Question about Volumetric Analysis

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In summary, the conversation involved using a given formula to calculate the meq of H2SO4 reacted, which was found to be equal to 5.205 meq of salt. However, the n-factor of the salt was calculated to be 10, which was deemed incorrect as it did not align with the expected result of 76.87% purity. The problem was identified as potentially being related to the n-factor being calculated for a redox reaction, and a balanced equation was provided to help guide the calculation towards the correct answer.
  • #1
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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  • #2
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:
 
  • #3
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?
 

Related to Question about Volumetric Analysis

1. What is volumetric analysis?

Volumetric analysis is a quantitative analytical method used in chemistry to determine the concentration of a substance in a solution. It involves measuring the volume of a solution of known concentration that is required to react completely with a known volume of the solution being analyzed.

2. What are some common applications of volumetric analysis?

Volumetric analysis is commonly used in various industries, such as pharmaceuticals, food and beverage, and environmental testing, to determine the concentration of substances in solutions. It is also used in research and development to study chemical reactions and in quality control to ensure the accuracy of product formulations.

3. How is volumetric analysis performed?

Volumetric analysis involves using a burette to dispense a known volume of a solution of known concentration, called the titrant, into a flask containing the solution being analyzed, called the analyte. The titrant is added gradually until the reaction between the two solutions is complete, as indicated by a color change or the appearance of a precipitate. The volume of titrant used is then used to calculate the concentration of the analyte.

4. What are the advantages of using volumetric analysis?

Volumetric analysis is a relatively simple and inexpensive method that can provide accurate results. It also allows for the analysis of multiple samples at once, making it efficient for high-throughput testing. Additionally, it is a non-destructive method, meaning the sample can be recovered and used for further testing or analysis.

5. What are some sources of error in volumetric analysis?

Some common sources of error in volumetric analysis include human error in reading volumes or making calculations, impurities in the solutions being used, and environmental factors such as temperature and humidity. It is important to carefully follow proper techniques and use high-quality equipment to minimize these sources of error.

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