Calculate the equilibrium concentration of c2o42 - in a 0.20 m solution of oxalic acid.

[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