Standard solutions and titration
Volumetric analysis: standard solutions and the properties of primary standards, preparing a standard solution, standardising sodium hydroxide, titration glassware and how to rinse it, reading a burette, concordant titres, titration calculations including dilution steps, and the errors that affect results.
A good titration in a school lab can find the concentration of an acid to within about 0.2%. That puts a burette, a pipette and a few drops of indicator in the same league as instruments costing thousands of dollars. The catch is that every step has to be done exactly right, and the whole result rests on one solution whose concentration you know for certain.
The language of titration
Volumetric analysis finds the amount of a substance by measuring the volume of a solution of known concentration that reacts exactly with it. In an acid–base titration:
- The titrant is the solution in the burette. The analyte is the solution in the conical flask.
- An aliquot is an accurately measured volume of solution, delivered by a pipette.
- The titre is the volume of titrant delivered from the burette.
- The equivalence point is when the acid and base have been mixed in exactly the mole ratio of the equation. The end point is when the indicator changes colour. A well-chosen indicator makes the two match closely (6I).
- Concordant titres agree within 0.10 mL (some schools use 0.05 mL). The rough titration is never included in the average.
Standard solutions and primary standards
A standard solution is one whose concentration is accurately known. You can make one directly only from a primary standard: a substance so pure and stable that its mass tells you exactly how many moles you have. A primary standard should:
- be available in very high purity (99.9% or better) with a known formula
- be stable in air: not absorb water (non-hygroscopic) or carbon dioxide, and not lose water of crystallisation
- have a high molar mass, so weighing errors are a small fraction of the mass
- dissolve readily in water and react completely and quickly in the titration.
| Good primary standards | Not primary standards, and why |
|---|---|
| anhydrous sodium carbonate, Na₂CO₃ (dried at about 270 °C first), for standardising acids | NaOH: absorbs water and reacts with CO₂ from the air, so its mass is never exactly NaOH |
| potassium hydrogen phthalate, KHC₈H₄O₄ ("KHP", M = 204.22 g/mol), for standardising bases | HCl: a solution of a gas, which escapes from concentrated acid, so its concentration drifts |
| oxalic acid dihydrate, H₂C₂O₄·2H₂O, a diprotic acid for standardising bases | Na₂CO₃·10H₂O (washing soda crystals): loses water of crystallisation in air |
Solutions such as NaOH and HCl are therefore made up to an approximate concentration and then standardised: titrated against a primary standard to find their exact concentration. A solution standardised this way is called a secondary standard. NaOH solutions slowly absorb CO₂, so they should be standardised on the day of use.
Glassware and how to rinse it
Rinsing is the step students most often lose marks on. The rule: rinse anything that must deliver an exact concentration with the solution it will contain; rinse anything that only holds a known number of moles with distilled water.
| Glassware | Final rinse with | Why |
|---|---|---|
| burette | the titrant | water left inside would dilute the titrant, making titres too large |
| pipette | the solution to be pipetted | water left inside would dilute the aliquot, so it contains fewer moles |
| conical flask | distilled water (it can stay wet) | water doesn't change the moles in the aliquot; rinsing with the analyte would add extra moles |
| volumetric flask | distilled water | it is filled with water to the mark anyway |
Pipettes are filled with a pipette filler (never by mouth) until the meniscus sits on the line, and allowed to drain under gravity, touching the tip against the inside of the flask. The small drop left in the tip is part of the calibration: don't blow it out. Burettes are read at eye level, at the bottom of the meniscus, to the nearest 0.05 mL (two decimal places, the second being 0 or 5).
Titration technique
- Rinse and fill the burette with titrant, run some through to fill the tip with no air bubble, and record the initial reading.
- Pipette an aliquot of the analyte into a conical flask and add 2–3 drops of indicator. Stand the flask on a white tile.
- Do a rough titration quickly to find the approximate titre.
- For accurate titrations, run in titrant quickly to about 1 mL before the rough end point, then add dropwise, swirling constantly and rinsing the inside of the flask with distilled water from a wash bottle.
- Stop at the first permanent colour change (for phenolphthalein, a pale pink that lasts 30 seconds). Record the final reading.
- Repeat until you have three concordant titres. Average them.
- Do a rough titration with the +1.00 mL button and record the end point. Then do accurate trials, switching to single drops near the rough titre, until the sim reports three concordant titres.
- Read the calculation the sim shows. Is your concentration within 2% of the true value?
- Try overshooting on purpose in one trial. What does the sim say, and why does this titre spoil concordance?
- Turn on "Burette rinsed with water only" and repeat. Does the calculated concentration come out too high or too low? Explain why using the moles of NaOH.
- Turn that off and turn on "Flask rinsed with the acid". Which way is the error now, and why?
Titration calculations
Every titration calculation follows the same chain: known solution → moles → mole ratio → moles of unknown → concentration, using $n = cV$ with $V$ in litres.
