Question

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Answer 1

CH₃NH₃Br <====> CH₃NH₃⁺ + Br^-

CH₃NH₃⁺ + H₂O <===> CH₃NH₂ + H₃O⁺

ICE table:

                        CH₃NH₃⁺              CH₃NH₂                   H₃O⁺

initial                  0.12                       0                            0

change    -x +x    +x

equilibrium (0.12 - x) x                            x

pK_b = -logk_b = 3.35

pK_a = 14 - 3.35 = 10.65

K_a = antilog(-pK_a) = 2.23 x 10^-11

K_a = x²/0.12     (0.12 >> x)

2.23 x 10^-11 x 0.12 = x²

x = 1.63 x 10^-6 = [H⁺]

pH = -log[H⁺]

pH = 5.78