The following buffer was prepared: 50. mL 2.0 M methylamine (CH 3 NH 2 ) solutio
ID: 493832 • Letter: T
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.
A. What is the pH of the resulting solution after adding 200 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
B.What is the pH of the resulting solution after adding 50 mL of 1.0 M KOH to the original buffer solution?
a)11.12 b) 10.64 c) 12.32 d) 0.48 e) 5.32
C. What is the pH of the resulting solution after adding 100 mL of 1.0 M KOH to the original buffer solution?
a) 11.12 b) 10.64 c) 12.32 d) 0.48 e) 5.32
Explanation / Answer
A) option d) 0.48
millimoles of CH3NH2 = 50 x 2 = 100
millimoles of CH3NH3Cl = 50 x 2 = 100
millimoles of HCl = 200 x 1 = 200
millimoles of HCl > base
strong acid reamins = 200 - 100 / (200 + 50 + 50)
= 0.33 M
pH = -log [H+] = -log 0.33
pH = 0.48
B) option a) 11.12
millimoles of KOH = 50 x 0.1 = 50
pOH = pKb + log [100 - 50 / 100 + 50]
= 3.36 + log [50 / 150]
= 2.88
pH = 11.12
C.) option C)
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