Question

Part A

What is the pH of a 0.215 M ammonia solution?

Express your answer numerically to two decimal places.

Hints

pH =	4.64

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

Part B

What is the percent ionization of ammonia at this concentration?

Express your answer with the appropriate units.

Hints

% ionization =

SubmitMy AnswersGive Up

Answer 1

A)
NH3 dissociates as:

NH3 +H2O     ----->     NH4+   +   OH-
0.215                   0         0
0.215-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 ((1.8*10^-5)*0.215) = 1.967*10^-3

since c is much greater than x, our assumption is correct
so, x = 1.967*10^-3 M



so.[OH-] = x = 1.967*10^-3 M


use:
pOH = -log [OH-]
= -log (1.967*10^-3)
= 2.70


use:
PH = 14 - pOH
= 14 - 2.70
= 11.30
Answer: 11.30

B)
% ionization = x*100/initial concnetration
= (1.967*10^-3 * 100) / 0.215
= 0.915 %
Answer: 0.915 %