Question

C Give Up? ssignment Score: Resources 267/300 Feedback Resume O Attempt 32 uestion 3 of 3 The reaction of the strong acid HCl with the weak base CH,NH, is HCl(aq)CHNH, (aq) » CH,NH3 (aq) + CI (aq) To compute the pH of the resulting solution if 28 mL of 0.41 M acid is mixed with 66 mL of 0.75 M base we need to start with the stoichiometry. Let's do the stoich in steps: How many moles of base? number of moles: 0.0495 How many moles of acid? number of moles:0.01148 What is the limiting reactant? limiting reactant: HCl

Answer 1

no of moles of HCl   = molarity * volume in L

                                 = 0.41*0.028   = 0.01148moles

no of moles of CH3NH2 = molarity * volume in L

                                         = 0.75*0.066   = 0.0495moles

HCl(aq) + CH3NH2(aq) ------------------> CH3NH3^+ (aq) + Cl^- (aq)

1 moles of CH3NH2 react with 1 mole of HCl

0.0495 moles of CH3Nh3 react with 0.0495moles of HCl is required

HCl is limiting reactant

PKb of CH3NH2   = 3.38

               HCl(aq) + CH3NH2(aq) ------------------> CH3NH3^+ (aq) + Cl^- (aq)

I              0.01148      0.0495                                      0

C             -0.01148    -0.01148                                 0.01148

E                0                0.03802                               0.01148

             POH   = Pkb + log[CH3NH3^+]/[CH3NH2]

                         = 3.38 + log(0.01148/0.03802)

                         = 3.38 -0.52

                         = 2.86

           PH   = 14-POH

                   = 14-2.86

                     = 11.14 >>>>answer