Question
To investigate how much kale would supply the daily recommended intake of iron a student:
1 weighed 79.6g of kale leaves and blended with 500\(cm^3\) of water
2 boiled, filtered and cooled
3 pipetted 10.0\(cm^3\) of the filtrate into 20.0\(cm^3\) of 2.00mol \(dm^{-3}\) sulfuric acid in a flask
4 titrated with 0.00100mol \(dm^{-3}\) potassium manganate(VII).
The reaction taking place is:
\(5Fe^{2+}(aq) + MnO_4^-(aq) + 8H^+ (aq) → 5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_2O(l)\)
(a) All species are almost colourless except for \(MnO_4^-\), which has an intense purple colour, though the kale extract is coloured by the chlorophyll present.
(i) State the colour change at the end point.
From:
To:
(ii) Outline how the addition of distilled water to the 10.0\(cm^3\) aliquot before titration will affect the titrant volume at the end point.
(b) State the class of errors that always affect results in a particular direction.
(c) The end point occurred when 3.1 ± 0.1\(cm^3\) of the titrant had been added.
(i) Calculate the percentage uncertainty associated with the titre.
(ii) Suggest one procedural modification which would reduce the percentage uncertainty for a single titration, other than using a burette with greater precision.
(iii) The solution in the titration flask contained \(8.66 × 10^{-4}\) g of iron. Determine, to three significant figures, the percentage of iron, by mass, in the kale leaves.
(d) The value obtained is about 30 times greater than published values for the percentage of iron in kale. Suggest one reason, other than human error, why there might be such a large discrepancy.
Answer/Explanation
Answer:
(a) (i) green to purple
OR
green to brown
OR
green to purple-green
(ii) none / no effect
(b) systematic
(c) (i)
(ii) using more dilute potassium manganate(VII)
OR
using more dilute titrant
OR
larger aliquot/volume of filtrate
(iii) ALTERNATIVE 1
mass Fe in the 79.6 g kale 〈〈= \(8.66×10^{−4} \times \frac{500}{10.0}\)〉〉
= 0.0433 «g»
percent by mass 〈〈=\(\frac{0.0433}{49.6} \times 100\)〉〉
= 0.0544«%»
ALTERNATIVE 2
mass of kale in titration flask 〈〈=\(79.6 \times \frac{10.0}{500}\)〉〉
= 1.592 «g»
percent by mass 〈〈=\(\frac{8.66 \times 10^{-4}}{1.592} \times 100\) 〉〉
= 0.0544«%»
(d) other substances in the leaves «as well as iron» react with the manganate(VII)
«ion»
OR
kale modified to have more iron/Fe
OR
iron/Fe in water/pipes/container used for boiling
OR
manganate(VII) oxidized/reacted with other ions/substances/metals
OR
manganate(VII) concentration changes over time