Titrations for Paper 3

AS · 14 min

Almost every Paper 3 has a titration, and it carries a large share of the marks: for accurate, concordant titres, for a correctly laid out results table, and for the calculation that follows. This note covers the technique exactly as Cambridge expects it, the three families of titration in the syllabus (acid–alkali, potassium manganate(VII) and iodine–thiosulfate), how to choose an indicator, how to process results, and the calculation patterns, including the multi-step ones that make up the hardest questions. It builds on the mole calculations in Volumes of gases and concentrations of solutions.

Apparatus and technique

apparatus	use	precision
burette (50 cm³)	delivers the titrant; read before and after	read to the nearest 0.05 cm³
pipette (10.0 or 25.0 cm³) with filler	measures an exact volume of the solution in the flask	tolerance marked on the pipette (about ±0.06 cm³ for a 25 cm³ grade B pipette)
volumetric (graduated) flask (250 cm³)	makes up a standard solution or dilution accurately	one-mark flask, fill to the line
conical flask	holds the pipetted solution and indicator; swirled during titration	not a measuring device
white tile	placed under the flask to see the colour change clearly

Method

Carrying out a titration

Rinse the burette with a little of the solution it will contain (not with water, which would dilute it). Rinse the pipette with the solution it will measure. The conical flask may be rinsed with distilled water: adding water does not change the number of moles in it.
Fill the burette using a funnel, with the tap closed; run a little out so that the jet below the tap is full and has no air bubble. Remove the funnel before reading (drips from it change the reading).
Pipette the solution into the conical flask using a pipette filler, filling until the bottom of the meniscus is on the line at eye level. Let it drain and touch the tip on the side of the flask; do not blow out the last drop.
Add two or three drops of indicator.
Read the burette at eye level, at the bottom of the meniscus, to the nearest 0.05 cm³.
Do a rough titration: add the titrant quickly while swirling until the colour changes. This tells you approximately where the end-point is.
Do accurate titrations: add quickly to about 1 cm³ before the rough end-point, then dropwise, swirling, until the indicator just changes colour permanently. Wash the inside of the flask with a little distilled water near the end-point to bring any drops into the solution.
Repeat until you have concordant results: two accurate titres within 0.10 cm³ of each other.
Calculate the mean titre from the concordant results only. Do not include the rough titre.

Definition

The end-point is the point at which the indicator changes colour. The equivalence point is when the reactants have reacted in exactly the ratio shown by the equation. A good indicator makes these coincide.

Recording titration results

Cambridge awards marks for the layout. Use a single table with headings and units, burette readings to two decimal places ending in 0 or 5, and a clearly marked rough titre.

	rough	1	2	3
final burette reading / cm³	25.20	24.70	49.05	24.45
initial burette reading / cm³	0.10	0.00	24.70	0.00
titre / cm³	25.10	24.70	24.35	24.45

Concordant titres: 24.35 and 24.45 (within 0.10 cm³). Titre 1 is 0.25–0.35 cm³ away from these, so it is not used.

\text{mean titre} = \frac{24.35 + 24.45}{2} = 24.40\ \text{cm}^3

Watch out

Every burette reading must be recorded to two decimal places: write 0.00, not 0 or 0.0; 24.70, not 24.7.
The second decimal place can only be 0 or 5 (you read to the nearest 0.05 cm³).
Never average all titres. Use only those within 0.10 cm³ of each other, and say which you used (for example by ticking them).
Do not round the mean titre in a way that loses precision: the mean of 24.35 and 24.40 is 24.375, quoted as 24.38 (or 24.375).

