Question

A solution is prepared by dissolving 0.1 mol of ethanoic acid (CH3COOH) in 500ml of water at 25 C. pKa of ethanoic acid is 4.7.

What is the pH of the solution?

I'm assuming the equation is pH = pKa + log[Base]/[Acid] but can't get it to work?
Nope. The HH equation is for buffers. You have an acid listed but no base to use. This is just the ionization of an acid. CH3COONa = HAc
.........HAc ==> H^+ + Ac^-
I........0.2.....0......0
C.......-x.......x......x
E.......0.2-x....x......x

Substitute the E line into the Ka expression for Ka and solve for x, then convert to pH.

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