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.