Acid–alkali titrations

Choosing an indicator

The indicator must change colour over a pH range that falls within the steep part (vertical section) of the titration curve.

indicator	pH range of colour change	colour in acid	colour in alkali	use for
methyl orange	2.9–4.6	red	yellow	strong acid with strong or weak base
bromophenol blue	3.0–4.5	yellow	blue	strong acid with strong or weak base
thymol blue (base range)	8.0–9.6	yellow	blue	weak or strong acid with strong base
thymolphthalein	9.3–10.5	colourless	blue	weak or strong acid with strong base

(These are the indicators listed in the syllabus. Phenolphthalein, colourless in acid and pink in alkali, range about 8.3–10, is used in the same situations as thymolphthalein.)

Key result

Strong acid + strong base: the steep section covers about pH 3–11, so any of the four works.
Strong acid + weak base (e.g. $\ce{HCl}$ and $\ce{NH3}$ ): steep section in the acidic region; use methyl orange or bromophenol blue.
Weak acid + strong base (e.g. $\ce{CH3COOH}$ and $\ce{NaOH}$ ): steep section in the alkaline region; use thymol blue or thymolphthalein.
Weak acid + weak base: no steep section; no indicator gives a sharp end-point.

The colour at the end-point is described in the direction of the titration. Adding hydrochloric acid from a burette to sodium hydroxide with methyl orange: yellow to orange (the end-point is the first permanent orange, not red, which means excess acid has been added).

Potassium manganate(VII) titrations

Acidified potassium manganate(VII), $\ce{KMnO4}$ , is a strong oxidising agent. The purple $\ce{MnO4-}$ ion is reduced to the almost colourless $\ce{Mn^2+}$ ion:

\ce{MnO4-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}

Key result

Self-indicating: no indicator is added. The $\ce{KMnO4}$ goes in the burette. While the other reagent is in excess, each drop is decolourised. The end-point is the first permanent pale pink colour in the flask (one drop of excess $\ce{MnO4-}$ ).
The flask must contain excess dilute sulfuric acid (the reaction uses $\ce{H+}$ ). Hydrochloric acid is not used, because $\ce{MnO4-}$ would oxidise $\ce{Cl-}$ to chlorine. Without enough acid, a brown precipitate of $\ce{MnO2}$ forms.
Because the solution is so dark, read the burette at the top of the meniscus (consistently).

The three reactions in the syllabus:

reducing agent	half-equation	overall equation	ratio $\ce{MnO4-}$ : reducing agent
iron(II)	$\ce{Fe^2+ -> Fe^3+ + e-}$	$\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$	1 : 5
hydrogen peroxide	$\ce{H2O2 -> O2 + 2H+ + 2e-}$	$\ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O}$	2 : 5
ethanedioate	$\ce{C2O4^2- -> 2CO2 + 2e-}$	$\ce{2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O}$	2 : 5

The ethanedioate (oxalate) titration must be warmed to about 60 °C because the reaction is slow at room temperature. It speeds up as it goes, because the $\ce{Mn^2+}$ formed catalyses it (autocatalysis).

Iodine–thiosulfate titrations

These titrations find the amount of an oxidising agent. The oxidising agent is added to excess potassium iodide, which it oxidises to iodine. The iodine released is then titrated with standard sodium thiosulfate:

\ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}

Key result

Thiosulfate goes in the burette. The brown iodine colour fades to pale yellow as the end-point approaches.
Then add a few drops of starch: the solution turns blue-black. Continue dropwise until the blue-black colour just disappears, leaving a colourless solution. That is the end-point.
Starch is added near the end, not at the start, because a large amount of iodine forms a starch complex that releases iodine only slowly, blurring the end-point.
Key ratio: $\ce{I2} : \ce{S2O3^2-} = 1 : 2$ .

