Aspirin (acetylsalicylic acid) has Ka = 3.0 × 10⁻⁴ — and the ICE table you will build in this lesson predicts exactly how much H⁺ aspirin releases in your stomach, why it irritates the gastric lining, and why enteric-coated tablets dissolve in the intestine instead.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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:
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⁺].
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
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].
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 |
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:
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.
"pH = pKa for any weak acid solution." — Incorrect. pH = pKa only at the half-equivalence point of a titration, when [A⁻] = [HA]. For a pure weak acid solution, pH = −log(√(Ka × c)) which is approximately pKa/2 + log(c)/2 — always lower than pKa.
"I don't need to check the 5% rule if Ka is small." — Incorrect. Ka must be small relative to c. For a very dilute solution (c = 0.001 mol/L) with Ka = 1.0 × 10⁻⁴, Ka/c = 0.10 = 10% — the assumption fails even though Ka appears small in absolute terms.
"Ka × Kb = Kw can be used for any acid and base." — Incorrect. This only applies to a specific conjugate pair (HA and its conjugate base A⁻). Ka(HCl) × Kb(NH₃) ≠ Kw — HCl and NH₃ are not a conjugate pair.
"A salt formed by neutralisation always has pH = 7." — Incorrect. Only salts from strong acid + strong base give pH = 7. Salts from weak acid + strong base (e.g. CH₃COONa) are basic; salts from strong acid + weak base (e.g. NH₄Cl) are acidic.
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:
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) ✓
ANSWER: [H⁺] = 1.40 × 10⁻³ mol/L; pH = 2.85. Degree of ionisation = 0.93% — simplifying assumption valid.
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.
ANSWER: (a) pH = 2.67. (b) Kb(CH₃COO⁻) = 5.56 × 10⁻¹⁰. (c) pH = 9.07. (d) Claim incorrect — acetate ion is the conjugate base of a weak acid; it hydrolyses to produce OH⁻, giving a basic solution. pH = 7 requires a salt of a strong acid + strong base (e.g. NaCl).
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)).
ANSWER: (a) [H⁺] = 1.66 × 10⁻³ mol/L; Ka = 5.70 × 10⁻⁵. (b) α = 3.32% < 5% — assumption was valid; pH error only 0.01 units. (c) n(HX)rem = 1.50 × 10⁻³ mol; n(X⁻) = 5.00 × 10⁻⁴ mol; pKa = 4.244; pH = 3.77. (d) Error: ICE table assumes [A⁻] = 0 — invalid when X⁻ is present from neutralisation; the solution is a buffer; Henderson-Hasselbalch is required; ICE method ignores common ion suppression and gives erroneously low pH.
Minimum facts and procedures for Lesson 9.
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 |
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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)
Return to your prediction about aspirin vs ibuprofen. You can now calculate this precisely:
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 for an accurate result. Quadratic gives x = 0.01133 mol/L (~3% smaller). Option A is wrong — the assumption must always be checked. Option C uses an incorrect formula (Ka/c has units mol/L, not mol/L). Option D has wrong ICE table sign — H⁺ is produced (+x), not consumed.
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 it as strong (wrong). Option B confuses pKa with pH — equal only at half-equivalence point. Option D uses Ka × c instead of √(Ka × c) — dimensional error.
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 — serious conceptual error. Option C applies ICE table to total volume — still ignores common ion effect of CH₃COO⁻. Option D incorrectly manipulates pKa.
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. Option A took −log[OH⁻] as pH directly. Option B is the same error restated. Option D is wrong — 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 Ka = [H⁺]²/c (approximation — omits c − [H⁺] denominator; gives same answer here since α = 0.63% << 5%, but the exact expression should always be used in the reverse calculation). Option C is 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 the highly specific (and physically unrealistic) case where [H⁺][A⁻] = [H⁺][HA] simultaneously — which cannot occur in a simple weak acid equilibrium. 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 (2.00 × 10⁻⁴ mol in 40 mL total). 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, even though the buffer contains undissociated HA. The Henderson-Hasselbalch equation accounts for this suppression mathematically through the log(n(A⁻)/n(HA)) term, which here gives log(0.5) = −0.30 — the pH is 0.30 units lower than pKa = 3.85, not as low as the pure acid's pH of 2.79.