0.5105 g of KHP (KHC₈H₄O₄, M = 204.22 g/mol, a monoprotic acid) was dissolved in water and titrated with NaOH, needing 24.35 mL to reach the end point. Find the concentration of the NaOH.
20.00 mL aliquots of a sulfuric acid solution were titrated with 0.1150 mol/L NaOH. Titres: 23.45 (rough), 22.80, 22.90, 22.85 and 23.20 mL. Find the concentration of the acid.
25.00 mL of white vinegar was diluted to 250.0 mL in a volumetric flask. 20.00 mL aliquots of the diluted vinegar needed an average of 16.42 mL of 0.1027 mol/L NaOH. Find the concentration of ethanoic acid in the vinegar in mol/L and in g per 100 mL.
Errors and how they affect the result
For each error, work out whether the titre is too large or too small, then follow it through the calculation. When the titrant is the known solution and you're finding the analyte's concentration, a titre that is too large gives a concentration that is too high.
| Error (NaOH titrant, finding [acid]) | Effect on titre | Calculated [acid] |
|---|---|---|
| burette rinsed with water only | NaOH diluted, so more needed: larger | too high |
| pipette rinsed with water only | fewer moles of acid in the flask: smaller | too low |
| conical flask rinsed with the acid | extra acid in the flask: larger | too high |
| conical flask wet with distilled water | no change in moles | no effect |
| air bubble in the burette tip that comes out during titration | reading change larger than the volume delivered | too high |
| overshooting the end point | larger | too high |
| NaOH standardised days earlier (absorbed CO₂) | actual [OH⁻] lower: larger | too high |
These are systematic errors: they push every result the same way, so repeating doesn't fix them, although it does reduce the effect of random errors such as judging the end point colour or reading the meniscus. Concordant titres show precision (reliability), not accuracy; you can be precisely wrong if a systematic error is present.
Test yourself
Which property is not required of a primary standard?
It must dissolve in water for an aqueous titration; solubility in ethanol is irrelevant.
A student rinsed the pipette with distilled water before taking an aliquot of the acid. The calculated acid concentration will be
The water in the pipette dilutes the aliquot, so less base is needed but the calculation still assumes the original concentration.
Titres of 24.10, 23.65, 23.70 and 23.60 mL were recorded (the first was rough). The average titre to use is
Exclude the rough titre and average the three concordant values: (23.65 + 23.70 + 23.60) ÷ 3 = 23.65 mL.
1.060 g of anhydrous Na₂CO₃ (M = 105.99 g/mol) is dissolved and made up to 250.0 mL. 25.00 mL aliquots need 21.30 mL of HCl. The concentration of HCl is
n(Na₂CO₃) in 25.00 mL = 0.0100 × 0.1 = 1.000 × 10⁻³ mol; n(HCl) = 2.000 × 10⁻³ mol; c = 2.000 × 10⁻³ ÷ 0.02130 = 0.0939 mol/L.
Label the titration equipment.
How should each item be rinsed just before use?
A solution whose concentration is accurately known is a solution. An accurately measured volume delivered by a pipette is an . The volume delivered from the burette is the . The point where the reactants have been mixed in the exact mole ratio is the point. Titres that agree within 0.10 mL are called .
A student was asked to find the concentration of a sodium hydroxide solution. They weighed 1.326 g of anhydrous sodium carbonate, dissolved it and made it up to 250.0 mL. They used this to standardise a hydrochloric acid solution: 20.00 mL aliquots of the carbonate solution needed an average of 18.95 mL of HCl (methyl orange indicator). Then 20.00 mL aliquots of the NaOH needed an average of 22.40 mL of the same HCl. (a) Calculate the concentration of the NaOH. (b) Explain why the student did not simply weigh out solid NaOH to make a standard solution.
(a) n(Na₂CO₃) = 1.326 ÷ 105.99 = 0.01251 mol; c = 0.01251 ÷ 0.2500 = 0.05004 mol/L.
In each 20.00 mL aliquot: n(Na₂CO₃) = 0.05004 × 0.02000 = 1.001 × 10⁻³ mol.
Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g), so n(HCl) = 2 × 1.001 × 10⁻³ = 2.002 × 10⁻³ mol.
c(HCl) = 2.002 × 10⁻³ ÷ 0.01895 = 0.1056 mol/L.
In the second titration, n(HCl) = 0.1056 × 0.02240 = 2.366 × 10⁻³ mol. HCl + NaOH react 1 : 1, so n(NaOH) = 2.366 × 10⁻³ mol in 20.00 mL.
c(NaOH) = 2.366 × 10⁻³ ÷ 0.02000 = 0.1183 mol/L.
(b) Solid NaOH is not a primary standard. It is hygroscopic, absorbing water from the air, and it reacts with atmospheric CO₂ to form Na₂CO₃. A weighed mass of NaOH therefore contains an unknown amount of water and carbonate, so its true number of moles can't be calculated from its mass. It must be standardised against a primary standard instead.
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