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Topic: Titration of Acetic Acid and Sodium Hydroxide  (Read 4978 times)

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Offline Randomc

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Titration of Acetic Acid and Sodium Hydroxide
« on: March 20, 2014, 07:02:26 PM »
Hi guys,

I have this titration question which I can't solve.

25 mL of a 0.095M of acetic acid was diluted to 100mL with deionized water and titrated with 0.101M of NaOH.

The equation for the reaction occurring is CH3COOH (aq) + NaOH (aq) -> CH3COONa (aq) + H2O (aq)

a) Calculate the number of moles of sodium acetate formed at equivalence point. (I used around 23.1 mL of NaOH)
b) Calculate the concentration of sodium acetate at equivalence point.
c) Then using the equation pH = 1/2pKw + 1/2pKa + 1/2logc, calculate the pH at the equivalence point. (my pH comes out to about 1.86 here which is totally off the expected value)

Offline Borek

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #1 on: March 20, 2014, 07:16:40 PM »
c) Then using the equation pH = 1/2pKw + 1/2pKa + 1/2logc, calculate the pH at the equivalence point. (my pH comes out to about 1.86 here which is totally off the expected value)

That's not what I got, list all values you used.
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Offline Randomc

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #2 on: March 20, 2014, 07:33:41 PM »
1/2 (10^-14) + 1/2 (4.76) + 1/2 log 0.047M.

Comes up to around 1.72. I think that there's something wrong with the calculation of the concentration of sodium acetate.

a) 0.095M * 0.0231L = 2.1945 * 10^-3 mol of sodium acetate
b) conc of CH3COONa at equivalence point = (2.1945 * 10^-3 mol)/vol of CH3COONa
    conc of CH3COONa at equivalence point = (2.1945 * 10^-3 mol)/0.0467L
    conc of CH3COONa at equivalence point = 0.047M

vol of CH3COONa = Vol of acetic acid + Vol of NaOH
vol of CH3COONa = 0.025L + (2.1945*10^-3)/0.101M
vol of CH3COONa at equivalence point = 0.0467L

I am not sure if this makes sense... this is the first introductory chemistry module I have taken in a few years and I have found it pretty difficult to wrap my head around concepts like pH, pKa and the like.

Offline Borek

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #3 on: March 20, 2014, 07:55:13 PM »
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Offline Randomc

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #4 on: March 20, 2014, 08:00:30 PM »
My tutor says to assume pKw = 1.0 * 10^-14

Offline Borek

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #5 on: March 21, 2014, 03:54:42 AM »
My tutor says to assume pKw = 1.0 * 10^-14

This is not answer to the question "what is pKw"?  ;) And you are probably wrong - I strongly doubt you quoted your tutor correctly.

Do you understand the difference between Kw and pKw?
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Offline Randomc

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #6 on: March 21, 2014, 05:26:15 AM »
Oh... I get it now. No wonder I've been getting such a low pH value! I got mixed up with the two. Just another question - How do I calculate the total concentration of the following:

What is the total concentration (mol/L) of 3.4g of sodium acetate mixed with 30mL of 1.0M acetic acid?

Offline Borek

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Re: Titration of Acetic Acid and Sodium Hydroxide
« Reply #7 on: March 21, 2014, 05:41:36 AM »
What is the total concentration (mol/L) of 3.4g of sodium acetate mixed with 30mL of 1.0M acetic acid?

This is ambiguous, as it doesn't state total concentration of what.

But in the context of buffers we often say things like "0.1 M buffer" and it is intended to mean "sum of concentrations of HAcetate and Acetate- is 0.1 M".
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