Titration of a triprotic acid with strong base NaOH

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SUMMARY

The discussion focuses on the titration of a 0.307g sample of an unknown triprotic acid using 35.2 mL of 0.106 M NaOH to determine its molar mass. Participants clarify that the calculation must consider the third equivalence point due to the triprotic nature of the acid. The correct approach involves calculating the moles of NaOH used and relating it to the moles of the triprotic acid to find the molar mass. The final calculation yields a molar mass based on the stoichiometry of the reaction.

PREREQUISITES
  • Understanding of triprotic acids and their equivalence points
  • Knowledge of titration techniques and calculations
  • Familiarity with molarity and molar mass concepts
  • Proficiency in stoichiometry and chemical equations
NEXT STEPS
  • Learn how to calculate moles from molarity and volume
  • Study the concept of equivalence points in titrations
  • Explore the stoichiometry of triprotic acid-base reactions
  • Review examples of titration calculations for different acid types
USEFUL FOR

Chemistry students, educators, and laboratory technicians involved in acid-base titration experiments and those seeking to understand the properties of triprotic acids.

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
 
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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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