Question
7. At pH 7.4 a weak organic acid with a pKa of 6.4 would be 60% ionized
% Ionization =[(H^+)/(acid)]*100 = 3.98 x 10^-8
HA ==> H^+ + A^
K = (H^+)(A^-)/(HA)
K = (H^+)(A^-)/[HA-(H^+)] Plug in 3.98E-7
% Ionization =[(H^+)/(acid)]*100 = 3.98 x 10^-8
HA ==> H^+ + A^
K = (H^+)(A^-)/(HA)
K = (H^+)(A^-)/[HA-(H^+)] Plug in 3.98E-7
Answers
DrBob222
I don't get 60%. Is that your answer? An answer from your notes?