Common reactions that release the iodine:

oxidising agent	equation	moles of $\ce{S2O3^2-}$ per mole of oxidising agent
iodate(V)	$\ce{IO3- + 5I- + 6H+ -> 3I2 + 3H2O}$	6
copper(II)	$\ce{2Cu^2+ + 4I- -> 2CuI + I2}$	1
chlorate(I) (bleach)	$\ce{ClO- + 2I- + 2H+ -> Cl- + I2 + H2O}$	2
hydrogen peroxide	$\ce{H2O2 + 2I- + 2H+ -> I2 + 2H2O}$	2

The calculation method

Method

Titration calculations, including multi-step ones

Write the balanced equation(s). For redox titrations, build them from half-equations.
Calculate moles of the titrant: $n = \dfrac{c \times V}{1000}$ , with $V$ the mean titre.
Use the ratio to find moles of the substance in the flask (the pipetted portion). For iodine–thiosulfate titrations, chain the ratios: thiosulfate to iodine to oxidising agent.
Scale up to the whole sample if a portion was taken from a volumetric flask (e.g. $\times \dfrac{250}{25.0} = \times 10$ ).
Convert to what is asked: concentration, molar mass, percentage by mass, $x$ in a formula, or relative atomic mass.
Give the answer to the right number of significant figures (normally 3 or 4; see below) with units.

On significant figures: the syllabus says a calculated value should have the same number of significant figures as, or one more than, the least precise data. A titre of 24.40 cm³ (4 s.f.) with a concentration of 0.100 mol dm⁻³ (3 s.f.) allows an answer to 3 or 4 s.f.

Worked examples

Acid–alkali: processing results

Using the results table above, $25.0\ \text{cm}^3$ of sodium hydroxide solution was titrated with $0.100\ \text{mol dm}^{-3}$ hydrochloric acid. Calculate the concentration of the sodium hydroxide.

Solution

Mean titre (concordant results 2 and 3) $= 24.40\ \text{cm}^3$ .

$n(\ce{HCl}) = \dfrac{0.100 \times 24.40}{1000} = 2.440 \times 10^{-3}\ \text{mol}$

$\ce{NaOH + HCl -> NaCl + H2O}$ ( $1 : 1$ ): $n(\ce{NaOH}) = 2.440 \times 10^{-3}\ \text{mol}$ in $25.0\ \text{cm}^3$ .

c(\ce{NaOH}) = \frac{2.440 \times 10^{-3}}{0.0250} = 0.0976\ \text{mol dm}^{-3}

Standardising potassium manganate(VII)

$0.1675\ \text{g}$ of sodium ethanedioate, $\ce{Na2C2O4}$ , is dissolved in excess dilute sulfuric acid, warmed to 60 °C, and titrated with potassium manganate(VII) solution. The titre is $25.00\ \text{cm}^3$ . Calculate the concentration of the $\ce{KMnO4}$ . ( $M(\ce{Na2C2O4}) = 134.0\ \text{g mol}^{-1}$ )

Solution

$n(\ce{C2O4^2-}) = \dfrac{0.1675}{134.0} = 1.250 \times 10^{-3}\ \text{mol}$

$\ce{2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O}$

$n(\ce{MnO4-}) = \dfrac{2}{5} \times 1.250 \times 10^{-3} = 5.000 \times 10^{-4}\ \text{mol}$

c = \frac{5.000 \times 10^{-4}}{0.02500} = 0.0200\ \text{mol dm}^{-3}

Iron in an iron tablet

One iron tablet (containing iron(II) sulfate) is dissolved in dilute sulfuric acid and made up to $100\ \text{cm}^3$ . A $25.0\ \text{cm}^3$ portion needs $11.20\ \text{cm}^3$ of $0.00500\ \text{mol dm}^{-3}$ $\ce{KMnO4}$ . Calculate the mass of iron, in mg, in one tablet. ( $A_r$ : Fe 55.8)

Solution

$n(\ce{MnO4-}) = \dfrac{0.00500 \times 11.20}{1000} = 5.600 \times 10^{-5}\ \text{mol}$

$\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$ : $n(\ce{Fe^2+}) = 5 \times 5.600 \times 10^{-5} = 2.800 \times 10^{-4}\ \text{mol}$ in $25.0\ \text{cm}^3$ .

In $100\ \text{cm}^3$ : $\times 4 = 1.120 \times 10^{-3}\ \text{mol}$ .

