Question

Part A Find the [OH−] of a 0.45 M aniline (C6H5NH2) solution. (The value of Kb for aniline (C6H5NH2) is 3.9×10−10.) Express your answer to two significant figures and include the appropriate units. [OH−] = SubmitMy AnswersGive Up Part B Find the pH of a 0.45 M aniline (C6H5NH2) solution. Express your answer using two decimal places.

Answer 1

A)
C6H5NH2 dissociates as:

C6H5NH2 +H2O     ----->     C6H5NH3+   +   OH-
0.45                   0         0
0.45-x                 x         x


Kb = [C6H5NH3+][OH-]/[C6H5NH2]
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 ((3.9*10^-10)*0.45) = 1.325*10^-5

since c is much greater than x, our assumption is correct
so, x = 1.325*10^-5 M



so,
[OH-] = x = 1.325*10^-5 M

Answer: 1.3*10^-5 M

B)
use:
pOH = -log [OH-]
= -log (1.325*10^-5)
= 4.88


use:
PH = 14 - pOH
= 14 - 4.88
= 9.12
Answer: 9.12