 × 10⁻³
(d) Ranking conjugate bases: Largest Kb = strongest base.
S⁻ (1.5 × 10⁻³) > R⁻ (4.8 × 10⁻⁷) > Q⁻ (1.2 × 10⁻¹⁰) > P⁻ (7.7 × 10⁻¹³)
The ranking of conjugate bases is the exact inverse of the ranking of parent acids ✓
ANSWER: (a) P > Q > R > S. (b) pKa: P = 1.89; Q = 4.08; R = 7.68; S = 11.17. (c) Kb: P⁻ = 7.7 × 10⁻¹³; Q⁻ = 1.2 × 10⁻¹⁰; R⁻ = 4.8 × 10⁻⁷; S⁻ = 1.5 × 10⁻³. (d) Conjugate base ranking S⁻ > R⁻ > Q⁻ > P⁻ — perfect inversion of acid strength ranking.
Carbonic acid (H₂CO₃) has Ka1 = 4.3 × 10⁻⁷ and Ka2 = 4.7 × 10⁻¹¹.
(a) Write both ionisation equations with correct arrow notation. (b) Calculate the pH of a 0.0150 mol/L H₂CO₃ solution using Ka1 only, and justify this simplification. (c) Calculate the ratio Ka1/Ka2 and explain what it tells you. (d) Calculate Kb for CO₃²⁻ and predict whether 0.100 mol/L Na₂CO₃ is acidic, basic, or neutral.
GIVEN: Ka1(H₂CO₃) = 4.3 × 10⁻⁷; Ka2 = 4.7 × 10⁻¹¹; c = 0.0150 mol/L FIND: Equations, pH, ratio, Kb, Na₂CO₃ prediction
(a) Ionisation equations: Both Ka ≪ 1 → both use ⇌
First: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) Ka1 = 4.3 × 10⁻⁷
Second: HCO₃⁻(aq) ⇌ H⁺(aq) + CO₃²⁻(aq) Ka2 = 4.7 × 10⁻¹¹
(b) pH using Ka1:
Check assumption: Ka1/c = 4.3 × 10⁻⁷ / 0.0150 = 2.87 × 10⁻⁵ ≪ 0.0025 ✓
x = √(Ka1 × c) = √(4.3 × 10⁻⁷ × 0.0150) = √(6.45 × 10⁻⁹) = 8.03 × 10⁻⁵ mol/L
Verify: 8.03 × 10⁻⁵ / 0.0150 = 0.54% ≪ 5% ✓
pH = −log(8.03 × 10⁻⁵) = 5 − log(8.03) = 5 − 0.905 = 4.10
Justification: Ka2 = 4.7 × 10⁻¹¹ ≪ Ka1 = 4.3 × 10⁻⁷. The second ionisation contributes negligibly — including it would change pH by <0.001 units.
(c) Ka1/Ka2 ratio:
Ka1/Ka2 = (4.3 × 10⁻⁷) / (4.7 × 10⁻¹¹) = 9.1 × 10³ ≈ 10⁴
The first ionisation is approximately 10,000 times more favourable than the second. For every 10,000 H⁺ from the first ionisation, only ~1 additional H⁺ comes from the second — contributing <0.01% of total [H⁺].
(d) Kb for CO₃²⁻ and Na₂CO₃ prediction:
CO₃²⁻ is the conjugate base of HCO₃⁻ (second ionisation) → Ka for this pair = Ka2 = 4.7 × 10⁻¹¹
Kb(CO₃²⁻) = Kw / Ka2 = (1.0 × 10⁻¹⁴) / (4.7 × 10⁻¹¹) = 2.1 × 10⁻⁴ — a moderately strong weak base
Na₂CO₃ → Na⁺ (neutral spectator) + CO₃²⁻ (basic ion). CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻
Salt of strong base (NaOH) + weak acid (H₂CO₃) → basic solution (pH > 7) ✓
ANSWER: (a) H₂CO₃ ⇌ H⁺ + HCO₃⁻ (Ka1); HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (Ka2). (b) pH = 4.10; justified because Ka2/Ka1 ≈ 10⁻⁴. (c) Ka1/Ka2 ≈ 10⁴. (d) Kb(CO₃²⁻) = 2.1 × 10⁻⁴; Na₂CO₃ solution is basic.
