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Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on quantitative buffer calculations: derivation of the buffer equation from Ka, calculation of buffer pH from [HA] and [A⁻] (or equivalently the moles thereof in the same volume), calculation of pH change on addition of acid or base, calculation of required component ratio for a target pH, and choice of weak acid for a target buffer pH (refer to the official OCR H432 specification document for exact wording). The full Henderson-Hasselbalch form pH=pKa+log10([A−]/[HA]) is the workhorse equation throughout.
This lesson converts the qualitative buffer story of lesson 7 into a calculation routine. The single starting equation is the rearrangement [H+]=Ka⋅[HA]/[A−], derivable in two lines from the dissociation Ka. Three classes of question are common: (i) pH of a buffer of given composition; (ii) pH after adding strong acid or base to an existing buffer; (iii) the inverse — what ratio of components is needed for a target pH. All three reduce to the same algebra; the bookkeeping is what catches students out. The half-neutralisation identity pH=pKa when [HA]=[A−] is the most exam-tested result in the topic and the experimental route to measuring pKa of an unknown weak acid.
Key Equations: [H+]=Ka⋅[A−][HA](direct form) pH=pKa+log10[HA][A−](Henderson-Hasselbalch) Half-neutralisation: [HA]=[A−]⇒pH=pKa. Since both components share the same volume, [HA] : [A⁻] = moles HA : moles A⁻ — you can use moles directly in the ratio.
Start from the acid dissociation constant of the weak acid HA:
Ka=[HA][H+][A−]
Rearrange for [H+]:
[H+]=Ka⋅[A−][HA]
Take −log10 of both sides:
pH=−log10Ka−log10[A−][HA]=pKa+log10[HA][A−]
This is the Henderson-Hasselbalch equation (Lawrence Henderson, 1908; Karl Hasselbalch, 1916). Both forms are equivalent; pick whichever is more convenient for the numerical problem at hand. The direct [H+] form is often faster; the logarithmic form makes the pH = pKa structure transparent.
The buffer-pH method uses the initial concentrations of HA and A⁻ as if they were the equilibrium concentrations. This is justified because in a buffer the equilibrium HA⇌H++A− is heavily suppressed (Le Chatelier — the large pre-existing A⁻ concentration pushes the equilibrium far to the left), so the additional dissociation from "initial" → "equilibrium" is negligible at the few-percent level. The approximation is accurate to within ~1 % for typical OCR-A-Level buffer concentrations (0.05 to 1.0 mol dm⁻³) and ratios within 10:1.
flowchart TD
A["Buffer with HA and A-"] --> B{"Acid or base added?"}
B -- "Acid (H+)" --> C["H+ + A- -> HA"]
B -- "Base (OH-)" --> D["HA + OH- -> A- + H2O"]
B -- "Neither" --> E["Use initial [HA] and [A-]"]
C --> F["Update: [HA] up by added; [A-] down by added"]
D --> G["Update: [HA] down by added; [A-] up by added"]
E --> H["pH = pKa + log10([A-]/[HA])"]
F --> H
G --> H
Calculate the pH of a buffer made by dissolving 0.100 mol CH₃COOH and 0.100 mol CH₃COONa in 1.00 dm³ of water. Ka=1.74×10−5 mol dm⁻³.
[H+]=Ka⋅[A−][HA]=1.74×10−5×0.1000.100=1.74×10−5 mol dm−3 pH=−log10(1.74×10−5)=4.76
Result: when [HA]=[A−], the ratio is 1, [H+]=Ka, and pH=pKa. This is the half-neutralisation identity — central to titration curve analysis in lesson 9.
Calculate the pH of a buffer containing 0.200 mol dm⁻³ CH₃COOH and 0.100 mol dm⁻³ CH₃COONa.
[H+]=1.74×10−5×0.1000.200=3.48×10−5 mol dm−3 pH=−log10(3.48×10−5)=4.46(2 dp)
More HA than A⁻ means more dissociation, more H⁺, lower pH. The pH has dropped 0.30 units from the 1:1 case — exactly log10(2), the log of the doubled HA / A⁻ ratio.
Calculate the pH of a buffer containing 0.100 mol dm⁻³ CH₃COOH and 0.300 mol dm⁻³ CH₃COONa.
