Question

20) Determine the pH of a 0.188 M NH3 solution at 25°C. The K of NH3 is 1.76 x 10-5. 20) B) 12.656 A) 5.480 C) 8.520 D) 11.260 E) 2.740

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)

= 2.740

use:

PH = 14 - pOH

= 14 - 2.740

= 11.260

Answer: D