Question

Determine the pH of a 0.188 M NH₃ solution at 25°C. The Kb of NH₃ is 1.76x10^-5.

Answer 1

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.188 0 0

0.188-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.76*10^-5)*0.188) = 1.819*10^-3

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

so, x = 1.819*10^-3 M

So, [OH-] = x = 1.819*10^-3 M

use:

pOH = -log [OH-]

= -log (1.819*10^-3)

= 2.7402

use:

PH = 14 - pOH

= 14 - 2.7402

= 11.2598

Answer: 11.26