I read in a biology course that you can determine the concentration of oxygen in water by adding some $\ce{NaOH}$ and then a small amount of $\ce{MgCl2}$. They say that a dark coloured precipitate is formed. I can't see a way of writing this reaction. Can anyone help me?
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2$\begingroup$ Probably MnCl2 - manganese chloride not magnesium. $\endgroup$– MithoronCommented Mar 17, 2015 at 11:45
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$\begingroup$ It is definitely not magnesium, but manganese. $\endgroup$– Klaus-Dieter WarzechaCommented Mar 17, 2015 at 15:34
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1 Answer
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This is the old Winkler titration, published as
Die Bestimmung des im Wasser gelösten Sauerstoffes by Ludwig Wilhelm Winkler in Ber. Dtsch. Chem. Ges., 1888, 21, 2843-2854 (SRC)
How does it work?
- Add $\ce{MnSO4}$ to your water sample, then $\ce{NaOH}$ and $\ce{KI}$.
Under alkaline conditions, oxygen dissolved in your water sample oxidizes $\ce{Mn(II)}$ to $\ce{Mn(IV)}$. \[\ce{2Mn^{2+} + O2 + 4 OH- -> MnO(OH)2 v}\] This is the brownish precipitate mentioned in the question.
- Now, the sample is acidified (Winkler used hydrochloric acid) and the solution becomes yellowish due to the iodine formed. Under these conditions, $\ce{Mn(IV)}$ oxidizes iodide:
\[\ce{Mn(IV) + 2 I- -> Mn^{2+} + I2}\]
- The amount of iodine formed, which repesents the aount of oxygen in the water sample, can be determined by titration with thiosulfate solution.
answered Mar 17, 2015 at 15:33