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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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    $\begingroup$ Probably MnCl2 - manganese chloride not magnesium. $\endgroup$ – Mithoron Mar 17 '15 at 11:45
  • $\begingroup$ It is definitely not magnesium, but manganese. $\endgroup$ – Klaus-Dieter Warzecha Mar 17 '15 at 15:34
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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?

  1. 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.

  1. 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:

\[\ce{Mn(IV) + 2 I- -> Mn^{2+} + I2}\]

  1. The amount of iodine formed, which repesents the aount of oxygen in the water sample, can be determined by titration with thiosulfate solution.
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