Question

A 1.00 M solution of NH4Cl has a pH of 4.63. What is the percent ionization of this acid? a. 0.0023% b. 0.28% c. 4.6% d. 9.3% e. 36%

Answer 1

Kb of NH3 is 1.8*10^-5

use:

Ka = Kw/Kb

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

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ dissociates as:

NH4+ -----> H+ + NH3

1 0 0

1-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*1) = 2.357*10^-5

since c is much greater than x, our assumption is correct

so, x = 2.357*10^-5 M

% dissociation = (x*100)/c

= 2.357*10^-5*100/1

= 0.0023 %

Answer: a