A solotuion contains 0.60 mg/mL Mn2+. What minimum mass of KIO4 must be added to 5.00 mL of the solution in order to completely oxidize the Mn2+ to MnO4-?

Question

A solotuion contains 0.60 mg/mL Mn2+. What minimum mass of KIO4 must be added to 5.00 mL of the solution in order to completely oxidize the Mn2+ to MnO4-?

Answer 1

This is a redox reaction, so you should write and balance both oxidation and reduction half reactions:

 

Mn²⁺ + 4H₂O --> MnO4^- + 5e- + 8H⁺ [x2]
IO₄^- + 2e- + 2H⁺ --> IO₃^- + H₂O  [x5]

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2Mn²⁺ + 8H₂O + 5IO₄^- + 10e- + 10H⁺ ==> 2MnO₄^- + 10e- + 16H⁺ +5IO₃^- + 5H₂O

 

2Mn²⁺ + 3H₂O + 5IO₄^- ==> 2MnO₄^- + 6H⁺ + 5IO₃^-   BALANCED REDOX EQUATION (w/o spectator ions)

moles Mn²⁺ = 0.60 mg/ml x 5.00 ml x 24.3 mmole/mg = 0.1235 mmoles Mn²⁺

moles IO₄^- needed = 0.1235 mmoles Mn²⁺ x 5 mmoles IO₄^-/2 mmoles Mn²⁺ = 0.3088 mmoles IO₄^- needed

Since the IO₄^- is in the form of KIO₄, we need 0.3088 mmoles of KIO₄ to completely oxidize the Mn²⁺

Mass of KIO₄ needed = 0.3088 mg x 230 mg/mmole = 71.0 mg needed