Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: zeroh6x on August 07, 2006, 11:14:58 AM
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Calculate the amounts of acid and base forms of the buffer needed to prepare 2 L of a 0.1 M acetate buffer, pH 5.2.
my answer:Given that the concentration of the buffer is 2 x 0.1 = 0.2 molar
The concentration of the acid
pH = 5.2
H+ = - antilog 5.2
acid = 6.3 x 10 ^ -6 mol/dm3
ratio of salt by acid = 0.2
Concentration of the salt
0.2 = salt / 6.3 x 10 ^ -6
Salt = 0.2 x 6.3 x 10 ^-6
salt = 1.26 x 10 ^ -6 mol/dm3
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I'm not sure, but I think the problem asks you to calculate the weight of sodium acetate and acetic acid needed to prepare 2L of a buffer whose pH is 5.2 and where the molarity of sodium acetate is 0.1M.
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Maybe you should consider the Hendersson-Hasselbalch equation.
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Maybe you should consider the Hendersson-Hasselbalch equation.
Yes, I agree, but I think you still have to consider [base]=0.1M. Am I wrong?
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[CH3COO-] = 0,1 M yes.
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[CH3COO-] = 0,1 M yes.
Great. So, that's why I think the question is about weights (otherwise we've already solved half of it before beginning!) or, very uncommon, about moles.
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Yes, I agree, but I think you still have to consider [base]=0.1M. Am I wrong?
I think so. zeroh6x asked this question already earlier.
http://www.chemicalforums.com/index.php?topic=9928.msg47164#msg47164
0.1M buffer means you either start from 0.1M acetic acid solution and you add solid NaOH till pH reaches required value, or you mix 0.1M acetic acid with 0.1M sodium acetate to reach required pH. In both cases neither acid nor base concentration in the final buffer solution is 0.1M.
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if i have to make this buffer i would say that at the end [Ac-] + [HAc] = 0,1 M
An example of how to make an 100 mL 0,2 M acetate buffer pH = 5,0:
http://www.sonoma.edu/users/h/hanesda/B324/BUFFERS_lab.html
In my calculation i always end up with 2 equations which i solve with direct substitution. In this case:
[Ac-]/[HAc] = 10^(5,2-4,74) and [Ac-] + [HAc] = 0,1 M
then i would substitue [HAc] = 0,1 - [Ac-] and solve for [Ac-] in the other equation.
I also use this to calculate my Tris-HAc-buffers.
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if i have to make this buffer i would say that at the end [Ac-] + [HAc] = 0,2 M
So it will be 0.2M buffer, not 0.1M. Look at the border cases - when you go for higher and higher pH (or lower and lower ones) you end with 0.2M solution of either pure acid or pure salt. This is defnitely not something you will call 0.1M solution.
An example of how to make an 100 mL 0,2 M acetate buffer pH = 5,0:
http://www.sonoma.edu/users/h/hanesda/B324/BUFFERS_lab.html
Even this example contradicts what you have proposed. Read the last phrase:
A 0.2 M acetate buffer contains a total of 0.2 mole of acetate per liter. Some of the total acetate is in the conjugate acid form (HOAc), and some is in the conjugate base form (OAc-).
You see? That's 0.2M and we are asked to prepare 0.1M solution.
In my calculation i always end up with 2 equations which i solve with direct substitution. In this case:
[Ac-]/[HAc] = 10^(5,2-4,74) and [Ac-] + [HAc] = 0,2 M
then i would substitue [HAc] = 0,2 - [Ac-] and solve for [Ac-] in the other equation.
Which is exactly what I have answered yesterday, please read my post from the other thread.
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i modified my reply to [HAc] + [Ac-] = 0.1 for the problem in this topic. The sum equal to 0,2 M is in the problem in the reference.
And i've read your reply in the other topic.