Asked by ryan
7. At pH 7.4 a weak organic acid with a pKa of 6.4 would be ionized at what percent?
Answers
Answered by
DrBob222
%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.
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.
Answered by
Anonymous
I got 90% too
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