Topics

Standardising solutions and collecting quality data

Why standardising is named in the criteria

Merit requires the collection of quality data which includes standardising the standard solution(s). It is written into the criterion, not implied.
The reason is arithmetic: the titrant's concentration is a direct multiplier in every result you calculate.

$n(\text{analyte}) = c(\text{titrant}) \times V(\text{titre}) \times \text{mole ratio}$

A titrant 3% weaker than assumed makes every result 3% low, uniformly. Concordant titres will not reveal it, and repeating will not remove it.

Primary standards

A primary standard can be weighed out to make a solution of accurately known concentration. It must be pure, stable in air, of high molar mass, and soluble.

Substance	Primary standard?	Used to standardise
Anhydrous sodium carbonate, Na₂CO₃	Yes	acids
Potassium hydrogen phthalate (KHP)	Yes	bases
Potassium iodate, KIO₃	Yes	thiosulfate (via liberated iodine)
Sodium oxalate, Na₂C₂O₄	Yes	permanganate
Ammonium iron(II) sulfate (Mohr's salt)	Yes	permanganate, dichromate
Sodium hydroxide	No	absorbs H₂O and CO₂
Sodium thiosulfate	No	slowly decomposes; bacteria attack it
Potassium permanganate	No	oxidises traces of organic matter; decomposes in light
Iodine solution	No	iodine sublimes; iodide is air-oxidised

Note how many of the useful redox titrants are not primary standards. Permanganate, thiosulfate and iodine all need standardising, and all three are common choices for this standard.

Standardising permanganate against sodium oxalate

Sodium oxalate is weighed accurately, dissolved, acidified and warmed to about 60 °C, then titrated with the permanganate:

$2\text{MnO}_4^-(aq) + 5\text{C}_2\text{O}_4^{2-}(aq) + 16\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l)$

Warming is required because the reaction is slow at room temperature — the first few drops decolourise only sluggishly until enough Mn²⁺ has formed to autocatalyse it.
Justifying that warming step in relation to the reaction occurring is exactly the kind of Excellence-level explanation the criteria want.

Controlling significant variables

In a trend investigation, everything except the independent variable must be held constant, or a difference between samples could be caused by something other than the trend.

Variable	Why it must be controlled
Titrant concentration	one standardised batch throughout; a change mid-investigation makes early and late samples incomparable
Sample mass or volume	the same pipette or the same weighed mass for every sample
Time between preparation and titration	many analytes decompose; a varying delay creates a false trend
Temperature	affects both solution volumes and reaction rates
Acid concentration	redox half equations consume H⁺; too little acid changes the products
Endpoint judgement	the same person, the same criterion, the same white tile
Sample homogeneity	blend or mix so each aliquot represents the whole sample

The two that most often produce a false trend are titrant drift and varying delay before titration, because both change systematically over the course of an investigation whose samples are usually analysed in order.

Sufficient data

At least five values of the independent variable, so a trend can be distinguished from scatter.
Three concordant titres at each value, so each point has a reliable mean.
Concordant means agreeing within 0.05 mL (or your school's stated tolerance), and only the concordant ones are averaged.
Report all your titres, including the outliers, with a comment. Deleting data is not permitted; explaining it is.

Worked ExampleStandardising a thiosulfate solution

A student standardises sodium thiosulfate solution using potassium iodate as a primary standard. 0.1783 g of KIO₃ was dissolved and made up to 250.0 mL. A 25.0 mL aliquot was acidified and excess potassium iodide added; the liberated iodine required 24.15 mL of the thiosulfate solution. Calculate the concentration of the thiosulfate. (M(KIO₃) = 214.00 g mol⁻¹)

$\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}$ $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$

Step 1 — Moles of iodate weighed out

$n(\text{KIO}_3) = \frac{m}{M} = \frac{0.1783}{214.00} = 8.33178... \times 10^{-4} \text{ mol}$

Step 2 — Concentration of the iodate solution

$c = \frac{n}{V} = \frac{8.33178... \times 10^{-4}}{0.2500} = 3.33271... \times 10^{-3} \text{ mol L}^{-1}$

Step 3 — Moles of iodate in the aliquot

$n = c \times V = 3.33271... \times 10^{-3} \times 0.0250 = 8.33178... \times 10^{-5} \text{ mol}$

Step 4 — Moles of iodine liberated

From the first equation, the ratio IO₃⁻ : I₂ is 1 : 3:

$n(\text{I}_2) = 3 \times 8.33178... \times 10^{-5} = 2.49953... \times 10^{-4} \text{ mol}$

Note that the iodide is in excess — it is not limiting, so it does not enter the calculation.

Step 5 — Moles of thiosulfate used

From the second equation, the ratio I₂ : S₂O₃²⁻ is 1 : 2:

$n(\text{S}_2\text{O}_3^{2-}) = 2 \times 2.49953... \times 10^{-4} = 4.99907... \times 10^{-4} \text{ mol}$

Step 6 — Concentration of the thiosulfate

$c = \frac{n}{V} = \frac{4.99907... \times 10^{-4}}{0.02415} = 0.0207001... \text{ mol L}^{-1}$

Step 7 — Significant figures

The measurements were: mass 0.1783 g (4 sf), titre 24.15 mL (4 sf), aliquot 25.0 mL (3 sf), flask 250.0 mL (4 sf). The limiting measurement is the aliquot at 3 sf.

$c(\text{S}_2\text{O}_3^{2-}) = 0.0207 \text{ mol L}^{-1}$

Answer: the thiosulfate solution is 0.0207 mol L⁻¹. This value, not the nominal concentration on the bottle, must be used in every subsequent calculation.