Acid-Base Titrations

Spec Mapping — OCR H432 Module 2.1.4 — Acids, covering acid-base titration technique, standard solution preparation, indicator selection, titre and concentration calculations, percentage purity, percentage uncertainty propagation, and back titration. This lesson is PAG 2 (acid-base titration) anchor, and the quantitative arithmetic introduced here will be re-examined in Module 5.1.3 buffer/Ka work and in Module 3 enthalpy calorimetry (refer to the official OCR H432 specification document for exact wording).

Acid-base titration is the single technique most often examined in OCR H432 quantitative chemistry and the most frequently reported in industrial QA labs (pharmaceutical purity, food acidity, water hardness). This lesson takes the qualitative neutralisation framework of the previous lesson and turns it into a precision-measurement protocol that can determine a solution's concentration to within roughly 0.2 %. You will learn how to prepare a standard solution from a primary standard, run a titration to concordant titres, choose an indicator using the pH-curve framework, propagate apparatus uncertainty, and tackle back titrations for slow- or two-stage reactions. The lesson also re-emphasises why strong-vs-weak matters operationally — it dictates indicator choice — and lays groundwork for the Module 5.1.3 quantitative pH treatment that comes a year later.

Key Equation: $n = c \cdot V$n=c⋅V where $n$n is amount in moles, $c$c is concentration in mol dm⁻³, and $V$V is volume in dm³. With $V$V in cm³: $n = (c \cdot V)/1000$n=(c⋅V)/1000.

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What a titration measures, and why

A titration is a quantitative neutralisation in which a solution of accurately known concentration (the titrant, traditionally in the burette) is added in measured volumes to a known volume of analyte (in the conical flask) until the reaction is just complete. The volume of titrant required — the titre — is read from the burette and used with the balanced equation to compute the moles of analyte.

Two terms must be held apart:

• Equivalence point — the theoretical point at which stoichiometrically equivalent moles of acid and base have been combined.
• End point — the observed point at which the indicator changes colour.

A well-chosen indicator makes the end point coincide with the equivalence point to within a fraction of a drop, which corresponds to about ±0.05 cm³ on a 50 cm³ burette — better than 0.2 % of a 25 cm³ titre. Acid-base titration is therefore one of the highest-precision benchtop techniques in chemistry, beaten in absolute accuracy only by gravimetry and a handful of instrumental analyses.

Key Definitions:

• Primary standard — a pure, stable, non-hygroscopic solid of accurately known formula and high $M_r$Mr​ used to prepare standard solutions by mass.
• Standard solution — a solution of accurately known concentration prepared from a primary standard or by titration against one.
• Titre — the volume of titrant delivered from the burette to reach the end point.
• Concordant titres — repeat titres agreeing to within ±0.10 cm³.

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Preparing a standard solution from a primary standard

To make a 250.0 cm³ standard solution at a known concentration $c$c, you need a primary standard with the following properties:

• Pure to a high standard (≥99.9 % typical).
• Stable in air — does not absorb moisture or CO₂, does not decompose on storage.
• Non-hygroscopic — does not gain mass from atmospheric moisture during weighing.
• High relative molecular mass — minimises the percentage uncertainty in the weighed mass.

Common primary standards: anhydrous sodium carbonate (Na₂CO₃, $M_r = 106.0$Mr​=106.0), potassium hydrogen phthalate (KHP, $M_r = 204.2$Mr​=204.2), oxalic acid dihydrate ($\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$H2​C2​O4​⋅2H2​O, $M_r = 126.1$Mr​=126.1). Sodium hydroxide is not a primary standard — it is hygroscopic, absorbs atmospheric CO₂ to form Na₂CO₃, and cannot be weighed accurately.

Procedure (Na₂CO₃ example, 0.100 mol dm⁻³ × 250.0 cm³)

1. Required moles: $n = c \cdot V = 0.100 \times 0.2500 = 0.02500$n=c⋅V=0.100×0.2500=0.02500 mol.
2. Required mass: $m = n \cdot M_r = 0.02500 \times 106.0 = 2.650$m=n⋅Mr​=0.02500×106.0=2.650 g.
3. Weigh by difference on an analytical balance to ±0.001 g (record empty weighing bottle, transfer, reweigh; difference is the transferred mass).
4. Dissolve in ~100 cm³ distilled water in a beaker; rinse the weighing bottle three times and add washings (quantitative transfer).
5. Transfer the solution to a 250.0 cm³ volumetric flask via a funnel; rinse beaker and funnel three times.
6. Add distilled water until the meniscus reaches the graduation mark (use a dropper for the last 1 cm³).
7. Stopper and invert at least ten times to homogenise.

Quantitative transfer matters: leaving 1 % of the solute in the original beaker creates a 1 % systematic error that cannot be corrected by careful titration technique.

