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This question is about ethanoic acid (HOOCCH3) and the ethanoate ion (−OOCCH3) - AQA - A-Level Chemistry - Question 1 - 2021 - Paper 3

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This question is about ethanoic acid (HOOCCH3) and the ethanoate ion (−OOCCH3). 1. Ethanoic acid reacts with propane-1,3-diol (HOCH2CH(CH3)CH2OH) to form a polyeste... show full transcript

Worked Solution & Example Answer:This question is about ethanoic acid (HOOCCH3) and the ethanoate ion (−OOCCH3) - AQA - A-Level Chemistry - Question 1 - 2021 - Paper 3

Step 1

Draw the repeating unit of this polyester.

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Answer

The repeating unit of the polyester formed from the reaction between ethanoic acid and propane-1,3-diol can be represented as follows:

        O
        ||
H2C - C - O - C - H
        |   |
     H   H

In this structure, the ester bond (–O–) links the two monomer units.

Step 2

Explain why polyesters are biodegradable but polyalkenes are not biodegradable.

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Answer

Polyesters are biodegradable because they can be broken down by microorganisms due to the presence of polar functional groups, such as esters, which are more susceptible to hydrolysis. In contrast, polyalkenes are made up of non-polar hydrocarbon chains, which resist microbial attack and are generally more stable in the environment, making them non-biodegradable.

Step 3

Calculate the concentration, in mol dm−3, of the potassium manganate(VII) solution.

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Answer

First, calculate the moles of sodium ethanoate:

  1. Number of moles of Na2C2O4:

a. Moles = mass / molar mass

a. Moles = 162 g / 134 g/mol = 1.209 mol

  1. Concentration of Na2C2O4 solution:

Concentration = moles / volume

Volume = 0.250 dm³

Concentration = 1.209 mol / 0.250 dm³ = 4.836 mol/dm³

  1. From the balanced equation, 2 moles of MnO4− react with 5 moles of C2O4^2−. Therefore:

Moles of MnO4− used = (2/5) * moles of C2O4^2−

4.836 mol/dm³ * 0.025 dm³ = 0.1209 mol

Thus, moles of MnO4− = (2/5) * 0.1209 mol = 0.04836 mol

  1. Volume of MnO4− used in the titration:

Volume = 23.85 cm³ = 0.02385 dm³

  1. Concentration of MnO4− solution:

Concentration = moles / volume = 0.04836 mol / 0.02385 dm³ = 2.023 mol/dm³.

Step 4

Drawing the meniscus in Figure 1.

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Answer

When drawing the meniscus in Figure 1, ensure that the bottom of the curve touches the graduation mark and does not cross it.

Step 5

Safety precautions when filling the burette with potassium manganate(VII) solution.

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Answer

  1. Ensure to wear gloves and goggles to protect skin and eyes from harmful chemicals.
  2. Always use a funnel when pouring the solution to prevent spills and minimize exposure.

Step 6

Safety precautions when dissolving the solid sodium ethanoate in water.

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Answer

  1. Always wear gloves to avoid skin contact with the corrosive solid.
  2. Work under a fume hood to avoid inhalation of dust.

Step 7

State the colour change seen at the end point of each titration.

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Answer

The colour change observed at the end point of the titration of potassium manganate(VII) is from deep purple to pale pink.

Step 8

Practical steps needed before recording the initial burette reading.

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Answer

  1. Ensure the burette is filled with the solution without any air bubbles.
  2. Allow the tap to fill the space below the tap.

Step 9

Explain why the replacement of water ligands by ethanoate ions is favourable.

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Answer

The replacement of water ligands by ethanoate ions is favourable due to the following reasons:

  1. Enthalpy change: The formation of more stable bonds with ethanoate ions releases energy, contributing to a negative enthalpy change.
  2. Entropy change: The displacement of six water molecules by a more complex ligand like ethanoate increases the disorder (entropy) of the system.
  3. The overall free-energy change of the reaction can be expressed as:

ΔG=ΔHTΔS\Delta G = \Delta H - T \Delta S

Where a negative enthalpy change and a positive entropy change will lead to a negative free energy change, indicating a spontaneous reaction.

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