Mass of Fe $= 1.120 \times 10^{-3} \times 55.8 = 0.0625\ \text{g} = 62.5\ \text{mg}$ .

Exam-style: standardising thiosulfate with potassium iodate

$0.214\ \text{g}$ of potassium iodate(V), $\ce{KIO3}$ , is dissolved and made up to $250\ \text{cm}^3$ . A $25.0\ \text{cm}^3$ portion is added to excess potassium iodide and dilute sulfuric acid. The iodine released needs $24.00\ \text{cm}^3$ of sodium thiosulfate solution. Calculate the concentration of the thiosulfate. ( $M(\ce{KIO3}) = 214.0\ \text{g mol}^{-1}$ )

Solution

$n(\ce{KIO3})$ in flask $= \dfrac{0.214}{214.0} = 1.000 \times 10^{-3}\ \text{mol}$ ; in $25.0\ \text{cm}^3$ : $1.000 \times 10^{-4}\ \text{mol}$ .

$\ce{IO3- + 5I- + 6H+ -> 3I2 + 3H2O}$ : $n(\ce{I2}) = 3.000 \times 10^{-4}\ \text{mol}$

$\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}$ : $n(\ce{S2O3^2-}) = 6.000 \times 10^{-4}\ \text{mol}$

c = \frac{6.000 \times 10^{-4}}{0.02400} = 0.0250\ \text{mol dm}^{-3}

Chain of ratios: 1 $\ce{IO3-}$ $\to$ 3 $\ce{I2}$ $\to$ 6 $\ce{S2O3^2-}$ .

Exam-hard: copper in brass

$1.00\ \text{g}$ of brass (an alloy of copper and zinc) is dissolved in nitric acid. The solution is neutralised, made up to $250\ \text{cm}^3$ , and $25.0\ \text{cm}^3$ portions are added to excess potassium iodide. The iodine released needs a mean titre of $19.70\ \text{cm}^3$ of $0.0500\ \text{mol dm}^{-3}$ sodium thiosulfate.

\ce{2Cu^2+(aq) + 4I-(aq) -> 2CuI(s) + I2(aq)}

(a) Calculate the percentage by mass of copper in the brass.

(b) Zinc ions do not react with iodide. Explain why this matters for the method.

(c) The off-white precipitate of CuI makes the end-point hard to see. Suggest why starch is still added only near the end-point.

( $A_r$ : Cu 63.5)

Solution

(a) $n(\ce{S2O3^2-}) = \dfrac{0.0500 \times 19.70}{1000} = 9.850 \times 10^{-4}\ \text{mol}$

$n(\ce{I2}) = 4.925 \times 10^{-4}\ \text{mol}$ ; $n(\ce{Cu^2+}) = 2 \times 4.925 \times 10^{-4} = 9.850 \times 10^{-4}\ \text{mol}$ (ratio $\ce{Cu^2+} : \ce{S2O3^2-} = 1 : 1$ ).

In $250\ \text{cm}^3$ : $9.850 \times 10^{-3}\ \text{mol}$ ; mass $= 9.850 \times 10^{-3} \times 63.5 = 0.625\ \text{g}$ .

Percentage $= \dfrac{0.625}{1.00} \times 100 = 62.5\%$ .

(b) Only copper(II) releases iodine, so the titre measures copper alone; zinc does not interfere. If another ion in the alloy also oxidised iodide, the titre would be too large and the copper content overestimated.

(c) If starch is added while a lot of iodine is present, iodine becomes held in the blue-black starch complex and is released only slowly, so the colour lingers and the end-point is overshot or unclear. Adding starch when the solution is pale yellow means only a little iodine remains, giving a sharp change from blue-black to (off-)white.

