Calculating percentage ionization of a weak acid!?

Alright. LOTS of number are coming your way. But try to follow along.

Take this general expression

HA <------> H + A

This is a weak acid, so it will be in equilibrium between the ionized product and the original acid.

So we know this thing ionized and will have a Ka of (7.95 x 10^ -5). I got this from the pKa.

Next, pH is going to be equal to 7.4, which means [H]=(4x10^ -8)

With that, let's go back to our general expression and fill in the values we know

HA <-------> H + A

HA-X X X

where X is the moles. So looking at a couple of equations

Ka equations ----------> 7.94 x (10^-5)= (X^2)/(HA-X)

Henderson-Has equation ----> 7.4 = 4.1 + log (A/HA)

Rewritten Henderson Has Eq-----> 3.3=log(X/HA)

Look. 2 equations, 2 unknowns! What you'll want too do is solve for X, because this is you're A, which is your ion. Then divide this by the total volume. This will give you the molarity of your ion.

Then divide the molarity of your ion to the initial molarity of your acid. Multiply by 100 and voila!!!

------To the guy below me!----------

This actually suggests it'll be a weak acid. If it were a strong acid, the pH would be acidic.

Ionization Of Weak Acids

It can not be a weak acid only since a solution of pH 7.4 is slightly basic. So at least you must have a mixture of this acid and a salt of strong base with this acid

with Henderson -hasselbach, you have

pH =pK + log ([salt]/[acid] ]

7.4=4.1 +log([salt]/[acid]

log([salt]/[acid]=7.4-4.1=3.3

[salt]/[acid]=10^3.3=2000