A food scientist develops a sports drink containing citric acid (Ka1 = 7.4 × 10⁻⁴, Ka2 = 1.7 × 10⁻⁵, Ka3 = 4.0 × 10⁻⁷) at 0.0200 mol/L and malic acid (Ka = 3.5 × 10⁻⁴) at 0.0150 mol/L. (a) Calculate the pH contribution from citric acid alone and malic acid alone. (b) Rank citric acid (Ka1), malic acid, acetic acid (Ka = 1.8 × 10⁻⁵), and carbonic acid (Ka1 = 4.3 × 10⁻⁷) by decreasing acid strength with pKa values. (c) The target pH range is 3.0–3.5. Assess whether 0.0200 mol/L citric acid alone achieves this target. (d) Explain, using Ka values and degree of ionisation, why the drink would need a dental health warning if consumed daily, and why an acetic acid formulation at the same concentration would be less concerning.
GIVEN: Citric Ka1 = 7.4 × 10⁻⁴, c = 0.0200 mol/L; Malic Ka = 3.5 × 10⁻⁴, c = 0.0150 mol/L
(a) pH of citric acid — assumption check:
Ka1/c = 7.4 × 10⁻⁴ / 0.0200 = 3.7% — <5% at first glance, but verify after calculation.
x = √(7.4 × 10⁻⁴ × 0.0200) = √(1.48 × 10⁻⁵) = 3.85 × 10⁻³
Check: 3.85 × 10⁻³ / 0.0200 = 19.2% > 5% — assumption INVALID → use quadratic.
x² + 7.4 × 10⁻⁴ x − 1.48 × 10⁻⁵ = 0
x = (−7.4 × 10⁻⁴ + √((7.4 × 10⁻⁴)² + 4 × 1.48 × 10⁻⁵)) / 2 = 3.50 × 10⁻³ mol/L
pH(citric) = −log(3.50 × 10⁻³) = 2.46
Malic acid (quadratic similarly required): x = 2.12 × 10⁻³ mol/L → pH(malic) = 2.67
(b) pKa ranking:
Citric Ka1 = 7.4 × 10⁻⁴ → pKa = 3.13; Malic Ka = 3.5 × 10⁻⁴ → pKa = 3.46
Acetic Ka = 1.8 × 10⁻⁵ → pKa = 4.74; Carbonic Ka1 = 4.3 × 10⁻⁷ → pKa = 6.37
Ranking (decreasing strength): Citric (3.13) > Malic (3.46) > Acetic (4.74) > Carbonic (6.37)
(c) Assess target pH:
Citric acid alone at 0.0200 mol/L gives pH = 2.46. This is below the target range of 3.0–3.5. The concentration must be reduced ~8× to achieve pH 3.0.
(d) Dental warning and acetic acid comparison:
Citric acid at pH 2.46 is well below the enamel dissolution threshold of pH 5.5. Ka1(citric) = 7.4 × 10⁻⁴ gives degree of ionisation = 17.5% — high ionisation producing [H⁺] = 3.50 × 10⁻³ mol/L. Dental warning is warranted.
Acetic acid at same concentration (Ka = 1.8 × 10⁻⁵, 0.0200 mol/L): [H⁺] = √(1.8 × 10⁻⁵ × 0.0200) = 6.0 × 10⁻⁴ mol/L. pH = 3.22. Degree of ionisation = 3.0%. The Ka is ~41× smaller — [H⁺] is ~5.8× lower. pH 3.22 still <5.5, so erosion still occurs, but significantly less aggressive.