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Apparatus and uncertainty budget

Apparatus	Function	Tolerance	% uncertainty (typical reading)
25.00 cm³ volumetric pipette	Deliver fixed analyte volume	±0.06 cm³	0.24 %
250.0 cm³ volumetric flask	Prepare standard solution	±0.30 cm³	0.12 %
50.00 cm³ burette	Deliver variable titrant volume	±0.05 cm³ per reading	0.40 % for 25 cm³ titre (two readings → ±0.10 cm³)
Analytical balance	Weigh primary standard	±0.001 g per reading	0.04 % for 2.650 g mass (two readings → ±0.002 g)
100 cm³ conical flask	Hold analyte during titration	—	—
Wash bottle, white tile, funnel, droppers	Support equipment	—	—

Two-reading rule: any apparatus read twice (burette start + end, balance empty + full) contributes 2 × tolerance of absolute uncertainty.

Percentage uncertainty: $\% \text{u} = \dfrac{\text{absolute uncertainty}}{\text{measured value}} \times 100$%u=measured valueabsolute uncertainty​×100. For a burette titre of 22.50 cm³: $\% \text{u} = (0.10 / 22.50) \times 100 = 0.44\,\%$%u=(0.10/22.50)×100=0.44%.

To minimise percentage uncertainty, maximise the titre within the burette range — choose analyte concentration and aliquot volume so the typical titre is 20–35 cm³. A 5 cm³ titre would carry 2.0 % uncertainty (unacceptable); a 45 cm³ titre risks running off the bottom of the burette and forcing a refill mid-titration.

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Titration procedure (PAG 2 — acid-base titration)

1. Rinse pipette with the analyte (typically the base or unknown). Pipette 25.0 cm³ into a clean (but not necessarily dry) conical flask. The flask only needs to be rinsed with distilled water — extra solute would change the moles present.
2. Add 3–4 drops of suitable indicator chosen by the acid-base strength combination (see indicator section).
3. Rinse burette with the titrant (typically the acid of known concentration). Fill the burette above the zero mark, then run titrant out through the tap into a waste beaker until the meniscus sits at or just below 0.00 cm³ and the jet is filled (no air bubble below the tap).
4. Initial burette reading to 2 d.p. (with the last digit a 0 or a 5 — the burette is graduated to 0.1 cm³ and you estimate the half-division).
5. Rough titre: add titrant rapidly while swirling until the indicator changes — record the rough titre. This locates the end-point.
6. Accurate titres: refill burette. Pipette a fresh 25.0 cm³ aliquot. Add titrant to within 1–2 cm³ of the rough end-point quickly, then dropwise (with continuous swirling) until the indicator changes colour permanently. Record the titre.
7. Repeat until you have at least two titres concordant to within ±0.10 cm³. Discard the rough titre and any obvious outlier.
8. Mean titre = average of the concordant titres only. Report to 2 d.p.

Why these rinses?

• Pipette / burette rinsed with their own solution — residual water would dilute the solution and reduce the effective concentration delivered, introducing a positive or negative systematic bias.
• Conical flask rinsed with distilled water only — adding more analyte would change the moles being titrated (and hence the titre); adding distilled water only changes the dilution, which doesn't matter because the indicator responds to [H⁺], and the end-point depends on the moles of each reactant, not on total volume.

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Indicator selection — the pH-curve framework

An acid-base indicator (HIn) is itself a weak acid whose protonated form (HIn) and deprotonated form (In⁻) have different colours:

$\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-, \quad K_{\text{In}} = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}$HIn⇌H++In−,KIn​=[HIn][H+][In−]​

The colour change is sharp over roughly $\text{p}K_{\text{In}} \pm 1$pKIn​±1, which defines the indicator's pH range — the narrow band over which the eye perceives the colour shift.

Indicator	Acid colour	Alkali colour	pH range
Methyl orange	Red	Yellow	3.1 – 4.4
Bromothymol blue	Yellow	Blue	6.0 – 7.6
Phenolphthalein	Colourless	Pink	8.3 – 10.0
Methyl red	Red	Yellow	4.4 – 6.2

For a titration to work, the indicator's pH range must fall within the vertical (steep) portion of the pH-vs-volume curve at the equivalence point.

Indicator pH bands: Methyl orange 3.1–4.4 Bromothymol 6.0–7.6 Phenolphthalein 8.3–10.0

The fourth combination (weak acid + weak base) has no usable indicator because the pH change at equivalence is gradual rather than vertical. A pH meter is the only way to locate the equivalence point precisely.

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Calculations — the three-step framework

Step 1: Moles of the known reagent

$n_1 = \frac{c_1 \cdot V_1}{1000}$n1​=1000c1​⋅V1​​ (V in cm³)

Step 2: Apply the stoichiometric ratio

$\frac{n_2}{n_1} = \frac{\text{coeff}_2}{\text{coeff}_1}$n1​n2​​=coeff1​coeff2​​

Step 3: Calculate the unknown

For unknown concentration: $c_2 = \dfrac{n_2 \cdot 1000}{V_2}$c2​=V2​n2​⋅1000​.

For percentage purity: $\% \text{purity} = \dfrac{m_{\text{pure, calculated}}}{m_{\text{sample, weighed}}} \times 100$%purity=msample, weighed​mpure, calculated​​×100.

