In 1971, pharmacologist John Vane measured the Ka of aspirin (acetylsalicylic acid) as 3.0 × 10⁻⁴ — ten times larger than ibuprofen's Ka of 1.3 × 10⁻⁵. That difference determines which drug ionises more in the stomach, which is absorbed faster in the intestine, and is why the ICE table method exists: for any weak acid, [H⁺] depends on both Ka and concentration simultaneously, and you cannot find it without solving the equilibrium properly. Vane won the Nobel Prize in 1982.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A pharmacology researcher is comparing two pain relief tablets. Tablet A contains 500 mg of aspirin (acetylsalicylic acid, Ka = 3.0 × 10⁻⁴, M = 180 g/mol). Tablet B contains 500 mg of ibuprofen (Ka = 1.3 × 10⁻⁵, M = 206 g/mol). Both dissolve in 200 mL of stomach fluid (approximately 0.1 mol/L HCl, pH ≈ 1).
The researcher wants to predict which tablet releases more H⁺ ions and therefore causes more gastric irritation. Before you read on, write down your prediction: which tablet do you think releases more H⁺? Is it the one with the higher Ka, the higher mass, or the higher molar concentration in stomach fluid? Will either tablet significantly change the pH of stomach acid? You will return to this prediction at the end of the lesson.
For a strong acid, [H⁺] = concentration because ionisation is complete — no equilibrium to solve. For a weak acid, most molecules remain intact at equilibrium, and the fraction that ionise depends on both Ka and concentration — which is why a systematic method (the ICE table) is required to find the actual [H⁺].
The ICE table is the systematic framework for weak acid calculations. ICE = Initial, Change, Equilibrium — three rows that track concentrations of every species from start through to equilibrium. For a weak acid HA at initial concentration c:
| Stage | [HA] | [H⁺] | [A⁻] |
|---|---|---|---|
| I (Initial) | c | 0 | 0 |
| C (Change) | −x | +x | +x |
| E (Equilibrium) | c − x | x | x |
Substituting E-row into Ka = [H⁺][A⁻]/[HA] gives: Ka = x²/(c − x). Solving for x gives [H⁺] at equilibrium → pH = −log(x).
For 0.1 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵), using [H⁺] = c (the strong acid shortcut) gives pH = 1.0. The correct ICE method gives pH ≈ 2.87 — a difference of nearly 2 pH units, corresponding to a factor of ~74 in [H⁺].
ICE table framework for weak acids — I row: [HA] = c, [H⁺] = 0, [A⁻] = 0. C row: −x, +x, +x. E row: (c−x), x, x. Ka = x²/(c−x). Solve for x = [H⁺], then pH = −log(x). Never use [H⁺] = c for a weak acid — this ignores the equilibrium and overcalculates [H⁺] by up to 100×.
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Quick Check — Card 1
For 0.100 mol/L acetic acid (Ka = 1.8 × 10⁻⁵), which entry belongs in the E (Equilibrium) row for [HA]?
We just saw the ICE table framework — Ka = x²/(c−x) where x = [H⁺] at equilibrium. That raises a question: Do you always need to solve the full quadratic, or is there a shortcut — and how do you know when it's safe? This card answers it → if Ka/c < 0.0025 (degree of ionisation < 5%), use x = √(Ka × c); otherwise solve the full quadratic.
Solving Ka = x²/(c − x) exactly requires the quadratic formula — but for most weak acid problems a simplifying assumption reduces this to a one-step square root calculation, and knowing when the assumption is valid versus when the quadratic is necessary is itself an HSC exam skill.
The simplifying assumption: if x << c, then c − x ≈ c, and Ka ≈ x²/c → x ≈ √(Ka × c).
The assumption is valid when degree of ionisation < 5%. A faster check: if Ka/c < 0.0025, the assumption is valid (since x/c = √(Ka/c)).
When the assumption fails (Ka/c ≥ 0.0025), solve the full quadratic: x² + Ka·x − Ka·c = 0, taking the positive root only:
x = (−Ka + √(Ka² + 4·Ka·c)) / 2
Simplifying assumption: if Ka/c < 0.0025, use x = √(Ka × c) for [H⁺] (saves solving a quadratic). Always verify AFTER: degree of ionisation = (x/c) × 100% must be < 5%. If it fails, use the quadratic: x = (−Ka + √(Ka² + 4Ka·c))/2. Never write "assumption valid" without actually checking.
