A patient in diabetic ketoacidosis has a blood pH of 7.10 — just 0.30 units below normal. Without the bicarbonate buffer system, that same metabolic acid load would drive blood pH to around 6.80, causing cardiac arrest within minutes. The buffer system bought the patient enough time to reach hospital.
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A biochemistry student adds 0.001 mol of HCl to two beakers, each containing 100 mL of solution. Beaker A contains pure water (pH 7.00). Beaker B contains a mixture of 0.100 mol/L acetic acid and 0.100 mol/L sodium acetate (pH 4.74). After adding the HCl: Beaker A drops from pH 7.00 to pH 2.00 — a change of 5 pH units. Beaker B drops from pH 4.74 to pH 4.66 — a change of only 0.08 pH units. Same amount of acid, same volume, dramatically different pH responses.
Before reading on: What is present in Beaker B that is not present in Beaker A, and how does it resist the pH change? What would happen to Beaker B's pH if you added NaOH instead of HCl? What do you think limits a buffer's ability to resist pH changes?
By the end of this lesson you will be able to:
A buffer is one of the most misunderstood concepts in Module 6 — students frequently confuse it with a neutralising agent, a diluting agent, or a substance that simply has a particular pH, and none of these misconceptions survives a precise examination of what a buffer actually does.
A buffer is a solution that resists significant changes in pH when small amounts of strong acid or strong base are added, or when the solution is diluted with a small amount of water. The key word is "resists" — a buffer does not prevent pH changes entirely, but it dramatically limits them compared to an unbuffered solution. A buffer consists of a weak acid (HA) and its conjugate base (A⁻) present simultaneously in comparable concentrations.
The weak acid component provides resistance to base addition — when OH⁻ is added, HA reacts with it: HA + OH⁻ → A⁻ + H₂O. The conjugate base component provides resistance to acid addition — when H⁺ is added, A⁻ reacts with it: A⁻ + H⁺ → HA. Both reactions consume the added species before it can significantly change [H⁺].
Three things a buffer is NOT: (1) a neutral solution — buffers can be acidic or basic depending on their pKa (the acetic acid/acetate buffer has pH 4.74 when [HA] = [A⁻] — fully functional, but acidic); (2) a neutralising agent — it absorbs added acid or base into the equilibrium, it does not neutralise them to completion; (3) a substance with unlimited capacity — every buffer has finite capacity determined by the moles of HA and A⁻ present.
| Property | Buffer solution | Pure water | Strong acid solution |
|---|---|---|---|
| pH stability | Resists change on small acid/base addition | pH changes dramatically | pH changes dramatically |
| Components required | Weak acid + conjugate base (comparable concentrations) | H₂O only | One species only |
| pH determined by | pKa + ratio [A⁻]/[HA] (Henderson-Hasselbalch) | Kw only | Concentration of acid/base |
| Response to added H⁺ | A⁻ + H⁺ → HA (consumes H⁺, minor pH change) | pH drops dramatically | pH drops dramatically |
| Response to added OH⁻ | HA + OH⁻ → A⁻ + H₂O (consumes OH⁻, minor pH change) | pH rises dramatically | pH rises dramatically |
The two equations that describe buffer action are the most important equations in this lesson — understanding each one at the molecular level, rather than just memorising it, is what makes buffer theory accessible under exam pressure.
Consider a buffer made from acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵) and sodium acetate (CH₃COONa). In solution, CH₃COOH (weak acid component) and CH₃COO⁻ (conjugate base component) coexist in equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻.
When a small amount of strong acid (H⁺) is added: H⁺ + CH₃COO⁻ → CH₃COOH. The added H⁺ is consumed by reacting with the abundant CH₃COO⁻, converting it to CH₃COOH. [CH₃COO⁻] decreases slightly and [CH₃COOH] increases slightly — because both were present in large amounts, neither changes significantly in relative terms. The log([A⁻]/[HA]) term in Henderson-Hasselbalch changes only slightly — pH changes by a small amount.
When a small amount of strong base (OH⁻) is added: OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O. The added OH⁻ is consumed by reacting with the abundant CH₃COOH. [CH₃COOH] decreases slightly and [CH₃COO⁻] increases slightly — again, both were present in large amounts, so the ratio changes only modestly — pH changes by a small amount.
Mathematical reason for buffering: pH = pKa + log([A⁻]/[HA]). Adding a small amount of H⁺ or OH⁻ changes the numerator and denominator of the ratio by equal molar amounts — but because both are large, the relative change in the ratio is small. The logarithm compresses this further into a tiny pH change.