Exam tip

Marks are given for accuracy: your titre is compared with the supervisor's value. Careful dropwise addition near the end-point is worth more than speed.
Marks are also given for concordance (two titres within 0.10 cm³) and for recording: a single table, headings with units (/ cm³), all readings to 0.05 cm³ and two decimal places.
Show every step of the calculation. If you make an early arithmetic slip, later marks can still be awarded "error carried forward".
Learn the end-point colours exactly: methyl orange yellow to orange (acid added to alkali); $\ce{KMnO4}$ colourless to pale pink; thiosulfate with starch blue-black to colourless.
When asked why a rough titration is done: to find the approximate end-point so that the accurate titrations can be done quickly and then dropwise near the end-point.

Practical skills

Making a standard solution: weigh the solid accurately by difference (weigh the weighing bottle full, tip the solid into a beaker, reweigh the bottle; the difference is the mass transferred). Dissolve in distilled water in a beaker, transfer to a 250 cm³ volumetric flask with a funnel, rinsing the beaker, rod and funnel into the flask. Add distilled water until the bottom of the meniscus is on the line, then stopper and invert several times to mix. Common errors: not rinsing the beaker (solute lost, concentration too low), overshooting the line, not mixing thoroughly, and reading the meniscus from above or below eye level.

Summary

Rinse the burette and pipette with the solutions they will hold; the conical flask with distilled water.
Read the burette to 0.05 cm³, at eye level, and record to two decimal places.
Do a rough titration, then accurate ones until two are within 0.10 cm³; average the concordant titres only.
Indicators: methyl orange or bromophenol blue for strong acid–weak base; thymol blue or thymolphthalein for weak acid–strong base; any for strong–strong.
$\ce{KMnO4}$ : self-indicating, colourless to pale pink, needs excess dilute $\ce{H2SO4}$ ; ratios $\ce{MnO4-} : \ce{Fe^2+} = 1 : 5$ , $\ce{MnO4-} : \ce{H2O2}$ or $\ce{C2O4^2-} = 2 : 5$ .
Iodine–thiosulfate: $\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}$ ; add starch near the end (blue-black to colourless).
Calculation: moles of titrant, ratio, scale up, convert, correct significant figures.

Practice

Question

Explain why the burette is rinsed with the solution it will contain, but the conical flask is rinsed with distilled water.
Titres of 23.65, 23.10, 23.25 and 23.15 cm³ were obtained (the first is the rough). Identify the concordant titres and calculate the mean titre.
Suggest a suitable indicator for the titration of ethanoic acid with sodium hydroxide, and state the colour change seen at the end-point if the alkali is in the burette.
Why is no indicator needed in a potassium manganate(VII) titration? What colour change is seen at the end-point?
$25.0\ \text{cm}^3$ of $0.0500\ \text{mol dm}^{-3}$ sulfuric acid is neutralised by $20.85\ \text{cm}^3$ of potassium hydroxide solution. Calculate the concentration of the potassium hydroxide.
$10.0\ \text{cm}^3$ of hydrogen peroxide solution is diluted to $250\ \text{cm}^3$ . A $25.0\ \text{cm}^3$ portion, acidified, needs $20.40\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\ce{KMnO4}$ . Calculate the concentration of the original hydrogen peroxide.
Explain why hydrochloric acid is not used to acidify potassium manganate(VII) titrations, and what is seen if too little sulfuric acid is added.
$1.00\ \text{g}$ of impure iron(II) ammonium sulfate, $\ce{(NH4)2Fe(SO4)2.6H2O}$ ( $M = 392.0$ ), is dissolved in dilute sulfuric acid and titrated with $0.0200\ \text{mol dm}^{-3}$ $\ce{KMnO4}$ ; the titre is $24.80\ \text{cm}^3$ . Calculate the percentage purity.
A student adds starch at the start of an iodine–thiosulfate titration. Explain the effect on the end-point.
$2.50\ \text{g}$ of a hydrated iron(II) sulfate, $\ce{FeSO4.xH2O}$ , is dissolved and made up to $250\ \text{cm}^3$ . $25.0\ \text{cm}^3$ portions, acidified, need a mean titre of $18.00\ \text{cm}^3$ of $0.0100\ \text{mol dm}^{-3}$ $\ce{KMnO4}$ . Calculate $x$ . ( $M$ : $\ce{FeSO4}$ 151.9, $\ce{H2O}$ 18.0)