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Quick Check — Card 2
For 0.0100 mol/L weak acid with Ka = 1.0 × 10⁻⁴, what is Ka/c and is the simplifying assumption valid?
We just saw the ICE table and simplifying assumption for weak acids. That raises a question: How do you handle a weak base — does the same ICE approach apply? This card answers it → identical ICE table but products are BH⁺ and OH⁻; use Kb instead of Ka; then convert: [OH⁻] → pOH = −log[OH⁻] → pH = 14 − pOH.
Weak base pH calculations follow exactly the same ICE table logic as weak acid calculations — the only differences are that the equilibrium produces OH⁻ instead of H⁺, the equilibrium constant is Kb instead of Ka, and the final step requires converting [OH⁻] to pH via the pOH route.
For a weak base B at initial concentration c: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq). Kb = [BH⁺][OH⁻]/[B].
| Stage | [B] | [BH⁺] | [OH⁻] |
|---|---|---|---|
| I (Initial) | c | 0 | 0 |
| C (Change) | −x | +x | +x |
| E (Equilibrium) | c − x | x | x |
If Kb/c < 0.0025: x = √(Kb × c) = [OH⁻]. Then: pOH = −log[OH⁻] → pH = 14 − pOH.
The conjugate pair relationship Ka × Kb = Kw = 1.0 × 10⁻¹⁴ connects acid and base calculations. Given Ka for HA, find Kb for A⁻: Kb = Kw/Ka. For example, CH₃COO⁻ (conjugate base of acetic acid, Ka = 1.8 × 10⁻⁵): Kb = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰.
| Weak base | Kb | c (mol/L) | [OH⁻] = √(Kb × c) | pOH | pH |
|---|---|---|---|---|---|
| NH₃ | 1.8 × 10⁻⁵ | 0.10 | 1.34 × 10⁻³ | 2.87 | 11.13 |
| CH₃COO⁻ | 5.6 × 10⁻¹⁰ | 0.10 | 7.48 × 10⁻⁶ | 5.13 | 8.87 |
| CO₃²⁻ | 2.1 × 10⁻⁴ | 0.050 | 3.24 × 10⁻³ | 2.49 | 11.51 |
Weak base ICE table: B + H₂O ⇌ BH⁺ + OH⁻; same structure as weak acid but products differ. Route: [OH⁻] = √(Kb × c) (if Kb/c < 0.0025) → pOH = −log[OH⁻] → pH = 14 − pOH. Ka × Kb = Kw only for conjugate pairs (HA / A⁻). Never write pH = −log[OH⁻] — that gives acidic pH for a basic solution.
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Quick Check — Card 3
NH₃ has Kb = 1.8 × 10⁻⁵. For 0.10 mol/L NH₃, [OH⁻] = 1.34 × 10⁻³ mol/L. What is the correct pH?
We just saw the ICE table gives Ka when you know pH is unknown — but can you run it in reverse? That raises a question: If you measure pH of a known-concentration weak acid in the lab, how do you find Ka? This card answers it → [H⁺] = 10⁻ᵖᴴ; [A⁻] = [H⁺]; [HA] = c − [H⁺]; Ka = [H⁺]²/(c − [H⁺]).
Every Ka value in every data table was originally determined from a pH measurement — and being able to reverse the calculation (from measured pH back to Ka) is both a practical laboratory skill and a frequently examined HSC calculation type.
If the pH of a weak acid solution of known concentration is measured, Ka can be calculated by reversing the ICE table:
Reverse ICE (Ka from pH): [H⁺] = 10⁻ᵖᴴ → [A⁻] = [H⁺] (1:1 stoichiometry) → [HA]eq = c − [H⁺] → Ka = [H⁺]²/(c − [H⁺]). Always use (c − [H⁺]) in the denominator, not just c. Degree of ionisation α = ([H⁺]/c) × 100%; if < 5%, the simplifying assumption would have been valid. Ka ≠ pH.
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Quick Check — Card 4
A 0.100 mol/L weak acid has pH 3.00. What is [H⁺] and the correct expression for Ka?
