Question

Calculate the pH of a solution prepared by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer that is comprised of 0.25 M CH3NH2 and 0.25 M CH3NH3Cl. Kb of CH3NH2 = 6.8 x 10-5.

Answer 1

before reaction,
moles of CH3NH2 = M*V =0.25 M * 80 mL = 20 mmol
moles of CH3NH3+ = M*V =0.25 M * 80 mL = 20 mmol

moles of added H+ = M*V = 0.100 M * 20.0 mL = 2 mmol
This H+ will react with CH3NH2 to form additional CH3NH3+

After reaction,
moles of CH3NH2 = 20 -2 = 18 mmol
moles of CH3NH3+ = 20 + 2 = 22 mmol

pkb = - log Kb
= - log (6.8*10^-5)
= 4.17

use:
pOH = pKb + log {[CH3NH3+]/[CH3NH2]}
since volume is same for CH3NH3+ and CH3NH2,
we can use number of moles instead of concentration

pOH = pKb + log {[CH3NH3+]/[CH3NH2]}
= 4.17 + log (22/18)
= 4.26

pH = 14 - pOH
= 14 - 4.26
= 9.74

Answer: 9.74