a) Calculate the percent ionization of 0.125M lactic acid (Ka=1.4x10^-4) b) Calculate the percent ionization of 0.125M lactic acid in a solution containing 0.00075 M sodium lactate.

Answer 1

Lactic acid = HL, sodium lactate = NaL.

..........HL ==> H^+ + L^-
I...... 0.125....0.....0
C.........-x.....x.....x
E......0.125-x...x.....x

Substitute the E line into Ka expression and solve for x = (H^+), the
% ion = [((H^+)/(HL)]*100 = ?

for b part, plug into Ka expression the following concns:
Remember NaL is 100% ionized in solution.
.........NaL ==> Na^+ + L^-
I......7.5E-4.....0......0
C.....-7.5E-4....7.5E-4..7.5E-4
so final concns are
(H^+) = x
(L^-) = x from HL + 7.5E-4 from NaL to make x + 7.5E-4
(HL) = 0.125-x

These are plugged into the Ka expression and solve for x = (H^+), then
%ion = [(H^+)/(HL)]*100 = ?