Volumetric analysis and titration: VCE Chemistry Unit 2

What this dot point is asking

VCAA wants you to describe the principles of volumetric analysis, to perform acid-base and redox titrations (including the choice of indicator based on the equivalence-point pH), to use primary and secondary standards, and to do the stoichiometric calculations that turn a titre into a concentration or mass, including back-titration.

The answer

What a titration measures

A titration determines the concentration (or amount) of an analyte by reacting it with a known volume of a solution of known concentration (a standard solution) until the reaction is just complete (the equivalence point), signalled by a colour change at the endpoint.

Apparatus: a burette delivers the standard (the titrant); a pipette delivers a known aliquot of the analyte solution to a conical flask; an indicator changes colour at (or near) the equivalence point.

Replicate titres should agree to within about 0.10 mL; the average of concordant titres (usually three) is the mean titre used in calculations.

Primary and secondary standards

A primary standard is a substance you can weigh directly and trust as the basis of an accurate solution. Criteria:

• High purity (more than 99.9%).
• Stable in air (does not absorb water, react with CO2, oxidise on standing).
• High molar mass (small mass errors are diluted out).
• Soluble in water and reacts in a known stoichiometry.

Common primary standards: anhydrous sodium carbonate (Na2CO3), oxalic acid dihydrate (H2C2O4.2H2O), potassium hydrogen phthalate (KHP), AgNO3.

A secondary standard is a solution whose concentration was determined by titration against a primary standard (or another secondary standard already traced to one). NaOH and HCl are usually used as secondary standards because they absorb CO2 / are not pure as supplied.

Acid-base titrations and indicator choice

The shape of the titration curve depends on whether the acid and base are strong or weak.

Indicator end-point ranges:

• Methyl orange: pH 3.1 to 4.4, red to yellow.
• Methyl red: pH 4.2 to 6.3, red to yellow.
• Bromothymol blue: pH 6.0 to 7.6, yellow to blue.
• Phenolphthalein: pH 8.2 to 10.0, colourless to pink.

The chosen indicator must change colour within the steep vertical section of the titration curve around the equivalence point.

Redox titrations

Same workflow with a redox reaction instead of acid-base. The standard may be itself coloured (KMnO4 is intensely purple) and acts as its own indicator: the first persistent pink colour beyond the analyte signals the endpoint. Iodine titrations use starch (deep blue with I2). Acidified KMnO4 is used to titrate Fe^2+ and other reductants.

The stoichiometric calculation

The unbreakable workflow:

1. Calculate moles of titrant from c x V (use the mean titre and volumes in litres).
2. Convert to moles of analyte using the mole ratio from the balanced equation.
3. Calculate concentration or mass of the analyte from c = n / V (analyte aliquot volume) or m = n x M.
4. If the analyte is in a diluted aliquot, scale back up to the original sample concentration using the dilution factor.

Back-titration

Used when:

• The analyte reacts too slowly with a titrant for a direct titration.
• The analyte is volatile or insoluble in the usual solvent.
• An exact endpoint is hard to see.

Procedure:

1. Add a known excess of a standard reagent that reacts completely with the analyte.
2. Titrate the unreacted excess against a second standard.
3. Moles of analyte = (moles of first reagent added) - (moles of first reagent left over) all corrected by the mole ratio with the analyte.

Classic example: an antacid tablet (calcium carbonate) dissolved in excess HCl, with the leftover HCl back-titrated against NaOH. The titration of CaCO3 directly is impractical because the solid reacts slowly and produces CO2 bubbles that disrupt the endpoint.

