Question

The following buffer was prepared:

50. mL 2.0 M methylamine (CH3NH2) solution and 50. mL of 2.0 M methylamine hydrochloride (CH3NH3Cl) solution. The Kb for methylamine is 4.4 x 10-4.

What is the pH of the resulting solution after adding 100 ml of 1.0 M HCl to the original buffer solution?

a) 11.12 b) 10.64 c) 12.32 d) 0.48 e) 5.32

Answer 1

50. mL 2.0 M methylamine (CH3NH2) solution and 50. mL of 2.0 M methylamine hydrochloride (CH3NH3Cl) solution. The Kb for methylamine is 4.4 x 10-4.

What is the pH of the resulting solution after adding 100 ml of 1.0 M HCl to the original buffer solution?

mmol of base = MV = 50*2 = 100 mmol of base

mmol of conjugate = MV = 50*2 = 100 mmol of conjugate

pKb = -log(Kb) = -log(4.4*10^-4) = 3.36

pOH = pKb + log(BH+/B)

pOH = 3.36 + log(BH+/B)

after adding:

mmol of H+ = 100*1 = 100 mmol of H+

mmol of base = MV = 50*2 = 100 mmol of base --> reacts with 100 mmol of acid

mmol of conjguate formed = 100 mmol

mmol of conjugate original = 100 mmol

total mmol of cojnujgate BH+ = 100+100 = 200 mmol

Vtotal = 50+50 +100 = 200 mL

[BH+] = 200/200 = 1 M

now

BH+ <--> B + H+

Ka = [B][H+]/[BH+]

Ka = Kw/KB = (10^-14)/(10^-3.36) = 2.29086*10^-11

2.29086*10^-11 = x*x/(1-x)

x = H+ = 4.78*10^-6

pH = -log(4.78*10^-6)

pH = 5.320