Chemistry question - Acids and bases?

Question

Ammonia Solution (0.8M) has a pH of 11.00. What is the pKb of Ammonia?

Answer 1

NH3 + H2O  ---->  NH4+ + OH-  

Kb = [NH4+] x [OH-] / [NH3]

pH + pOH = 14

At pH = 11  === > pOH = 3  ====> [OH-] = 10^ -3 M

[NH4+] = 10^ -3 M  ( equal to [OH-]

[NH3] = 0.8 - 10^ -3

= 0.799

Kb = [NH4+] x [OH-] / [NH3]

= ( 10^ -3 ) x ( 10^-3 ) / ( 0.799 )

= 1.251 x 10^ -6

pKb = - log Kb = - log ( 1.251 x 10^ -6 )

= ( - 0.1306) + 6

= 5.8694