First, balance the titration equation.
g As2O3 = 0.1322 x 0.9916 = ??
moles As2O3 = g/molarmass
Using the coefficients in the balanced equation to find moles KMnO4.
mols As2O3 x (xxmoles KMnO4/yy moles As2O3).
M KMnO4 = moles KMnO4/L KMnO4
Suppose that the KMnO4 (aq) were standardized by reaction with As2O3.
As2O3 + MnO4- + H+ +H2O yields H3AsO4 + Mn2+ (not balanced).
If a 0.1322-g sample that is 99.16 % As2O3 by mass had been used in the titration, how many milliliters of the 0.02140 M KMnO4 (aq) would have been required?
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