The Henderson-Hasselbalch equation is not a separate formula to memorise in isolation — it is the Ka expression rearranged to solve for pH directly, and deriving it from Ka in two algebraic steps demonstrates that it is simply the equilibrium expression in a more convenient form.
Derivation: Start with Ka = [H⁺][A⁻]/[HA]. Rearrange: [H⁺] = Ka × [HA]/[A⁻]. Take −log of both sides: −log[H⁺] = −log(Ka) − log([HA]/[A⁻]). pH = pKa − log([HA]/[A⁻]) = pKa + log([A⁻]/[HA]).
Three key cases: (1) When [A⁻] = [HA]: log(1) = 0 → pH = pKa. This is the half-equivalence point, and it is where buffer capacity is maximised. (2) When [A⁻] > [HA]: log > 0 → pH > pKa — more basic than pKa. (3) When [A⁻] < [HA]: log < 0 → pH < pKa — more acidic than pKa.
Buffer capacity: A buffer is effective within approximately one pH unit on either side of pKa — the effective buffering range is approximately pKa ± 1. Outside this range, one component is present in such low relative concentration that it cannot adequately resist pH changes.
| [A⁻]/[HA] ratio | log([A⁻]/[HA]) | pH relative to pKa | Buffering effectiveness |
|---|---|---|---|
| 10:1 | +1.00 | pH = pKa + 1.00 | Weakening — A⁻ mostly consumed if more H⁺ added |
| 2:1 | +0.30 | pH = pKa + 0.30 | Moderate — still effective |
| 1:1 | 0 | pH = pKa | Maximum buffer capacity |
| 1:2 | −0.30 | pH = pKa − 0.30 | Moderate — still effective |
| 1:10 | −1.00 | pH = pKa − 1.00 | Weakening — HA mostly consumed if more OH⁻ added |
A buffer solution is not found ready-made in nature — it is deliberately prepared by combining specific amounts of a weak acid and its conjugate base, and the quantitative control of this ratio is precisely what gives a buffer its target pH.
Method 1 — Direct mixing of weak acid and conjugate base salt. Choose a weak acid with a pKa close to the target pH (within ±1 unit). Calculate the required [A⁻]/[HA] ratio using Henderson-Hasselbalch: [A⁻]/[HA] = 10^(pH − pKa). Dissolve the calculated masses of weak acid and its sodium (or potassium) salt in water.
Example: to prepare an acetate buffer at pH 5.00 using CH₃COOH (Ka = 1.8 × 10⁻⁵, pKa = 4.74). Desired ratio = 10^(5.00 − 4.74) = 10^0.26 = 1.82. For a 0.100 mol/L total buffer: [CH₃COO⁻] + [CH₃COOH] = 0.100. Solving: [CH₃COO⁻] = 0.0645 mol/L; [CH₃COOH] = 0.0355 mol/L.
Method 2 — Partial neutralisation of weak acid with strong base. Add a calculated volume of strong base to a weak acid solution, neutralising the required fraction. Since HA + OH⁻ → A⁻ + H₂O (complete), moles of A⁻ formed = moles of OH⁻ added, and moles of HA remaining = initial moles − moles of OH⁻. Apply Henderson-Hasselbalch to the n(A⁻)/n(HA) ratio. This is the most common method in HSC questions and the method by which buffers form during a titration before the equivalence point.
The biological and environmental significance of buffer systems is not incidental to their chemistry — it is the reason the chemistry matters, and understanding the H₂CO₃/HCO₃⁻ blood buffer at the molecular level connects every calculation in this lesson to a real system where the margin between normal function and death is 0.3 pH units.
The blood buffer system. Normal blood pH is 7.35–7.45. The primary buffer is the carbonate system: H₂CO₃(aq)/HCO₃⁻(aq), pKa = 6.10. At pH 7.40: 7.40 = 6.10 + log([HCO₃⁻]/[H₂CO₃]) → [HCO₃⁻]/[H₂CO₃] = 20:1. Blood operates far from equal concentrations of the two components — yet it is maintained by continuously replenishing H₂CO₃ from dissolved CO₂: CO₂(aq) + H₂O ⇌ H₂CO₃. Breathing rate regulates CO₂ concentration — faster breathing expels more CO₂, reducing [H₂CO₃] and raising pH (respiratory alkalosis). Slower breathing accumulates CO₂, lowering pH (respiratory acidosis).
