Question

A buffer solution is prepared by dissolving 0.250 mol of solid methylamine hydrochloride (CH₃NH₃Cl) in 1.00 L of 1.10 M methylamine (CH₃NH₂). The Kb for methylamine is 4.40x10^-4. (Assume the final volume of the solution is 1.00 L).

A) Write a balanced molecular, total ionic, and net ionic equation for the completion reaction that occurs when NaOH is added to this buffer.

B) What is the pH of the solution when 100.0 mL of 1.15 M NaOH is added? (show work)

Answer 1

no of moles of CH₃NH₃Cl   = 0.25moles

no of moles of CH₃NH₂       = molarity * volume in L

                                             = 1.1*1   = 1.1moles

Pkb   = -logKb

         = -log4.4*10^-4    = 3.3565

POH = PKb + log[CH3NH3Cl]/[CH3NH2]

           = 3.3565 + log0.25/1.1

          = 3.3565-0.6434   = 2.7131

PH   = 14-POH

          = 14-2.7131   = 11.2869

A.    CH3NH3Br --------------> CH3NH3^+ (aq) + Br^- (aq)

       NaOH(aq) --------------> Na^+ (aq) + OH^- (aq)

CH3NH3Br(aq) + NaOH(aq) ------------------> CH3NH3OH(aq) + NaBr(aq)

CH3NH3^+(aq) +Br^-(aq) + Na^+(aq) +OH^-(aq) ------------------> CH3NH3OH(aq) + Na^+(aq)+Br^-(aq)

removal of spectator ions to get net ionic equation

        CH3NH3^+(aq) + OH^-(aq) ---------> CH3NH3OH(aq)

B. no of moles of NaOH = molarity * volume in L

                                        = 1.15*0.1 = 0.115 moles

no of moles of CH3NH3Br after addition of NaOH = 0.25-0.115   = 0.135 moles

no of moles of CH3NH2 after addition of NaOH       = 1.1+0.115   = 1.215moles

    POH = PKb + log[CH3NH3Br]/[CH3NH2]

              = 3.3565 + log0.135/1.21

            = 3.3565-0.9525   = 2.404

PH   = 14-POH

           = 14-2.404 = 11.596