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!
Dissolved Oxygen in Water Titration - Sodium Thiosulphate
Moderators: Xen, expert, ChenBeier
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
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
Remember safety first! Check MSDS and consult with professionals before performing risky experiments.
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
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.
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.
Remember safety first! Check MSDS and consult with professionals before performing risky experiments.