Consider a solution prepared by dissolving 0.35 mol of CH 3 NH 3 + Cl - (methyla
ID: 493477 • Letter: C
Question
Consider a solution prepared by dissolving 0.35 mol of CH3NH3+Cl- (methylamine hydrochloride) in 1.00 L of 1.1 M CH3NH2 (methylamine). If 10 mL of 0.10 M HCl is added to this buffer solution, the pH of the solution will ________ slightly because the HCl reacts with the _________ present in the solution to produce a _______.
A. increase; CH3NH2 weak acid
B. increase; CH3NH3+Cl-; weak base
C. decrease, CH3NH2; weak acid
D. decrease; CH3NH3+Cl-; weak base
E. increase; CH3NH2; weak base
Explanation / Answer
moles of CH3NH3+Cl- = 0.35
moles of CH3NH2 = 1 x 1.1 = 1.1
pOH = pKb + log [salt / base]
= 3.30 + log [0.35 / 1.1]
= 2.803
pH = 11.197
if moles of HCl added = 10 x 0.1 / 1000 = 0.001
pOH = pKb + log [salt + C / base - C]
= 3.30 + log [0.35 + 0.001 / 1.1 - 0.001]
= 2.804
pH = 11.196
so pH decrease
CH3NH2 + HCl --------------> CH3NH3+
so
option C) is correct
C. decrease, CH3NH2; weak acid