When metabolic acids add H⁺ to blood: H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O. The H₂CO₃ formed decomposes to CO₂, which is expelled by the lungs — effectively removing the added H⁺ as gaseous CO₂. This makes blood buffering an "open" system that regenerates capacity far beyond what a closed solution buffer could provide.
Ocean buffering and ocean acidification. Ocean pH is maintained at ~8.1 by the carbonate system: CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. Increased atmospheric CO₂ from fossil fuel combustion dissolves in ocean water, pushing the equilibrium right and increasing [H⁺] — lowering ocean pH. Since the industrial revolution, average ocean pH has fallen from ~8.2 to ~8.1 — a shift of 0.1 units corresponding to a 26% increase in [H⁺]. Organisms that build calcium carbonate shells are vulnerable: CaCO₃(s) + H⁺ → Ca²⁺(aq) + HCO₃⁻(aq). Increased [H⁺] accelerates shell dissolution and increases shell formation energy cost, threatening marine ecosystems.
Enzyme function. Most enzymes have an optimal pH range within which their active site geometry and ionisation state support catalysis. Outside this range, the ionisation of acidic and basic amino acid side chains changes, altering binding affinity and catalytic rate. For example, pepsin (stomach protease) is optimally active at pH 1.5–2.5; trypsin (intestinal protease) at pH 7.5–8.5. Intracellular phosphate buffers (H₂PO₄⁻/HPO₄²⁻, pKa = 7.21) maintain cytoplasmic pH within the range where metabolic enzymes function optimally.
"A buffer prevents pH changes." A buffer resists, not prevents. There is always a small pH change — the buffer minimises it. "Prevents" earns no credit in an explanation; "resists" is the required term.
"A buffer must have pH 7." Buffer pH is determined by pKa + log([A⁻]/[HA]). The acetate buffer operates at pH ~4.74 — acidic, not neutral. Blood buffer operates at 7.35–7.45 — basic, not exactly neutral.
"The HA component reacts with added H⁺." Wrong — the conjugate base A⁻ reacts with added H⁺ (A⁻ + H⁺ → HA). The weak acid HA reacts with added OH⁻ (HA + OH⁻ → A⁻ + H₂O). Many students reverse these roles.
"Equal volumes of weak acid and strong base makes a buffer." If concentrations are equal, this is the equivalence point — all HA is consumed, no buffer exists. Only if the strong base is added in less than stoichiometric amount does a buffer form.
"Use pH = −log(Ka) to find buffer pH." This is wrong for all solutions. For a buffer, use Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). pH = pKa only when [A⁻] = [HA] — a special case, not the general formula.
GIVEN: [CH₃COOH] = 0.150 mol/L; [CH₃COO⁻] = 0.200 mol/L; Ka = 1.8 × 10⁻⁵; V = 500 mL = 0.500 L; V(HCl) = 5.00 mL = 0.00500 L; c(HCl) = 1.00 mol/L
FIND: (a) pH; (b) required ratio for pH 5.20; (c) new pH after adding HCl
METHOD and ANSWER (a) — Buffer pH:
pKa = −log(1.8 × 10⁻⁵) = 5 − log(1.8) = 5 − 0.255 = 4.74
pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.200/0.150) = 4.74 + log(1.333) = 4.74 + 0.125 = 4.87
METHOD and ANSWER (b) — Required ratio for pH 5.20:
pH = pKa + log([A⁻]/[HA]) → log([A⁻]/[HA]) = pH − pKa = 5.20 − 4.74 = 0.46
[A⁻]/[HA] = 10^0.46 = 2.88 — approximately 3 parts acetate to 1 part acetic acid
METHOD and ANSWER (c) — pH after adding HCl:
n(CH₃COOH) = 0.150 × 0.500 = 0.0750 mol
n(CH₃COO⁻) = 0.200 × 0.500 = 0.100 mol
n(HCl) = 1.00 × 0.00500 = 5.00 × 10⁻³ mol H⁺ added
H⁺ reacts with CH₃COO⁻: CH₃COO⁻ + H⁺ → CH₃COOH
n(CH₃COO⁻) after = 0.100 − 5.00 × 10⁻³ = 0.0950 mol
n(CH₃COOH) after = 0.0750 + 5.00 × 10⁻³ = 0.0800 mol
pH = 4.74 + log(0.0950/0.0800) = 4.74 + log(1.188) = 4.74 + 0.075 = 4.81
Change in pH = 4.81 − 4.87 = −0.06 pH units. Compare: same HCl in pure water at pH 7.00 → [H⁺] = 5.00 × 10⁻³/0.505 = 9.9 × 10⁻³ mol/L → pH 2.00 — a change of 5.00 pH units. The buffer reduced the pH change by a factor of ~83.