Worked example 1 — concentration of NaOH against H₂SO₄

Q: 25.0 cm³ of a sodium hydroxide solution of unknown concentration is titrated against 0.100 mol dm⁻³ sulfuric acid. The mean titre is 22.50 cm³. Calculate the concentration of NaOH.

Equation: $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$2NaOH+H2​SO4​→Na2​SO4​+2H2​O.

• $n(\text{H}_2\text{SO}_4) = 0.100 \times 22.50 / 1000 = 2.250 \times 10^{-3}$n(H2​SO4​)=0.100×22.50/1000=2.250×10−3 mol.
• $n(\text{NaOH}) = 2 \times 2.250 \times 10^{-3} = 4.500 \times 10^{-3}$n(NaOH)=2×2.250×10−3=4.500×10−3 mol.
• $c(\text{NaOH}) = 4.500 \times 10^{-3} \times 1000 / 25.0 = 0.180$c(NaOH)=4.500×10−3×1000/25.0=0.180 mol dm⁻³.

Worked example 2 — percentage purity of impure Na₂CO₃

Q: 1.25 g of impure sodium carbonate is dissolved and made up to 250.0 cm³ in a volumetric flask. 25.0 cm³ of this solution requires a mean titre of 22.80 cm³ of 0.100 mol dm⁻³ HCl. Calculate the percentage purity.

Equation: $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$Na2​CO3​+2HCl→2NaCl+H2​O+CO2​.

• $n(\text{HCl}) = 0.100 \times 22.80 / 1000 = 2.280 \times 10^{-3}$n(HCl)=0.100×22.80/1000=2.280×10−3 mol.
• $n(\text{Na}_2\text{CO}_3 \text{ in 25.0 cm}^3) = 2.280 \times 10^{-3} / 2 = 1.140 \times 10^{-3}$n(Na2​CO3​ in 25.0 cm3)=2.280×10−3/2=1.140×10−3 mol.
• $n(\text{Na}_2\text{CO}_3 \text{ in 250 cm}^3) = 1.140 \times 10^{-3} \times 10 = 1.140 \times 10^{-2}$n(Na2​CO3​ in 250 cm3)=1.140×10−3×10=1.140×10−2 mol.
• $m(\text{pure Na}_2\text{CO}_3) = 1.140 \times 10^{-2} \times 106.0 = 1.2084$m(pure Na2​CO3​)=1.140×10−2×106.0=1.2084 g.
• $\% \text{purity} = (1.2084 / 1.25) \times 100 = 96.7\,\%$%purity=(1.2084/1.25)×100=96.7%.

Worked example 3 — diprotic vs monoprotic stoichiometry trap

Q: 20.0 cm³ of 0.250 mol dm⁻³ H₂SO₄ requires 25.0 cm³ of NaOH. Calculate [NaOH].

• $n(\text{H}_2\text{SO}_4) = 0.250 \times 20.0 / 1000 = 5.00 \times 10^{-3}$n(H2​SO4​)=0.250×20.0/1000=5.00×10−3 mol.
• $n(\text{NaOH}) = 2 \times 5.00 \times 10^{-3} = 1.00 \times 10^{-2}$n(NaOH)=2×5.00×10−3=1.00×10−2 mol (2:1 ratio).
• $c(\text{NaOH}) = 1.00 \times 10^{-2} \times 1000 / 25.0 = 0.400$c(NaOH)=1.00×10−2×1000/25.0=0.400 mol dm⁻³.

Forgetting the 2:1 ratio gives 0.200 mol dm⁻³ — half the correct value, and a frequent OCR mark-scheme deduction.

Worked example 4 — back titration to find CaCO₃ in eggshell

Q: 0.500 g of crushed eggshell is dissolved in 50.00 cm³ of 1.000 mol dm⁻³ HCl. The excess acid is titrated against 0.500 mol dm⁻³ NaOH, requiring 25.20 cm³. Calculate the percentage of CaCO₃ in the eggshell.

• $n(\text{HCl initial}) = 1.000 \times 50.00 / 1000 = 5.000 \times 10^{-2}$n(HCl initial)=1.000×50.00/1000=5.000×10−2 mol.
• $n(\text{NaOH}) = 0.500 \times 25.20 / 1000 = 1.260 \times 10^{-2}$n(NaOH)=0.500×25.20/1000=1.260×10−2 mol.
• $n(\text{HCl excess}) = n(\text{NaOH}) = 1.260 \times 10^{-2}$n(HCl excess)=n(NaOH)=1.260×10−2 mol (1:1).
• $n(\text{HCl reacted with CaCO}_3) = 5.000 \times 10^{-2} - 1.260 \times 10^{-2} = 3.740 \times 10^{-2}$n(HCl reacted with CaCO3​)=5.000×10−2−1.260×10−2=3.740×10−2 mol.
• $n(\text{CaCO}_3) = 3.740 \times 10^{-2} / 2 = 1.870 \times 10^{-2}$n(CaCO3​)=3.740×10−2/2=1.870×10−2 mol (CaCO₃ + 2HCl → ..).
• $m(\text{CaCO}_3) = 1.870 \times 10^{-2} \times 100.1 = 1.872$m(CaCO3​)=1.870×10−2×100.1=1.872 g.