Posted by Marcus on Friday, April 26, 2013 at 8:24pm.
To illustrate the calculation, assume that 21.95 mL of a thiosulfate solution is required to reduce the iodine liberated by 20.85 mL of 0.0200M KIO3 and that 18.63 mL of the same thiosulfate solution reacts with the iodine replaced by 20.00 mL of an unknown solution of sodium hypochlorite.
(This is in the sample calculation given in my lab manual, for reference.)
And the question I'm given in my lab manuel is:
See the question in your lab manual. Suppose that you determine that the molarity of your sodium thiosulfate solution is 0.100 M. Suppose that 10.02 mL of the same thiosulfate solution reacts with the iodine replaced by 20.00 mL of an unknown solution of sodium hypochlorite. Calculate the weight, in grams, of available Cl in a liter of the unknown solution. Enter your answer as a decimal number with 3 decimal places.
This is what I did.
Since M thio = 0.100,
2 x M OCl x 0.0200L = 0.100 M x .01002L
M OCl = .02505
0.02505 mol NaOCl x 1 mole Cl x 35.45 g Cl x 1000 mg = 888.0225 mg Cl/L =888.023 ppm
But it says the answer is incorrect. So what would I be doing wrong here?
Chemistry - DrBob222, Saturday, April 27, 2013 at 6:26pm
I would look at the following. I think your value of 0.02505 M (mols/L) for NaOCl is correct.
I believe available chlorine is calculated as Cl2. Therefore, I would multiply by 70.90 instead of 35.45 and you have changed that to mg which I would not do since the problem asks for grams and not mg. Let me know how this turns out and/or shed any new information about exactly what is meant by available chlorine.
Chemistry - Marcus, Sunday, April 28, 2013 at 8:11pm
Turns out all the values were right such as .02505 M and 35.45 g Cl etc, I just didn't need to multiply by 1000 and leave it in grams as you said.
Chemistry - DrBob222, Monday, April 29, 2013 at 12:48pm
Many thanks. The literature is not clear on how to report "available chlorine"; i.e., as Cl, Cl2, or as some suggest NaOCl.
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