Chemistry: Titration of of a base to solve for the mass of an unknown acid.

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SUMMARY

The discussion centers on the titration of a 0.4630g unknown monoprotic acid using 0.1060M NaOH, with an equivalence point volume of 28.70mL. The calculation for moles of NaOH used is confirmed as 3.04E-3 mol. The equivalents of the unknown acid are equal to the moles of NaOH due to the 1:1 stoichiometry, resulting in 3.04mmol of acid. The equivalent mass of the unknown acid is calculated as 161.03 g/mol, derived from dividing the mass of the acid by the moles.

PREREQUISITES
  • Understanding of titration principles and stoichiometry
  • Knowledge of molarity and its calculations
  • Familiarity with monoprotic acids and their properties
  • Ability to interpret and use pKa values in acid-base chemistry
NEXT STEPS
  • Study the concept of equivalence point in titrations
  • Learn how to calculate pKa from given acid dissociation constants
  • Explore the relationship between moles, mass, and molar mass in chemical reactions
  • Review balanced chemical equations for acid-base reactions
USEFUL FOR

Chemistry students, educators, and anyone involved in laboratory work related to acid-base titrations and stoichiometric calculations.

Hemolymph
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A student titrated a .4630g of unknown monoprotic acid with .1060M NaOH. The equivalance point volume is 28.70mL.
a) Calculate the number of moles of NaOH used.
Im pretty sure I got this one
I did .1060 M X .02870L to equal 3.04E-3 mol
b)how many equivalents of unknown acid were titrated? no clue

c) Determine the equivalent mass of the unknown acid
Which is the moles times MM


There is also a table given of acids and there pKa and equivalent masses. If I can figure out what Pka is I should be fine.
 
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Hemolymph said:
A student titrated a .4630g of unknown monoprotic acid with .1060M NaOH. The equivalance point volume is 28.70mL.
a) Calculate the number of moles of NaOH used.
Im pretty sure I got this one
I did .1060 M X .02870L to equal 3.04E-3 mol
b)how many equivalents of unknown acid were titrated? no clue

c) Determine the equivalent mass of the unknown acid
Which is the moles times MM


There is also a table given of acids and there pKa and equivalent masses. If I can figure out what Pka is I should be fine.

If the unknown acid is HA, can you write a balanced equation for the titration?
 
would it be

NaOH + HA= H_2_O+ NaA?
 
Yes, so if you have 3.04 mmol of NaOH, how many moles of acid do you have?
 
Last edited:
Well since its one to one it would be the same so 3.04mmol of acid. So would that just be the answer for part b? If so I did not think it could be that simple.
 
Part b is more about the recognition that it's 1:1, rather than an actual number.
 
Would part c be .4630g/ .00304moles? Which I got to be 161.03
 

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