[pmart491]

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]

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]

OK

Simple stoichiometry. Na⁺ is just a spectator.