Calculate the pH of a 0.033 M solution of sodium acrylate (NaC3H3O2).?

Answer 1

pKa of C3H3O2H or H2C=CH-COOH is required.

Answer 2

Lancenigo di Villorba (TV), Italy

THE ANSWER IS pH GREATER THAN 11.

BRIEF CALCULATION
Once I verified if this salt's concentration obeys to the numerical limitations of the Ostwald's ionic equilibria

0.10 >= (Ka / Cs)^0.5 = (5.5E-5 / 0.033)^0.5 = 0.02

I can proceeds applying the simplest formula of Ostwald

|OH-| = (Kw * Cs / Ka)^0.5 = (1E-14 * 0.033 / 5.5E-5)^0.5 =
= 2.4-3 M

hence

|H+| = Kw / |OH-|= 1E-14 / 2.4E-3 = 4.2E-12 M

pH = 0 - LOG(|H+|) = 11.4

CORRECT PATHWAY

I hope this helps you.