Question

0.1 mole of CH3NH2 (kb = 5 x 10-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre , then calculate resulting hydrogen ion concentration in the solution .

Answer 1

after the reaction: 0.02 mol CH3NH2 and 0.08 mol CH3NH3

by Henderson-Hasselbalch Equation: 

pH = pKa + log (base/acid) 

pKb = - log 5 x 10⁻⁴
= 3.30103 

pKa = 14 - 3.30103
= 10.69897 

pH = 10.69897 + log (0.02 / 0.08)

pH = 10.69897 + (-0.60206) 

pH = 10.09691 

Antilog the pH to get the H+ conc. which will be = 8 x 10⁻¹¹ M

Answer 2

CH3NH2 + HCl = CH3NH3+ + Cl-

Initial 0.1 0.08 -

At Eq. 0.02 - 0.08

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

= 5*10^-4 * 0.02/0.08

= 1.25 * 10^-4

[H+] = Kw/[OH-]

= 10^-14 * 4/ 5*10^-4

= 8 * 10^-11