ANSWERS: (a) pH = 4.87. (b) [A⁻]/[HA] = 2.88 (3 parts acetate to 1 part acetic acid). (c) pH after HCl = 4.81; ΔpH = −0.06 units (vs −5.00 units in pure water).
GIVEN: V(NaOH) = 30.0 mL = 0.0300 L; c(NaOH) = 0.200 mol/L; V(NaH₂PO₄) = 50.0 mL = 0.0500 L; c(NaH₂PO₄) = 0.200 mol/L; Ka₂ = 6.2 × 10⁻⁸
FIND: (a) buffer components; (b) pH; (c) maximum NaOH capacity; (d) biological suitability
(a) Buffer components:
NaH₂PO₄ provides H₂PO₄⁻ — the weak acid (Ka₂ = 6.2 × 10⁻⁸; ionisation: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻)
When NaOH added: H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O → HPO₄²⁻ is the conjugate base
Buffer pair: H₂PO₄⁻ (weak acid) / HPO₄²⁻ (conjugate base)
(b) Buffer pH:
n(H₂PO₄⁻) initial = 0.200 × 0.0500 = 0.0100 mol
n(OH⁻) = 0.200 × 0.0300 = 6.00 × 10⁻³ mol
H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O (complete)
n(H₂PO₄⁻) remaining = 0.0100 − 6.00 × 10⁻³ = 4.00 × 10⁻³ mol
n(HPO₄²⁻) formed = 6.00 × 10⁻³ mol
pKa₂ = −log(6.2 × 10⁻⁸) = 8 − log(6.2) = 8 − 0.792 = 7.21
pH = 7.21 + log(6.00 × 10⁻³ / 4.00 × 10⁻³) = 7.21 + log(1.50) = 7.21 + 0.176 = 7.39 ✓ (within physiological range)
(c) Maximum NaOH capacity:
The buffer fails when all H₂PO₄⁻ is consumed. Remaining n(H₂PO₄⁻) = 4.00 × 10⁻³ mol.
Maximum additional NaOH = 4.00 × 10⁻³ mol
Maximum HCl that can be absorbed = n(HPO₄²⁻) = 6.00 × 10⁻³ mol (before HPO₄²⁻ exhausted)
(d) Why phosphate buffer suits biological use:
pKa₂(H₂PO₄⁻) = 7.21 is within 0.2 pH units of physiological pH (7.35–7.45). Effective buffering range = pKa ± 1 = 6.21–8.21, which comfortably spans physiological pH. At pH 7.39 (from our calculation), [HPO₄²⁻]/[H₂PO₄⁻] = 1.50 — both components present in comparable concentrations, providing capacity to resist both acid and base loads. Additionally, phosphate is a naturally occurring biological molecule, non-toxic at physiological concentrations, and participates directly in metabolic pathways (ATP/ADP cycle).
ANSWERS: (a) Weak acid = H₂PO₄⁻; conjugate base = HPO₄²⁻. (b) pH = 7.39. (c) Maximum NaOH = 4.00 × 10⁻³ mol; maximum HCl = 6.00 × 10⁻³ mol. (d) pKa₂ = 7.21 close to physiological pH; effective range 6.21–8.21; both components in comparable concentrations; non-toxic and naturally present.
GIVEN: pH = 7.10 (measured); [HCO₃⁻] = 6.0 mmol/L; [H₂CO₃] = 0.60 mmol/L (compensated); [H₂CO₃] = 1.20 mmol/L (normal, uncompensated); pKa = 6.10
FIND: (a) Verify pH; (b) explain HCO₃⁻ fall; (c) respiratory compensation; (d) uncompensated pH and reason compensation is insufficient
(a) Verify blood pH:
pH = pKa + log([HCO₃⁻]/[H₂CO₃]) = 6.10 + log(6.0/0.60) = 6.10 + log(10) = 6.10 + 1.00 = 7.10 ✓
Calculated pH exactly matches measured value — confirms the Henderson-Hasselbalch equation applies to this clinical scenario.
