Where Did I Go Wrong in Calculating the n-Factor in Volumetric Analysis?

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SUMMARY

The discussion centers on the calculation of the n-factor in volumetric analysis, specifically regarding Rochelle salt (NaKC4H4O6.4H2O). The user calculated the meq of H2SO4 reacted as 5.205 but encountered discrepancies in the n-factor, which they calculated as 10. This value was deemed incorrect, as it led to an erroneous weight of 0.1467g and a purity percentage of 15.3%, while the expected answer was 76.87%. The correct interpretation of the reaction and n-factor is crucial for accurate results in volumetric analysis.

PREREQUISITES
  • Understanding of volumetric analysis and equivalent weight calculations.
  • Familiarity with Rochelle salt and its chemical composition.
  • Knowledge of redox reactions and their balancing.
  • Proficiency in calculating milliequivalents (meq) in titration.
NEXT STEPS
  • Review the concept of n-factors in acid-base and redox reactions.
  • Learn how to balance chemical equations involving complex salts like Rochelle salt.
  • Investigate the calculation of purity percentages in volumetric analysis.
  • Explore the implications of incorrect n-factor calculations on analytical results.
USEFUL FOR

Chemistry students, laboratory technicians, and professionals involved in analytical chemistry and volumetric analysis will benefit from this discussion.

subhradeep mahata
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Homework Statement
0.9546 g of a sample of Rochelle salt (NaKC4H4O6.4H2O), on ignition, gave NaKCO3, which was treated with 41.72 mL of 0.1307 N H2SO4. The unreacted H2SO4 was then neutralized by 1.91 mL of 0.1297 N NaOH. Find the percentage purity of the salt in the sample. Use equivalent concept.
Relevant Equations
meq of H2SO4 reacted = total meq - unreacted meq=meq of salt
unreacted meq of H2SO4 = meq of NaOH
Using the given formula, meq of H2SO4 reacted= meq of salt = 5.205
The math is correct and I have verified it.
Now, I am pretty sure that I am making a mistake in calculating the n- factor of the salt.
According to the question, C4+6=4C+4
The n- factor is coming out to be 10, which is pretty weird (now, I might be wrong here, maybe there's nothing weird in it. Please also tell me about it's limits)
Now, on doing further calculations, we get the weight as 0.1467g, and the percentage purity is coming out to be 15.3%.
But the answer according to my book is 76.87%.
Can you tell me where I went wrong?
 
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subhradeep mahata said:
the math is correct and I have verified it
Good. So how many mole of Rochelle stuff have reacted ? How many grams is that ?

subhradeep mahata said:
According to the question, C4+6=4C+4
Don't see it in the problem statement ... :rolleyes:
subhradeep mahata said:
0.1467g
check and verify that too !:wink:
 
0.1467g is exactly 1/5 of the right answer. I suspect the problem is with your n-factor of 10. I don't know how you got that, and frankly I don't care, as I was never taught n-factors. But it does look as if you are calculating it for the redox reaction, on the assumption that all the carbons in the Rochelle salt end up as carbonate. This is not true. (Where would the extra Na and K come from?)
Can you write a balanced equation for the reaction? Assuming no K or Na is created or destroyed, it must be something like
NaKC4H4O6.4H2O → NaKCO3 + stuff...
i.e. 1 mole of Rochelle salt gives 1 mole of carbonate.
You correctly calculated the meq of carbonate; can you now proceed to the right answer?
 

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