How are concentrations and amounts found accurately by titration and analysis?
Redox titrations using potassium manganate(VII) and iodine-thiosulfate, the calculations involved, colorimetry for coloured ions, and the planning and evaluation of quantitative analysis with attention to uncertainty.
A CCEA A-Level Chemistry answer on quantitative analysis, covering redox titrations with potassium manganate(VII) and iodine-thiosulfate, the associated calculations, colorimetry for coloured ions, and the planning and evaluation of analysis with uncertainty.
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What this dot point is asking
CCEA wants you to carry out and calculate redox titrations (manganate(VII) and iodine-thiosulfate), use colorimetry for coloured ions, and plan and evaluate quantitative analysis with attention to uncertainty.
Redox titrations in outline
A redox titration finds an unknown amount of an oxidising or reducing agent by reacting it with a standard solution of the other in a known mole ratio taken from the balanced ionic equation. The key skill is the mole calculation: find the moles of the standard from , scale by the ratio from the equation, then divide by the volume to get the unknown concentration. CCEA examines two systems in detail, manganate(VII) and iodine-thiosulfate, each with its own way of detecting the endpoint.
Manganate(VII) titrations
The manganate(VII) is placed in the burette and run into the acidified iron(II) solution. While any iron(II) remains, each drop of purple manganate(VII) is reduced to almost colourless and the colour vanishes on swirling. Once all the iron(II) has reacted, the next drop is not reduced and the solution turns a permanent pale pink, which is the endpoint. The acid must be sulfuric: hydrochloric acid would be oxidised to chlorine by the manganate(VII), consuming extra oxidant and giving a falsely high titre, while nitric acid is itself an oxidising agent and would interfere. The amount of acid must be in excess so the medium stays strongly acidic, otherwise brown forms instead of the colourless .
Iodine-thiosulfate titrations
This is an example of an indirect (back) titration. The oxidising agent itself may have no convenient colour change, so it is first reacted with excess potassium iodide; the iodine set free is exactly equivalent to the oxidising agent, and the iodine is then measured by titration against thiosulfate. For example, copper(II) liberates iodine: , so each mole of frees half a mole of and needs one mole of thiosulfate. As the brown iodine colour fades to pale straw the endpoint is close; only then is starch added, turning the mixture blue-black, and the endpoint is the single drop that makes the blue-black just disappear.
Colorimetry
The colorimeter passes light of a chosen colour (a filter is selected so the solution absorbs strongly, usually the complementary colour of the solution) through the sample, and the detector reads the absorbance. A series of standards of known concentration is measured to plot a calibration curve of absorbance against concentration, which is a straight line through the origin at low concentrations (the Beer-Lambert relationship). The absorbance of the unknown is then read against this line. Colorimetry suits coloured transition-metal ions and is faster and uses less sample than a titration.
Planning and uncertainty
Good quantitative analysis controls uncertainty. The percentage uncertainty in a measuring step is the absolute uncertainty divided by the reading, so using larger volumes and masses lowers the percentage uncertainty. A burette read to at each end gives on a titre, so a titre of carries a smaller percentage uncertainty than one of . Repeat until titres are concordant (agreeing within ), and mean only the concordant results, discarding the rough trial. The overall percentage uncertainty of the experiment is found by adding the percentage uncertainties of each apparatus reading, and the largest contributor is usually the burette.
Examples in context
Manganate(VII) titrations measure the iron content of an ore or of an iron tablet: the tablet is dissolved in sulfuric acid and titrated, and the self-indicating endpoint avoids any added indicator. Iodine-thiosulfate titrations measure the chlorine content of pool water or the copper content of an alloy (copper(II) liberates iodine from iodide). CCEA practical assessment rewards quoting the percentage uncertainty in each measuring step and identifying the largest contributor, usually the burette reading.
Try this
Q1. State why manganate(VII) needs no separate indicator. [1 mark]
- Cue. It is self-indicating; the first permanent pink colour marks the endpoint.
Q2. Write the equation for the reaction of iodine with thiosulfate. [1 mark]
- Cue. .
Q3. of manganate(VII) reacts with iron(II) in a ratio. Calculate the moles of iron(II). [2 marks]
- Cue. ; .
Exam-style practice questions
Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
CCEA 20195 marksA sample of acidified iron(II) sulfate solution required of potassium manganate(VII) for complete reaction. Determine the concentration of the iron(II) ions.Show worked answer →
The redox equation is , so the ratio is to .
Moles of manganate(VII): .
Moles of iron(II): .
Concentration of iron(II): .
Markers reward (1) the correct ratio from the balanced equation, (2) moles of manganate, (3) scaling to moles of iron(II), (4) dividing by the iron volume, (5) the concentration. No indicator is needed because manganate(VII) is self-indicating (the first permanent pink marks the endpoint).
CCEA 20214 marksDescribe how an iodine-thiosulfate titration is used to find the concentration of an oxidising agent, and explain the role of starch indicator.Show worked answer →
The oxidising agent is reacted with excess potassium iodide, which it oxidises to iodine: the amount of iodine liberated is equivalent to the amount of oxidising agent.
The liberated iodine is then titrated with standard sodium thiosulfate: .
As the endpoint nears, the solution turns from brown to pale straw. Starch is added near the endpoint; it forms a deep blue-black complex with the remaining iodine. The endpoint is when this blue-black colour just disappears. Starch is added late so the large amount of iodine does not bind it irreversibly.
Markers reward (1) oxidising agent liberates iodine from excess iodide, (2) titrate iodine with thiosulfate, (3) starch gives a blue-black colour, (4) endpoint when the blue-black just disappears, with starch added near the end.
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Sources & how we know this
- CCEA GCE Chemistry specification — CCEA (2016)