Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: pmart491 on May 06, 2008, 04:45:34 PM
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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?
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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?
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OK :)
Simple stoichiometry. Na+ is just a spectator.