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  1. Problems
  2. /A 25.0 mL sample of old ether is shaken with excess acidified potassium iodide…
HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Alcohols, Phenols and EthersNumerical

A 25.0 mL sample of old ether is shaken with excess acidified potassium iodide…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A 25.0 mL25.0\ \text{mL}25.0 mL sample of old ether is shaken with excess acidified potassium iodide. The peroxide ROOH\text{ROOH}ROOH oxidises iodide to iodine: ROOH+2I−+2H+→ROH+H2O+I2\text{ROOH} + 2 \text{I}^{-} + 2 \text{H}^{+} \to \text{ROH} + \text{H}_{2}\text{O} + \text{I}_{2}ROOH+2I−+2H+→ROH+H2​O+I2​. The iodine is titrated with 0.0100 mol/L0.0100\ \text{mol/L}0.0100 mol/L sodium thiosulphate, and 4.60 mL4.60\ \text{mL}4.60 mL is needed. Two thiosulphate ions react with each I2\text{I}_2I2​. Find the amount of thiosulphate in micromoles, the amount of peroxide in micromoles, and the peroxide in mg per litre of ether, expressed as hydrogen peroxide (molar mass 34.015 g/mol34.015\ \text{g/mol}34.015 g/mol).

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micromol
micromol
mg/L
Need a reminder? The key ideaShowHide

One mole of an ideal gas occupies RTP\dfrac{RT}{P}PRT​: 24.46 L24.46\ \text{L}24.46 L at 25 ∘C25\ ^\circ\text{C}25 ∘C and 101.325 kPa101.325\ \text{kPa}101.325 kPa. The lower flammable limit of ether vapour is 1.9 %1.9\ \%1.9 % by volume. A titration gives moles of thiosulphate, and two thiosulphate ions react with each I2\text{I}_2I2​. An activating group directs to the ortho and para carbons.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. 25.0 mL25.0\ \text{mL}25.0 mL of ether, 4.60 mL4.60\ \text{mL}4.60 mL of 0.0100 mol/L0.0100\ \text{mol/L}0.0100 mol/L thiosulphate.

Step 1. Thiosulphate. 0.0100×4.60=0.0460 mmol=46.0 μmol0.0100 \times 4.60 = 0.0460\ \text{mmol} = 46.0\ \mu\text{mol}0.0100×4.60=0.0460 mmol=46.0 μmol.

Step 2. Iodine and peroxide. I2=46.02=23.0 μmol\text{I}_2 = \dfrac{46.0}{2} = 23.0\ \mu\text{mol}I2​=246.0​=23.0 μmol. One ROOH gives one I2\text{I}_2I2​, so the peroxide is 23.0 μmol23.0\ \mu\text{mol}23.0 μmol.

Step 3. As hydrogen peroxide. 23.0 μmol×34.015 g/mol=782 μg=0.782 mg23.0\ \mu\text{mol} \times 34.015\ \text{g/mol} = 782\ \mu\text{g} = 0.782\ \text{mg}23.0 μmol×34.015 g/mol=782 μg=0.782 mg in 25.0 mL25.0\ \text{mL}25.0 mL. In a litre: 0.782×100025.0=31.3 mg/L0.782 \times \dfrac{1000}{25.0} = 31.3\ \text{mg/L}0.782×25.01000​=31.3 mg/L.

Check. A reading of this size means peroxide is present in the ether. Many laboratory safety guides call a level of about 100 mg/L100\ \text{mg/L}100 mg/L or more unsafe to distil. This sample should be treated before any distillation.

Answer: 46.0 μmol46.0\ \mu\text{mol}46.0 μmol, 23.0 μmol23.0\ \mu\text{mol}23.0 μmol and 31.3 mg/L31.3\ \text{mg/L}31.3 mg/L.

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