Is the N Factor of Oxalic Acid in Titration Correctly Applied?

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SUMMARY

The discussion centers on the titration of oxalic acid with sodium hydroxide (NaOH) and the confusion surrounding the calculation of the n-factor. Participants assert that the n-factor of oxalic acid is 2, leading to a calculated amount of 4 grams of NaOH, while the official answer suggests only 2 grams. The ambiguity arises from the presence of two possible reactions during titration, which can yield different endpoints. The use of phenolphthalein as an indicator is also debated, given its pKa of approximately 9, which may not accurately reflect the endpoints of oxalic acid titration.

PREREQUISITES
  • Understanding of titration principles and stoichiometry
  • Knowledge of oxalic acid's dissociation and pKa values (pKa1 = 1.27, pKa2 = 4.27)
  • Familiarity with the role of indicators in acid-base titrations, specifically phenolphthalein
  • Basic chemistry knowledge regarding diprotic acids and their neutralization reactions
NEXT STEPS
  • Research the titration of diprotic acids and their n-factors
  • Learn about the use of phenolphthalein in acid-base titrations and its limitations
  • Explore the concept of equivalence points in titrations and how they relate to pH changes
  • Investigate the calculation of moles and grams in titration scenarios
USEFUL FOR

Chemistry students, educators, and anyone involved in laboratory titration experiments, particularly those focusing on acid-base reactions and stoichiometry.

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Homework Statement
[" 50mL of "0.5M" Oxalic Acid is needed to neutralize "25mL" of sodium hydroxide "],[" solution.The amount of "NaOH" in "50mL" of the given sodium hydroxide solution is "]
Relevant Equations
Equation moles
I got a answer by equation the moles i took the n factor of oxalic acid as 2 and the naoh solution came out to be 2M but i got an answer of 4grams but the answer given is 2 grams am i correct ??
 

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It's a bit ambiguous because there are two possible reactions:
OH- + 1/2 H2Ox → H2O + Ox2-
OH- + H2Ox → H2O + HOx-
I've written them that way because you are adding the oxalic acid to the NaOH (by the way, there's no such thing as naoh). You will first form disodium oxalate, then monosodium hydrogen oxalate. The titration will have two endpoints.
Can you estimate the pH at these endpoints? To make it a bit simpler, just try to calculate the pH of a 1M solution of Na2Ox, and a 1M solution of NaHOx. Which do you think best corresponds to "neutralisation" (if that term is not better defined in the problem)?
hint: for oxalic acid, pKa1 = 1.27, pKa2 = 4.27 (This question is assuming a lot of knowledge. How many marks do you get?)
 
mjc123 said:
It's a bit ambiguous because there are two possible reactions:
OH- + 1/2 H2Ox → H2O + Ox2-
OH- + H2Ox → H2O + HOx-
I've written them that way because you are adding the oxalic acid to the NaOH (by the way, there's no such thing as naoh). You will first form disodium oxalate, then monosodium hydrogen oxalate. The titration will have two endpoints.
Can you estimate the pH at these endpoints? To make it a bit simpler, just try to calculate the pH of a 1M solution of Na2Ox, and a 1M solution of NaHOx. Which do you think best corresponds to "neutralisation" (if that term is not better defined in the problem)?
hint: for oxalic acid, pKa1 = 1.27, pKa2 = 4.27 (This question is assuming a lot of knowledge. How many marks do you get?)
this is a high school grade question but why can't we simply equate the milliequivallents of both oxalic acid and NaOH
 
Titrations with oxalic acid are most usually done if I remember right by adding alkali to the acid, because you traditionally use as indicator phenolphthalein whose alkaline form is coloured (has pKa about 9) well above the oxalic acid pK's, so both -COOH groups have been titrated when you see the pink colour appear.

I get 4g; the student would have to set out reasoning in more detail for us to identify any misconception or mistake.
 
epenguin said:
Titrations with oxalic acid are most usually done if I remember right by adding alkali to the acid, because you traditionally use as indicator phenolphthalein whose alkaline form is coloured (has pKa about 9) well above the oxalic acid pK's, so both -COOH groups have been titrated when you see the pink colour appear.

I get 4g; the student would have to set out reasoning in more detail for us to identify any misconception or mistake.
ya i am also getting 4g hence the answer given must be wrong thanks
 
What is the source of the question?
 
epenguin said:
What is the source of the question?
its a sample question for jee mains its an enterance exam in india
 
Is 2g the official answer? I don't think I would be able to enter there.
However I hope we shall get a second opinion.
 
Something is definitely wrong, 2 g is amount of NaOH that reacted, not amount of NaOH in 50 mL of the solution.
 
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  • #10
Which equivalence point? Are the two of them close-enough for using phenolphthalein as the only indication? My guess is both hydrogens from the oxalic would be strong enough ionized in solution for that one indicator to be good enough; but my judgement could be wrong (unless I review the theory thoroughly).
 
  • #11
oxalic_acid_titrated_with_sodium_hydroxide.png
 
  • #12
Borek has given a graphical clarification about the endpoint pH's for the titration so that surely helps. If you want to go to both endpoints, then this will be titrating for BOTH hydrogen ions from the oxalic acid. Notice that the more endpoint to pH 2.8 is not going to be very sharp; maybe this means you would want to take just the first enpoint, for which phenolphthalein would be the suitable endpoint (for ONE hydrogen ion).
 
  • #13
On the second thought the problem is this question is poorly defined. It doesn't specify what it means by "needed to neutralize". If taken literally (and that would be my approach), oxalic acid is diprotic. If it is about titration, only the first end point is sharp enough to be usable. That gives two possible answers that differ by exactly 100%.

sodium_hydroxide_titrated_with_oxalic_acid.png
 
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