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Topic: [Help] Finding Molar weight of Unknown Diprotic Acid  (Read 3675 times)

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Offline satoshi1

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[Help] Finding Molar weight of Unknown Diprotic Acid
« on: April 05, 2015, 06:41:11 PM »
My pka1= 2.14 pka2= 5.99
[NaOH] = 0.10402M
Second equivalence pt. at 28 mL.
First equivalence pt. at 14 mL.
1st half at 7 mL
2nd half at 21 mL.
Mass of unknown diprotic acid is 0.71g.
Titrating 25 mL of unknown diprotic acid.   How do I determine the molar weight of the unknown diprotic acid with the second equivalent pt. and stoichiometry?

The titration Curve:
https://docs.google.com/spreadsheets/d/1trThaHTfx2AZ5LVxiWqbpJlZPxezmBRq0YbAFN01ryk/pubchart?oid=1774542269&format=image
« Last Edit: April 05, 2015, 09:12:03 PM by satoshi1 »

Offline Borek

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #1 on: April 06, 2015, 03:50:12 AM »
You have to show your attempts at solving the question to receive help. This is a forum policy.

You are given mass of the acid. Would knowing number of moles help calculate molar mass?
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Offline satoshi1

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #2 on: April 06, 2015, 05:31:48 AM »
I kind of figured that I needed to show my work.  I just have no idea how to write it out.

Well I got the M of NaOH and I have the volume of it at the second equivalence pt.  I get 0.00291 moles.  Since I use 2NaOH for every H2X, I get 0.001455 mole of H2X.  Now I simply divide the grams of the unknown and I get 487.97 g/mol.  I still don't know how to show this better. 

Offline Borek

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #3 on: April 06, 2015, 06:58:35 AM »
Now I simply divide the grams of the unknown and I get 487.97 g/mol.

Surprisingly high IMHO, in a typical school lab I would expect something well below 200 g/mol.

Mass of unknown diprotic acid is 0.71g.
Titrating 25 mL of unknown diprotic acid.

25 mL of what solution? Does it contain all 0.71 g, or just a fraction?

Quote
I still don't know how to show this better.

That was good enough :)
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Offline satoshi1

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #4 on: April 06, 2015, 07:17:28 AM »
Now I simply divide the grams of the unknown and I get 487.97 g/mol.

Surprisingly high IMHO, in a typical school lab I would expect something well below 200 g/mol.

Mass of unknown diprotic acid is 0.71g.
Titrating 25 mL of unknown diprotic acid.

25 mL of what solution? Does it contain all 0.71 g, or just a fraction?

Quote
I still don't know how to show this better.

That was good enough :)

Yes it is surprisingly high.  The unknown are well below 200g/mol.
25 mL of the unknown diprotic acid.  That is how much you take out of the volumetric flask after dissolving 0.71g in distilled water.  25mL out of a 100 mL volumetric flask used to make the solution.

Offline Borek

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #5 on: April 06, 2015, 08:03:53 AM »
25 mL of the unknown diprotic acid.  That is how much you take out of the volumetric flask after dissolving 0.71g in distilled water.  25mL out of a 100 mL volumetric flask used to make the solution.

Now try to answer the question I asked:

25 mL of what solution? Does it contain all 0.71 g, or just a fraction?
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Offline satoshi1

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #6 on: April 06, 2015, 05:08:58 PM »
25 mL of the unknown diprotic acid.  That is how much you take out of the volumetric flask after dissolving 0.71g in distilled water.  25mL out of a 100 mL volumetric flask used to make the solution.

Now try to answer the question I asked:

25 mL of what solution? Does it contain all 0.71 g, or just a fraction?

I found that it was 121.99 g/mol when I diluted it.  (0.71g/100mL)(25mL)=?g  This allowed me to get a much lower molar mass.

Offline Borek

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Re: [Help] Finding Molar weight of Unknown Diprotic Acid
« Reply #7 on: April 06, 2015, 06:16:03 PM »
I found that it was 121.99 g/mol when I diluted it.  (0.71g/100mL)(25mL)=?g

Much better.
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