Calculating Molarity in Redox Reactions: NaCl Solution Example

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Discussion Overview

The discussion revolves around calculating the molarity of a NaCl solution involved in a redox reaction with KMnO4. Participants explore the stoichiometry of the reaction, the balancing of half-reactions, and the application of molarity calculations in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a homework problem involving the molarity of a NaCl solution reacting with KMnO4 and asks how to find the moles of NaCl from the half-reaction.
  • Another participant suggests that the problem is straightforward and emphasizes the need to find the full balanced reaction equation first.
  • A participant provides a balanced reaction equation and calculates the moles of KMnO4 used, expressing confusion about how to derive the moles of NaCl from the half-reaction.
  • One participant concludes that there are 5 times as many moles of NaCl as there are moles of KMnO4 based on the stoichiometry of the reaction.
  • Another participant questions the ratio of NaCl to KMnO4, suggesting that the relationship is not 1:1 and seeks clarification on the stoichiometric coefficients from the balanced equation.
  • Further clarification is sought regarding the necessity of balancing the reaction equation and the meaning behind the coefficients used.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the stoichiometry of the reaction and the process of balancing equations. There is no consensus on the clarity of the steps involved in deriving the molarity of NaCl from the reaction.

Contextual Notes

Some participants express uncertainty about the balancing of the reaction and the implications of stoichiometric coefficients, indicating potential gaps in understanding the underlying principles of redox reactions.

pmart491
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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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