When 25.00 mL of unknown were passed through a Jones redactor, molybdate ion (MoO42–) was converted into Mo3+. The filtrate required 16.3 mL of 0.01033 M KMnO4 to reach the purple end point.
3 MnO4– + 5 Mo3+ + 4 H+ → 3 Mn2+ + 5 MoO22+ + 2 H2O
If a blank required 0.04 mL, what is the molarity of molybdate in the unknown?
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