Question

what is the Ka for a 0.1994M propionic acid solution if pH is 2.795?

CH3CH2CO2H(aq)+H2O(l)<->CH3CH2CO2-(aq)+H3O(aq)

someone please help.. I have a final in two days.

Answer 1

Call propionic acid HPr.

HPr ==> H^+ + Pr^-

Ka = (H^+)(Pr^-)/(HPr)

initial:
HPr = 0.1994 M
H^+ 0
Pr^- = 0

final:
pH = 2.795. Convert to (H^+) by
pH = -log(H^+) = ??
Pr^- = same as H^+.
HPr = 0.1994-(H^+)

Substitute into Ka expression and solve for Ka.