When
NaC2H3O2 is in solution, it would act as a base since it ionizes into sodium
ions and acetate ions, the acetate ions would interact with water with the
following reaction:
C2H3O2- + H2O <---> C2H3O2H + OH-
So, from the reaction, hydroxide ions are formed
giving it basic properties. We calculate as follows:
[NaC2H3O2] = [Na+] = [C2H3O2-] = 0.250 m
By the ICE table:
[C2H3O2-] C2H3O2H OH-
I
0.250
0 0
C
-x
+x +x
------------------------------------------------------------
E 0.250
- x x
x
1.80×10^−5 = x^2 / (0.250 - x)
Solving for x, we have
x = 0.002112
pOH = -log 0.002112
pOH = 2.68
pH = 14 - 2.68 = 11.32
Therefore, the pH of the solution is about 11.32.
