Question

21. Calculate the pH of the following aqueous solutions. Choose your answer from the given pH ranges. 0.1 M methylamine (pKb = 3.36) A) pH 6.00–8.99 C) pH 9.00–10.99 | E) pH 3.00–5.99 B) pH 0.00–2.99 D) pH 11.00–14.00
please help!!

Answer 1

use:

pKb = -log Kb

3.36= -log Kb

Kb = 4.365*10^-4

CH3NH2 dissociates as:

CH3NH2 +H2O -----> CH3NH3+ + OH-

0.1 0 0

0.1-x x x

Kb = [CH3NH3+][OH-]/[CH3NH2]

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 ((4.365*10^-4)*0.1) = 6.607*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)

4.365*10^-4 = x^2/(0.1-x)

4.365*10^-5 - 4.365*10^-4 *x = x^2

x^2 + 4.365*10^-4 *x-4.365*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 4.365*10^-4

c = -4.365*10^-5

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 = 1.748*10^-4

roots are :

x = 6.392*10^-3 and x = -6.829*10^-3

since x can't be negative, the possible value of x is

x = 6.392*10^-3

So, [OH-] = x = 6.392*10^-3 M

use:

pOH = -log [OH-]

= -log (6.392*10^-3)

= 2.1943

use:

PH = 14 - pOH

= 14 - 2.1943

= 11.8057

Answer: D