How Does Winkler's Method Calculate Dissolved Oxygen in Water?

[Tyto alba]

We were measuring the amount of O2 dissolved in water by Winkler's method.
Here's a link to the procedure and reactions: http://file:///C:/Users/SANJUKTA/Desktop/Downloads/Capture.PNG http://dropcanvas.com/6su7m/1 (The same is attached to this post)
We had to use this formula: c(O2) (mg/L)=(V1×N×E×1000)/V4
where
V1 = Vol of titrant Na2S2O3
V4 =Vol of analyte
E =Equ wgt of O2
N =Normality of Na2S2O3 sol

Now I've got two questions:

1. How do I calculate Equivalent weight of O₂? Is it from the 2nd reaction where n=2 as O₂ is being reduced to O^2-?
2. How's this formula V₁S₁=V₂S₂ going to calculate the conc. of O2 straight way when Na₂S₂O₃ reacts with I₂ and not O₂? Through this equation aren't we supposed to relate only the Volume and strength of chemicals that react ? O₂ and N₂S₂O₃ are not directly reacting!

 

[MoreKnowledge]

We were measuring the amount of O2 dissolved in water by Winkler's method.
Here's a link to the procedure and reactions: http://file:///C:/Users/SANJUKTA/Desktop/Downloads/Capture.PNG http://dropcanvas.com/6su7m/1 (The same is attached to this post)
We had to use this formula: c(O2) (mg/L)=(V1×N×E×1000)/V4
where
V1 = Vol of titrant Na2S2O3
V4 =Vol of analyte
E =Equ wgt of O2
N =Normality of Na2S2O3 sol

Now I've got two questions:

1. How do I calculate Equivalent weight of O₂? Is it from the 2nd reaction where n=2 as O₂ is being reduced to O^2-?
2. How's this formula V₁S₁=V₂S₂ going to calculate the conc. of O2 straight way when Na₂S₂O₃ reacts with I₂ and not O₂? Through this equation aren't we supposed to relate only the Volume and strength of chemicals that react ? O₂ and N₂S₂O₃ are not directly reacting!

Hello. Some more information would have helped me form a proper answer to your questions, but with what has been given I have tried my best. I have tried to include explanations so that each step makes sense. Hopefully it isn't too lengthy though.

To answer your 1st question, the equivalent weight of O2 is the mass of a mole of O2. You can find the mass per mole of O2 by first finding the mass per mole of O and multiplying it by 2. The mass per mole of O is equal to the atomic weight of oxygen as grams. Oxygen has an atomic weight (average atomic mass) of 16.00 a.m.u. (atomic mass units). The mass per mole of oxygen is thus 16.00 grams per mole. So, the mass per mole of O2 is 32.00 g / mol, and the equivalent weight is 32.00 grams.

In your 2nd question, you mention that it is I2 that reacts with Na2S2O3, and not O2. You are right. However, the amount of I2 that is available to react with the Na2S2O3 is determined by the amount of O2 present in the water sample. By the equations in the lab procedures you provided, every molecule of oxygen (O2) reacts to free two molecules of iodine (2 I2). This iodine is what is causing the golden color of the solution. When Na2S2O3 is added to the solution, 2 units of it react with each molecule of iodine and effectively work to eliminate the golden color. So, four units of Na2S2O3 are needed for every molecule of O2. And therefore, four moles of Na2S2O3 are needed for every mole of O2.

Now, by determining the number of units of Na2S2O3 in the titrant solution via the normality, you can determine the amount of O2 gas in the water. The normality is the number of equivalent weights of solute per liter of solution. The equivalent weight of the solute is the grams per mole of the solute. So, the normality is equivalent to the number of moles of solute per liter of solution. If there are X moles of Na2S2O3 per liter of titrant solution, then there are X/1000 moles of Na2S2O3 per milliliter of titrant solution. If, for example, you added 5 mL of titrant solution to make the sample clear, you added 5X/1000 moles of Na2S2O3 to the solution. In this example, because there are 4 moles of Na2S2O3 per 1 mole of oxygen, there would be 5X/4000 moles of oxygen present in the water. Using the equivalent weight of a molecule of oxygen, you can find the mass of that number of moles.

Now you must find the volume of the water which this amount of oxygen is found in. In the procedures you included, you take 100 mL of the sample and then titrate it with Na2S2O3. However, you must remember that the sample now is taken from more than just the sampled water, but also includes part of the 2 mL of manganese sulfate, 2 mL of NaOH + KI reagent, and 2 mL of sulfuric acid. The 100 mL sample is taken from a 306 mL solution. 300 mL of the solution is the water sample, and so ~98% of the solution is the sample. Therefore, the mass of oxygen you determined is found in ~0.98 mL of water, not 100 mL. You must then convert the mass of oxygen per milliliter to be # grams per 100 mL, or an appropriate equivalent such as # of milligrams per liter.

Criticism of my answer is very welcome, as I am learning chemistry myself and people make mistakes! I believe that this way of calculating the amount of oxygen in a sample of water makes sense. Hopefully it helps!

Best regards