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Topic: Problem of the week - 26/11/2012  (Read 8281 times)

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Offline Borek

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Problem of the week - 26/11/2012
« on: November 26, 2012, 12:37:56 PM »
Yesterday someone asked whether pH of the mixture prepared by mixing pH 2.8 and pH 5.1 solutions will be (2.8+5.1)/2=3.95. The answer is simple - no, or more complicated - generally no, but you can design such an experiment.

Let's start with something simpler. Let's say we have a 100 mM acetate buffer of pH 4.00. We want to mix it in 1:1 volume ratio with another acetate buffer of pH 5.00. What must be the concentration of the other buffer, so that the final solution has pH that is an average of the initial pHs of both solutions?
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Offline XGen

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Re: Problem of the week - 26/11/2012
« Reply #1 on: November 26, 2012, 08:20:36 PM »
Sorry for posting it so early in the week, but I really thought this problem was interesting :3



Acetic acid has pKa 4.74, so using the Henderson-Hasselbach equation:

4 = 4.74 + log (A-/HA)
A-/HA = 0.18

Therefore, the ratio of acetate to acetic acid in the 0.1M buffer is 0.18, and we can say the concentration of acetate is 0.015 M and the concentration of acetic acid is 0.085 M.

Applying the same equation to the buffer of pH 4.5, we get a ratio of 0.58, and with the buffer of pH 5 we get a ratio of 1.82. If we assume 1 liter of both buffers, and the concentration of the pH 5 buffer to be x, we can write:

(0.015 + 1.82x)/(0.085 + x) = 0.58

For which we can solve and get x ~ 0.0277, meaning the concentration of the pH 5 buffer is 0.0277 M.

Offline Rutherford

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Re: Problem of the week - 26/11/2012
« Reply #2 on: November 28, 2012, 12:43:07 PM »
I got that the concentration is 0.075M and I checked it. Should be okay if I didn't mistake in both solving and testing.

Offline Borek

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Re: Problem of the week - 26/11/2012
« Reply #3 on: December 03, 2012, 05:09:04 AM »
I got that the concentration is 0.075M

I got 0.076 M, close enough.
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