(b) Why [HCO₃⁻] has fallen:
In DKA, ketone bodies (e.g. acetoacetic acid) dissociate in blood, releasing H⁺:
HA ⇌ H⁺ + A⁻ (ketone bodies, weak acids)
The released H⁺ reacts with bicarbonate via the buffer reaction:
H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O
Each mole of metabolic H⁺ produced consumes one mole of HCO₃⁻, producing CO₂ (exhaled). As ketone body production exceeds the rate of HCO₃⁻ replenishment by the kidneys, [HCO₃⁻] falls progressively — from 24 mmol/L to 6.0 mmol/L in this patient. This depletion of the conjugate base component destroys buffer capacity.
(c) Respiratory compensation — significance of reduced [H₂CO₃]:
The patient's reduced [H₂CO₃] (0.60 mmol/L vs normal 1.20 mmol/L) indicates respiratory compensation — the body is breathing faster and deeper (Kussmaul breathing) to expel CO₂.
Expelling CO₂ removes the acid component from the buffer equilibrium: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻
Reducing [CO₂(aq)] shifts the equilibrium to the left (Le Chatelier's principle), decreasing [H⁺] and raising pH toward normal. The respiratory system compensates for metabolic acidosis by lowering [H₂CO₃] — increasing the [HCO₃⁻]/[H₂CO₃] ratio and raising pH. This is the body applying Le Chatelier's principle to maintain homeostasis.
(d) Uncompensated pH and why compensation is insufficient:
Without respiratory compensation: [H₂CO₃] = 1.20 mmol/L (normal); [HCO₃⁻] = 6.0 mmol/L
pH = 6.10 + log(6.0/1.20) = 6.10 + log(5.0) = 6.10 + 0.699 = 6.80
Without respiratory compensation, blood pH would be 6.80 — severely life-threatening. Respiratory compensation raised pH from 6.80 to 7.10 (+0.30 units) by halving [H₂CO₃] through CO₂ expulsion.
Why compensation is insufficient to normalise pH (7.35–7.45): [HCO₃⁻] has fallen from 24 to 6.0 mmol/L — even if respiratory compensation reduced [H₂CO₃] to near zero, the ratio [HCO₃⁻]/[H₂CO₃] cannot reach the required ~20:1 because the numerator ([HCO₃⁻] = 6.0) is the limiting factor. The kidneys replenish HCO₃⁻ slowly (over hours to days) while respiration acts over minutes. Without insulin to stop ketone body production and IV fluids to restore [HCO₃⁻], the buffer system continues to fail.
ANSWERS: (a) pH = 6.10 + log(6.0/0.60) = 7.10 ✓. (b) Ketone body H⁺ + HCO₃⁻ → H₂CO₃ → CO₂ + H₂O; consumes HCO₃⁻ faster than replaced; [HCO₃⁻] falls from 24 to 6.0 mmol/L. (c) Kussmaul breathing expels CO₂ → [H₂CO₃] ↓ → [HCO₃⁻]/[H₂CO₃] ratio ↑ → pH rises (Le Chatelier). (d) Uncompensated pH = 6.10 + log(5.0) = 6.80; compensation raised pH to 7.10 (+0.30 units); insufficient because [HCO₃⁻] = 6.0 mmol/L is too depleted to restore the 20:1 ratio regardless of [H₂CO₃] reduction — renal HCO₃⁻ replenishment and insulin are required.
📷 Image slot — Buffer mechanism two beakers showing pH response to HCl addition (pure water vs acetate buffer)
📷 Image slot — Blood buffer H₂CO₃/HCO₃⁻ equilibrium with CO₂/lung connection and pH 7.35–7.45 range
Write these by hand for retention.
Use the Henderson-Hasselbalch equation to solve each scenario. Show all working.
Scenario 1: A formate buffer is prepared by mixing 0.120 mol/L HCOOH (Ka = 1.77 × 10⁻⁴) and 0.080 mol/L HCOONa. Calculate the pH of this buffer and determine what volume of 1.00 mol/L NaOH must be added to 250 mL of this buffer to raise its pH by exactly 0.20 units. Assume volume change is negligible.
Scenario 2: A student needs to prepare a buffer at pH 7.20 using the phosphate pair H₂PO₄⁻/HPO₄²⁻ (Ka₂ = 6.2 × 10⁻⁸). She has 100 mL of 0.500 mol/L NaH₂PO₄. What volume of 0.500 mol/L NaOH must be added to achieve the desired pH? Why is this buffer suitable for simulating intracellular conditions?
Answer each question in full sentences, referencing the Henderson-Hasselbalch equation where relevant.
