Posted by Nina on .
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.

chemistry 
DrBob222,
Lactic acid = HL, sodium lactate = NaL.
..........HL ==> H^+ + L^
I...... 0.125....0.....0
C.........x.....x.....x
E......0.125x...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......7.5E4.....0......0
C.....7.5E4....7.5E4..7.5E4
so final concns are
(H^+) = x
(L^) = x from HL + 7.5E4 from NaL to make x + 7.5E4
(HL) = 0.125x
These are plugged into the Ka expression and solve for x = (H^+), then
%ion = [(H^+)/(HL)]*100 = ?