Question

Methylamine is a weak base with pKb = 3.36. Calculate the pH of a 0.010 M solution of methylamine in water.

A)3.36

B)7.00

C)10.64

D)11.26

E)12.05

Which answer is correct and show all work.

Answer 1

we have below equation to be used:

pKb = -log Kb

3.36-log Kb

log Kb = -3.36

K = 10^(-3.36)

Kb = 4.365*10^-4

Lets write the dissociation equation of CH3NH2

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

1*10^-2 0 0

1*10^-2-x x x

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

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 ((4.365*10^-4)*1*10^-2) = 2.089*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/(1*10^-2-x)

4.365*10^-6 - 4.365*10^-4 *x = x^2

x^2 + 4.365*10^-4 *x-4.365*10^-6 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 4.365*10^-4

c = -4.365*10^-6

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 1.765*10^-5

putting value of d, solution can be written as:

x = {-4.365*10^-4 + √(1.765*10^-5)}/2

x = {-4.365*10^-4 - √(1.765*10^-5)}/2

solutions are :

x = 1.882*10^-3 and x = -2.319*10^-3

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

x = 1.882*10^-3

So, [OH-] = x = 1.882*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.882*10^-3)

= 2.73

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.73

= 11.27

Answer: D