Posted by **amy** on Tuesday, April 10, 2007 at 4:49pm.

in analyzing household bleach. I have a flask that contains 10.00 mL diluted bleach + 0.4g KI + 20 mL distilled water and 20 drops of 2M HCl. To that I titrated 30.60 mL Na2S2O3 and I need to calculate the number of moles of NaOCl reacted. This is what I was thinking..

Calculate volume Na2S2O3 used * molarity Na2S2O3 * balanced equation ratio. So it would look like (.03060L Na2s2O3) * (7.67*10-4 moles/L Na2S2O3)* 1 mol NOCl/2 mol Na2S2O3)

Does this look correct??

Yes, I think so; however, perhaps for the wrong reason. Perhaps you got lucky.

You titrated I2 with the S2O3^= and you omitted one factor involving I2.

2H^+ + OCl^- + 2I^- ==> I2 + Cl^- + H2O

and the titration step of

2S2O3^= + I2 ==> S4O6^= + 2I^-

so

vol S2O3^= * molarity S2O3^= x (1 mol I2/2 mols S2O3)*(1 mol OCl^-/1 mol I2) = mols OCl^-

The molarity of 7.67 x 10^-4 seems small to me but I guess you know the molarity of the thiosulfate.

check my work.

Thank you Dr.Bob. This is how I came up with the molarity of S2O3.

We used 30.60 mL of .0250M Na2S2O3 so I did .03060 L * .0250 mol/1 L. Is that not correct?

No. If you used 0.025 mol/L Na2S2O3, then 0.025 M is what goes in for molarity. You have calcualted mols Na2S2O3 which is ok but then you don't multiply by 0.03060 again.

mols NaOCl = 0.03060*0.0250*1/2 = ??

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