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?

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.

Sure! To calculate the approximate pH of lemon juice, which is about 5% citric acid by mass, we can use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa value of the acid and the ratio of the concentrations of the acid and its conjugate base.

First, we need to determine the concentration of the acid and its conjugate base in the lemon juice solution. Since we know that lemon juice is about 5% citric acid by mass, we can assume that the citric acid is the only acid in the solution. Therefore, we can calculate the concentration of citric acid by assuming that the density of lemon juice is similar to that of water (1 g/mL).

Here's how we can set it up:
1. Assume that we have 100 g of lemon juice (since density is similar to water, 100 mL = 100 g).
2. The mass of citric acid in the lemon juice is 5% of 100 g, which is 5 g.
3. The molar mass of citric acid is approximately 192.14 g/mol (C6H8O7).
4. Calculate the number of moles of citric acid: 5 g / 192.14 g/mol = 0.026 moles.

Now that we have the concentration of citric acid, we can calculate the concentrations of the acid and its conjugate base. Since citric acid has three ionizable hydrogens and three corresponding pKa values, we have three possible protonation states.

For the first protonation stage (pKa1 = 3.13):
5. Write the equilibrium reaction for the dissociation of citric acid: C6H8O7 ⇄ C6H7O7^- + H+
6. Since we assume that citric acid is the only acid present and all of it dissociates, the concentration of the citrate ion (C6H7O7^-) is equal to the concentration of the citric acid (0.026 moles).
7. The concentration of H+ is unknown at this stage.

For the second protonation stage (pKa2 = 4.76):
8. Write the equilibrium reaction for the dissociation of the citrate ion: C6H7O7^- ⇄ C6H6O7^2- + H+
9. The concentration of the citrate ion is now the concentration of the citric acid from the first stage (0.026 moles).
10. The concentration of H+ is unknown at this stage.

For the third protonation stage (pKa3 = 6.40):
11. Write the equilibrium reaction for the dissociation of the second-stage citrate ion: C6H6O7^2- ⇄ C6H5O7^3- + H+
12. The concentration of the second-stage citrate ion is now the concentration of the citric acid from the second stage (0.026 moles).
13. The concentration of H+ is unknown at this stage.

Using the Henderson-Hasselbalch equation, we can express the pH in terms of pKa and the concentrations of the acid and its conjugate base:
pH = pKa + log10 ([base] / [acid])

Please note that this is an approximate calculation, as the Henderson-Hasselbalch equation assumes ideal conditions and neglects the influence of other ions or buffers, which may be present in lemon juice.