For the datasheet found at Chemistry Libretext,
#K_(a1)=5.6*10^(-2)# and #K_(a2)=1.5*10^(-4)# [1]
for the dissociation of the first and second proton when oxalic acid #"C"_2 "H"_2 "O"_4# dissolves in water at #25^"o" "C"# (#298*"K"#).
Construct the #"RICE"# table (in moles per liter, #mol*dm^(-3)#, or equivalently #"M"#) for the dissociation of the first oxalic proton. Let the increase in #"H"^+(aq)# concentration be #x*mol*dm^(-3)#.
#color(grey)R" " "C"_2 "H"_2 "O"_4 (aq)\rightleftharpoons "C"_2 "H" "O"_4^(-)(aq) + "H"^+(aq)#
#color(grey)(I)" " 0.20#
#color(grey)(C)" " " "-x " " +x" "+x#
#color(grey)(E)" "0.20-x " "x" "x#
By definition,
#K_(a1)=(["C"_2 "H" "O"_4^(-)(aq)] ["H"^+(aq)])/(["C"_2 "H"_2 "O"_4 (aq)])=5.6*10^(-2)#
Simplifying the expression shall give a quadratic equation about #x#, solving for #x# gives #["C"_2 "H" "O"_4^(-)]=x=8.13*10^(-2)*mol*dm^(-3)#
(discard the negative solution since concentrations shall always be greater or equal to zero).
Now construct a second #"RICE"# table, for the dissociation of the second oxalic proton from the amphiprotic #"C"_2 "H" "O"_4^(-)(aq)# ion. Let the change in #"C"_2 "O"_4^(2-)(aq)# be #+y*mol*dm^(-3)#. None of these species was present in the initial solution. (#K_w# is neglectable) Thus the initial concentration of both #"C"_2 "H" "O"_4 ^(-)# and #"H"^+# shall be the same as that at the equilibrium position of the first ionization reaction.
#color(grey)R" " "C"_2 "H" "O"_4 ^(-)(aq)\rightleftharpoons "C"_2 "O"_4^(2-)(aq) + "H"^+(aq)#
#color(grey)(I)" " 8.13*10^(-2) " "0" "8.13*10^(-2)#
#color(grey)(C)" " " "-y " " +y" "+y#
#color(grey)(E)" "8.13*10^(-2)-y " "y" " 8.13*10^(-2)+y#
It is reasonable to assume that
a. #8.13 *10^(-2)-y~~8.13 *10^(-2)#,
b. #8.13 * 10^(-2)+y ~~8.13 * 10^(-2)# , and
c. The dissociation of #"C" _2 "H" "O"_4^(-)(aq )# #
pose neglectable influence on the equilibrium position of the first reaction
Given the fact that #K_(a1)<<K_(a2)#
Thus
#K_(a2)=(["C"_2"O"_4^(2-)(aq)] ["H"^+(aq)])/(["C"_2 "H" "O"_4 ^(-)(aq)])=1.5*10^(-4)~~((cancel(8.13 *10^(-2)))*y)/(cancel(8.13 *10^(-2)))#
Hence #["C"_2"O"_4^(2-)(aq)]=y~~K_(a2)=1.5\cdot 10^(-4)]\cdot mol\cdot dm^(-3)#
References:
[1] E1: Acid Dissociation Constants at 25°C, Chemistry LibreText, https://chem.libretexts.org/Reference/Reference_Tables/Equilibrium_Constants/E1%3A_Acid_Dissociation_Constants_at_25%C2%B0C
