Question

A) Calculate the pH of a 0.0116 M aqueous solution of methylamine (CH₃NH₂, Kb = 4.2×10^-4) and the equilibrium concentrations of the weak base and its conjugate acid.

pH	= ?
[CH3NH2]equilibrium	= ?	M
[CH3NH3+ ]equilibrium	= ?	M

B)Calculate the pH of a 0.0115 M aqueous solution of nitrous acid (HNO₂, Ka= 4.6×10^-4) and the equilibrium concentrations of the weak acid and its conjugate base.

pH	= ?
[HNO2]equilibrium	= ?	M
[NO2- ]equilibrium	= ?	M

C) Calculate the pH of a 0.0251 M aqueous solution of caffeine (C₈H₁₀N₄O₂, Kb= 4.1×10^-4).

Answer 1

                  CH3-NH2(aq) + H2O (l) --------------> CH3-NH3^+ (aq) + OH^- (aq)

I                   0.0116                                                   0                           0

C                   -x                                                          +x                        +x

E                0.0116-x                                                   +x                         +x

                           Kb    = [CH3-NH3^+][OH^-]/[CH3-NH2]

                          4.2*10^-4   = x*x/(0.0116-x)

                          4.2*10^-4*(0.0116-x) = x^2

                               x   = 0.002

                      [CH3-NH2]   = 0.0116-x = 0.0116-0.002   = 0.0096M

                       [CH3-NH3^+]   = x        = 0.002M

                       [OH^-]               = x         = 0.002M

                          POH   = -log[OH^-]

                                       = -log0.002

                                        = 2.6989

                            PH    = 14-POH

                                     = 14-2.6989   = 11.3

B.

                HNO2(aq) +H2O---------> H3O^+ (aq)   + NO2^- (aq)

I              0.0115                               0                    0

C             -x                                     +x                  +x

E            0.0115-x                               +x                   +x

                    Ka     =   [H^+][NO2^-]/[HNO2]

                      4.6*10^-4        = x*x/(0.0115-x)

                     4.6*10^-4 *(0.0115-x)   = x^2

                       x   = 0.002

              [HNO2]   = 0.0115-x   = 0.0115-0.002   = 0.0095M

              [H3O^+]         = x              = 0.002M

             [NO2^-]         =x            = 0.002M

              PH   = -log[H3O^]

                         = -log0.002   = 2.6989

C.  

                  C₈H₁₀N₄O₂ (aq) +H2O(l) -----------> C₈H₁₀N₄O₂H^+(aq) +OH^-(aq)

I                      0.0251                                                0                              0

C                      -x                                                      +x                            +x

E                  0.0251-x                                                +x                            +x

                                Kb     =    [C₈H₁₀N₄O₂H^+][OH^-]/[C₈H₁₀N₄O₂]

                                  4.1*10^-4   = x*x/(0.0251-x)

                                  4.1*10^-4 *(0.0251-x) = x^2

                                x   = 0.003

                      [C₈H₁₀N₄O₂]    = 0.0251-x = 0.0251- 0.003    = 0.0221M

                      [[C₈H₁₀N₄O₂H^+]   = x    = 0.003M

                    [OH^-]                       = x     = 0.003M

                            POH   = -log[OH^-]

                                       = -log0.003

                                        = 2.5228

                            PH    = 14-POH

                                     = 14-2.5228   = 11.4772 >>>>answer