If 20.0 mL of 0.42 M acetic acid (CH

Question

3COOH) is added to 10.0
mL of 0.42 M sodium acetate (CH
3COONa) solution, what is the
pH of the resulting solution? (K
= 1.8 x 10-5)

Answer 1

I notice that your Ka is missing units is this deliberate???

start from the equation

HAc -> H+ + Ac-

so

Ka=[H+][Ac-]/[HAc]

at the start concentration acid
20 ml x 0.42 M/30 ml
=0.28 M

at the start concentration acetate
10 ml x 0.42 M/30 ml
=0.14 M

if H+ concentration at equilibrium = x

at equilibrium

acid concentration
0.28 M - x

acetate concentration
0.14 M +x

substitute

Ka=[H+][Ac-]/[HAc]

Ka=[x][0.14 + x]/[0.28-x]=1.8 x 10-5

solve the resulting quadratic for x

or

assume x is small wrt 0.14

[x][0.14]/[0.28]=1.8 x 10-5

find x.

and pH=-log(x)

Answer 2

Great, thanks so much! :)
And yes Ka doesn't have any units since K is an equilibrium constant.