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

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.