Question

53. What is the pH of a 0.4283 M aqueous solution of ammonia? K_b (NH₃) = 1.8x10^-5

Answer 1

Lets write the dissociation equation of NH3

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

0.4283 0 0

0.4283-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*0.4283) = 2.777*10^-3

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

so, x = 2.777*10^-3 M

so.[OH-] = x = 2.777*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.777*10^-3)

= 2.55

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.56

= 11.44

Answer: 11.44