Calculate the percent ionization of benzoic acid at the following concentrations?

Question

0.00032 M

I keep doing it but i think i am messing up with the qudratic equation

Answer 1

Benzoic acid is a weak acid, with the formula C6H5COOH, and a Ka value of 6.3 x 10^-5

Benzoic acid dissociates slightly, because it is a weak acid.
R: C6H5COOH ↔ H+ + C6H5COO-
I: .00032M...........0..........0
C: -x..................+x......+x
E: .00032-x...........x.........x

Write the equilibrium expression from the reaction written above.

Ka=[H+][C6H5COO-]
.....------------------------
..........[C6H5COOH]

Substitute in the equilibrium values.

6.3 x 10^-5= [x][x]
.................-----------
.................[.00032-x]

The quadratic equation I got was x^2 + 6.3 x 10^-5x -2.016 x 10^-8 =0
x= 1.1 x 10^-4

Percent ionization= amount ionized/initial concentration x 100

1.1 x 10^-4/.00032 x 100

Percent ionized= 34%

I'm not sure this is the right answer, but this is what I got. Don't know if this answer is too high, but in a way it makes sense because the lower the concentration, the greater percent ionization.

Hope this helps somewhat!! Sorry if it's completely wrong...