Part 1: A patient with severe pneumonia cannot expel CO₂ efficiently. Explain, using the Henderson-Hasselbalch equation and Le Chatelier's principle, how this will affect blood pH. Name the clinical condition and state whether it is acidosis or alkalosis.
Part 2: Atmospheric CO₂ has increased from 280 ppm (pre-industrial) to 420 ppm (current), and ocean pH has fallen from 8.2 to 8.1. (a) Write the equilibrium showing how CO₂ dissolves in ocean water and produces H⁺. (b) Use the Henderson-Hasselbalch equation to explain qualitatively why increasing [H₂CO₃] lowers ocean pH. (c) Explain why organisms with CaCO₃ shells are particularly vulnerable.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
1. A buffer contains 0.080 mol/L benzoic acid (C₆H₅COOH, Ka = 6.5 × 10⁻⁵) and 0.120 mol/L sodium benzoate. 0.010 mol HCl is added to 1.00 L of this buffer. Which response correctly calculates the initial pH and the new pH after adding HCl?
2. Blood has a normal [HCO₃⁻]/[H₂CO₃] ratio of approximately 20:1 and pKa(H₂CO₃) = 6.10. A patient hyperventilates, halving [H₂CO₃] while [HCO₃⁻] remains constant. What is the new blood pH and clinical classification?
3. A student prepares a buffer by mixing 50.0 mL of 0.200 mol/L NH₃ (Kb = 1.8 × 10⁻⁵) with 50.0 mL of 0.200 mol/L NH₄Cl. Which correctly identifies the buffer components and calculates the pH?
4. A buffer is prepared by dissolving 0.050 mol CH₃COOH and 0.050 mol CH₃COONa in 1.00 L of solution (Ka = 1.8 × 10⁻⁵). What is the maximum number of moles of NaOH that can be added before the buffer fails?
5. Increased atmospheric CO₂ dissolves in ocean water via: CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺. Using the Henderson-Hasselbalch equation for the ocean carbonate buffer, which prediction is most directly supported?
Q6. A student claims: "Any solution that contains both an acid and a base is a buffer." Using a specific counterexample, explain why this claim is incorrect. Then write the two molecular equations that describe how a CH₃COOH/CH₃COO⁻ buffer resists pH change when (i) HCl is added and (ii) NaOH is added. (4 marks)
Q7. A student adds 20.0 mL of 0.300 mol/L NaOH to 50.0 mL of 0.300 mol/L lactic acid (CH₃CH(OH)COOH, Ka = 1.38 × 10⁻⁴). (a) Show that a buffer forms and identify its components. (b) Calculate the pH of the resulting buffer using the Henderson-Hasselbalch equation. (c) Calculate the pH if an additional 2.00 mL of 0.300 mol/L NaOH is added to the buffer in (b). (5 marks)
Q8. Blood pH is maintained at 7.35–7.45 primarily by the H₂CO₃/HCO₃⁻ buffer system (pKa = 6.10). (a) Calculate the normal [HCO₃⁻]/[H₂CO₃] ratio at blood pH 7.40. (b) During intense exercise, lactic acid enters the bloodstream, lowering blood pH to 7.30. Using the Henderson-Hasselbalch equation, calculate the new [HCO₃⁻]/[H₂CO₃] ratio and identify whether this represents acidosis or alkalosis. (c) Explain at the molecular level why blood pH at 7.30 is classified as acidosis even though the solution is still basic, and describe two physiological mechanisms the body uses to restore pH to the normal range. (7 marks)
Return to your Think First response. Can you now explain the beaker experiment with quantitative precision?
Q1: A — pKa = −log(6.5 × 10⁻⁵) = 5 − log(6.5) = 5 − 0.813 = 4.19. pH = 4.19 + log(0.120/0.080) = 4.19 + 0.176 = 4.37. Adding 0.010 mol H⁺ to 1.00 L: H⁺ reacts with C₆H₅COO⁻ (conjugate base), not the acid component. n(C₆H₅COO⁻) after = 0.110; n(C₆H₅COOH) after = 0.090. New pH = 4.19 + log(0.110/0.090) = 4.28. ΔpH = −0.09. Option B incorrectly has H⁺ reacting with the acid (producing H₂). Option C incorrectly states the initial pH equals pKa (only true when [A⁻] = [HA]). Option D incorrectly claims a small HCl addition destroys the buffer — 0.010 mol << 0.120 mol C₆H₅COO⁻.