ANSWER: (a) Citric pH = 2.46; malic pH = 2.67 (both require quadratic). (b) Citric (3.13) > Malic (3.46) > Acetic (4.74) > Carbonic (6.37). (c) pH 2.46 is below target 3.0–3.5; concentration must be reduced ~8×. (d) Dental warning warranted — 17.5% ionisation, pH 2.46 well below 5.5 threshold. Acetic acid gives pH 3.22 and 3.0% ionisation — less concerning but still below threshold.
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Use the data below to complete the table, then answer the questions.
| Acid | Ka | pKa (calculate) | Acid strength rank (1=strongest) | Kb of conjugate base | Conjugate base strength rank (1=strongest) |
|---|---|---|---|---|---|
| HF | 6.8 × 10⁻⁴ | ||||
| CH₃COOH | 1.8 × 10⁻⁵ | ||||
| HCN | 6.2 × 10⁻¹⁰ | ||||
| HNO₂ | 4.5 × 10⁻⁴ | ||||
| H₂CO₃ | 4.3 × 10⁻⁷ |
(i) Which conjugate base is the strongest base? Calculate its pKb. (ii) A student claims HF (Ka = 6.8 × 10⁻⁴) and HNO₂ (Ka = 4.5 × 10⁻⁴) are "practically the same strength." Evaluate this claim. (iii) Why does the acid with the highest Ka have the lowest Kb for its conjugate base?
Each student response below contains at least one error. Identify the error and write the correct reasoning.
Student 1: "Acid A has pKa = 3.2 and acid B has pKa = 7.5. Therefore acid A is weaker because 3.2 < 7.5."
Student 2: "Ka(CH₃COOH) = 1.8 × 10⁻⁵. Using Ka × Kb = Kw: Kb(NH₃) = Kw/Ka(CH₃COOH) = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰."
Student 3: "For H₃PO₄: Ka = [H⁺][PO₄³⁻]/[H₃PO₄] — this gives the overall Ka for phosphoric acid losing all three protons."
Student 4: "I compared the acidity of HNO₂ and CH₃COOH by measuring the pH of 0.1 mol/L solutions and got pH 2.17 and pH 2.87. Since pH(HNO₂) < pH(CH₃COOH), HNO₂ is a stronger acid — but only because its concentration is higher."
Student 5: "To calculate the pH of 0.050 mol/L H₃PO₄, I set up three ICE tables — one for each Ka — and summed up the H⁺ from all three. My answer was pH 1.63."
1. Four acids have pKa values: W = 2.1; X = 5.8; Y = 9.3; Z = 3.7. Which correctly ranks them from strongest to weakest?
2. Phosphoric acid (H₃PO₄) has Ka1 = 7.5 × 10⁻³, Ka2 = 6.2 × 10⁻⁸, Ka3 = 4.8 × 10⁻¹³. A student calculates pH of 0.100 mol/L H₃PO₄ using all three Ka values and summing H⁺ contributions. A second student uses only Ka1. Which approach is correct, and approximately what pH does the correct approach give?
3. A student measures the pH of 0.100 mol/L solutions of two unknown acids: Acid M gives pH 2.15; Acid N gives pH 3.22. The student concludes that Acid M is a strong acid because it has a lower pH. Evaluate this conclusion.
4. HCN has Ka = 6.2 × 10⁻¹⁰. What is the Kb of CN⁻ (the conjugate base of HCN), and what does this value tell you about a solution of NaCN in water?
5. A food technologist compares cola (pH 3.2, containing H₃PO₄ and H₂CO₃) and sparkling water (pH 4.8, containing only H₂CO₃) for their relative risk of dental erosion. The technologist states: "Sparkling water is safe for teeth because its pH is above 4.0." Which statement best evaluates this claim?
