Dichromate and Iodine Titrations – Solutions and Titration (Senior 5 Chemistry) | Ganzaa

Solutions and Titration

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Senior 5 · Chemistry

Solutions and Titration

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Lesson 4 of 5

Dichromate and Iodine Titrations

25 min

Two more oxidising agents, two different end points

Potassium manganate(VII) is not the only oxidising titrant. Potassium dichromate(VI) is a primary standard, which manganate(VII) is not, and it is still the basis of the chemical oxygen demand (COD) test that water laboratories use to measure organic pollution. Iodine is a much gentler oxidant, and iodide ions are a mild reducing agent; between them they let you measure vitamin C in juice, chlorine in bleach and copper in an ore. The price is that neither titration can show its own end point clearly: you need an indicator.

Definition

Dichromate titration: a redox titration with acidified potassium dichromate(VI), $\mathrm{Cr_2O_7^{2-}}$ (orange), which is reduced to $\mathrm{Cr^{3+}}$ (green).

Iodimetry: direct titration of a reducing agent with a standard iodine solution, e.g. vitamin C.

Iodometry: indirect determination of an oxidising agent: the oxidant reacts with excess iodide to release iodine, which is then titrated with standard sodium thiosulfate.

Starch indicator: forms an intense blue-black complex with iodine (as $\mathrm{I_3^{-}}$ ); the colour appears or disappears at the end point.

Dichromate(VI) titrations

$\mathrm{Cr_2O_7^{2-}(aq) + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)}$

With iron(II), which gives one electron each, the ratio is 1 : 6:

$\mathrm{Cr_2O_7^{2-} + 6Fe^{2+} + 14H^{+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$

Feature	$\mathrm{KMnO_4}$	$\mathrm{K_2Cr_2O_7}$
Primary standard?	no	yes
Electrons gained per ion	5	6
Indicator	none (self-indicating)	sodium diphenylamine sulfonate: green → violet-blue
Hydrochloric acid allowed?	no, oxidises $\mathrm{Cl^{-}}$	yes in dilute acid (a weaker oxidant)
Stability of solution	decomposes slowly in light	stable for years

Phosphoric acid is usually added with the indicator: it forms a colourless complex with $\mathrm{Fe^{3+}}$ , which sharpens the end point.

Iodine titrations

Iodine is only slightly soluble in water, so "iodine solution" is iodine dissolved in potassium iodide, where it exists as the brown $\mathrm{I_3^{-}}$ ion. The key titration reaction is with thiosulfate:

$\mathrm{I_2(aq) + 2S_2O_3^{2-}(aq) \rightarrow 2I^{-}(aq) + S_4O_6^{2-}(aq)}$

Thiosulfate is oxidised to tetrathionate; two thiosulfate ions give up one electron each. Iodometry uses this reaction to measure an oxidant indirectly:

Oxidant	Reaction with excess iodide	mol $\mathrm{S_2O_3^{2-}}$ per mol oxidant
iodate(V)	$\mathrm{IO_3^{-} + 5I^{-} + 6H^{+} \rightarrow 3I_2 + 3H_2O}$	6
copper(II)	$\mathrm{2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2}$	1
chlorate(I) (bleach)	$\mathrm{ClO^{-} + 2I^{-} + 2H^{+} \rightarrow I_2 + Cl^{-} + H_2O}$	2
hydrogen peroxide	$\mathrm{H_2O_2 + 2I^{-} + 2H^{+} \rightarrow I_2 + 2H_2O}$	2

Figure 11.5: Colours in an iodometric titration. Starch is added only near the end point.

Why add starch late? With a lot of iodine present, the starch–iodine complex forms so strongly that iodine is released only slowly, and the end point is overshot or blurred. Adding it when the solution is pale straw-yellow gives a sharp change from blue-black to colourless.

Iodimetry works the other way: standard iodine is run from the burette into the reducing agent with starch already present, and the end point is the first permanent blue-black. Vitamin C (ascorbic acid, $\mathrm{C_6H_8O_6}$ ) reacts 1 : 1: $\mathrm{C_6H_8O_6 + I_2 \rightarrow C_6H_6O_6 + 2H^{+} + 2I^{-}}$ .

Worked example

Problem. 25.00 cm³ of an iron(II) solution needs 20.15 cm³ of 0.01667 mol dm⁻³ $\mathrm{K_2Cr_2O_7}$ . Calculate $[\mathrm{Fe^{2+}}]$ .

Solution.

$n(\mathrm{Cr_2O_7^{2-}}) = 0.01667 \times 0.02015 = 3.359 \times 10^{-4}\,\mathrm{mol} \qquad n(\mathrm{Fe^{2+}}) = 6 \times 3.359 \times 10^{-4} = 2.015 \times 10^{-3}\,\mathrm{mol}$

$[\mathrm{Fe^{2+}}] = \frac{2.015 \times 10^{-3}}{0.02500} = 0.0806\,\mathrm{mol\,dm^{-3}}$

Worked example

Problem. (a) 0.1070 g of potassium iodate(V) (M = 214.00) is dissolved, excess KI and dilute sulfuric acid are added, and the iodine released needs 30.00 cm³ of sodium thiosulfate. Find the thiosulfate concentration. (b) 10.00 cm³ of household bleach is diluted to 250.0 cm³; 25.00 cm³ of this, with excess KI and acid, needs 16.10 cm³ of the same thiosulfate. Find the concentration of $\mathrm{ClO^{-}}$ in the bleach and the mass of NaClO per dm³ (M = 74.44).

Solution.

