What is the [H+] in a 4M solution of butanoic acid, where the Ka of butanoic acid is 1.48 x 10^-5? Butanoic acid partially dissociates forming an equilibrium:
H(C4H7O2) <-> C4H7O2- + H+
How is this problem done?
2006-08-25 11:19:02 · 2 answers · asked by RED MIST! 5 in Science & Mathematics ➔ Chemistry
The Ka is graphically equal to [H+] [Bu-] / [HBu].
The [HBu] = 4M - [Bu-] , and ignoring the H+ from the water in the solution, [Bu-] = [H+]. We now have an equation in one unknown, if we express everything in terms of [H+]:
1.48 x 10^-5 = [H+][H+] / (4-[H+])
If you assume that [H+] is much smaller than 4, this is simplified even further to 1.48 x 10^-5 = [H+]^2 / 4
This comes out to about 8 x 10^-3 M, so it is indeed much smaller than 4 and much greater than 10^-7, so our assumptions check out, and we're done.
2006-08-25 11:34:34 · answer #1 · answered by Mr. E 5 · 0⤊ 0⤋
In your own words to a question of mine...
"May you get an education in this wonderful country of ours and utilize your OWN thoughts...."
God Bless You! Oh btw, anymore contacts from you will be considered harrassment and will be reported immediately.
.
2006-08-27 12:41:21 · answer #2 · answered by blaze 4 · 0⤊ 0⤋