Molar Mass of Diprotic

  • Thread starter Valenti
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  • #1
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Hey, I've been having some trouble with my Lab report. After doing a titration of an unknown diprotic acid we had to find the molar mass of the acid and then find out which acid it is by comparing it to a list of given acids with molar masses. But when I run through the calculations my answer is off by quite a bit to list (lowest molar mass there being 104)

1. Homework Statement

Determine the molar mass of H2X based on the number of moles of NaOH used in the titration.

Concentration of NaOH = 0.2M
Mass of unknown = 0.0987 g
Volume used (first equivalence point) = 10.49 mL
Volume used (second equivalence point) = 20.31 mL

Endpoints were found via 1st and 2nd derivative
The 2nd equivalence point is steeper so it will be used for the calculation

Homework Equations


n=m/M
n=cv

The Attempt at a Solution



n=cv
Moles of NaOH = 0.2(20.31/1000)
= 0.004 mols
H2X : NaOH = 1:2

Mols of H2x = 0.002 mols

M=m/n
0.0987/0.002
M= 49.35g
 

Answers and Replies

  • #2
symbolipoint
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Reviewing your given experimental information,
Dissolved 0.0987 grams of H2X and titrated to second equivalence point volume 20.31 mL of 0.2 M NaOH

0.2(moles/liter)*0.02031*liters=0.004062 moles of NaOH
which means that
0.0987 grams of H2X contains 0.004062 moles of H+1 .

Can you use this to find the formula weight for H2X ?
 
Last edited:
  • #3
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Reviewing your given experimental information,
Dissolved 0.0987 grams of H2X and titrated to second equivalence point volume 20.31 mL of 0.2 M NaOH

0.2(moles/liter)*0.02031*liters=0.04062 moles of NaOH
which means that
0.0987 grams of H2X contains 0.04062 moles of H+1 .

Can you use this to find the formula weight for H2X ?

using M=m/n I get formula weight to be 24.29g/mol
 
  • #4
Borek
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I got 48.6 g/mol (without rounding intermediate results, which you should never do). Your number is in a correct ballpark (that is, correctly calculated from the given data). No idea what and where went wrong, but it is not the calculation to blame.
 
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  • #5
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I got 48.6 g/mol (without rounding intermediate results, which you should never do). Your number is in a correct ballpark (that is, correctly calculated from the given data). No idea what and where went wrong, but it is not the calculation to blame.

When doing the same calculation for the first endpoint (10.24 mL used) I get a molar mass of 98g/mol. Should it matter which endpoint I use for the calculation?
 
  • #6
Borek
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When doing the same calculation for the first endpoint (10.24 mL used) I get a molar mass of 98g/mol.

At the first endpoint what you have neutralized was a monoprotic acid, not a diprotic.
 
  • #7
symbolipoint
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First, or Second endpoint to use - depends on how fast pH changes with change in titrant volume. You still need to be sure titration is complete enough so you can decide if you have one or two (or maybe more) endpoints.
 
  • #8
Borek
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First, or Second endpoint to use - depends on how fast pH changes

Yes, one should choose the endpoint which is easier to detect, at the same time they can't be treated the same way as they mean different stoichiometry of the neutralization, which is apparently what OP missed.
 

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