Question

a. Calculate the pH of a 0.26M methylamine solution? (weak base)

b. Consider the following three solutions of equal concentration. Rank the three solutions from most basic to least basic based on their Kb: (a) aniline 4.3 x 10^-10, (b) methylamine5.0× 10^–4, (c) caffeine 4.1 x 10^-4.

Answer 1

Solution :-

Part a) Calculating the pH of 0.26 M methylamine

Kb of methyl amine = 5.0*10^-4

Reaction equation with water and ICE table

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

0.26 M                                  0                       0

-x                                           +x                      +x

0.26-x                                   x                          x

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

5.0*10^-4 = [x][x]/[0.26-x]

Since the kb is small therefore we can neglect the x from denominator

5.0*10^-4 = [x][x]/[0.26]

5.0*10^-4 = x^2

1.3*10^-4 = x^2

Taking square root on both sides we get

0.0114 M= x = [OH-]

Now lets calculate the pOH

pOH= -log[OH-]

pOH= -log [0.0114]

pOH= 1.94

pH + pOH = 14

pH = 14 – pOH

pH= 14 – 1.94

pH= 12.06

part b) given kb values are as follows

aniline = 4.3*10^-10

methylamine = 5.0*10^-4

caffeine = 4.1*10^-4

when the kb value is larger than means it is stronger base and vice versa

so among the given compounds methylamine is having higher kb value, secondly, caffeine has a higher kb value than aniline

so the strongest base to weak base order is as follows

methylamine > caffeine > aniline