Didn't I answer this earlier?
You have 0.2M KI and it has been diluted. So the initial concn is
0.2 x [(25 mL/(25 + 48+1+1+25)]
You have 0.4M Na2S2O3 so the initial concn is (before any reaction) is
0.4 x (1 mL/total ML)
My prelab is telling me to find the initial concentrations of I(^-) and S208(^2-) given the following reaction:
25 mL of 0.2 M KI + 48.00 mL 0.2 M KNO3 + 1 mL 0.4 M Na2S2O3 + 1 mL starch + 25 mL 0.2 M (NH4)2S2O8 + 1 drop EDTA
How do I calculate the concentration of I^-1 when I am given the molality of KI? Is the molality of I^-1 a percentage of this number of the total? Same goes for S2O8(^-2) in (NH4)2S2O8.
Thanks!!
1 answer