Every soft drink you consume contains carbonic acid — and the Ka values of its two successive ionisation steps explain exactly why carbonated drinks erode tooth enamel at pH 3.5 but not at pH 5.5, and why the second ionisation barely contributes to acidity at all.
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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?
By the end of this lesson you will be able to:
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 |
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.
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.
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.
Ka values are not just textbook constants — they are the analytical foundation for industrial quality control, pharmaceutical verification, and environmental monitoring, all of which depend on the relationship between Ka, pH, and concentration to make quantitative decisions about real substances.
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.
"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.
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.
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 for Ka1 only: Ka2 = 4.7 × 10⁻¹¹ ≪ Ka1 = 4.3 × 10⁻⁷. The second ionisation starts from HCO₃⁻ present at only ~8 × 10⁻⁵ mol/L, with an extremely small Ka2. The additional [H⁺] from the second step is negligible — 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⁺]. This large ratio confirms that using Ka1 alone introduces negligible error.
(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) ✓ This is why Na₂CO₃ (washing soda) is used as a cleaning agent.
ANSWER: (a) H₂CO₃ ⇌ H⁺ + HCO₃⁻ (Ka1); HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (Ka2). (b) pH = 4.10; justified because Ka2/Ka1 ≈ 10⁻⁴ — second ionisation contributes <0.01% of [H⁺]. (c) Ka1/Ka2 ≈ 10⁴ — first step 10,000× more favourable. (d) Kb(CO₃²⁻) = 2.1 × 10⁻⁴; Na₂CO₃ solution is basic — CO₃²⁻ is conjugate base of a weak acid and hydrolyses to produce OH⁻.
GIVEN: Citric acid Ka1 = 7.4 × 10⁻⁴, c = 0.0200 mol/L; Malic acid Ka = 3.5 × 10⁻⁴, c = 0.0150 mol/L FIND: pH(citric), pH(malic), pKa ranking, target pH assessment, dental risk comparison
(a) pH of citric acid — assumption check:
Ka1/c = 7.4 × 10⁻⁴ / 0.0200 = 0.037 = 3.7% — looks < 5%, 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
= (−7.4 × 10⁻⁴ + √(5.476 × 10⁻⁷ + 5.92 × 10⁻⁵)) / 2 = (−7.4 × 10⁻⁴ + √(5.975 × 10⁻⁵)) / 2
= (−7.4 × 10⁻⁴ + 7.730 × 10⁻³) / 2 = 7.0 × 10⁻³ / 2 = 3.50 × 10⁻³ mol/L
pH(citric) = −log(3.50 × 10⁻³) = 3 − log(3.50) = 3 − 0.544 = 2.46
Malic acid: x = √(3.5 × 10⁻⁴ × 0.0150) = √(5.25 × 10⁻⁶) = 2.29 × 10⁻³; check: 15.3% > 5% → quadratic.
x = (−3.5 × 10⁻⁴ + √(1.225 × 10⁻⁷ + 2.10 × 10⁻⁵)) / 2 = (−3.5 × 10⁻⁴ + √(2.112 × 10⁻⁵)) / 2 = (−3.5 × 10⁻⁴ + 4.596 × 10⁻³) / 2 = 2.12 × 10⁻³ mol/L
pH(malic) = −log(2.12 × 10⁻³) = 3 − log(2.12) = 3 − 0.326 = 2.67
(b) pKa ranking:
Citric Ka1 = 7.4 × 10⁻⁴ → pKa = 4 − log(7.4) = 4 − 0.869 = 3.13
Malic Ka = 3.5 × 10⁻⁴ → pKa = 4 − log(3.5) = 4 − 0.544 = 3.46
Acetic Ka = 1.8 × 10⁻⁵ → pKa = 5 − log(1.8) = 5 − 0.255 = 4.74
Carbonic Ka1 = 4.3 × 10⁻⁷ → pKa = 7 − log(4.3) = 7 − 0.633 = 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 to achieve pH 3.0. At pH 3.0, [H⁺] = 1.0 × 10⁻³ mol/L. Using Ka1 = x²/(c−x): c = x²/Ka1 + x = (1.0 × 10⁻³)²/(7.4 × 10⁻⁴) + 1.0 × 10⁻³ ≈ 1.35 × 10⁻³ + 1.0 × 10⁻³ = 2.35 × 10⁻³ mol/L. The target pH of 3.0 requires approximately 0.0024 mol/L citric acid — about 8× less than 0.0200 mol/L. The formulation requires significant citric acid reduction to meet the dental safety target.
(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 = 3.50 × 10⁻³/0.0200 = 17.5% — high ionisation producing [H⁺] = 3.50 × 10⁻³ mol/L. Sustained daily exposure at pH 2.46 provides 20+ minutes of acid attack per consumption, well below the protective pH 5.5 threshold. 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) = √(3.6 × 10⁻⁷) = 6.0 × 10⁻⁴ mol/L. pH = −log(6.0 × 10⁻⁴) = 3.22. Degree of ionisation = 6.0 × 10⁻⁴/0.0200 = 3.0%. The Ka is ~41× smaller than citric Ka1 — [H⁺] from acetic acid is 3.50 × 10⁻³/6.0 × 10⁻⁴ ≈ 5.8× lower. pH 3.22 is still below 5.5, so enamel erosion still occurs — but the erosion rate is significantly lower. A dental caution still applies, but the risk is substantially less than with citric acid.
