A) 0.02 M Ch3COOH (weak acid, Ka = 1.8 x 10^-5)

B ) 0.02 M NH3 (weak base, Kb = 1.8 x10^-5)

For these am I supposed to use the Ka and Kb and divide by Kw or do I just multiply the molarity by the Ka/Kb?

This post was edited by Braxton11 on Nov 11 2013 03:55pm

First thing to recognize is that weak acids and bases don't dissociate completely. then use an ice table

definition Ka = ([H+][A-]) /[HA]

CH3COOH (aq) -> H+ (aq) + CH3COO- (aq)

I 0.02 0 0
C -x +x +x
E 0.02 -x x x

Approximated here and eliminated this term since I don't have a calculator

1.8*10^-5 = x^2 / (0.02-x)
x = 6*10^-4

pH = -log[H+]
pH = -log(6*10^-4)
pH ~ 3.22