We just saw how to extract Ka from a measured pH using the reverse ICE calculation. That raises a question: What happens when you add a strong base to a weak acid but don't neutralise all of it — which method applies? This card answers it → partial neutralisation creates a buffer; use Henderson-Hasselbalch: pH = pKa + log(n(A⁻)/n(HA)); at half-equivalence pH = pKa.
When a weak acid is partially neutralised by a strong base — before the equivalence point — the solution contains significant amounts of both the weak acid and its conjugate base simultaneously, and the pH calculation must account for this mixture using the Henderson-Hasselbalch equation.
When NaOH is added to HA before the equivalence point: HA + OH⁻ → A⁻ + H₂O (goes to completion — NaOH is strong). After the reaction, some HA remains and A⁻ has been formed. The solution contains both HA and A⁻ — this is a buffer solution (developed fully in L13).
pH is calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). In the same volume, [A⁻]/[HA] = n(A⁻)/n(HA) (volumes cancel). At the half-equivalence point — exactly half the acid neutralised — n(A⁻) = n(HA), so log(1) = 0 and pH = pKa.
Partial neutralisation creates a buffer (HA + A⁻ both present). Henderson-Hasselbalch: pH = pKa + log(n(A⁻)/n(HA)) — use moles directly (volumes cancel in ratio). At the half-equivalence point n(A⁻) = n(HA) → log(1) = 0 → pH = pKa. Never use the plain ICE table for a solution containing both HA and A⁻ — it ignores the common ion effect.
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Quick Check — Card 5
25.0 mL of 0.100 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵, pKa = 4.74) is mixed with 12.5 mL of 0.100 mol/L NaOH. What is the pH?
Worked Example 1 — ICE Table pH Calculation with Assumption Check
Calculate the pH of 0.150 mol/L propanoic acid (CH₃CH₂COOH, Ka = 1.3 × 10⁻⁵) at 25°C. Check whether the simplifying assumption is valid.
GIVEN: c = 0.150 mol/L, Ka = 1.3 × 10⁻⁵. FIND: pH.
Pre-check assumption: Ka/c = (1.3 × 10⁻⁵)/0.150 = 8.7 × 10⁻⁵ = 0.0087% << 5% → assumption likely valid. Proceed with square root.
ICE table:
| Stage | [CH₃CH₂COOH] | [H⁺] | [CH₃CH₂COO⁻] |
|---|---|---|---|
| I | 0.150 | 0 | 0 |
| C | −x | +x | +x |
| E | 0.150 − x | x | x |
Apply simplifying assumption: x = √(Ka × c) = √(1.3 × 10⁻⁵ × 0.150) = √(1.95 × 10⁻⁶) = 1.396 × 10⁻³ mol/L.
Verify: degree of ionisation = (1.396 × 10⁻³/0.150) × 100% = 0.93% < 5% ✓ Assumption valid.
Calculate pH: pH = −log(1.396 × 10⁻³) = −log(1.396) + 3 = −0.145 + 3 = 2.85.
Sanity check: pH 2.85 < 7 ✓ (acid). pH 2.85 > 1.0 (not treating it as strong acid) ✓
Worked Example 2 — Multi-Part: Ka, Kb, Conjugate Base pH, and Claim Evaluation
Acetic acid has Ka = 1.8 × 10⁻⁵. (a) Calculate the pH of 0.250 mol/L CH₃COOH. (b) Calculate Kb for the acetate ion (CH₃COO⁻). (c) Calculate the pH of 0.250 mol/L sodium acetate (CH₃COONa). (d) Evaluate the claim: "The pH of 0.250 mol/L sodium acetate should be 7 because it is a salt formed by neutralisation."
GIVEN: Ka(CH₃COOH) = 1.8 × 10⁻⁵; c = 0.250 mol/L for both solutions.
METHOD (a — acetic acid pH): Check: Ka/c = 1.8 × 10⁻⁵/0.250 = 7.2 × 10⁻⁵ << 0.0025 ✓. x = √(1.8 × 10⁻⁵ × 0.250) = √(4.5 × 10⁻⁶) = 2.121 × 10⁻³ mol/L. Verify: 2.121 × 10⁻³/0.250 = 0.85% < 5% ✓. pH = −log(2.121 × 10⁻³) = 2.67.
