Chemistry Forum

I have been carrying out an in house calibration of our Dissolved Oxygen Meter. To do this we follow the blue book method 'Dissolved Oxygen in Natural and Waste Waters 1979, Tritimetric Method A'

We have to standardise the 0.0125M sodium thiosulphate solution against potassium iodate solution. We carried this out and found our sodium thiosulphate solution to be 0.0118M.

We then carried out the rest of the titrations using this 0.0118M solution. I believed the calulations at the end of the experiment took the actual value of the sodium thiosulphate solution in to consideration.

How wrong I was! We now have a situation where our values are out from the readings from the DO probe. I am sure this is because our sodium thiosulphate was not standardised to the 0.0125M it stated, but to 0.0118M

Is there a way I can convert/calculate my results with the value 0.0118M rather than the expected 0.0125M?

I would rather not have to re-do the titrations due to time contraints but I do realise this might have to be my final option!

Without knowing details of your titration method it’s hard to answer this question. Apparently the method releases iodine from potassium or sodium iodide and then titrated with sodium thiosulfate. In any case, the amount of sodium thiosulfate should be x4 times equimolar to oxygen according to this formal equation:

4 Na₂S₂O₃ + O₂ + 2 H₂O = 2 Na₂S₄O₆ + 4 NaOH

It is formal means that actual reaction takes place between O₂ and I(-) and then I₂ titrated with thiosulfate:

4 KI + O₂ + 4 HCl = 2 I₂ + 2 H₂O + 4 KCl

and subsequently:

2 Na₂S₂O₃ + I₂ = Na₂S₄O₆ + 2 NaI

Basically one mol O₂ produces two mols of I₂ that will react with four mols of Na₂S₂O₃

I have noted down the reactions that are occuring:

precipitation of white manganous hydroxide

Mn++ + 2OH- = Mn(OH)₂

Oxidation of the white manganous hydroxide by dissolved oxygen

4MN(OH)₂ + O₂ = 4MnO(OH) 2H₂O

brown basic oxide dissolves in sulphuric acid forming unstable manganic ions

MnO(OH) + 3H+ = Mn+++ + 2H₂O

manganic ions release free iodine from iodide

2Mn+++ + 2I- = "Mn++ +I₂

liberated iodine is titrated with thiosulphate

I₂ + 2S₂O₃-- = S₄O₆-- +2I-

nitrite interferes in the determination by liberating iodine from iodide ion and by catalysing the reaction between iodide and atmospheric oxygen during the titration

nitite is destroyed initially by reaction with azide ion in acid solution added in the procedure

N₃- + NO₂- + 2H+ = N₂ + N₂O + H₂O

We are told that 1ml of 0.0125 = 0.1mg dissolved oxygen.

We also have the following equation for the calculation of the dissolved oxygen

DO = V₁ x 0.1 x 1000 x F / V₂

V₁ = volume of sodium thio used
V₂ = volume of sample titrated
F= 1.017

If anyone can help with my first problem with all the information I have supplied I would be very grateful

Thanks

No matter how you capture oxygen, your final formal process is as follows:
4 Na₂S₂O₃ + O₂ + 2 H₂O = 2 Na₂S₄O₆ + 4 NaOH

and you can use it for calculations: four mols of sodium thiosulfate per one mol of oxygen.

Molar concentrations of the reagents