Determine the pH of a 0.045mol dm–3 solution of sodium ethanoate (ethanoic acid: pKa=4.75).

Sodium ethanoate is NaOEt and the OEt^- will be hydrolyzed.

..........EOt^- + HOH ==> EotH + OH^-
I.......0.045..............0.......0
C.........-x...............x.......x
E......0.045-x.............x........x

Kb for OEt = (Kw/Ka for t) = (x)(x)/(0.045-x)

Ka is 1.77E-5. Solve for x = (OH^-) and convert to pH.