METHOD (b — Kb for CH₃COO⁻): Ka × Kb = Kw → Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰.
METHOD (c — sodium acetate pH): CH₃COONa → Na⁺ (neutral spectator) + CH₃COO⁻ (weak base). ICE for CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻: Check: Kb/c = 5.56 × 10⁻¹⁰/0.250 = 2.22 × 10⁻⁹ << 0.0025 ✓. x = √(5.56 × 10⁻¹⁰ × 0.250) = √(1.39 × 10⁻¹⁰) = 1.179 × 10⁻⁵ mol/L = [OH⁻]. Verify: 1.179 × 10⁻⁵/0.250 = 0.0047% < 5% ✓. pOH = −log(1.179 × 10⁻⁵) = 4.93. pH = 14.00 − 4.93 = 9.07.
METHOD (d — evaluate claim): The student's claim is incorrect. Sodium acetate is the salt of a strong base (NaOH) and a weak acid (CH₃COOH). A neutral solution (pH = 7) results only from a salt of a strong acid + strong base. The acetate ion is the conjugate base of a weak acid — it has a significant tendency to accept H⁺ from water (Kb = 5.56 × 10⁻¹⁰), producing OH⁻ and making the solution basic (pH = 9.07). "Neutralisation" refers to the process, not the product.
Worked Example 3 — Band 6: Ka from pH, Quadratic Check, and Partial Neutralisation
A 0.0500 mol/L solution of weak monoprotic acid HX is measured at 25°C and found to have pH 2.78. (a) Calculate [H⁺] and the Ka of HX. (b) Check whether the simplifying assumption would have been valid for this calculation. (c) 40.0 mL of 0.0500 mol/L HX is mixed with 10.0 mL of 0.0500 mol/L NaOH. Calculate the pH of the resulting mixture. (d) A student attempts part (c) using x = √(Ka × c(HX)). Identify the conceptual error and explain the correct method.
GIVEN: c = 0.0500 mol/L, pH = 2.78; part (c): 40.0 mL HX + 10.0 mL NaOH, both 0.0500 mol/L.
METHOD (a — [H⁺] and Ka): [H⁺] = 10⁻²·⁷⁸ = 10⁻³ × 10⁰·²² = 1.0 × 10⁻³ × 1.660 = 1.66 × 10⁻³ mol/L. Ka = [H⁺]²/(c − [H⁺]) = (1.66 × 10⁻³)²/(0.0500 − 1.66 × 10⁻³) = (2.756 × 10⁻⁶)/(0.04834) = 5.70 × 10⁻⁵.
METHOD (b — check assumption): α = ([H⁺]/c) × 100% = (1.66 × 10⁻³/0.0500) × 100% = 3.32% < 5% ✓. The simplifying assumption would have been valid — x = √(5.70 × 10⁻⁵ × 0.0500) = √(2.85 × 10⁻⁶) = 1.688 × 10⁻³ → pH = 2.77. Error of only 0.01 pH units vs measured 2.78. ✓
METHOD (c — partial neutralisation): n(HX) = 0.0500 × 0.0400 = 2.00 × 10⁻³ mol. n(OH⁻) = 0.0500 × 0.0100 = 5.00 × 10⁻⁴ mol. After HA + OH⁻ → A⁻ + H₂O (complete): n(HX) remaining = 2.00 × 10⁻³ − 5.00 × 10⁻⁴ = 1.50 × 10⁻³ mol. n(X⁻) formed = 5.00 × 10⁻⁴ mol. pKa = −log(5.70 × 10⁻⁵) = 4.244. pH = pKa + log(n(X⁻)/n(HX)) = 4.244 + log(5.00 × 10⁻⁴/1.50 × 10⁻³) = 4.244 + log(0.333) = 4.244 − 0.477 = 3.77.
METHOD (d — conceptual error): The student's approach uses x = √(Ka × c(HX)) — the ICE table method for a pure weak acid with [A⁻] = 0 initially. This is wrong because the solution after partial neutralisation contains significant X⁻ (5.00 × 10⁻⁴ mol) simultaneously with HX. The ICE table assumes [A⁻] = 0 at the start — completely invalid here. This is a buffer solution. The X⁻ already present suppresses further ionisation of HX (common ion effect), making [H⁺] lower than the ICE table predicts. The correct method is Henderson-Hasselbalch: pH = pKa + log(n(X⁻)/n(HX)).
