In: Chemistry
The value of Ka in water at 25 °C for propionic acid (C2H5CO2H) is 3.4 x 10-5 M. What is the pH of a 0.020 M aqueous solution of propionic acid?
proponoic acid is a week acid
construct the ICE table
C2H5CO2H + H2O <------> C2H5CO2- + H3O+
I 0.02 0 0
C -x +x +x
E 0.02-x +x +x
now dissociation constant
Ka = [C2H5CO2-][H3O+] / [C2H5CO2H]
3.4 x 10-5 = [x][x] / [0.02-x]
x2 + x 3.4 * 10-5 - 6.8 x 10-7 = 0
solve the quadratic equation to get x value
x = 0.0008078 M = [H3O+]
now
pH = -log[H3O+] = -log[0.0008078] = 3.09