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Oxalic acid , HOOCCOOH (aq) , is a diprotic acid with Ka1= 0.0560 and Ka2=0.0145. Determine the [-OOCCOO-] in a solution with a...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
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)
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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