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(a) CH₃COOH is a weak acid — it only partially ionises. Assuming [H⁺] = 0.1 mol/L treats it as if it fully ionises like HCl, massively overestimating [H⁺] by a factor of ~74. (b) The ICE table: I row has [HA]=0.1, [H⁺]=0, [A⁻]=0. C row has changes of −x, +x, +x. E row gives [H⁺]=x. Substituting into Ka = x²/(0.1−x) and solving gives x ≈ 0.00134, so pH = −log(0.00134) ≈ 2.87.
For each weak acid or base, set up a full ICE table, check the simplifying assumption, and calculate pH. Show all working.
| Species | Concentration | Ka or Kb | Ka/c check | [H⁺] or [OH⁻] | pH |
|---|---|---|---|---|---|
| HNO₂ (nitrous acid) | 0.100 mol/L | Ka = 4.5 × 10⁻⁴ | |||
| HCN (hydrocyanic acid) | 0.0500 mol/L | Ka = 6.2 × 10⁻¹⁰ | |||
| NH₃ (ammonia) | 0.200 mol/L | Kb = 1.8 × 10⁻⁵ | |||
| CH₃COOH (acetic acid) | 0.0100 mol/L | Ka = 1.8 × 10⁻⁵ |
1. A student sets up an ICE table for 0.200 mol/L hydrofluoric acid (HF, Ka = 6.8 × 10⁻⁴) and obtains x = 0.0117 mol/L using the simplifying assumption, without checking the validity. What error has the student made?
2. The Ka of benzoic acid (C₆H₅COOH) is 6.5 × 10⁻⁵. Calculate the pH of 0.100 mol/L benzoic acid. Which answer and method are correct?
3. 25.0 mL of 0.100 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵) is mixed with 12.5 mL of 0.100 mol/L NaOH. A student calculates pH using x = √(Ka × 0.100). What is wrong and what is the correct pH?
4. Ammonia has Kb = 1.8 × 10⁻⁵. What is the pH of 0.050 mol/L NH₃?
5. A 0.050 mol/L solution of weak acid HA has a measured pH of 3.50 at 25°C. Which of the following correctly calculates Ka?
Question 6. Calculate the pH of each of the following at 25°C. Show a full ICE table and assumption check for each.
(a) 0.0800 mol/L chloroacetic acid (ClCH₂COOH, Ka = 1.4 × 10⁻³) (b) 0.500 mol/L sodium propanoate (CH₃CH₂COONa; Ka(CH₃CH₂COOH) = 1.3 × 10⁻⁵)
Question 7. A 0.200 mol/L solution of weak acid HB has a measured pH of 2.52 at 25°C.
(a) Calculate [H⁺] and Ka of HB. (b) Calculate the degree of ionisation. Would the simplifying assumption have been valid? (c) A student claims "Ka of HB = 10⁻²·⁵² because Ka equals [H⁺] in a weak acid solution." Identify and correct the error in this reasoning.
Question 8. Lactic acid (HC₃H₅O₃, Ka = 1.4 × 10⁻⁴) builds up in muscles during intense exercise. Consider a 0.0200 mol/L solution of lactic acid in muscle fluid at 25°C.
(a) Determine whether the simplifying assumption is valid for this system. If not, use the quadratic. Calculate [H⁺] and pH. (3 marks) (b) 30.0 mL of 0.0200 mol/L lactic acid is mixed with 10.0 mL of 0.0200 mol/L NaOH. Calculate the pH of the resulting buffer mixture. (2 marks) (c) Explain why the pH you calculated in (b) is higher than the pH in (a), even though more acid is present than in (b)'s buffer. Use the concept of the common ion effect in your answer. (2 marks)
Q1: B — x = √(6.8 × 10⁻⁴ × 0.200) = √(1.36 × 10⁻⁴) = 0.01166 mol/L. Check: 0.01166/0.200 = 5.83% > 5% — assumption marginally invalid. The student must check after calculating x and apply the quadratic. Quadratic gives x = 0.01133 mol/L (~3% smaller). Option A: always must check. Option C: Ka/c has units mol/L (wrong formula). Option D: wrong sign — H⁺ is produced (+x).
