Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: valeg96 on 22/02/2014 13:55:29
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how do I calculate the pH of a solution of citric acid? If i wanted to make it 2M, 1M, 0.5M and 0.1M what would their pH be? From http://en.wikipedia.org/wiki/Citric_acid (http://en.wikipedia.org/wiki/Citric_acid)
pKa1 = 3.14
pKa2 = 4.75
pKa3 = 6.39
What's the procedure?
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This sounds like homework, so I don't want to give too much away, but:
for the reaction H3A <=> H2A– + H+
10–pKa = Ka = [H2A–]*[H+]/[H3A]
The values in square brackets are concentrations in molarity. Since one H+ and one H2A– is produced by the consumption of one H3A, you can plug in the initial concentration of H3A as whatever molarity you want (M), assume that there is no appreciable concentration of H+ or H2A–
so the calculation becomes
10–pKa = x2/[M–x]
solve for x, which is then the concentration of H+, so pH = –log(x).
I have made the assumption that the citric acid is only deprotonated once. You can do the calculation taking all pKa values into account, but that is a LOT more work, and ends up giving a negligibly different answer.
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It's not homework, I'm preparing various solutions to prepare a scale of a vegetal indicator I extracted some time ago. I calculated them long ago actually, and i got 1.57 for 1M, 2 for 0.5M and 2.57 for 0.1M, but I couldn't remember the procedure, and consequently if it was correct.
Edit: damn repetitions.