Can someone help me figure out the pH of these 2 solutions?

Question

would it be impossible b/c they're both a base?

1) 0.10 M CH3COOH and 0.20 M NaCH3COO

2) 0.10 M CH3COOH and 0.10 M NaCH3COO

Answer 1

pH = pKa + log (A-/HA)

A- is the conjugate base (without the proton) - NaCH3COO-

HA is the acid, CH3COOH.

pKa of acetic acid is 4.75

For first case

pH = 4.75 + log(.2/.1)

= 4.75 + log(2)

= 5.05

For second case

pH = 4.75 + log(.1/.1)

= 4.75

Answer 2

No, the first is the structural formula for acetic acid.  It is an "oxyacid", since the hydrogen that ionizes is from the OH group bonded to the carbon, and that the carbon is "stabilized" by a double-bonded oxygen, which allows it to carry a negative charge as the acetate.  

In either combo, the acetate (as Na-acetate) is totally ionized.  This controls the H+, since it depresses the tendency of the acetic acid to ionize.  In either case, you use the Ka eqtn:



    [Acetate-][H+]/[Acetic acid] = Ka =1.8x1-5

Substitute the conc of the acetate and the acid in these and compute the H+.  I trust you can find pH from H+.