Question

Calculate:

a) Ka and pKa of a 0.010M solution of a weak acid that is 4% ionized.

b) Kb and pKb of a 0.010M solution of a weak base whose pH = 10.10.

Please show all steps.

Answer 1

a)

% dissociation = (x*100)/c

4= x*100/0.01

x = 4*10^-4

Lets write the dissociation equation of HA

HA -----> H+ + A-

1*10^-2 0 0

1*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 4*10^-4*4*10^-4/(0.01-4*10^-4)

Ka = 1.667*10^-5

we have below equation to be used:

pKa = -log Ka

= -log (1.667*10^-5)

= 4.7782

Ka = 1.67*10^-5

pKa = 4.78

b)

we have below equation to be used:

pH = -log [H+]

10.1 = -log [H+]

log [H+] = -10.1

[H+] = 10^(-10.1)

[H+] = 7.943*10^-11 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(7.943*10^-11)

[OH-] = 1.259*10^-4 M

Lets write the dissociation equation of B

B +H2O -----> BH+ + OH-

1*10^-2 0 0

1*10^-2-x x x

Kb = [BH+][OH-]/[B]

Kb = x*x/(c-x)

Kb = 1.259*10^-4*1.259*10^-4/(0.01-1.259*10^-4)

Kb = 1.605*10^-6

we have below equation to be used:

pKb = -log Kb

= -log (1.605*10^-6)

= 5.7945

Kb = 1.605*10^-6

pKb = 5.79