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 = ??
The correct calculation for the moles of NaOCl reacted would be:
mols NaOCl = 0.03060 L * 0.0250 mol/L * 1 mol NaOCl / 2 mol Na2S2O3
Make sure to use the correct molarity for Na2S2O3, which in this case is 0.0250 mol/L. By multiplying the volume of Na2S2O3 used (0.03060 L) by the molarity of Na2S2O3 (0.0250 mol/L) and then dividing by the balanced equation ratio (1 mol NaOCl / 2 mol Na2S2O3), you will get the moles of NaOCl reacted.
Please note that this assumes the chemical reactions mentioned in the question are balanced correctly and accurately reflect the reaction taking place in the flask.