Question

What is the equilibrium concentration of H3O+ in a 0.20 M solution of oxalic acid?

Polyprotic acids contain more than one dissociable proton. Each dissociation step has its own acid-dissociation constant, Ka1, Ka2, etc. For example, a diprotic acid H2A reacts as follows:
H2A(aq)+H2O(l)⇌H3O+(aq)+HA−(aq)
Ka1=[H3O+][HA−][H2A]
HA−(aq)+H2O(l)⇌H3O+(aq)+A2−(aq)
Ka2=[H3O+][A2−][HA−]
In general, Ka2 = [A2−] for a solution of a weak diprotic acid because [H3O+]≈[HA−].
Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1=5.9×10−2, Ka2=6.4×10−5.

Answer 1

See below.

Explanation:

If the concentration of the acid is 0.2 then we can find the `H_3O^+` in total by using the two different `K_a`'s.

Also, I call Oxalic acid as `[HA_c]` and the oxalate ion as `[A_c^-]`, although this is often used for acetic acid. It was simpler than to write out the entire formula...

Remember:

`K_a=([H_3O^+] times [A_c^-])/([HA_c])`

So in the first dissociation:

`5.9 times 10^-2=([H_3O^+] times [A_c^-])/([0.2])`

Hence we can say the following:

`0.118=[H_3O^+]^2`

As the `[H_3O^+]` ion and the respective anion must exist in a 1:1 ratio in the solution.

So:
`0.1086=[H_3O^+]= [A_c^-]`

Now the oxalate ion will continue to dissociate, and we know that this is the anion- so we plug in the `[A_c^-]` found in the first dissociation as the acid in the second dissociation (aka the term in the denominator).

`6.4 times 10^-5=([H_3O^+] times [B_c^-])/[0.1086]`

`6.95 times 10^-6=[H_3O^+]^2`

`0.002637=[H_3O^+]`

Then we add the concentrations from the dissociations:
`0.002637+0.1086= 0.111237 mol dm^-3` (using rounded answers, the real value would be: `0.1112645035 moldm^-3`
of `H_3O^+`.