Posted by jack on Wednesday, November 24, 2010 at 12:44pm.
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-
so
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
find x
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.
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