Question

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook). Ka=1.3*10^-5 Part A 0.250 M . Express your answer using two significant figures. Part B 7.72×10−2 M . Express your answer using two significant figures. Part C 1.92×10−2 M .

Answer 1

i am hoping you know how to construct ICE table

expression for Ka = [C2H5COO-][H3O+] / [C2H5COOH]

1.3 x 10^-5 = [x][x] / [0.25 - x]

x² + x1.3 * 10^-5 - 3.25 *10^-6 =0

solve the quadratic equation

x = 0.00179 M = [H3O+]

now percentahe of ionozation = (concentration of acid at equilibrium / intial concentration of Acid ) x 100

= (0.00179 M / 0.25) x 100

= 0.71%

PartB

1.3 x 10^-5 = [x][x] / [0.0772 - x]

x² + x1.3 * 10^-5 - 0.10036 * 10^-5 =0

solve the quadratic equation

x = 0.00099 M = [H3O+]

now percentahe of ionozation = (concentration of acid at equilibrium / intial concentration of Acid ) x 100

= (0.00099 M / 0.0772) x 100

= 1.3%

Part C

1.3 x 10^-5 = [x][x] / [0.0192 - x]

x² + x1.3 * 10^-5 - 2.496 *10^-7 = 0

solve the quadratic equation

x = 0.000493 M = [H3O+]

now percentahe of ionozation = (concentration of acid at equilibrium / intial concentration of Acid ) x 100

= (0.000493 M / (0.0192) x 100

= 2.6%