November 29, 2021, 12:12:06 PM
Forum Rules: Read This Before Posting


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

0 Members and 1 Guest are viewing this topic.

Offline pmart491

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
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?

Offline pmart491

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
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?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26863
  • Mole Snacks: +1742/-403
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Redox Reaction/Molarity problem
« Reply #2 on: May 06, 2008, 05:22:21 PM »
OK :)

Simple stoichiometry. Na+ is just a spectator.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Sponsored Links