You titrated a 22.00 mL solution of 0.0300 M oxalic acid with
freshly prepared solution of KMnO4. If it took 48.99 mL of this
solution, what is the molarity of the KMnO4?
0.022 L of 0.300 M oxalic acid
(0.022)(0.03) = 0.00066 mol oxalic acid
0.0006 mol oxalic acid x (2 mol KMnO4)/(5 mol oxalic acid)=
0.000264 mol KMnO4
0.000264 mol/0.04899 L = 0.05388855 M KMnO4
= 0.005388 M KMnO4
where did I go wrong?
Homework Answers
You did perfectly
correct.nothing wrong.except rounding off answer
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