Answer:
[C2O2−4] =1.5⋅10−4⋅mol⋅dm−3
Explanation:
For the datasheet found at Chemistry Libretext,
Ka1=5.6⋅10−2 and Ka2=1.5⋅10−4 [1]
for the separation of the primary and second nucleon once oxalic corrosive C2H2O4 breaks up in water at 25oC (298⋅K).
Build the RICE table (in moles per l, mol⋅dm−3, or identically M) for the separation of the primary oxalic nucleon. offer the growth access H+(aq) fixation be x⋅mol⋅dm−3.
R C2H2O4(aq)⇌C2HO−4(aq)+H+(aq)
I 0.20
C −x +x
E 0.20−x x
By definition,
Ka1=[C2HO−4(aq)][H+(aq)][C2H2O4(aq)]=5.6⋅10−2
Improving the articulation can provides a quadratic condition regarding x, sinking for x provides [C2HO−4]=x=8.13⋅10−2⋅mol⋅dm−3
(dispose of the negative arrangement since fixations can faithfully be additional noteworthy or appreciate zero).
Presently build a second RICE table, for the separation of the second oxalic nucleon from the amphoteric C2HO−4(aq) particle. offer the modification access C2O2−4(aq) be +y⋅mol⋅dm−3. None of those species was out there within the underlying arrangement. (Kw is neglectable) therefore the underlying centralization of each C2HO−4 and H+ are going to be appreciate that at the harmony position of the first ionization response.
R C2HO−4(aq)⇌C2O2−4(aq)+H+(aq)
I 8.13⋅10−2 zero eight.13⋅10−2
C −y +y
E 8.13⋅10−2−y y eight.13⋅10−2+y
It is smart to just accept that
a. 8.13⋅10−2−y≈8.13⋅10−2,
b. 8.13⋅10−2+y≈8.13⋅10−2 , and
c. The separation of C2HO−4(aq)
Accordingly
Ka2=[C2O2−4(aq)][H+(aq)][C2HO−4(aq)]=1.5⋅10−4≈(8.13⋅10−2 )⋅y8.13⋅10−2
Consequently [C2O2−4(aq)]=y≈Ka2=1.5⋅10−4]⋅mol⋅dm−3
