Solving for pH of citric acid?

Question

Could I get some help on this problem?

Citric acid (H3C6H5O7) is a triprotic acid with Ka1 = 8.4 10-4,

Ka2 = 1.8 10-5, and Ka3 = 4.0 10-6.

Calculate the pH of 0.23 M citric acid.

Thank you in advance!

Answer 1

The citric acid chemical reaction is

HCitric + H2O ----> Ci- + H3O+

The Ka1 expression is

Ka1 = {H+} {Ci-} / {HCi} since you don't know the {H+} you set it equal to x so the expression becomes

Ka1 = [x][x] / [HCi]

x^2 = Ka1 * [HCi]

x= square root of (Ka1 x [HCi] )

If you solve x for each Ka value you get

x= 0.0139 for Ka1

x= 0.00203 for Ka2

x= 9.59x10^-4 for Ka 3

x= [H+] so the pH is the -log of [H+]

The pH will be 1.85, 2.69, and 3.01 for each Ka value respectively.