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

  • Thread starter Hemolymph
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
445
5
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?
 
  • #3
30
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would it be

NaOH + HA= H_2_O+ NaA?
 
  • #4
445
5
Yes, so if you have 3.04 mmol of NaOH, how many moles of acid do you have?
 
Last edited:
  • #5
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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.
 
  • #6
445
5
Part b is more about the recognition that it's 1:1, rather than an actual number.
 
  • #7
30
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Would part c be .4630g/ .00304moles? Which I got to be 161.03
 

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