Question

you have 250.0mL of a solution containing 0.150M trimethylamine,,(CH3)3Nand 0.125 M trimethylaminium nitrate(CH3)3NHNO3, DETERMINE the PH OF THIS solution.. What i will the PH of this solution be after the addition of 25.0 mL of 1.00M sodium hydroxide solution? The ionization constant for ammonia is 1.8*10^-5

Answer 1

no of moles of (CH3)3N   = molarity *volume in L

                                         = 0.15*0.25   = 0.0375 moles

no of moles of (CH3)3NHNO3 = molarity *volume in L

                                                    = 0.125*0.25   = 0.03125 moles

kb of ,(CH3)3N   = 6.5*10^-5

PKb   = -logKb

             = -log6.5*10^-5

                = 4.187

POH   = PKb + log[,(CH3)3NHNO3]/[,(CH3)3N]

           = 4.187 + log0.03125/0.0375

           = 4.187 -0.07918   = 4.1

PH   = 14-POH

           = 14-4.1   = 9.9

no of moles of NaOH   =molarity *volume in L

                                      = 1*0.025   = 0.025 moles

no of moles of (CH3)3N after addition of 0.025 moles of NaOH   = 0.0375+0.025   = 0.0625 moles

no of moles of (CH3)3NHNO3 after addition of 0.025 moles of NaOH   = 0.03125-0.025 = 0.00625mole

POH   = PKb + log[,(CH3)3NHNO3]/[,(CH3)3N]

           = 4.187 + log0.00625/0.0625

           = 4.187 -1 = 3.187

PH   = 14-POH

           = 14-3.187 = 10.813