(a) $n(\mathrm{IO_3^{-}}) = \frac{0.1070}{214.00} = 5.000 \times 10^{-4}$ ; $n(\mathrm{I_2}) = 3 \times 5.000 \times 10^{-4} = 1.500 \times 10^{-3}$ ; $n(\mathrm{S_2O_3^{2-}}) = 2 \times 1.500 \times 10^{-3} = 3.000 \times 10^{-3}\,\mathrm{mol}$ :

$c = \frac{3.000 \times 10^{-3}}{0.03000} = 0.1000\,\mathrm{mol\,dm^{-3}}$

(b) $n(\mathrm{S_2O_3^{2-}}) = 0.1000 \times 0.01610 = 1.610 \times 10^{-3}$ ; $n(\mathrm{I_2}) = n(\mathrm{ClO^{-}}) = 8.050 \times 10^{-4}\,\mathrm{mol}$ in 25.00 cm³, so $8.050 \times 10^{-3}\,\mathrm{mol}$ in the 10.00 cm³ of bleach:

$[\mathrm{ClO^{-}}] = \frac{8.050 \times 10^{-3}}{0.01000} = 0.805\,\mathrm{mol\,dm^{-3}} \qquad 0.805 \times 74.44 = 59.9\,\mathrm{g\,dm^{-3}}$

Answer: (a) 0.1000 mol dm⁻³; (b) 0.805 mol dm⁻³, about 60 g of NaClO per dm³ (6.0 % mass by volume).

Warning

Chromium(VI) compounds are toxic and carcinogenic. Use the smallest amounts possible, wear gloves and goggles, never handle the solid powder outside a fume cupboard, and collect all chromium waste for safe disposal; never pour it down the sink. First aid (CDC/NIOSH Pocket Guide, chromic acid and chromates): eyes, irrigate immediately; skin, wash with soap and flush immediately; swallowed, get medical attention immediately.

Never mix bleach with acid outside a small, controlled titration: the two release chlorine gas. Iodine stains and irritates skin.

Tip

In iodometry, follow the iodine: oxidant → I₂ → thiosulfate. Write the ratio for each arrow, then multiply. For $\mathrm{IO_3^{-}}$ : 1 → 3 → 6. For $\mathrm{Cu^{2+}}$ : 2 → 1 → 2, i.e. 1 : 1.

tmjc chemistry: Iodometric Titration. Watch the colour run from brown to pale yellow as thiosulfate is added, then the blue-black after starch, and the sudden change to colourless at the end point.

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Practice

Write the half-equation for the reduction of dichromate(VI) in acid and state the colour change.
Give two advantages of potassium dichromate(VI) over potassium manganate(VII) as a titrant.
Calculate the mass of $\mathrm{K_2Cr_2O_7}$ (M = 294.18) needed for 250.0 cm³ of 0.01667 mol dm⁻³ solution.
25.00 cm³ of orange juice needs 9.45 cm³ of 0.00500 mol dm⁻³ iodine with starch. Calculate the mass of vitamin C (M = 176.12) per 100 cm³ of juice.
1.248 g of a copper(II) salt is dissolved and made up to 250.0 cm³. 25.00 cm³ with excess KI releases iodine that needs 20.00 cm³ of 0.02500 mol dm⁻³ thiosulfate. Calculate the amount of $\mathrm{Cu^{2+}}$ in the sample and show that the salt could be $\mathrm{CuSO_4{\cdot}5H_2O}$ (M = 249.68).
Explain why starch is added near the end of an iodometric titration but at the start of an iodimetric one.
A student leaves the flask of liberated iodine standing in sunlight for an hour before titrating. Predict and explain the effect on the calculated amount of oxidant.

Show answersHide answers

$\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O}$ ; orange to green.
Any two: it is a primary standard; its solution is stable; it can be used in hydrochloric acid.
$0.01667 \times 0.2500 \times 294.18 = 1.226\,\mathrm{g}$ .
$n = 0.00500 \times 0.00945 = 4.725 \times 10^{-5}\,\mathrm{mol} = 8.32\,\mathrm{mg}$ in 25.00 cm³; 33.3 mg per 100 cm³.
$n(\mathrm{S_2O_3^{2-}}) = 5.000 \times 10^{-4} = n(\mathrm{Cu^{2+}})$ per 25.00 cm³; total $5.000 \times 10^{-3}\,\mathrm{mol}$ ; $M = \frac{1.248}{5.000 \times 10^{-3}} = 249.6\,\mathrm{g\,mol^{-1}}$ , matching $\mathrm{CuSO_4{\cdot}5H_2O}$ .
In iodometry iodine is present from the start, and a large amount would bind strongly to starch and blur the end point. In iodimetry no iodine is present until the end point, so starch can be there from the start to show the first excess.
Two errors act in opposite directions. Iodine is volatile, so some escapes and the titre falls; but in light, air slowly oxidises the excess iodide to more iodine, which raises it. The result is unreliable either way. Titrate the liberated iodine at once, in a stoppered flask, away from sunlight.

Sources and credits

First aid: NIOSH Pocket Guide to Chemical Hazards: Chromic acid and chromates, US Centers for Disease Control and Prevention.
Image: Potassium dichromate solutions.JPG by Leiem, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Iodometric Titration, tmjc chemistry, embedded from YouTube.
The text, Figure 11.5, the tables, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Two more oxidising agents, two different end points

2Dichromate(VI) titrations

3Iodine titrations

Two more oxidising agents, two different end points

Dichromate(VI) titrations

Iodine titrations