Question

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.

Part A

Part complete

What is the pH of a 0.100 M ammonia solution?

Express your answer numerically to two decimal places.

View Available Hint(s)

pH =

11.13

SubmitPrevious Answers

Correct

Part B

What is the percent ionization of ammonia at this concentration?

Express your answer with the appropriate units.

Answer 1

A)
NH3 dissociates as:

NH3 +H2O     ----->     NH4+   +   OH-
0.1                   0         0
0.1-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)*0.1) = 1.342*10^-3

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
1.8*10^-5 = x^2/(0.1-x)
1.8*10^-6 - 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -1.8*10^-6

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 7.2*10^-6

roots are :
x = 1.333*10^-3 and x = -1.351*10^-3

since x can't be negative, the possible value of x is
x = 1.333*10^-3

So, [OH-] = x = 1.333*10^-3 M


use:
pOH = -log [OH-]
= -log (1.333*10^-3)
= 2.87


use:
PH = 14 - pOH
= 14 - 2.8753
= 11.13
Answer: 11.13

B)

x = 1.333*10^-3

% dissociation = (x*100)/c
= 1.333*10^-3*100/0.1
= 1.33 %
Answer: 1.33 %