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

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^+#.