posted by jack .
The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is
__________. The Ka of acetic acid is 1.76 × 10-5.
You are missing units for your Ka.
I am sure that I have answered this one recently. Anyway start from the equation
HAc <-> H+ + Ac-
Ka = [H+][Ac-]/[HAc]
at the start we have 0.181 M HAc
and 0.172 M NaAc
at equilibrium we have
(0.181-x)M HAc (some has dissociated)
(0.172+x)M Ac- (we now have more Ac- due to the dissociation)
x M H+
so we can write Ka
(0.172-x)(x)/(0.181-x)= 1.76 × 10-5
you can either solve the quadratic or we can say that x is small wrt 0.172 and 0.181, hence
(0.172)(x)/(0.181)= 1.76 × 10-5
hence find the pH=-log(x)
sorry, answer found. no longer need help..
I hope the method above agrees with the answer you have found.