Question

Determine the percent ionization for formic acid at 25 degrees C.

a) 0.066M

b) 2.7x10^-4M

c) 1.95M

Answer 1

For a weak acid

HA <-------------> H+ (aq) + A-(aq)

and the degree of ionization x and Ka are related as

x = (Ka/C)^1/2

and Ka of formic acid = 1.77x10^-4

^a) C = 0.066M

Thus degree of ionization x = (1.77x10^-4 / 0.066)^1/2

= 5.17x10^-2

and percent ionization = degree of ionization x 100

= 5.17%

b) at C= 2.7x10^-4 M

degree of ionization = (1.77x10^-4 / 2.7x10^-4 )^1/2

=0.8096

and % ionisation = 80.96%

c) C=1.95 M

degree of ionisation =(1.77x10^-4 / 1.95)^1/2

= 9.25x10^-3

and percent ionistion = 0.925%