[H+]=1.74×10−5×0.3000.100=5.80×10−6 mol dm−3 pH=−log10(5.80×10−6)=5.24(2 dp)
More A⁻ pushes the dissociation equilibrium to the left, lowering [H⁺] and raising pH. The pH has risen log10(3)=0.48 units above pKa (4.76 + 0.48 = 5.24). ✓
Because both HA and A⁻ are dissolved in the same volume, the ratio [HA]/[A−] equals the ratio of moles of HA to moles of A⁻. You don't need to compute separate concentrations — the volume cancels.
A buffer contains 0.150 mol of methanoic acid (pKa=3.75) and 0.050 mol of sodium methanoate in 500 cm³ of solution. Calculate the pH.
Ka=10−3.75=1.78×10−4 mol dm−3 Ratio (mol HA : mol A−)=0.150:0.050=3:1 [H+]=Ka⋅nA−nHA=1.78×10−4×3=5.34×10−4 mol dm−3 pH=−log10(5.34×10−4)=3.27(2 dp)
The 500 cm³ volume never enters the calculation.
25.0 cm³ of 0.200 mol dm⁻³ NaOH is added to 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid. Calculate the pH.
Step 1 — moles before reaction: n(CH3COOH)=0.200×0.0500=0.0100 mol n(NaOH)=0.200×0.0250=0.00500 mol
Step 2 — react. NaOH consumes half the ethanoic acid, producing an equal amount of ethanoate: CH3COOH+NaOH→Na++CH3COO−+H2O n(CH3COOH)remaining=0.0100−0.00500=0.00500 mol n(CH3COO−)formed=0.00500 mol
Step 3 — Henderson-Hasselbalch. Ratio is 1:1. [H+]=Ka=1.74×10−5 mol dm−3 pH=4.76
This is the half-neutralisation point of the ethanoic acid titration curve (lesson 9). The pH equals the pKa of the weak acid being titrated — the central experimental route to measuring pKa.
A 1.00 dm³ buffer contains 0.200 mol of CH₃COOH and 0.150 mol of CH₃COO⁻. Calculate (a) the initial pH, and (b) the pH after adding 0.020 mol of HCl (assume no volume change).
(a) Initial pH: [H+]=Ka⋅0.1500.200=1.74×10−5×1.333=2.32×10−5 mol dm−3 pH=−log10(2.32×10−5)=4.63
(b) After 0.020 mol HCl: The added H⁺ reacts with A⁻ to form HA: CH3COO−+H+→CH3COOH n(CH3COO−)=0.150−0.020=0.130 mol n(CH3COOH)=0.200+0.020=0.220 mol [H+]=Ka⋅0.1300.220=1.74×10−5×1.692=2.94×10−5 mol dm−3 pH=−log10(2.94×10−5)=4.53(2 dp)
The pH dropped from 4.63 to 4.53 — a change of just 0.10 units. Compare with the same 0.020 mol HCl added to 1 dm³ of pure water: [H+] goes from 10−7 to 2×10−2 and pH crashes from 7.00 to 1.70 — a drop of 5.3 units. The buffer absorbed the acid 53× more effectively.
Starting again from the same initial 1.00 dm³ buffer (0.200 mol HA, 0.150 mol A⁻), add 0.020 mol NaOH.
The added OH⁻ reacts with HA to form A⁻ and water: CH3COOH+OH−→CH3COO−+H2O n(CH3COOH)=0.200−0.020=0.180 mol n(CH3COO−)=0.150+0.020=0.170 mol [H+]=Ka⋅0.1700.180=1.74×10−5×1.0588=1.84×10−5 mol dm−3 pH=−log10(1.84×10−5)=4.74(2 dp)
The pH rose from 4.63 to 4.74 — a change of just 0.11 units. Again, tiny compared to unbuffered water.
A chemist wants a buffer at pH = 5.00 using ethanoic acid (pKa=4.76). What ratio of [CH₃COONa] to [CH₃COOH] is required?
[H+]=10−5.00=1.00×10−5 mol dm−3 1.00×10−5=1.74×10−5×[A−][HA] [A−][HA]=1.74×10−51.00×10−5=0.575 [HA][A−]=0.5751=1.74
So [CH₃COONa] must be 1.74× [CH₃COOH]. A practical recipe: 0.100 mol dm⁻³ ethanoic acid + 0.174 mol dm⁻³ sodium ethanoate gives pH 5.00 with good buffer capacity.
