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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: bubblyporcupine on November 03, 2005, 10:52:30 AM

Title: pH and Buffers
Post by: bubblyporcupine on November 03, 2005, 10:52:30 AM

I'm told to make a 60mL buffer solution with a pH of 5.00 using  0.1M acetic acid and 0.1M sodium acetate, but how do I determine the volumes required to make the buffer?

Here's what I've done so far...

CH3COOH + CH3COONa + H2O<--> H3O+ + 2CH3COO- + Na+

and then I'm kinda stuck...

Title: Re:pH and Buffers
Post by: Borek on November 03, 2005, 11:03:58 AM

What you need is a Henderson-Hasselbalch equation (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch) to calculate ratio of acid and conjugate base and then some rather simple (although possibly tedious) stoichiometry/concentration calculations. This is one of the rare cases when reaction equation is of no help ;)

Title: Re:pH and Buffers
Post by: bubblyporcupine on November 03, 2005, 11:20:13 AM

booo! I was starting to like the reaction equation - it solved EVERYTHING! lol

so how's this for a start:

pH = pKa + log[conj. base] / [acid]

5.0 = 4.74 + log x / 0.1
5.0 = 4.74 + log(x) - log(0.1)
0.2553 = log(x)  - 1
1.2553 = log(x)
10^1.2553 = x
x=18.001

so 18 is the concenntration of my conjugate base?
 that seems a little high... lol


and then for my [H+],

[H+] = Ka[HA]/[A-]
[H+] = (1.8*E-5(0.1)/(18)
[h+] = 1.01e-7
TINY NUMBER lol - this doesn't look right to me. :-\

Title: Re:pH and Buffers
Post by: bubblyporcupine on November 03, 2005, 11:57:33 AM

or hang on - you said RATIO right?

so does this maybe look better? :
pH = pKa + log[ base] / [acid]
5.0 = 4.74 + log[base] / [acid]
0.2553 = log[base] / [acid]
10^0.2553 = [base] / [acid]
1.8 = [base] / [acid]

and then
1.8 = [base] / [acid] so:
1.8 = (60mL - x) / x
x= 21.43mL
so CH3COOH volume needed is 21.43mL

and for the sodium acetate: 60mL - x = 60mL - 21.413mL = 38.57mL

does that look ok?



 

Title: Re:pH and Buffers
Post by: Albert on November 03, 2005, 12:35:28 PM

Quote from: bubblyporcupine on November 03, 2005, 11:57:33 AM

so CH3COOH volume needed is 21.43mL

and for the sodium acetate: 60mL - x = 60mL - 21.413mL = 38.57mL

Now, you've got it. :)

Title: Re:pH and Buffers
Post by: bubblyporcupine on November 03, 2005, 12:46:43 PM

FANTASTIC!!    ;D

thanks guys!


now, if I were to find the pH at the acidic equivalence point, could I go like this:

CH3COOH + H2O -> CH3COO- + H3O
0.0357          -              0            0
-x                 -             +x         +x
0.0357 - x    -              x            x

1.8E-5 = x^2 / 0.0357-x

x = 7.93E-4
[H+] = x
pH = -log[H+]\
pH = 3.10

and use pOH to fond the basic equivalence point - does that look ok?