Q2: C — Benzoic acid is weak — ICE method required. Ka/c = 6.5 × 10⁻⁵/0.100 = 6.5 × 10⁻⁴ << 0.0025 ✓. x = √(6.5 × 10⁻⁵ × 0.100) = √(6.5 × 10⁻⁶) = 2.55 × 10⁻³. Verify: 2.55% < 5% ✓. pH = −log(2.55 × 10⁻³) = 2.59. Option A: treats as strong. Option B: confuses pKa with pH. Option D: uses Ka × c instead of √(Ka × c).
Q3: B — n(CH₃COOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. n(OH⁻) = 0.100 × 0.0125 = 1.25 × 10⁻³ mol. After reaction: n(CH₃COOH)rem = 1.25 × 10⁻³ mol; n(CH₃COO⁻) = 1.25 × 10⁻³ mol. Half-equivalence point: [A⁻]/[HA] = 1 → pH = pKa = −log(1.8 × 10⁻⁵) = 4.74. Option A: ignores NaOH. Option C: still ignores common ion effect of CH₃COO⁻.
Q4: C — [OH⁻] = √(1.8 × 10⁻⁵ × 0.050) = √(9.0 × 10⁻⁷) = 9.49 × 10⁻⁴ mol/L. pOH = −log(9.49 × 10⁻⁴) = 3.02. pH = 14.00 − 3.02 = 10.98. Options A and B: took −log[OH⁻] as pH directly (wrong). Option D: NH₃ is basic, not neutral.
Q5: B — [H⁺] = 10⁻³·⁵⁰ = 3.16 × 10⁻⁴ mol/L. Ka = [H⁺]²/(c − [H⁺]) = (3.16 × 10⁻⁴)²/(0.050 − 3.16 × 10⁻⁴) = (9.99 × 10⁻⁸)/0.04968 = 2.01 × 10⁻⁶. Option A: uses approximation Ka = [H⁺]²/c (omits c − [H⁺]). Option C: nonsensical. Option D: confuses Ka with [H⁺].
(a) 0.0800 mol/L chloroacetic acid (Ka = 1.4 × 10⁻³):
Check: Ka/c = 1.4 × 10⁻³/0.0800 = 0.0175 > 0.0025 ✗. x = √(1.4 × 10⁻³ × 0.0800) = √(1.12 × 10⁻⁴) = 1.058 × 10⁻² mol/L. Verify: (1.058 × 10⁻²/0.0800) × 100% = 13.2% > 5% ✗. Assumption fails — use quadratic.
x² + (1.4 × 10⁻³)x − (1.4 × 10⁻³)(0.0800) = 0 → x² + 1.4 × 10⁻³x − 1.12 × 10⁻⁴ = 0
x = (−1.4 × 10⁻³ + √((1.4 × 10⁻³)² + 4 × 1.12 × 10⁻⁴))/2 = (−1.4 × 10⁻³ + √(4.49 × 10⁻⁴))/2 = (−1.4 × 10⁻³ + 0.02119)/2 = 0.009895/2 = 9.79 × 10⁻³ mol/L
pH = −log(9.79 × 10⁻³) = 2.01
(b) 0.500 mol/L sodium propanoate (Ka(propanoic acid) = 1.3 × 10⁻⁵):
Kb(CH₃CH₂COO⁻) = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.3 × 10⁻⁵) = 7.69 × 10⁻¹⁰
Check: Kb/c = 7.69 × 10⁻¹⁰/0.500 = 1.54 × 10⁻⁹ << 0.0025 ✓
[OH⁻] = √(7.69 × 10⁻¹⁰ × 0.500) = √(3.85 × 10⁻¹⁰) = 1.962 × 10⁻⁵ mol/L
Verify: (1.962 × 10⁻⁵/0.500) × 100% = 0.0039% << 5% ✓
pOH = −log(1.962 × 10⁻⁵) = 4.707; pH = 14.00 − 4.707 = 9.29
(a) [H⁺] and Ka:
[H⁺] = 10⁻²·⁵² = 10⁻³ × 10⁰·⁴⁸ = 1.0 × 10⁻³ × 3.020 = 3.02 × 10⁻³ mol/L
Ka = [H⁺]²/(c − [H⁺]) = (3.02 × 10⁻³)²/(0.200 − 3.02 × 10⁻³) = (9.12 × 10⁻⁶)/(0.1970) = 4.63 × 10⁻⁵
(b) Degree of ionisation:
α = (3.02 × 10⁻³/0.200) × 100% = 1.51% < 5% ✓. The simplifying assumption would have been valid — the error from using x = √(Ka × c) would be less than 5% in [H⁺].