A 1.00 dm³ buffer initially contains 0.0500 mol HA and 0.0500 mol A⁻ (Ka=1.00×10−5, pKa = 5.00). 0.0100 mol of strong acid HCl is added (no volume change). Set out an ICE-style table for the reaction and calculate the new pH.
| Species | Initial / mol | Change / mol | Final / mol |
|---|---|---|---|
| HA | 0.0500 | +0.0100 (gains from A⁻) | 0.0600 |
| A⁻ | 0.0500 | −0.0100 (consumed by H⁺) | 0.0400 |
| H⁺ (added) | 0.0100 | −0.0100 (all consumed by A⁻) | ≈ 0 (extra) |
The "C" row codifies the stoichiometric step that the unprepared candidate omits. The "F" row is then the input to Henderson-Hasselbalch:
[H+]=Ka⋅n(A−)n(HA)=1.00×10−5×0.04000.0600=1.50×10−5 mol dm−3 pH=−log10(1.50×10−5)=4.82
The pH has dropped from 5.00 to 4.82 — a fall of 0.18 units for adding 0.0100 mol of strong acid to a 0.1 mol total buffer. Contrast with adding the same 0.0100 mol HCl to 1 dm³ of pure water, which would crash pH from 7.00 to 2.00 (a drop of 5 units). The buffer absorbs the perturbation almost 28× more effectively.
The discipline of writing the ICE table — even when not formally required — keeps the stoichiometric step visible and prevents the most common error: applying Henderson-Hasselbalch with the initial HA and A⁻ values, ignoring the added perturbation entirely.
You have 500 cm³ of 0.200 mol dm⁻³ ethanoic acid (pKa = 4.76). You wish to prepare a buffer at pH = 4.50 by adding solid sodium ethanoate. What mass of NaCH₃COO (Mr = 82.0) must you add?
Step 1 — required ratio. From Henderson-Hasselbalch, log10([A−]/[HA])=pH−pKa=4.50−4.76=−0.26, so [A−]/[HA]=10−0.26=0.550.
Step 2 — moles. n(HA) in the starting 500 cm³ = 0.200×0.500=0.100 mol. Required n(A−)=0.550×0.100=0.0550 mol.
Step 3 — mass. m(NaCH3COO)=n×Mr=0.0550×82.0=4.51 g.
Adding 4.51 g of sodium ethanoate to the 500 cm³ of 0.200 mol dm⁻³ ethanoic acid (assuming no volume change) produces a buffer at pH = 4.50. The inverse-design pattern — given a target pH, work out the component ratio, then quantify it — is heavily examined and is the basis of practical buffer recipes in biochemistry. A robust answer also flags buffer capacity: total component concentration is (0.100+0.0550)/0.500=0.310 mol dm⁻³, well above the 0.01 mol dm⁻³ minimum, so capacity is good.
You need a buffer at pH = 7.00. Which weak acid would you choose, and in what component ratio?
The closest convenient pKa to 7.0 is H₂PO₄⁻ with pKa2=7.21.
Ka=10−7.21=6.17×10−8 mol dm−3 [H+]=10−7.00=1.00×10−7 mol dm−3 [A−][HA]=Ka[H+]=6.17×10−81.00×10−7=1.62 [HA][A−]=0.617
A practical recipe: 0.050 mol NaH₂PO₄ (the HA) + 0.031 mol Na₂HPO₄ (the A⁻) in 1 dm³ gives pH 7.00. This is the formulation of one of the most widely used "phosphate-buffered saline" (PBS) variants in cell biology.
| Target pH | Suggested weak acid | pKa | Useful range |
|---|---|---|---|
| 3.5 – 4.5 | Methanoic acid (HCOOH) | 3.75 | 2.8 – 4.8 |
| 4.0 – 5.5 | Benzoic acid | 4.20 | 3.2 – 5.2 |
| 4.0 – 5.5 | Ethanoic acid | 4.76 | 3.8 – 5.8 |
| 6.2 – 8.2 | H₂PO₄⁻ / HPO₄²⁻ | 7.21 | 6.2 – 8.2 |
| 8.5 – 10.0 | NH₄⁺ / NH₃ (basic) | 9.25 | 8.3 – 10.3 |
The Henderson-Hasselbalch method assumes [HA] and [A⁻] are unchanged from their initial values. This is accurate when:
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