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Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on weak-acid pH calculations: using Ka with the equilibrium concentration of a weak monobasic acid to calculate pH, the two standard approximations ([H+]=[A−] and [HA]eq≈c), the validity criterion (< 5 % ionisation), reverse calculation of Ka from a measured pH, calculation of percentage dissociation, and the effect of dilution on degree of dissociation (Ostwald dilution law in qualitative form) (refer to the official OCR H432 specification document for exact wording). This is the practical computation lesson — it converts the Ka formalism set up in lesson 4 into pH values.
This lesson is the calculation engine for the rest of Module 5.1.3 — buffer derivation in lessons 7–8 uses the same Ka approximation, and the half-equivalence-point logic in lesson 9 (pH=pKa at half-neutralisation) is a direct consequence of the relations established here. The two-approximation framework — [H+]=[A−] (from water self-ionisation being negligible) and [HA]eq≈c (from < 5 % dissociation) — leads to a single tidy result: [H+]=Kac. Mark schemes reward candidates who show the approximations rather than producing the answer with no working. The 5 % validity check is part of the answer, not optional.
Key Equation: Ka=c−xx2≈cx2(weak acid, x << c) [H+]=x=Ka⋅c pH=−21log10(Ka⋅c)=21(pKa−log10c) Approximations valid when α=x/c<5%.
For a weak acid HA at initial concentration c, let x be the equilibrium concentration of H⁺ produced by dissociation:
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| Initial | c | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | c−x | x | x |
Substituting into Ka:
Ka=[HA][H+][A−]=c−xx⋅x=c−xx2
This is a quadratic in x. The full solution is x=2−Ka+Ka2+4Kac, but we almost never need it because two excellent approximations collapse the problem to a single square root.
The ICE table assumed all H⁺ comes from the acid's dissociation, so [H+]=[A−] exactly. This is only an approximation if you remember that water self-ionisation also contributes some H⁺ (about 10−7 mol dm⁻³). For any solution with pH less than ~6, the acid's [H+] (>10−6) dominates over the water's contribution (10−7) by at least a factor of 10, and we can safely write [H+]=[A−]. Approximation 1 fails only for extremely weak acids (Ka<10−10) at very low concentrations — beyond OCR.
If only a small fraction of HA dissociates, the molecular form is essentially undepleted, so c−x≈c. The conventional validity threshold is α=x/c<5%. Above this, the error in [H+] from the approximation exceeds about 2.5 % — small at A-Level precision but worth flagging.
With both approximations:
Ka≈cx2⇒x=Ka⋅c pH=−log10(Ka⋅c)=−21log10(Kac)=21(pKa−log10c)
You should work from first principles (Ka=x2/c, then , then pH = −log10) rather than memorise the compact pH = 21(pKa−log10c) formula. The reasoning matters as much as the answer in OCR mark schemes.
flowchart TD
A["Weak acid HA at concentration c with Ka"] --> B["ICE: [H+]=[A-]=x, [HA]=c-x"]
B --> C{"Is x < 5% of c?"}
C -- "Yes (typical case)" --> D["Approximation: Ka = x^2 / c"]
C -- "No (concentrated strong-weak)" --> E["Use full quadratic"]
D --> F["x = sqrt(Ka * c)"]
F --> G["pH = -log10(x)"]
G --> H["Check x/c < 5% (state this)"]
Calculate the pH of 0.100 mol dm⁻³ ethanoic acid. Ka=1.74×10−5 mol dm⁻³.
Step 1 — equilibrium and Ka: CH3COOH⇌H++CH3COO− Ka=[CH3COOH][H+][CH3COO−]=1.74×10−5
Step 2 — approximations: [H+]=[CH3COO−]=x; [CH3COOH]≈0.100.
Step 3 — substitute and solve: 1.74×10−5=0.100x2 x2=1.74×10−6 x=1.32×10−3 mol dm−3
Step 4 — pH: pH=−log10(1.32×10−3)=2.88(2 dp)
Step 5 — validity check: x/c=1.32×10−3/0.100=0.0132=1.32%. Well below 5 %, approximation safe.
