Titration of a triprotic acid with strong base NaOH

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Discussion Overview

The discussion revolves around the titration of a triprotic acid with a strong base (NaOH) to determine the molar mass of the acid. Participants explore the implications of reaching different equivalence points in the titration process and the calculations involved in determining the molar mass based on the titration data provided.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how to start the calculation for the molar mass after determining the grams of NaOH used.
  • Another participant questions which equivalence point is being referred to, noting the significance of this in the context of a triprotic acid.
  • A participant suggests calculating the number of moles of NaOH used and finding a relationship to the moles of triprotic acid to determine the molar mass.
  • There is a correction regarding the calculation of grams of NaOH, with a suggestion that the correct value should be around 0.149 g instead of the previously calculated 120.430 g.
  • One participant challenges the idea of subtracting from the total, indicating disagreement with that approach.
  • A later post introduces the possibility that the titration reaches the third equivalence point, raising further questions about the implications for the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, with multiple competing views on how to interpret the equivalence points and the calculations involved.

Contextual Notes

There are unresolved assumptions regarding the specific equivalence point being referenced and the calculations for grams and moles of NaOH, which may affect the determination of the molar mass of the acid.

xpatelsxownage
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Homework Statement


A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


Homework Equations





The Attempt at a Solution


I just don't know where to start with this question.
35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
I don't know where to go after this part
 
Last edited:
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A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.
Not "Its" equivalence point; WHICH equivalence point? The first, second, or third? The question is about a triprotic acid.
 
xpatelsxownage said:

Homework Statement


A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


Homework Equations





The Attempt at a Solution


I just don't know where to start with this question.
35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
I don't know where to go after this part

You don't need to calculate the number of grams of NaOH (which you did incorrectly). You set it up right but probably entered it in your calculator incorrectly. You should have gotten something like 0.149 g NaOH if this was what the question asked, which it wasn't.

Why don't you calculate the number of moles of NaOH used and find a relationship between that and the number of moles of triprotic acid. Once you have that, you will have the number of moles of acid and the mass. From that you can calculate formula weight.
 
xpatelsxownage said:

Homework Statement


A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


Homework Equations





The Attempt at a Solution


I just don't know where to start with this question.
35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
I don't know where to go after this part



It should equal to .149g NaOH
then I think you just subtract from the total.
 
yaho8888 said:
...then I think you just subtract from the total.

no, nope, nein, nyet.
 


A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 mL of 0.106 M NaOH. Calculate the molar mass of the acid.

What if the problem is to the Third equivalence point?
 

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