ANSWER: (a) Citric acid pH = 2.46 (quadratic required; degree of ionisation = 17.5%); malic acid pH = 2.67 (quadratic required). (b) Citric (pKa 3.13) > Malic (3.46) > Acetic (4.74) > Carbonic (6.37). (c) Citric alone gives pH 2.46 — below target range 3.0–3.5; concentration must be reduced ~8× to pH 3.0. (d) Citric acid at pH 2.46 is well below enamel dissolution threshold (5.5); 17.5% ionisation confirms sustained high [H⁺] — dental warning warranted. Acetic acid at same concentration gives pH 3.22 and only 3.0% ionisation — [H⁺] is 5.8× lower; less aggressive but still below 5.5, so dental caution still applies.
These are the key definitions, relationships, and rules for this lesson. Write them by hand for retention.
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 |
Questions: (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 using the Ka data and the difference in degree of ionisation at 0.1 mol/L. (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."
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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 from the calculated value. (d) The analyst notes that this sample is below pH 5.6. Using Ka values and the context of environmental monitoring, explain the significance of this finding and what further analysis you would recommend.
Return to your Think First response. Can you now answer all three questions with quantitative detail?
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. Option B reverses the direction — larger pKa is weaker, not stronger.
Q2: B — Second student correct; quadratic required; pH ≈ 1.62
Ka1 ≫ Ka2 ≫ Ka3 (ratios ~10⁵) — only Ka1 matters. Check assumption: √(Ka1 × c) = 0.0274 mol/L; degree = 27.4% > 5% → assumption invalid, quadratic required → x ≈ 0.0239 mol/L, pH = 1.62. Option C uses the approximation without checking — invalid here.
Q3: B — Both acids are weak; correct Ka values calculated from pH
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) = (7.08 × 10⁻³)²/(0.100 − 7.08 × 10⁻³) = 5.40 × 10⁻⁴; Ka(N) = 3.66 × 10⁻⁶ — both ≪ 1, both weak. M is the stronger weak acid.
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⁻ is the conjugate base of HCN (a weak acid) → CN⁻ has non-trivial Kb → CN⁻ + H₂O ⇌ HCN + OH⁻ → solution is basic. Option A uses the wrong Kb formula. Option D correctly calculates Kb but incorrectly predicts acidic (NaCN is basic, not acidic).
Q5: D — Enamel dissolution threshold is pH 5.5, not 4.0; sparkling water at pH 4.8 causes erosion
The technologist's threshold of pH 4.0 is incorrect — enamel dissolves below pH 5.5. 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. The claim that "pH above 4.0 is safe" is factually wrong by the dental chemistry criterion.
(a) Ranking by Ka (largest = strongest): 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 − log(3.2) = 3 − 0.505 = 2.50. pKa(L) = −log(1.4 × 10⁻¹¹) = 11 − log(1.4) = 11 − 0.146 = 10.85.
(c) Kb(M⁻) = Kw/Ka(M) = 1.0 × 10⁻¹⁴/5.0 × 10⁻⁵ = 2.0 × 10⁻¹⁰.
(d) Ka × Kb = Kw applies to each conjugate pair. Acid L has the smallest Ka (weakest acid) → its 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 (strongest) → conjugate base J⁻ has Kb = 1.0 × 10⁻¹⁴/3.2 × 10⁻³ = 3.1 × 10⁻¹². The inverse relationship between acid and conjugate base strength is a direct consequence of Ka × Kb = Kw being constant.
(a) Ionisation equations (all ⇌ — weak acid steps):
H₃PO₄(aq) ⇌ H⁺(aq) + H₂PO₄⁻(aq) Ka1 = 7.5 × 10⁻³
H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) Ka2 = 6.2 × 10⁻⁸
HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) Ka3 = 4.8 × 10⁻¹³
(b) pH using Ka1 only (Ka2 and Ka3 negligible):
Ka1/c = 7.5 × 10⁻³/0.0800 = 0.094 = 9.4% > 5% → assumption invalid at this Ka/c ratio → use quadratic.
x² + 7.5 × 10⁻³x − 6.0 × 10⁻⁴ = 0
x = (−7.5 × 10⁻³ + √(5.625 × 10⁻⁵ + 2.40 × 10⁻³))/2 = (−7.5 × 10⁻³ + √(2.456 × 10⁻³))/2
= (−7.5 × 10⁻³ + 0.04956)/2 = 0.04206/2 = 2.10 × 10⁻² mol/L
pH = −log(2.10 × 10⁻²) = 2 − log(2.10) = 2 − 0.322 = 1.68
(c) Kb for HPO₄²⁻:
HPO₄²⁻ is the conjugate base of H₂PO₄⁻ (second ionisation) → Ka for this pair = Ka2 = 6.2 × 10⁻⁸
Kb(HPO₄²⁻) = Kw/Ka2 = 1.0 × 10⁻¹⁴/6.2 × 10⁻⁸ = 1.6 × 10⁻⁷
Na₂HPO₄ → 2Na⁺ (neutral) + HPO₄²⁻. HPO₄²⁻ has Kb = 1.6 × 10⁻⁷ — it accepts H⁺ from water: HPO₄²⁻ + H₂O ⇌ H₂PO₄⁻ + OH⁻. The solution is basic (pH > 7). Na₂HPO₄ is the salt of a strong base (NaOH) and a polyprotic weak acid (H₃PO₄) — salt hydrolysis produces OH⁻, giving a basic solution.
