H2S + I2 = S + HI
I2 + 2Na2S203 = 2Na2S4O6+ NaI
- How to determine normality and molarity of I2 in this equation?
- Which step do we have to take to determine normality and molarity?
Thanks and regards,
prabhu
To Measure H2S in water, you need 0.1N I2 standard solution, titrate it till brown colour disappear, as a end point, if you feel uneasy to locate the endpoint add 1% starch solution as indicator near to endpoint to locate exactly the end point.
then calculate Normality or Molarity of the H2S or % composition by using formulas
N1V1 = N2V2
M1V1 = M2V2
H2S %= [Titer(ml) x 0.1 N of I2x eq wt of H2Sx 100]/ [wt of the sample grams taken x 1000]
For H2S analysis you can follow iodimetric titration as mentioned by Dr. Imran. For I2 the normality is computed as follows. Mol. wt: 127*2
So 254g I2 per litre gives 1 M solution. In this case eq.wt.: Mol.wt/2. So 127gI2 per 1litre refers to 1 N solution. Eq. wt. = Mol.wt/no. of electrons involved in 1 mole. Here no. of electrones involved : 2 as per the following reaction.
I2 + 2e = 2I-
I did a research project on radionuclides in groundwater. I had the groundwater analyzed for sulfide by ASTM 4500-S2¯ G:
http://www.standardmethods.org/store/ProductView.cfm?ProductID=200
This method uses an ion selective electrode. I don’t consider myself to be an expert in this area. This was the method recommended to me by our Inorganics Unit.
Yes, it is true that at low level concentrations (ppm and subppm levels) Ion selective electrode technique is the best choice.
Which step do we have to take to determine normality and molarity of iodine and sodium thiosulphate.
Let us consider the following reaction:
I2 + 2Na2S203 = 2Na2S4O6+ NaI, This reaction can be resolved into individual redox reactions.
I2 + 2e = 2I- ............ (1)
2S2O3-- = S4O6-- + 2e .......... (2)
Hence eq. of I2: Mol. wt./2 ( 2 : no. of electrons involved in 1 mole of I2)
Eq. wt. of Na2S2O3: Mol. wt./1 (1: no. of electrons involved per mole of S2O3--)
In this case mol. wt. and eq. wt are same, whereas in case I2, eq.wt. is half of mol. wt.
From this, for preparation of 1N or 1M solutions suitable wts can be taken.
I am pleased to see that you have provided such a quick response to the question Iasked . I thank you again for taking the time to respond to my question.
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