Calculate the pH of 0.0500 mol dm⁻³ HCOOH. Ka=1.60×10−4 mol dm⁻³.
x2=Ka⋅c=1.60×10−4×0.0500=8.00×10−6 x=2.83×10−3 mol dm−3 pH=−log10(2.83×10−3)=2.55(2 dp)
Check: x/c=2.83×10−3/0.0500=5.7%. Just above the 5 % threshold — flag this. A full quadratic gives 2.56 — a 0.01 unit correction, marginal at A-Level precision. State the boundary and proceed with the approximation; an OCR mark scheme would accept 2.55.
Calculate the pH of 0.250 mol dm⁻³ benzoic acid C₆H₅COOH. pKa=4.20.
Convert: Ka=10−4.20=6.31×10−5 mol dm⁻³.
x2=6.31×10−5×0.250=1.578×10−5 x=3.97×10−3 mol dm−3 pH=−log10(3.97×10−3)=2.40(2 dp)
Check: x/c=1.6%. Approximation safe.
Calculate the pH of 0.100 mol dm⁻³ HCN. Ka=4.9×10−10 mol dm⁻³.
x2=4.9×10−10×0.100=4.9×10−11 x=7.0×10−6 mol dm−3 pH=−log10(7.0×10−6)=5.15(2 dp)
Check: x/c=7.0×10−5 = 0.007 %, extraordinarily safe. A very weak acid like HCN barely moves the pH from 7.
Boundary case warning: for extremely weak acids (Ka<10−10) at very dilute concentrations, the water self-ionisation contribution (∼10−7 mol dm⁻³) becomes comparable with the acid's [H⁺], and approximation 1 breaks down. For 0.001 mol dm⁻³ HCN the calculation above gives x=7×10−7 mol dm⁻³ — only 7× the water value, so the simple formula begins to over-estimate the acidity. This case is beyond OCR examinable content but worth recognising.
The mirror-image problem: given pH and c, find Ka.
A 0.100 mol dm⁻³ solution of propanoic acid CH₃CH₂COOH has pH = 2.94 at 298 K. Calculate Ka and pKa.
[H+]=10−2.94=1.148×10−3 mol dm−3
Since 1.148×10−3≪0.100 (i.e. α≈1.1%), the approximation [HA]≈c holds. [A−]=[H+].
Ka=0.100(1.148×10−3)2=1.32×10−5 mol dm−3 pKa=−log10(1.32×10−5)=4.88
The value matches the literature pKa of propanoic acid (4.89) to two decimal places.
A weak acid has pKa=3.52. Calculate the pH of a 0.250 mol dm⁻³ solution.
Ka=10−3.52=3.02×10−4 mol dm−3 x=3.02×10−4×0.250=8.69×10−3 mol dm−3 pH=−log10(8.69×10−3)=2.06(2 dp)
Check: x/c=3.5%, just inside the approximation window.
The degree of dissociation α is the fraction of HA molecules that have shed a proton:
α=cx=cKa
This is the Ostwald dilution law (Friedrich Wilhelm Ostwald, 1888): α scales as 1/c, so dilution increases the percentage dissociated.
Calculate the percentage dissociation of 0.0100 mol dm⁻³ ethanoic acid (Ka=1.74×10−5).
x=1.74×10−5×0.0100=4.17×10−4 mol dm−3 α=0.01004.17×10−4=4.17%
At 10× lower concentration than worked example 1, ethanoic acid is now three times more dissociated (4.17 % vs 1.32 %). This is Le Chatelier: dilution shifts the equilibrium toward the side with more particles (right side, two particles per one on the left), so the dissociated fraction grows.
A 0.100 mol dm⁻³ ethanoic acid solution is diluted 10×. Calculate the pH of the diluted solution.
