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

Polyesters are biodegradable due to the presence of ester bonds in their structure, which can be hydrolyzed by microorganisms. The natural enzymes produced by these microorganisms break down the polyester into smaller, non-toxic molecules over time.

In contrast, polyalkenes such as polyethylene and polypropylene are highly stable and resistant to microbial attack because they lack functional groups that can undergo reactive interactions. Their long-chain hydrocarbon structure does not easily degrade, making them persist in the environment for centuries.

To calculate the concentration of the potassium manganate(VII) solution, we start from the titration reaction:

$2 MnO4− + 16 H+ + 5 C2O42− → 2 Mn2+ + 8 H2O$

From the balanced equation, we can see that 2 moles of potassium manganate(VII) react with 5 moles of sodium ethanoate.

1. Calculate moles of sodium ethanoate:

Given:

• Mass of Na2C2O4 = 162 g
• Molar mass (Mr) = 134.0 g/mol

Moles of Na2C2O4 = ( \frac{162 ext{ g}}{134.0 ext{ g/mol}} = 1.21 ext{ mol} )

2. Calculate concentration of the sodium ethanoate solution:

Concentration (C) = ( \frac{n}{V} = \frac{1.21 ext{ mol}}{0.250 ext{ dm}^3} = 4.84 ext{ mol dm}^{-3} )

3. Calculate the moles of MnO4− used in the titration:

From the titration: Volume of MnO4− = 23.85 cm³ = 0.02385 dm³

Using the molar ratio from the balanced equation, we can find the moles of MnO4−:

The concentration of KmnO4 = ( C_{MnO4} = \frac{(5/2) * 4.84 ext{ mol}}{0.02385 ext{ dm}^3} )

Calculating this will give us the concentration of the potassium manganate(VII) solution.

1. Ensure that the burette is closed and not leaking before filling it to avoid spillage.
2. Wear gloves to protect yourself from skin contact with potassium manganate(VII), as it is a harmful substance that can cause skin irritation.

1. Use a fume hood or ensure good ventilation to avoid inhalation of any dust or vapors.
2. Always add the solid sodium ethanoate to water slowly to prevent excessive exothermic reactions.

The color change at the end point of each titration is from purple (appearance of permanganate ion) to colorless (when all MnO4− has reacted).

1. Ensure there are no air bubbles in the burette tip, as they can affect the accuracy of the reading.
2. Allow the tap to fill space below the tap to avoid any errors in the initial reading.

The replacement of the water ligands by ethanoate ions is favourable because:

1. Enthalpy Change: The formation of new bonds with ethanoate ions typically releases energy, making the process exothermic and thus favourable in terms of enthalpy.
2. Entropy Change: The replacement of six bound water molecules with six ethanoate ions increases the disorder (entropy) of the system, which is favourable under the Second Law of Thermodynamics.
3. Free-energy Change: The Gibbs free-energy change ($\Delta G$) for the reaction can be calculated through the relationship: $\Delta G = \Delta H - T \Delta S$. A negative free-energy change indicates that the reaction is spontaneous. Here, the combination of favorable enthalpy and increased entropy typically results in a negative $\Delta G$, making the ligand substitution highly favorable.