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Topic: Redox Reaction/Molarity problem  (Read 3361 times)

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pmart491

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Redox Reaction/Molarity problem
« on: May 06, 2008, 04:45:34 PM »
What is the molarity of a NaCl solution if 18.3mL of the solution reacted with 13.6mL of 0.1M KMnO4 based on the following unbalanced redox reaction in an acidic solution?

Cl- + MnO4- -> Cl2 + Mn2+

Attempt
I did the two half reactions
Cl- -> Cl2
MnO4- -> Mn2+

and I got 10Cl- + 16H+ + 2MnO4- -> 5Cl2 + 2Mn2+ + 8H2O

Next I did 13.6mL KMnO4 x .1M = .00136moles KMnO4

I'm kind of having trouble about where to go next. How does the Na fit into the equation?

pmart491

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Re: Redox Reaction/Molarity problem
« Reply #1 on: May 06, 2008, 04:54:22 PM »
well i think i got it

based on the half reaction, there are 5 times as many Cl- ions as there are MnO4- ions.
therefore there are 5 times as many moles of NaCl than there are moles of KMnO4.
5 x .00136=.0068moles NaCl

.0068moles NaCl/.0186L NaCl=.372M NaCl

is this correct logic?

Borek

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Re: Redox Reaction/Molarity problem
« Reply #2 on: May 06, 2008, 05:22:21 PM »
OK

Simple stoichiometry. Na+ is just a spectator.
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