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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: pmart491 on May 06, 2008, 04:45:34 PM

Title: Redox Reaction/Molarity problem
Post by: pmart491 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?
Title: Re: Redox Reaction/Molarity problem
Post by: pmart491 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?
Title: Re: Redox Reaction/Molarity problem
Post by: Borek on May 06, 2008, 05:22:21 PM
OK :)

Simple stoichiometry. Na+ is just a spectator.