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

Vedi sotto.

#### Spiegazione:

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...

Ricorda:

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

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

Così:
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^+.