(a) Expected pH from Ka1 and concentration:
Ka1/c = 4.3 × 10⁻⁷/0.0080 = 5.4 × 10⁻⁵ ≪ 0.0025 ✓ (assumption valid)
x = √(4.3 × 10⁻⁷ × 0.0080) = √(3.44 × 10⁻⁹) = 5.87 × 10⁻⁵ mol/L
Verify: 5.87 × 10⁻⁵/0.0080 = 0.73% ≪ 5% ✓
pH(calculated) = −log(5.87 × 10⁻⁵) = 5 − log(5.87) = 5 − 0.769 = 4.23
Measured pH = 5.1. The measured pH is higher than calculated (5.1 > 4.23) — the water appears less acidic than expected for pure H₂CO₃.
(b) Ka(obs) from measured pH 5.1:
[H⁺] = 10⁻⁵·¹ = 7.94 × 10⁻⁶ mol/L
Ka(obs) = [H⁺]²/(c − [H⁺]) = (7.94 × 10⁻⁶)²/(0.0080 − 7.94 × 10⁻⁶) = 6.31 × 10⁻¹¹/7.99 × 10⁻³ = 7.9 × 10⁻⁹
(c) Comparison: Ka(obs) = 7.9 × 10⁻⁹ ≪ Ka1(lit.) = 4.3 × 10⁻⁷. The observed Ka is about 54× smaller than the literature Ka1. A plausible reason: the sample contains dissolved alkalinity (HCO₃⁻ from dissolved minerals) that partially neutralises the carbonic acid, reducing [H⁺] and raising the apparent pH. Other possibilities include the presence of bicarbonate buffering or measurement error in total dissolved CO₂.
(d) Environmental significance: pH 5.1 is below the natural threshold of pH 5.6 (pH of CO₂-saturated pure water in equilibrium with atmospheric CO₂). A water sample below pH 5.6 indicates the presence of additional acid beyond H₂CO₃ from atmospheric CO₂ — consistent with acid deposition (acid rain). The Ka values of H₂SO₄ and HNO₃ (both strong acids, complete ionisation) from combustion emissions explain how small concentrations of these acids can drive pH significantly below 5.6. Recommended further analysis: (1) titrate for total acidity to quantify all acid species; (2) measure sulfate and nitrate ion concentrations (ion chromatography) to identify acid rain contribution; (3) measure pH of a nearby non-impacted sample for baseline comparison.
Student 1 Error: Reversed the pKa ranking. Larger pKa = smaller Ka = weaker acid. Acid A (pKa 3.2) has Ka = 6.3 × 10⁻⁴; Acid B (pKa 7.5) has Ka = 3.2 × 10⁻⁸. Acid A 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 — CH₃COO⁻ is the conjugate base of CH₃COOH, and NH₄⁺ is the conjugate acid of NH₃. To find Kb(NH₃), use the relationship with its conjugate acid: Kb(NH₃) = Kw/Ka(NH₄⁺). Ka(NH₄⁺) = 5.6 × 10⁻¹⁰; Kb(NH₃) = 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 ionisation is a separate step with its own Ka. Correct Ka1 expression: Ka1 = [H⁺][H₂PO₄⁻]/[H₃PO₄]. The expression [H⁺][PO₄³⁻]/[H₃PO₄] would give the product of Ka1 × Ka2 × Ka3 — not a standard Ka.
Student 4 Error: Incorrectly attributed the lower pH of HNO₂ to higher concentration (both solutions are 0.1 mol/L — same concentration). The correct comparison uses Ka: Ka(HNO₂) = 4.5 × 10⁻⁴ > Ka(CH₃COOH) = 1.8 × 10⁻⁵. HNO₂ is a stronger acid with higher degree of ionisation at the same concentration → lower pH. The pH comparison is valid only because the concentrations are equal.
Student 5 Error: Unnecessarily set up three ICE tables. For H₃PO₄, Ka2/Ka1 ≈ 10⁻⁵ — the second ionisation contributes <0.001% of [H⁺]. Correct approach: use Ka1 only, set up one ICE table for H₃PO₄ ⇌ H⁺ + H₂PO₄⁻, check assumption (Ka1/c = 0.075 > 5% → quadratic required), solve for x, calculate pH.
Answer questions on Ka, pKa, comparing acid strengths and conjugate pair relationships before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–12.