Asked by Chemistry
in a 0.01 M solution of 1.4 butanedicarboxylic acid, (Ka1=2.9x10^-5, Ka2=5.3x10^-6), which species is present in the lowest concentration?
Answers
Answered by
DrBob222
Let's call 1,4-butanedicarboxylic acid simply H2B. Then
H2B ==> H^+ + HB^-
HB^- ==> H^+ + B^2-
k1 = (H^+)(HB^-)/(HB)
k2 = (H^+)(B^2-)/(HB^-)
Since both k1 and 2 are small, you know H2B will be the largest of all.
Since the first H to come off is from k1 (k1 is larger than k2 and that's how you can tell) and you can see that H^+ = HB^-. Therefore, both H^+ and HB^- must be smaller than H2B but larger than the B^2-. That makes B^2- the smallest. You also know what B^2- is. It = Ka2.
H2B ==> H^+ + HB^-
HB^- ==> H^+ + B^2-
k1 = (H^+)(HB^-)/(HB)
k2 = (H^+)(B^2-)/(HB^-)
Since both k1 and 2 are small, you know H2B will be the largest of all.
Since the first H to come off is from k1 (k1 is larger than k2 and that's how you can tell) and you can see that H^+ = HB^-. Therefore, both H^+ and HB^- must be smaller than H2B but larger than the B^2-. That makes B^2- the smallest. You also know what B^2- is. It = Ka2.
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