Asked by Amanda

Oxalic acid , HOOCCOOH (aq) , is a diprotic acid with Ka1= 0.0560 and Ka2=0.0145. Determine the [-OOCCOO-] in a solution with an initial concentration of 0.139 M oxalic acid.

HOOCCOOH(aq) = HOOCCOO-(aq) + H+(aq)
HOOCCOO-(aq) = -OOCCOO-(aq) + H+(aq)

I have tried a ICE diagram:

HOOCCOO^- <=> ^-OOCCOO^-+ H^+
I: 0......0.139......0
C: -X......+X.......+X
E: -X......0.139+x......X

K= 0.139+X *X / -X
0.0145 = 0.139+X *X /-X
X= 0.03823
[-OOCCOO-] = 0.139-.03823= 0.10077

Not sure where I went wrong. Please help

Amanda

Answers

Answered by DrBob222
Where did you get these Ka values. The literature gives values that are about 1000 between k1 and k2 but your values have a ratio of about 4 and not 1000.
Answered by Lynn
the Ka values are given in the question from the homework.

Ka1 = 0.0560, for the equation: HOOCCOOH(aq) = HOOCCOO-(aq) + H+(aq)

While Ka2 = 0.0145, for the equation:
HOOCCOO-(aq) = -OOCCOO-(aq) + H+(aq)

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