7. At pH 7.4 a weak organic acid with a pKa of 6.4 would be ionized at what percent?

Answer 1

%ionization =[(H^+)/(acid)]*100 = ??
We have the (H^+), or we can get it, from pH. I find 3.98 x 10^-8 but you should confirm that.
Now, what is (acid)?
acid we will call HA.
HA ==> H^+ + A^-
K = (H^+)(A^-)/(HA)
K = (H^+)(A^-)/[HA-(H^+)]
Plug in 3.98E-7 for K(from pKa = 6.4), plug in 3.98 x 10^-8 for (H^+) in the above equation (don't forget the H^+ in the denominator) and solve for HA.
Substitute for H^+ and HA in the first equation and solve for % ionization. I think the answer is about 90% or so but I didn't check my figures.

Answer 2

I got 90% too