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### Topic: Acetic acid Buffer calculation  (Read 5545 times)

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#### Lourdes

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##### Acetic acid Buffer calculation
« on: June 03, 2017, 02:35:28 PM »
Hi

I have another example here:

An acetic acid / acetate buffer with pH = 4 is to be produced. Acetic acid has a
PKS value of 4.75.

A) We prepare a sodium acetate solution with a concentration of 0.025 mol / L.
How much mg of 98.5% sodium acetate should be weighed to produce 200 mL of this solution?

Now this one I got right, about 410 mg

B) How much mL of the 0.025 molar solution must be added to 150 mL of a 0.1 molar solution
acetic acid are added to obtain a pH = 4 buffer?

I have now idea what to do here, I tried it several times but everytime I get 200 mL, the correct answer is 108 mL.

Can someone help me out with B)?

Thank you all very much in advance!

#### Borek

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##### Re: Acetic acid Buffer calculation
« Reply #1 on: June 03, 2017, 03:28:14 PM »
B) How much mL of the 0.025 molar solution must be added to 150 mL of a 0.1 molar solution
acetic acid are added to obtain a pH = 4 buffer?

Can't make much sense of this one. 0.025 molar solution of what? Sodium acetate?

If so, then yes, correct answer is around 108 mL. Show what you did.
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#### Lourdes

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##### Re: Acetic acid Buffer calculation
« Reply #2 on: June 03, 2017, 03:45:05 PM »
OK, it might take some time to type this all in so be patient for a sec. Thank you

#### Lourdes

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##### Re: Acetic acid Buffer calculation
« Reply #3 on: June 03, 2017, 03:55:33 PM »
Alright so:
c = n/V

For the sodium acetate

0,025moles/L = 0,005moles/V
V = 0,2 L = 200 mL

For acetic acid I calculated:
c =n/V
0,1moles/L = n/0,15L
0,015 moles = n

And then I calculated the new concentration of sodium acetate and acetic acid considering the new volume of 350 mL (100mL + 250mL)

The concentration for the acid is: 0,0428 moles/L and for the salt: 0,0142 moles/L
I put this in the Henderson-Hasselbach equation, which gave me a pH of around 4, which is consistent with what is requiered.

I am sensing a major mistake here, what did I do wrong?
Thanks!!

#### Borek

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##### Re: Acetic acid Buffer calculation
« Reply #4 on: June 03, 2017, 04:30:44 PM »
In general - I have no idea what you are doing.

For the sodium acetate

0,025moles/L = 0,005moles/V

Where did you got 0.005 moles from?

This can be solved pretty easily if you remember that ratio of concentrations equals ratio of numbers of moles (volume is identical, as we are talking about the same solution).
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#### Lourdes

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##### Re: Acetic acid Buffer calculation
« Reply #5 on: June 03, 2017, 04:46:20 PM »
Well I got n = 0,005 from the first part of this problem. 0,025 mol/L = n/0,2 L.

#### Borek

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##### Re: Acetic acid Buffer calculation
« Reply #6 on: June 03, 2017, 05:08:01 PM »
They are completely unrelated.
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#### Lourdes

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##### Re: Acetic acid Buffer calculation
« Reply #7 on: June 03, 2017, 05:29:20 PM »
Can you at least tell me how you managed to get the correct answer?

#### Borek

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##### Re: Acetic acid Buffer calculation
« Reply #8 on: June 04, 2017, 02:50:54 AM »
Can you at least tell me how you managed to get the correct answer?

I already gave you a hint, have you tried to follow it?
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#### Lourdes

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##### Re: Acetic acid Buffer calculation
« Reply #9 on: June 04, 2017, 05:51:21 AM »
OK, so you said that the ratio of the concentration equals the ratio of moles. So c1:c2 = n1:n2 When I calculate this I get 0,00375 for n(NaOH). Now I have n and c of NaOH which gives me 0,15 L which still is not the correct answer.

#### Borek

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##### Re: Acetic acid Buffer calculation
« Reply #10 on: June 04, 2017, 06:25:00 AM »
No idea what you are talking about, there is no NaOH in the problem.

No idea what is c1, c2, n1, n2.

Show your calculations, step by step, describing precisely what is what, as now we are just wasting time.
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#### Lourdes

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##### Re: Acetic acid Buffer calculation
« Reply #11 on: June 04, 2017, 06:41:55 AM »
Never mind, I am getting confused here even more
I'll just skip this problem.

#### Borek

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##### Re: Acetic acid Buffer calculation
« Reply #12 on: June 04, 2017, 06:48:13 AM »
Never mind, I am getting confused here even more

Mostly you are confusing yourself by being chaotic. Please approach the problem in a systematic way, listing what you know, listing necessary equations and taking notes of which variable means what. You are writing down only parts of your thoughts and they are impossible to follow.
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