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
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
See your post above.
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