mols KI = mols I^- = M x L = ?
Then (I^-) = mols/L of solution. Total L = 25 + 48 + 1 + 1 + 25
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!!
2 answers
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