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

a) 0.250

M .

b 7.74×10⁻²

M .

c 2.03×10⁻²

M .

Answer 1

a) Concentration of Propanoic acid = C2H5COOH = 0.250M = C

ka = 1.34x10^-5

percent ionization = = square root of Ka/C

                                      = square root of ( 1.34x10^-5/0.250)

percent ionization = = 7.32x10^-3

percent ionization = = = 7.3x10^-3 x100 = 0.73%.

b)

Concentration of C2H5COOH = C=7.74x10^-2M

percent ionization = = square root of ( 1.34x10^-5/7.74x10^-2)

percent ionization = = 1.316x10^-2

percent ionization = = 1.316x10^-2 x100 = 1.316%.

C)

Concentration of C2H5COOH = C= 2.03x10^-2M

percent ionization = = square root of ( 1.34x10^-5/2.03x10^-2)

percent ionization = = 2.57x10^-2

percent ionization = = 2.57x10^-2 x100 = 2.57%.