Question

A. 1.30M
B. .500M
C. .110M
D. 5.40*10^-2M

Answer 1

Ka of acetic acid is 1.8*10^-5
A) HA <=> H+ + A-

Ka = [H+]^2 / [HA] {As [H+] = [A-] }
1.8*10^-5 = [H+]^2 / 1.30
[H+] = 4.84*10^-3 M

Percent ionization = [ 4.84*10^-3 M] / 1.30 * 100
= 0.37 %


B) 1.8*10^-5 = [H+]^2 / 0.50
[H+] =3*10^-3 M

Percent ionization = [ 3*10^-3 M] / 0.50 * 100
= 0.6 %


C) 1.8*10^-5 = [H+]^2 / 0.110
[H+] = 1.41*10^-3 M

Percent ionization = [ 1.41*10^-3 M] / 0.110 * 100
= 1.28 %



D) 1.8*10^-5 = [H+]^2 / 5.40*10^-2
[H+] = 9.86*10^-4 M

Percent ionization = [ 9.86*10^-4 M] / 5.40*10^-2 * 100
= 1.83 %