Calculating Molarity in Redox Reactions: NaCl Solution Example

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The discussion revolves around calculating the molarity of a NaCl solution reacting with KMnO4 in a redox reaction. The balanced half-reactions are established, leading to the conclusion that there are 5 moles of Cl- for every mole of MnO4-. The participant calculates the moles of KMnO4 used and derives the moles of NaCl based on the stoichiometric ratio from the balanced equation. The final calculation yields a molarity of 0.372M for the NaCl solution. Understanding the necessity of balancing the reaction is emphasized as crucial for accurate stoichiometric calculations.
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Homework Statement


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- yields-> Cl2 + Mn2+


The Attempt at a Solution



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

Do I find the full half reactions (ie: the electrons gained/lost H2O, H+) and then do a molarity problem?
 
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Yes - this is in fact simple stoichiometric question, just based on the redox reaction. Find full balanced reaction equation first.
 
Ok I got 10Cl- + 16H+ + 2MnO4- -> 5Cl2 + 2Mn2+ + 8H2O

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

what i don't understand is how to find moles of NaCl through the half reaction
 
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
 
OK
 
pmart491 said:
Ok I got 10Cl- + 16H+ + 2MnO4- -> 5Cl2 + 2Mn2+ + 8H2O

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

what i don't understand is how to find moles of NaCl through the half reaction

pmart491 said:
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

I don't understand this. If we have V of NaCl is 18.3 mL, can we find M by using M=mole/V. So the ratio of NaCl and KMnO4 isn't 1:1 ?
 
Ratio is given by reaction equation. Take a look at stoichiometric coeffcients.
 
pmart491 said:
Ok I got 10Cl- + 16H+ + 2MnO4- -> 5Cl2 + 2Mn2+ + 8H2O

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

what i don't understand is how to find moles of NaCl through the half reaction

Borek said:
Ratio is given by reaction equation. Take a look at stoichiometric coeffcients.

Why do we have to do 10Cl + 16H ...?

Sorry, I just don't understand your step.
 
Because these coefficients where necessary to balance the equation. Do you know what it means to balance the equation and why the reaction equations needs to be balanced?
 
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