Question

What is the pH of a solution prepared from .250 mol NH3 dissolved in sufficient water to make 1.00 L of solution? (Kb= 1.8x10^-5)

A. 2.12

B. 2.67

C. 8.92

D. 11.33

E. 13.40

Answer 1

Lets write the dissociation equation of NH3

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

0.25 0 0

0.25-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)*0.25) = 2.121*10^-3

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

so, x = 2.121*10^-3 M

so.[OH-] = x = 2.121*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.121*10^-3)

= 2.67

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.67

= 11.33

Answer: 11.33