Question

Calculate the [H3O+] and pH of each polyprotic acid solution.

Dissociation Constants for Acids at 25 ∘C

Formula	Ka1	Ka2	Ka3
H2CO3	4.3×10−7	5.6×10−11
H3C6H5O7	7.4×10−4	1.7×10−5	4.0×10−7

PART A

0.130 M H3C6H5O7 H3O+ =

PART B

0.130 M H3C6H5O7 pH =

Answer 1

If the Ka or pKa values are not provided in the question ( they should be) then you have to source these from some reference list - I often find some slight differences depending on the source. So final answers may differ slightly between different contrbutors. But the calculation theory and methods should be the same.
From my reference list I find:
Ka1 H2CO3 = 4.3*10^-7, pKa = 6.37
Ka2 H2CO3 = 5.6*10^-11, pKa = 10.25

Because H2CO3 is a diprotic weak acid, we will make some assumptions that will not affect the end result, but will allow much easier calculation:
H2CO3 dissociates in two steps:
H2CO3 ↔ H+ + HCO3-
HCO3- ↔ H+ + CO3 2-

We will assume:
1) [H+] = [HCO3-]= [CO3 2-]
[H2CO3] = [HCO3-]

For first dissociation:
Ka = [H+]*[HCO3-] / [H2CO3]
Ka = [H+]²/[H2CO3]
4.3*10^-7 = [H+]²/ 0.130
[H+]² = 5.59*10^-8
[H+] = 2.364*10^-4

For second dissociation:
Ka = [H+]*[CO3 2-]/[HCO3-]
5.6*10^-11 = [H+]²/[HCO3-]
5.6*10^-11 * 0.130 = [H+]²
[H+]² = 7.28*10^-12
[H+] = 2.698*10^-6

Total [H+] ( or [H3O+] = (2.364*10^-4) +( 2.698*10^-6) = 2.3909 *10^-4 M

pH = -log 2.3909 *10^-4 =3.6214