Question

Calculate the hydroxide concentration of a 0.024 M methylamine solution. Kb is 4.2 x 10^-4

This problem has a second part in the problem below....

part 1 .What is the OH- concentration?

B)

Consider the same 0.024 M methylamine solution. Kb is 4.2 x 10^-4

part 2 -- what is the pH?

Answer 1

first construct the ICE table and then substitute the values in the K equation and then calculate the equilibrium values as follows

CH3NH2 - methyl amine Kb equation is as follows

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

I    0.024                                                       0                0

C    -x                                                            +x            +x

E (0.024-x)                                                  (x) (x)

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

=> 4.2*10-4 = x2 / (0.024-x)

=> 4.2*10-4 * (0.024-x) = x2

=> x = 0.003 M

[OH-] = 0.003

pOH = -log(OH-)

pOH = -log(0.003)

pOH = 2.52

pH = 14- pOH

pH = 14 - 2.52 => 11.48