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Topic: Degree of dissociation  (Read 7755 times)

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Offline busque00

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Degree of dissociation
« on: October 09, 2010, 06:13:38 AM »
Hey guys

I have a problem and I need your *delete me*
I'm quite confused and I don't know even what they're askin me about

1. An unkown organic acid was isolated from urine and then analyzed. 150mg sample was titrated with 0.05N NaOH (f=1.1). 3.6ml NaOH elevated the pH to 4.3, 8.0ml NaOH to 5.35. Calculate the dissocation constant if the equivalent weight of the acid is 300.


Offline Borek

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Re: Degree of dissociation
« Reply #1 on: October 09, 2010, 06:35:58 AM »
I can't see degree of dissociation in the question.

Do you know what dissociation constant is?

Can you calculate number of moles of acid from the given data?
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Offline busque00

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Re: Degree of dissociation
« Reply #2 on: October 09, 2010, 07:01:08 AM »
How can I calculate K of that acid if I don't have [H+]?

Offline Borek

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Re: Degree of dissociation
« Reply #3 on: October 09, 2010, 07:26:38 AM »
Concentration of H+ is given. What is pH definition?

But IMHO there is something wrong the question. Data given allows calculation of two different values of the pKa, values that differ by 0.2.

Do you have an idea (from your course) what f=1.1 is supposed to mean? I have never seen such a symbol used to give an information in such a context, and even ignoring fact that I have never seen a symbol, I have no idea what it could be - 0.05N NaOH is just a 0.05N NaOH, no need for additional parameters.
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Offline busque00

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Re: Degree of dissociation
« Reply #4 on: October 09, 2010, 08:01:27 AM »
f means factor:

so what I should do now is....

Offline Borek

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Re: Degree of dissociation
« Reply #5 on: October 09, 2010, 10:21:02 AM »
This is crazy, I have never seen such an approach, and I have read or at least browsed dozens of analytical chemistry books.

Assume acid was neutralized quantitatively by added base (just correct amount of base added using f), that will let you calculate concentration of acid and its conjugate base. Combine it with known concentration of H+ and you know everything required to calculate Ka.
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Offline busque00

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Re: Degree of dissociation
« Reply #6 on: October 09, 2010, 04:56:29 PM »
so calculate concentration is equal 0.495M? yes? what about [H+]? I don't understand it at all.
the answer should be K=3.3 x 10-5

Offline Borek

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Re: Degree of dissociation
« Reply #7 on: October 09, 2010, 06:46:11 PM »
No, concentration is not 0.495M, no idea how you got it.

To calculate [H+] use pH definition.

Can you write formula for Ka? Can you identify acid and conjugate base in the formula?
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Offline busque00

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Re: Degree of dissociation
« Reply #8 on: October 11, 2010, 04:57:31 AM »
so which pH should I use to calculate [H+]? 4.3 or 5.35?

ps. czy 300 to jest po prostu masa molowa? tj. 300 g/mol? bo nie mam jednostek i trochę jestem zdezorientowana?
« Last Edit: October 11, 2010, 05:12:46 AM by busque00 »

Offline Borek

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Re: Degree of dissociation
« Reply #9 on: October 11, 2010, 06:20:14 AM »
so which pH should I use to calculate [H+]? 4.3 or 5.35?

Both - you will get two values. In theory they should be identical, can be they will differ because of experimental errors.

Quote
ps. czy 300 to jest po prostu masa molowa? tj. 300 g/mol?

Tak to należy rozumieć.
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Offline busque00

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Re: Degree of dissociation
« Reply #10 on: October 11, 2010, 07:19:01 AM »
should I calculate C from the formula above (using this f) or just from C=n/V (C=m/M*V) ?


Offline Borek

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Re: Degree of dissociation
« Reply #11 on: October 11, 2010, 08:14:31 AM »
f is for NaOH solution, not for acid.

I wonder what you are going to use for V.
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