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       

Answer 1

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