Citric acid (pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40) is found in citrus fruit. Calculate the approximate pH of lemon juice, which is about 5% citric acid by mass.
Can someone set this one up for me?
2 answers
pH = .5 (pka1)
I don't know what the reviewer means by approximate; however, this is a complex problem because Ka1, Ka2, and Ka3 are so close together. So I would approximate to the limit.
First, I would calculate the approximate concn of the lemon juice. Discounting any difference between the density of citric acid from lemons and that of water, I would call 5% solution about 5 g citric acid in about 95 g water.
moles citric acid = about 5 g/molar mass and that divided by 0.095 L = molarity citric.
Then I would treat this as a monoprotic acid and ignore the second and third ionization constants. Actually, k2 and k1 are so close together that such is not very realistic but it makes it a lot simpler.
If we call citric acid H3C, then
H3C ==> H^+ + H2C^-
Ka = (H^+)(H2C^-)/(H3C) and we work it as a monoprotic acid. The approximate pH should be about 1.8 or so. With the additional ions from K2 it would be lower than this in real life.
First, I would calculate the approximate concn of the lemon juice. Discounting any difference between the density of citric acid from lemons and that of water, I would call 5% solution about 5 g citric acid in about 95 g water.
moles citric acid = about 5 g/molar mass and that divided by 0.095 L = molarity citric.
Then I would treat this as a monoprotic acid and ignore the second and third ionization constants. Actually, k2 and k1 are so close together that such is not very realistic but it makes it a lot simpler.
If we call citric acid H3C, then
H3C ==> H^+ + H2C^-
Ka = (H^+)(H2C^-)/(H3C) and we work it as a monoprotic acid. The approximate pH should be about 1.8 or so. With the additional ions from K2 it would be lower than this in real life.
2 answers