New c=0.0100 mol dm⁻³.
x=1.74×10−5×0.0100=4.17×10−4 mol dm−3 pH=−log10(4.17×10−4)=3.38(2 dp)
Compare with the undiluted pH = 2.88 (worked example 1). The 10× dilution raises pH by only 0.50 units, not the 1.00 unit you would see for a strong acid. The reason: weak-acid dissociation becomes more extensive on dilution (Ostwald), partially compensating for the reduced concentration. The general result is pH rises by 0.50 per 10× dilution of a weak acid (and by 1.00 per 10× dilution of a strong acid).
Compare the percent dissociation of ethanoic acid at (a) 1.00 mol dm⁻³ and (b) 0.00100 mol dm⁻³. Ka=1.74×10−5.
(a) x=1.74×10−5×1.00=4.17×10−3 mol dm⁻³; α=0.42%.
(b) x=1.74×10−5×0.00100=1.32×10−4 mol dm⁻³; α=13.2%.
The low-concentration solution is 30× more percent-dissociated, but [H+] is actually lower (1.32×10−4 vs 4.17×10−3). At (b), α=13% is well above the 5 % threshold — the quadratic gives a more accurate α=12.4% and pH = 3.90 vs the approximation's 3.88. Worth flagging in mark-schemed answers.
Calculate the pH of 0.00500 mol dm⁻³ HF (Ka=5.6×10−4 mol dm⁻³). State whether the standard weak-acid approximation is valid.
Approximation attempt: x=5.6×10−4×0.00500=2.80×10−6=1.67×10−3 mol dm⁻³. Validity check: α=1.67×10−3/0.00500=33.5% — vastly above the 5 % threshold. Approximation invalid.
Full quadratic: x2+Kax−Kac=0, i.e. x2+5.6×10−4x−2.80×10−6=0. Quadratic formula:
x=2−5.6×10−4+(5.6×10−4)2+4(2.80×10−6)=2−5.6×10−4+1.151×10−5 x=2−5.6×10−4+3.39×10−3=1.41×10−3 mol dm−3 pH=−log10(1.41×10−3)=2.85(2 dp)
The naive approximation gave pH = 2.78 — a 0.07 unit error from the rigorous 2.85. Genuine α=1.41×10−3/0.00500=28.2%.
This is the cleanest A-Level illustration of where the standard approximation breaks: a "stronger weak" acid (HF, Ka in the 10−4 range) at low concentration (c<0.01 mol dm⁻³) produces α in the 30–50 % range, well outside the 5 % window. Note also the structural lesson: as c falls, the fraction dissociated rises but the absolute [H⁺] still falls, so pH rises. The two effects coexist and are sometimes confused.
The validity rule "approximation OK when α<5%" is sometimes also written as "approximation OK when c>100Ka." The two are algebraically identical: if α=Ka/c=0.05, then Ka/c=0.0025, i.e. c/Ka=400 ≈ 100 with rounding. Apply the test mentally to each of the following four cases and decide approximation-vs-quadratic:
| Acid | Ka / mol dm⁻³ | c / mol dm⁻³ | c/Ka | αapprox | Use quadratic? |
|---|---|---|---|---|---|
| CH₃COOH | 1.7×10−5 | 0.100 | 5.9×103 | 1.3 % | No |
| HCOOH | 1.6×10−4 | 0.0500 | 313 | 5.7 % | Borderline — flag |
| HF | 5.6×10−4 | 0.0100 | 17.9 | 23.7 % | Yes |
| Cl₃CCOOH | 0.23 | 0.500 | 2.17 | 67.8 % | Yes — essentially strong |
The decision tree gives you an immediate diagnostic without arithmetic. For typical OCR exam concentrations of carboxylic acids (~0.01–1 mol dm⁻³), the c/Ka ratio is in the thousands, so the approximation is safe; for stronger weak acids (HF and chlorinated derivatives) or for very dilute solutions, the ratio drops and the quadratic is mandatory.
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