Question 6. A student is given four unknown acids with the following Ka values: Acid J: 3.2 × 10⁻³; Acid K: 8.7 × 10⁻⁷; Acid L: 1.4 × 10⁻¹¹; Acid M: 5.0 × 10⁻⁵. (a) Rank the acids from strongest to weakest. (b) Calculate pKa for Acid J and Acid L. (c) Calculate Kb for the conjugate base of Acid M. (d) Explain why the conjugate base of Acid L is a stronger base than the conjugate base of Acid J.
Question 7. Phosphoric acid (H₃PO₄) is a triprotic acid. Ka1 = 7.5 × 10⁻³; Ka2 = 6.2 × 10⁻⁸; Ka3 = 4.8 × 10⁻¹³. (a) Write the three ionisation equations with correct arrow notation. (b) Calculate the pH of a 0.0800 mol/L H₃PO₄ solution, showing all steps including an assumption check. (c) Calculate Kb for HPO₄²⁻ and predict whether a 0.100 mol/L Na₂HPO₄ solution is acidic, basic, or neutral, giving reasons.
Question 8. A water quality analyst measures the pH of a water sample as 5.1 and determines the total dissolved CO₂ concentration is 0.0080 mol/L. (a) Using Ka1(H₂CO₃) = 4.3 × 10⁻⁷, calculate the expected pH of this H₂CO₃ concentration and compare it to the measured pH. (b) The analyst also measures Ka from the pH data: Ka(obs) = [H⁺]²/(c − [H⁺]). Calculate Ka(obs) using the measured pH 5.1. (c) Compare Ka(obs) to the literature Ka1 value and suggest one reason why the measured pH might differ. (d) The analyst notes this sample is below pH 5.6. Using Ka values and the context of environmental monitoring, explain the significance and what further analysis you would recommend.
MC Q1: A — W > Z > X > Y
Smaller pKa = stronger acid. Ranking by pKa: W (2.1) < Z (3.7) < X (5.8) < Y (9.3). Reading from strongest to weakest: W > Z > X > Y.
MC Q2: B — Second student correct; quadratic required; pH ≈ 1.62
Ka1 ≫ Ka2 ≫ Ka3 (ratios ~10⁵) — only Ka1 matters. Check assumption: degree = 27.4% > 5% → quadratic required → x ≈ 0.0239 mol/L, pH = 1.62.
MC Q3: B — Both acids are weak
A strong 0.100 mol/L acid gives pH = 1.00. Acid M at pH 2.15 has [H⁺] = 7.08 × 10⁻³ < 0.100 → partial ionisation → weak acid. Ka(M) = 5.40 × 10⁻⁴; Ka(N) = 3.66 × 10⁻⁶ — both ≪ 1, both weak.
MC Q4: C — Kb(CN⁻) = 1.6 × 10⁻⁵; NaCN solution is basic
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵. CN⁻ + H₂O ⇌ HCN + OH⁻ → solution is basic.
MC Q5: D — Enamel dissolution threshold is pH 5.5, not 4.0
Sparkling water at pH 4.8 is below 5.5 and does cause enamel erosion, though far less aggressively than cola at pH 3.2.
Q6 — Sample Answer:
(a) J (3.2 × 10⁻³) > M (5.0 × 10⁻⁵) > K (8.7 × 10⁻⁷) > L (1.4 × 10⁻¹¹)
(b) pKa(J) = −log(3.2 × 10⁻³) = 3 − 0.505 = 2.50. pKa(L) = −log(1.4 × 10⁻¹¹) = 11 − 0.146 = 10.85.
(c) Kb(M⁻) = Kw/Ka(M) = 1.0 × 10⁻¹⁴/5.0 × 10⁻⁵ = 2.0 × 10⁻¹⁰.
(d) Acid L has the smallest Ka (weakest acid) → conjugate base L⁻ has the largest Kb = Kw/Ka(L) = 1.0 × 10⁻¹⁴/1.4 × 10⁻¹¹ = 7.1 × 10⁻⁴. Acid J has the largest Ka → conjugate base J⁻ has Kb = 3.1 × 10⁻¹². The inverse relationship is a direct consequence of Ka × Kb = Kw being constant.