Examples in context

Example 1. Tartaric-acid determination in Yarra Valley wine. Yarra Valley wineries titrate finished wine to verify titratable acidity is within the $5.0$ to $7.0 \, \text{g/L}$ range expected for chardonnay. A $10.0 \, \text{mL}$ aliquot of wine is diluted with $50 \, \text{mL}$ of recently boiled water (to remove $\text{CO}_2$) and titrated against $0.100 \, \text{mol/L}$ $\text{NaOH}$ using phenolphthalein. Average titre $= 8.45 \, \text{mL}$. Tartaric acid is diprotic: $\text{H}_2 \text{C}_4 \text{H}_4 \text{O}_6 + 2 \text{NaOH} \to \text{Na}_2 \text{C}_4 \text{H}_4 \text{O}_6 + 2 \text{H}_2 \text{O}$. Moles NaOH $= 8.45 \times 10^{-4}$; moles tartaric acid $= 4.225 \times 10^{-4}$ in $10 \, \text{mL}$; mass per L $= 4.225 \times 10^{-2} \times 150 = 6.34 \, \text{g/L}$, within specification.

Example 2. Back-titration of antacid at the WEHI pharmacology lab. Pharmacists at WEHI determine $\text{CaCO}_3$ content in an antacid tablet via back-titration. A $0.5000 \, \text{g}$ tablet is dissolved in $50.0 \, \text{mL}$ of $0.500 \, \text{mol/L}$ $\text{HCl}$ (an excess). The unreacted acid is then titrated against $0.250 \, \text{mol/L}$ $\text{NaOH}$, requiring $35.2 \, \text{mL}$. Initial moles HCl $= 0.0250$; moles NaOH $= 8.80 \times 10^{-3}$; moles HCl reacted with $\text{CaCO}_3 = 0.0250 - 8.80 \times 10^{-3} = 0.0162$ mol. By $\text{CaCO}_3 + 2 \text{HCl} \to \text{CaCl}_2 + \text{H}_2 \text{O} + \text{CO}_2$, moles $\text{CaCO}_3 = 0.00810$; mass $= 0.00810 \times 100.1 = 0.811 \, \text{g}$, but tablet is only $0.5 \, \text{g}$... so $\text{CaCO}_3$ accounts for the entire neutralising capacity (suggesting the listed mass is the tablet's active content).

Try this

Q1. State three features that make a substance a suitable primary standard. [3 marks]

• Cue. High purity (over $99.9\%$); high molar mass (reduces weighing error); stable in air (does not absorb water or react); examples: $\text{Na}_2 \text{CO}_3$, KHP, oxalic acid dihydrate.

Q2. A $25.00 \, \text{mL}$ aliquot of vinegar is diluted to $250 \, \text{mL}$ in a volumetric flask. $25.00 \, \text{mL}$ aliquots of the diluted solution are titrated against $0.100 \, \text{mol/L}$ NaOH, average titre $20.50 \, \text{mL}$. Calculate the concentration of ethanoic acid in the original vinegar in g/L. [4 marks]

• Cue. $n(\text{NaOH}) = 2.05 \times 10^{-3}$ mol; $n(\text{CH}_3 \text{COOH})$ in $25 \, \text{mL}$ diluted $= 2.05 \times 10^{-3}$; scale by $\times 10$ for diluted-to-undiluted; concentration $= 0.0205 \times 10 / 0.025 = 8.20 \, \text{mol/L}$ undiluted; mass $= 8.20 \times 60.0 / 1 = 49.2 \, \text{g/L}$ (this is roughly $50 \, \text{g/L}$, typical of household vinegar).

Q3. A back-titration determines $\text{NH}_3$ in a fertiliser. (a) Outline the procedure. (b) $1.000 \, \text{g}$ of fertiliser releases $\text{NH}_3$ absorbed in $50.0 \, \text{mL}$ of $0.500 \, \text{mol/L}$ HCl. Back-titration needs $32.5 \, \text{mL}$ of $0.250 \, \text{mol/L}$ NaOH. Calculate $\%$ nitrogen. (c) Identify the indicator. [2+2+2 marks]

• Cue. (a) Excess standard HCl + sample, back-titrate excess. (b) $n(\text{HCl}) = 0.0250$; $n(\text{NaOH}) = 8.125 \times 10^{-3}$; $n(\text{NH}_3) = 0.01688$; mass N $= 0.01688 \times 14.0 = 0.236 \, \text{g}$; $\%$N $= 23.6\%$. (c) Methyl orange (titrating strong acid with strong base, but with $\text{NH}_4^+$ present in solution).