Q2: D — [H₂CO₃] is halved → new [H₂CO₃] = 0.5 × original. New ratio [HCO₃⁻]/[H₂CO₃] = 20/(0.5) = 40:1. pH = 6.10 + log(40) = 6.10 + 1.60 = 7.70. Since 7.70 > 7.45, this is respiratory alkalosis. Option A incorrectly uses ratio 20/2 = 10 — this would be if [HCO₃⁻] halved (it was [H₂CO₃] that halved). Option B uses the original ratio — incorrectly assuming no pH change. Option C halves the numerator instead of the denominator.
Q3: C — In NH₃/NH₄Cl, the weak acid is NH₄⁺ (proton donor: NH₄⁺ ⇌ H⁺ + NH₃). Ka(NH₄⁺) = Kw/Kb(NH₃) = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰. pKa = 9.26. Equal concentrations → log([NH₃]/[NH₄⁺]) = log(1) = 0 → pH = 9.26. Option A identifies NH₃ (not NH₄⁺) as the weak acid — wrong; NH₃ is the conjugate base. Option B uses an incorrect pKb approach in Henderson-Hasselbalch — the equation always uses pKa of the acid form. Option D is wrong — NH₃/NH₄Cl is a classic buffer.
Q4: B — The buffer fails when the weak acid (HA = CH₃COOH) component is entirely consumed. n(CH₃COOH) = 0.050 mol. Each mole of NaOH reacts with one mole of CH₃COOH: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. When 0.050 mol NaOH has been added, all CH₃COOH is consumed — no HA remains to resist further OH⁻ addition. Buffer fails at exactly 0.050 mol. Option A (0.025 mol) is the half-capacity point, not the failure point. Option C (0.100 mol) would require the conjugate base to also react, which it cannot do with NaOH. Option D is wrong — buffers always fail when one component is exhausted.
Q5: B — Henderson-Hasselbalch: pH = pKa + log([HCO₃⁻]/[H₂CO₃]). Increased atmospheric CO₂ → increased dissolved CO₂ → increased [H₂CO₃] (denominator). This decreases the [HCO₃⁻]/[H₂CO₃] ratio → log term decreases → pH falls. This is ocean acidification. Option A claims [HCO₃⁻] increases enough to raise pH — while some HCO₃⁻ does form, the net effect is acid addition that lowers pH. Option C has the effect backwards. Option D is wrong — the carbonate buffer can be overwhelmed by the scale of CO₂ increase.
Counterexample (1 mark): Mixing equal moles of HCl and NaOH gives NaCl(aq) — a neutral salt solution. This contains H⁺ (from HCl) and OH⁻ (from NaOH), both of which are technically "an acid and a base," yet NaCl has no buffering ability because no weak acid/conjugate base pair is present. A tiny addition of HCl would drop pH dramatically.
Correct definition (1 mark): A buffer requires a weak acid (HA) and its conjugate base (A⁻) present simultaneously in comparable concentrations. The strength of the acid is critical — a strong acid has a conjugate base so weak it cannot react with added H⁺.
(i) HCl added — equation (1 mark): H⁺ + CH₃COO⁻ → CH₃COOH — the conjugate base (A⁻) reacts with the added H⁺, consuming it before it can significantly change pH.
(ii) NaOH added — equation (1 mark): OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O — the weak acid (HA) reacts with the added OH⁻, consuming it before it can significantly change pH.
(a) Buffer formation (1 mark): n(lactic acid, HA) = 0.300 × 0.0500 = 0.0150 mol. n(NaOH) = 0.300 × 0.0200 = 6.00 × 10⁻³ mol. Since n(NaOH) < n(HA) (6.00 × 10⁻³ < 0.0150), a buffer forms. Buffer components: lactic acid (HA = CH₃CH(OH)COOH) and lactate ion (A⁻ = CH₃CH(OH)COO⁻).
(b) Buffer pH (2 marks): HA + OH⁻ → A⁻ + H₂O. n(A⁻) formed = 6.00 × 10⁻³ mol; n(HA) remaining = 0.0150 − 6.00 × 10⁻³ = 9.00 × 10⁻³ mol. pKa = −log(1.38 × 10⁻⁴) = 4 − log(1.38) = 4 − 0.140 = 3.86. pH = 3.86 + log(6.00 × 10⁻³ / 9.00 × 10⁻³) = 3.86 + log(0.667) = 3.86 − 0.176 = 3.68.
