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

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Discussion Overview

The discussion revolves around a titration experiment involving an unknown monoprotic acid and NaOH, focusing on calculations related to moles, equivalents, and equivalent mass. The scope includes mathematical reasoning and conceptual clarification regarding titration principles.

Discussion Character

  • Mathematical reasoning, Conceptual clarification

Main Points Raised

  • A student calculated the number of moles of NaOH used in the titration as 3.04E-3 mol based on the concentration and volume of NaOH.
  • Participants discussed how to determine the number of equivalents of the unknown acid, with some expressing uncertainty about this calculation.
  • There was a proposal that the equivalent mass of the unknown acid could be calculated as the mass of the acid divided by the moles of acid.
  • A balanced equation for the titration was suggested as NaOH + HA = H₂O + NaA.
  • It was noted that since the reaction is 1:1, the moles of acid would equal the moles of NaOH used.
  • One participant questioned whether the simplicity of the calculation for part b was appropriate, while another clarified that part b focuses on recognizing the 1:1 relationship rather than providing a numerical answer.
  • A calculation for part c was proposed as 0.4630g divided by 0.00304 moles, resulting in an equivalent mass of 161.03.

Areas of Agreement / Disagreement

Participants generally agree on the 1:1 stoichiometry of the titration reaction, but there is some uncertainty regarding the interpretation of part b and the implications of the calculations for part c.

Contextual Notes

Some assumptions regarding the definitions of equivalents and the nature of the unknown acid remain unaddressed, and there may be limitations in the provided data for calculating pKa or equivalent masses.

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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