Question

CH3NH2 is a weak base (Kb = 5.0 × 10–4) and so the salt, CH3NH3NO3, acts as a weak acid. What is the pH of a solution that is 0.0490 M in CH3NH3NO3 at 25 °C?

Answer 1

                           CH₃NH₃⁺ + H₂O         H₃O⁺ + CH₃NH₂

Initial                   0.049 M                       0              0

Change                   -x                              +x             +x

Equilibrium          (0.049-x)                    x               x

K_a = K_w/K_b = (10^-14)/(5 x 10^-4) = x²/(0.049-x)

2 x 10^-11 = x²/(0.049)

x = 9.89 x 10^-7 M

pH = -log(9.89 x 10^-7 M) = 6.00

pH = 6