(c) pH after additional NaOH (2 marks): n(NaOH) additional = 0.300 × 0.00200 = 6.00 × 10⁻⁴ mol. OH⁻ + HA → A⁻ + H₂O. n(HA) after = 9.00 × 10⁻³ − 6.00 × 10⁻⁴ = 8.40 × 10⁻³ mol. n(A⁻) after = 6.00 × 10⁻³ + 6.00 × 10⁻⁴ = 6.60 × 10⁻³ mol. pH = 3.86 + log(6.60 × 10⁻³ / 8.40 × 10⁻³) = 3.86 + log(0.786) = 3.86 − 0.105 = 3.76. ΔpH = +0.08 units (buffer resists the base addition effectively).
(a) Normal ratio at pH 7.40 (1 mark): 7.40 = 6.10 + log([HCO₃⁻]/[H₂CO₃]) → log = 1.30 → [HCO₃⁻]/[H₂CO₃] = 10^1.30 = 20:1.
(b) New ratio at pH 7.30 and classification (2 marks): 7.30 = 6.10 + log([HCO₃⁻]/[H₂CO₃]) → log = 1.20 → [HCO₃⁻]/[H₂CO₃] = 10^1.20 = 15.8:1. Since pH 7.30 < 7.35 (lower bound of normal range), this is metabolic acidosis. The ratio has fallen from 20:1 to 15.8:1 — lactic acid H⁺ has consumed HCO₃⁻, reducing the numerator.
(c) Molecular explanation and physiological mechanisms (4 marks): Although blood at pH 7.30 is still basic ([H⁺] = 5.01 × 10⁻⁸ mol/L; [OH⁻] = 2.0 × 10⁻⁷ mol/L — more OH⁻ than H⁺), acidosis is defined clinically relative to the normal range of 7.35–7.45. At pH 7.30, the increased [H⁺] (relative to normal pH 7.40) destabilises enzyme active sites, disrupts protein ionisation states, and impairs oxygen delivery by haemoglobin — all life-threatening even though the solution remains technically basic. Mechanism 1 — respiratory compensation: the brain detects the pH fall and signals increased breathing rate. Faster breathing expels more CO₂, shifting CO₂ + H₂O ⇌ H₂CO₃ to the left (Le Chatelier), reducing [H₂CO₃] and increasing [HCO₃⁻]/[H₂CO₃] → pH rises. Mechanism 2 — renal compensation: the kidneys increase HCO₃⁻ reabsorption and H⁺ excretion in urine over hours to days. This directly replenishes [HCO₃⁻], restoring the ratio toward 20:1 and returning blood pH to 7.35–7.45.
Scenario 1: pKa(HCOOH) = −log(1.77 × 10⁻⁴) = 4 − log(1.77) = 4 − 0.248 = 3.75. pH = 3.75 + log(0.080/0.120) = 3.75 + log(0.667) = 3.75 − 0.176 = 3.57. To raise pH by 0.20 to 3.77: log([A⁻]/[HA]) = 3.77 − 3.75 = 0.02 → [A⁻]/[HA] = 10^0.02 = 1.047. In 250 mL: n(HA) = 0.120 × 0.250 = 0.0300 mol; n(A⁻) = 0.080 × 0.250 = 0.0200 mol. After adding x mol NaOH: n(A⁻) = 0.0200 + x; n(HA) = 0.0300 − x. Ratio = (0.0200 + x)/(0.0300 − x) = 1.047. 0.0200 + x = 1.047(0.0300 − x) = 0.03141 − 1.047x. 2.047x = 0.01141 → x = 5.57 × 10⁻³ mol. Volume = 5.57 × 10⁻³/1.00 = 5.57 mL of 1.00 mol/L NaOH.
Scenario 2: pKa₂ = 7.21. pH target = 7.20. log([HPO₄²⁻]/[H₂PO₄⁻]) = 7.20 − 7.21 = −0.01 → ratio = 10^−0.01 = 0.977. n(NaH₂PO₄) = 0.500 × 0.100 = 0.0500 mol. Let x = n(NaOH). n(HPO₄²⁻) = x; n(H₂PO₄⁻) = 0.0500 − x. x/(0.0500 − x) = 0.977. x = 0.0489 − 0.977x → 1.977x = 0.0489 → x = 0.0247 mol. Volume = 0.0247/0.500 = 49.5 mL NaOH. Biological suitability: pKa₂ = 7.21 is within 0.01 pH units of the target; effective range 6.21–8.21 spans intracellular pH; phosphate is non-toxic and native to cells.
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