2. A procedure you want to duplicate calls for a 0.100 M sodium phosphate buffer
ID: 966345 • Letter: 2
Question
2. A procedure you want to duplicate calls for a 0.100 M sodium phosphate buffer, pH 7.0. You decide to prepare a liter of said buffer. The pKa’s of ortho-phosphoric acid are 2.12, 6.8 and 12.33 at 25 ºC. When you check the pH of your carefully prepared buffer, you find that it wrong. (The solution pH is actually 6.95.) Speculate on why this occurred. (Points will be deducted for blaming your crummy lab technique. Since everyone gets this result, you didn’t get it because you haven’t learned to use a pan balance.)
Explanation / Answer
First we need to know the concentrations that this buffer need to have. In this case, we want 1 L of a 0.1 M Buffer and a pH of 7 so, we'll use the pKa closest to this desired pH:
7 = 6.8 + log(A/HA)
7-6.8 = log(A/HA)
100.2 = A/HA
1.58 = A/HA
A = 1.58HA
With this we can know the concentrations of the species that forms the buffer:
[Bu] = [A] + [HA]
0.1 = 1.58HA + HA
0.1 = 2.58HA
HA = 0.0388 M
A = 1.58*0.0388 = 0.0613 M
This should be the concentrations for the buffer in order to get a pH of 7. Now if we got a pH of 6.95 this could happen:
6.95 = 6.8 + log(A/HA)
106.95-6.8 = A/HA
1.41 = A/HA
The ratio of A/HA is lower than the actual ratio needed to get a pH of 7 (1.58). So the most obvious thing that happens here is that the compounds used to prepare this buffer were not prepared correctly at the right concentrations, and that's why the pH is wrong.
Hope this helps
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