# Acetic acid is 3% dissociated in a 0.02M solution. Give the concentrations of each of the two ions, the acetic acid, and the Ka for acetic acid.

HC2H3O2 ==> H^+ + C2H3O2^-

So (H^+) = 0.02*0.003 = ??

(C2H3O2^-) = 0.02*0.003 = ??

Plug into Ka expression and solve for Ka.

Post your work if you get stuck.

I'm sorry, I don't understand why 0.003 & not 0.03

I got too happy with typing zeros. 3% is 0.03. You're right.

thank you

## You're welcome! It's easy to make mistakes while typing. Let's correct it using the correct value of 0.03 for the percent dissociation.

Here's the corrected calculation:

(H^+) = 0.02 * 0.03 = 0.0006 M

(C2H3O2^-) = 0.02 * 0.03 = 0.0006 M

Now that we have the concentrations of both ions, we can calculate the Ka for acetic acid.

The Ka expression for the ionization of acetic acid is:

Ka = (H^+) * (C2H3O2^-) / (HC2H3O2)

Substituting the given concentrations:

Ka = (0.0006) * (0.0006) / (0.02)

Now, you can calculate Ka using this equation.