Answers

Water left in the burette would dilute the titrant, lowering its concentration and giving a titre that is too large. Water in the conical flask does not change the number of moles of the pipetted substance, which is all that matters in the reaction.
Accurate titres 23.10, 23.25 and 23.15 cm³. Concordant (within 0.10): 23.10 and 23.15 (23.25 differs from 23.10 by 0.15). Mean $= (23.10 + 23.15) / 2 = 23.125 \approx 23.13\ \text{cm}^3$ .
Thymolphthalein (or thymol blue), since this is a weak acid–strong base titration with the steep section in the alkaline region. With thymolphthalein: colourless to (pale) blue.
$\ce{MnO4-}$ is intensely purple and its reduction product $\ce{Mn^2+}$ is almost colourless, so the first drop of excess manganate(VII) colours the solution. End-point: colourless to the first permanent pale pink.
$n(\ce{H2SO4}) = 0.0500 \times 25.0 / 1000 = 1.250 \times 10^{-3}\ \text{mol}$ . $\ce{H2SO4 + 2KOH -> K2SO4 + 2H2O}$ : $n(\ce{KOH}) = 2.500 \times 10^{-3}\ \text{mol}$ in 20.85 cm³. $c = 2.500 \times 10^{-3} / 0.02085 = 0.120\ \text{mol dm}^{-3}$ (0.1199).
$n(\ce{MnO4-}) = 0.0200 \times 20.40 / 1000 = 4.080 \times 10^{-4}$ . $n(\ce{H2O2}) = 2.5 \times 4.080 \times 10^{-4} = 1.020 \times 10^{-3}$ in 25.0 cm³; $1.020 \times 10^{-2}$ in 250 cm³, all from 10.0 cm³ of original. $c = 1.020 \times 10^{-2} / 0.0100 = 1.02\ \text{mol dm}^{-3}$ .
Manganate(VII) is a strong enough oxidising agent to oxidise chloride ions to chlorine, so some $\ce{MnO4-}$ would be used up by the acid and the titre would be too large. With too little acid, $\ce{MnO4-}$ is reduced only to manganese(IV) oxide, seen as a brown precipitate, and the ratio in the equation no longer applies.
$n(\ce{MnO4-}) = 0.0200 \times 24.80 / 1000 = 4.960 \times 10^{-4}$ . $n(\ce{Fe^2+}) = 5 \times 4.960 \times 10^{-4} = 2.480 \times 10^{-3}\ \text{mol}$ . Mass $= 2.480 \times 10^{-3} \times 392.0 = 0.972\ \text{g}$ . Purity $= 97.2\%$ .
With a lot of iodine present, the starch forms a blue-black complex that releases iodine only slowly. Near the end the blue-black colour fades gradually rather than disappearing sharply, so the end-point is difficult to judge and is usually overshot, giving a titre that is too large.
$n(\ce{MnO4-}) = 0.0100 \times 18.00 / 1000 = 1.800 \times 10^{-4}$ . $n(\ce{Fe^2+}) = 9.000 \times 10^{-4}$ in 25.0 cm³; $9.000 \times 10^{-3}$ in 250 cm³. $M = 2.50 / 9.000 \times 10^{-3} = 277.8\ \text{g mol}^{-1}$ . $151.9 + 18.0x = 277.8$ , $x = 6.99 \approx 7$ : $\ce{FeSO4.7H2O}$ .

Apparatus and technique

Recording titration results

Acid–alkali titrations

Choosing an indicator

Potassium manganate(VII) titrations

Iodine–thiosulfate titrations

The calculation method

Worked examples

Practice

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