Q7 — Sample Answer:
(a) H₃PO₄(aq) ⇌ H⁺(aq) + H₂PO₄⁻(aq) Ka1; H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) Ka2; HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) Ka3.
(b) Ka1/c = 7.5 × 10⁻³/0.0800 = 9.4% > 5% → quadratic required. x² + 7.5 × 10⁻³x − 6.0 × 10⁻⁴ = 0; x = 2.10 × 10⁻² mol/L; pH = −log(2.10 × 10⁻²) = 1.68.
(c) Kb(HPO₄²⁻) = Kw/Ka2 = 1.0 × 10⁻¹⁴/6.2 × 10⁻⁸ = 1.6 × 10⁻⁷. Na₂HPO₄ gives HPO₄²⁻ which hydrolyses: HPO₄²⁻ + H₂O ⇌ H₂PO₄⁻ + OH⁻. Solution is basic (pH > 7).
Q8 — Sample Answer:
(a) x = √(4.3 × 10⁻⁷ × 0.0080) = 5.87 × 10⁻⁵ mol/L; pH(calc) = 4.23. Measured pH = 5.1 is higher than calculated — water appears less acidic than expected for pure H₂CO₃.
(b) [H⁺] = 10⁻⁵·¹ = 7.94 × 10⁻⁶ mol/L; Ka(obs) = (7.94 × 10⁻⁶)²/(0.0080 − 7.94 × 10⁻⁶) = 7.9 × 10⁻⁹.
(c) Ka(obs) = 7.9 × 10⁻⁹ ≪ Ka1(lit.) = 4.3 × 10⁻⁷ (~54× smaller). Dissolved alkalinity (HCO₃⁻ from minerals) partially neutralises the carbonic acid, reducing [H⁺] and raising apparent pH.
(d) pH 5.1 is below the natural threshold of pH 5.6 — additional acid beyond atmospheric CO₂ is present, consistent with acid deposition. Recommended analysis: titrate for total acidity; measure sulfate/nitrate concentrations; compare with nearby non-impacted baseline sample.
Activity B — Spot & Fix Answers:
Student 1 Error: Reversed the pKa ranking. Larger pKa = smaller Ka = weaker acid. Acid A (pKa 3.2) is the stronger acid (smaller pKa). Correct: "Acid A (pKa 3.2) is stronger than Acid B (pKa 7.5)."
Student 2 Error: Applied Ka × Kb = Kw to a non-conjugate pair. CH₃COOH and NH₃ are not conjugate partners. To find Kb(NH₃), use Ka(NH₄⁺): Kb(NH₃) = Kw/Ka(NH₄⁺) = 1.0 × 10⁻¹⁴/5.6 × 10⁻¹⁰ = 1.8 × 10⁻⁵.
Student 3 Error: Treated all three proton removals as a single equilibrium. Each is a separate step. Correct Ka1: Ka1 = [H⁺][H₂PO₄⁻]/[H₃PO₄].
Student 4 Error: Incorrectly attributed the lower pH to higher concentration (both are 0.1 mol/L). Compare using Ka: Ka(HNO₂) = 4.5 × 10⁻⁴ > Ka(CH₃COOH) = 1.8 × 10⁻⁵ → HNO₂ is stronger at the same concentration.
Student 5 Error: Set up three ICE tables unnecessarily. Ka2/Ka1 ≈ 10⁻⁵ — use Ka1 only. One ICE table, check assumption (Ka1/c = 0.075 > 5% → quadratic required).
Return to your Think First. The 2014 BDA report explained by Ka: H₃PO₄ (Ka1 = 7.5 × 10⁻³) is 17,000× stronger than H₂CO₃ (Ka1 = 4.3 × 10⁻⁷). Now check your answers:
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