(c) Student's error:
The student claims Ka = [H⁺] = 10⁻²·⁵². This is incorrect. Ka is an equilibrium constant with units related to concentration — it equals [H⁺][A⁻]/[HA] = [H⁺]²/(c − [H⁺]), not [H⁺] itself. Ka describes the position of the equilibrium (the ratio of products to reactants); [H⁺] describes the actual concentration of H⁺ in a specific solution. They would only be numerically equal in a highly specific and physically unrealistic case. Correct Ka = 4.63 × 10⁻⁵ ≠ [H⁺] = 3.02 × 10⁻³.
(a) [H⁺] and pH of 0.0200 mol/L lactic acid:
Ka/c = 1.4 × 10⁻⁴/0.0200 = 7.0 × 10⁻³ > 0.0025 ✗. x = √(1.4 × 10⁻⁴ × 0.0200) = √(2.8 × 10⁻⁶) = 1.673 × 10⁻³ mol/L.
Verify: (1.673 × 10⁻³/0.0200) × 100% = 8.37% > 5% ✗. Assumption fails — use quadratic.
x² + (1.4 × 10⁻⁴)x − (1.4 × 10⁻⁴)(0.0200) = 0
x = (−1.4 × 10⁻⁴ + √((1.4 × 10⁻⁴)² + 4 × 1.4 × 10⁻⁴ × 0.0200))/2
= (−1.4 × 10⁻⁴ + √(1.96 × 10⁻⁸ + 1.12 × 10⁻⁵))/2 = (−1.4 × 10⁻⁴ + √(1.122 × 10⁻⁵))/2
= (−1.4 × 10⁻⁴ + 3.350 × 10⁻³)/2 = 3.210 × 10⁻³/2 = 1.605 × 10⁻³ mol/L
pH = −log(1.605 × 10⁻³) = 2.79
(b) pH of buffer (30.0 mL HA + 10.0 mL NaOH, both 0.0200 mol/L):
n(HA) = 0.0200 × 0.0300 = 6.00 × 10⁻⁴ mol; n(OH⁻) = 0.0200 × 0.0100 = 2.00 × 10⁻⁴ mol
After HA + OH⁻ → A⁻ + H₂O: n(HA)rem = 4.00 × 10⁻⁴ mol; n(A⁻) = 2.00 × 10⁻⁴ mol
pKa = −log(1.4 × 10⁻⁴) = 3.854
pH = 3.854 + log(2.00 × 10⁻⁴/4.00 × 10⁻⁴) = 3.854 + log(0.500) = 3.854 − 0.301 = 3.55
(c) Why pH(buffer) > pH(pure acid):
In the buffer (b), the solution contains both lactic acid HA and its conjugate base A⁻ (lactate). The A⁻ ion is already present at significant concentration. By Le Chatelier's principle, this product of the equilibrium HA ⇌ H⁺ + A⁻ suppresses the ionisation of HA — this is the common ion effect. The equilibrium shifts to the left, reducing [H⁺] compared to what pure HA alone would produce. As a result, the buffer has a higher pH (less acidic) than the pure weak acid at the same total concentration of HA. The Henderson-Hasselbalch equation accounts for this suppression mathematically through the log(n(A⁻)/n(HA)) term.
Return to your prediction about aspirin vs ibuprofen. Recall John Vane's 1971 measurements: Ka(aspirin) = 3.0 × 10⁻⁴, Ka(ibuprofen) = 1.3 × 10⁻⁵. You can now calculate which releases more H⁺ in the stomach using the ICE table method:
A 0.0400 mol/L solution of weak monoprotic acid HA has a measured pH of 3.20 at 25°C.
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