sample question obtained from class. cannot think clearly while not feeling well

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sample question obtained from class. cannot think clearly while not feeling well. little help please...

The Winkler method for Dissolved Oxygen involves two separate redox reactions;

one in the field and the other in the lab.

A 250 mL water sample was taken immediately downstream of the outflow from a food processing plant.

The sample was “fixed” on-site and transported back to the lab for the analysis of dissolved oxygen using the Winkler Method.

This required the addition of 2.23 mL of a 0.0250 M sodium thiosulfate solution to reach the end-point.

(i) For the initial redox reaction that occurs in the field. Identify the oxidant and reductant and write the associated redox half-reactions. Now, write the overall equation for the redox reaction

(ii) Similarly, for the final redox reaction that occurs in the lab. Identify the oxidant and reductant and write the associated redox half-reactions. Now, write the overall equation for the redox reaction

(iii) Determine the amount of DO in mg/L present in the original water sample.

(iv) Give reasons as to whether this level of DO would be able to support a healthy aquatic ecosystem.

Explanation / Answer

In wrinkler method: An excess of manganese(II) salt, iodide (I) and hydroxide (OH) ions is added to a water sample causing a white precipitate of Mn(OH)2 to form. The Mn(OH)2 so formed get oxidized with the oxygen dissolved in water to oxidize

2 MnSO4(s) + O2(aq) 2 MnO(OH)2(s) (brown ppt)

i) so the oxidant is Oxygen and reductant is Mn+2

Reduction half cell =

O2 --> 2O

Oxidation half reaction

Mn+2 --> Mn+4

ii) Mn(SO4)2 + 2 I(aq) Mn2+(aq) + I2(aq) + 2 SO42(aq)

2 S2O32(aq) + I2 S4O62(aq) + 2 I(aq)

iii) 2.23 mL of a 0.0250 M sodium thiosulfate

so moles of sodium thiosulphte used = molarity X volume = 2.23 X 0.025 =0.0557 millimoles

The general stoichiometry is

1 mole of O2 2 moles of MnO(OH)2 2 mole of I2 4 mole of S2O32

So for each mole of oxygen 4 moles of thiosulphate will be used

Or 4 moles ill be used for 1 mole of oxygen

so 0.0557 millimoles will be used for 0.0557 X 1 / 4 moles of oxygen = 0.0139millimoles must be present in 250 mL

so grams of O2 will be = 0.0139 X 10^-3 X 32 grams = 0.4448 X 10^-3 grams / 250mL

so in 1 L = 1.779 X 10^-3 grams

or 1 L will have = 1.779 mg

so concentration = 1.779ppm

iv) but most natural water systems require 5-6 parts per million to support a diverse population so this level of DO would not be able to support a healthy aquatic ecosystem.

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