# Deriving the equation for oxygen estimation by Winkler’s method

We were doing $\ce{O2}$ estimation by Winkler's method

The procedure and reactions involved in this experiment are:

1. Carefully fill a 300-mL glass stoppered bottle brim-full with sample water.
2. Immediately add 2mL of manganese sulfate to the collection bottle.
3. Add 2 mL of alkali-iodide reagent (NaOH + KI) in the same manner.
4. Stopper the bottle with care to be sure no air is introduced. Mix the sample by inverting several times.
5. Add 2 mL of concentrated sulfuric acid.Then mix well.
6. In a glass flask, titrate 100 mL of the sample with sodium thiosulfate to a pale straw color. Titrate by slowly dropping titrant solution from a calibrated pipette into the flask and continually stirring or swirling the sample water.
7. Add 2 mL of starch solution so a blue color forms.
8. Continue slowly titrating until the sample turns clear.

\begin{align}\ce{MnSO4 + 2NaOH &-> \underset{{white ppt.}}{Mn(OH)2} + Na2SO4}\\[0.6em] % \ce{2Mn(OH)2 + O2 &-> \underset{{orange-brown cloudy ppt.}}{2MnO(OH)2}}\\[0.6em] % \ce{2Na2SO4 + 2H2SO4 &-> 4 NaHSO4}\\[0.6em] % \ce{MnO(OH)2 + NaHSO4 + 2KI &-> \underset{{golden-brown colour}}{I2} + MnSO4 + K2SO4 \\&\quad\quad + 2Na2SO4 + 3H2O}\\[0.6em] % \ce{I2 + 2Na2S2O3 &-> 2NaI + Na2S4O6}\end{align}

And at the end of the experiment we needed to calculate the $\ce{O2}$ concentration in mg/l by the following equation:

$$c(\ce{O2})\ \mathrm{(mg/L)} = \frac{(V_1 \times N \times E \times 1000) }{[V_4 {(V_2-V_3)/V_2}]}$$

Here,

$V_1=$ Volume of titrant $\ce{Na2S2O3}$ (ml)

$V_2=$ Total volume of water sample (ml)

$V_3=$ Volume of ($\ce{MnSO4 + KI}$ solution) added (ml)

$V_4=$ Volume of analyte taken in titration

$E=$ Equivalent weight of $\ce{O2}$

$N=$ Normality of $\ce{Na2S2O3}$ solution

So how can I derive this equation?

Reference followed: Ghosh, K. C. & Manna, B. (2005) Practical Zoology

A different book though had a different formula :

Conc. of O2 (mg/l)= 0.2 x (Volume of sample water in ml) x (Strength of $\ce{Na2S2O3}$)

• Are you sure is $V_2 - V_3$? – Another.Chemist Dec 18 '16 at 7:53
• @Another.Chemist Yes I copied it from my book. – Tyto alba Dec 18 '16 at 16:53

[Concepts]

• Normality, N: The normality is the relation between the sustance equivalence and the liters of a solution. The sustance equivalents are referred to the charges per mole of a sustance. The number of equivalents (NOE) of a sustance is obtained as follows:

• Acid equal to number of hydrogens.
• Hidroxide equal to number of hydroxils.
• Salt equal to number of cations.

To determine the normality of a solution, these are the steps you must follow:

1. Weight Equivalent Gram, WEG: $$WEG=\frac{Molecular\ weight\ of\ solute}{NOE}$$
2. Weight Equivalent Number, WEN: $$WEN=\frac{Mass\ of\ solute}{WEG}$$
3. Normality, N: $$N = \frac{WEN}{L}$$ Doing some basic algebra, you find that: $$N=M\cdot NOE$$ Where M stands for Molarity.
• Dilution factors, DF: It is employed when the solution concetration of your sample is too concentrate. In order to determine the concentration of the substance you want to find, you have to dilute the initial solution. This is done through the dilution factors. I.e. you have 25 ml of a solution 10 M and you are asked to dilute it 100 times. To do so, you have to take an aliquot (A portion of a large whole) of 0.250 ml, pour into a volumetric flask of 25 ml and then fill it with destilled water to capacity. Then, you have a diluted solution of 0.1 M. In general, a dilution factor is expressed as: \begin{align} Dilution\ factor=\frac{Total\ volumen}{Sample\ volume} \end{align}

• Dilution: Dilution is the process of decreasing the concentration of a solute in solution, usually simply by mixing with more solvent. To dilute a solution means to add more solvent without the addition of more solute. This can be represented as: \begin{align} V_1\cdot M_1 &= V_2\cdot M_2\\ M_2&=\frac{V_1}{V_2}\cdot M_1 \end{align}

Now, if you put all together:

1. If you look at the first parenthesis at the right of the equa, having in mind that E is WEG and doing some basic algebra: \begin{align*} V_1\cdot N\cdot E\cdot 1000&=V_1(L)\cdot\frac{WEN}{L}\cdot WEG\cdot 1000\\ &=V_1(L)\cdot\frac{\frac{Mass\ of\ solute}{WEG}}{L}\cdot WEG\cdot 1000\\ &=V_1(L)\cdot\frac{Mass\ of\ solute}{WEG\cdot L}\cdot WEG\cdot 1000\\ &=V_1(L)\cdot\frac{Mass\ of\ solute}{L}\cdot 1000\\ &=V_1(mL)\cdot\frac{Mass\ of\ solute}{mL}\\ &=Mass\ of\ solute\cdot\frac{V_1(mL)}{mL}\\ &=Mass\ of\ solute \end{align*}
2. The square bracket is just a dilution factor.
3. If you put together the last two steps with the dilution section, you find a similarity.

[History]

If you are interested in the origins of this method, I recommend you the following paper:

• Winkler, L. "Die Bestimmung des in Wasser Gelösten Sauerstoffes". Berichte der Deutschen Chemischen Gesellschaft. (1888) 21, 2, 2843–2855.

I cannot give you a brief handout of it, my german is basic. But, I want to point out that in page 2845, you will find an equation similar than the indicate in dilution section.

[How to understand the Winkler method]

The Winkler method for the determination of dissolved $\ce{O2}$ involves adding an excess of ‘white’ $\ce{Mn(OH)2}$ which contains $\ce{Mn^{2+}}$ ions.

For every molecule of $\ce{O2}$ two molecules of $\ce{I2}$ are formed. The quantity of iodine present can be found though titration of the sample with sodium thiosulphate.

[Bibliography]

• How do I find out the equivalent weight of O2 here? How do I derive the number of equivalents (n) from the reactions? – Tyto alba Dec 23 '16 at 16:09
• Your 2nd reaction is a redox one. Therefore, the manganese is oxided by loosing two electrons and the oxigen is reduced by acepting 2 electrons. – Another.Chemist Dec 23 '16 at 17:23
• Then, look at weg formula. – Another.Chemist Dec 23 '16 at 19:08
• I get that. But as Na2S2O3 is reacting with I2 and not O2. I wonder how's this formula V1 S1 =V2 S2 going to calculate the conc of O2 present. – Tyto alba Dec 24 '16 at 17:57
• On your presented reactions, just go stoichiometrically backwards, having in mind that all reactions are redox. – Another.Chemist Dec 25 '16 at 0:40