Question

Calculate the percent ionization of 0.120 M lactic acid in a solution containing 7.0×10−3 M sodium lactate.

Answer 1

pKa = 3.86

Ka = 1.38 x 10^-4

Ka = [H+][Lactate]/[Lactic acid]

Lactic acid <===> Lactate   +   H+

0.120                 7 x 10^-3       0      Initial

(0.120-x)            (7 x 10^-3+x)   x     equilibrium

Feed values in Ka equation

1.38 x 10^-4 = (x)(7 x 10^-3 + x)/(0.120-x)

1.656 x 10^-5 - 1.38 x 10^-4x = 7 x 10^-3x + x^2

x^2 + 7.138 x 10^-3 - 1.656 x 10^-5 = 0

x = [H+] = 1.844 x 10^-3 M

percent ionization = (1.844 x 10^-3/0.120) x 100 = 1.54%