7. At pH 7.4 a weak organic acid with a pKa of 6.4 would be ionized at what percent?
%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.
I got 90% too
To determine the percentage of ionization of a weak acid at a specific pH, you need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]),
where pH is the solution's pH, pKa is the acid dissociation constant, [A-] is the concentration of the ionized form (conjugate base), and [HA] is the concentration of the non-ionized form (acid).
In this case, the pKa of the weak acid is given as 6.4, and the pH is 7.4.
Let's substitute the values into the equation and solve for [A-]/[HA]:
7.4 = 6.4 + log([A-]/[HA])
Rearranging the equation, we have:
log([A-]/[HA]) = 7.4 - 6.4
log([A-]/[HA]) = 1
Now, we need to convert the logarithmic equation into an exponential form:
10^log([A-]/[HA]) = 10^1
[A-]/[HA] = 10
This means that the ratio of [A-] to [HA] is 10.
To calculate the percentage of ionization, we need to consider the total amount of acid present before ionization. Assuming the initial concentration of the weak acid is 1M, we can set [HA] = 1M.
Therefore, [A-] = 10 * [HA] = 10 * 1M = 10M
To determine the percentage of ionization, divide the concentration of the ionized form by the initial concentration of the weak acid, and multiply by 100:
% ionization = ([A-] / [HA]) * 100
% ionization = (10M / 1M) * 100
% ionization = 1000%
Therefore, at pH 7.4, the weak organic acid with a pKa of 6.4 would be ionized at 1000%.