amount of dissolved oxygen

Question

In the year 2025, NASA was able to find another planet (XXV) which may be able to sustain life. In order to determine if the planet’s water can sustain an earth organism’s life, Winkler method was employed to determine dissolved oxygen content. A water sample from planet XXV was obtained and then treated with the usual reagents for Winkler titration. A 50.0 mL aliquot of this sample was titrated with the 0.010196M Na2S2O3 standard solution. It required 4.16 mL of the titrant to reach the starch endpoint.

A. Calculate the amount of dissolved oxygen in the sample in terms of ppm O2. (FW O2 = 32.00)

Answer 1

When you look at the reactions taking place in the Winkler method, it boils down to the fact that ONE MOLE of dissolved oxygen (O₂) reacts with TWO MOLES of Na₂S₂O₃ to produce the starch endpoint. Using this information, we can find the dissolved oxygen as follows ...

mols Na₂S₂O₃ used = 4.16 mls x 1 L / 1000 mls x 0.010196 mols / L = 4.242x10^-5 moles

mols O₂ present in 50.0 mls = 4.242x10^-5 mols Na₂S₂O₃ x 1 mol O₂ / 2 mols Na₂S₂O₃ = 2.121x10^-5 mols

mg O₂ in 50.0 mls = 2.121x10^-5 mols O₂ x 32.0 g / mol x 1000 mg / g = 0.6787 mg O₂ / 50.0 mls

Converting this to ppm O₂ (which is equivalent to mg / L), we have ...

0.6787 mg O₂ / 50.0 mls x 1000 mls / L = 13.6 mg / L = 13.6 ppm (3 sig. figs.)