Question

What is the pH of a 0.29 M solution of methylammonium chloride, CH₃NH₃Cl?
What is the concentration of methylamine in the solution?

Answer 1

CH3NH3Cl undergoes dissociation to form CH3NH2 and H+(aq) and the balanced chemical reaction is

------------ CH3NH3Cl(aq) -------> CH3NH2 + H+(aq) + Cl-(aq); Ka = 2.4x10^-11

init.conc: 0.29 M ------------------- 0 M, ------- 0 M

change: - 0.29y M, --------------- + 0.29 y M, + 0.29 y M

eqm.conc:0.29(1 - y)M, ---------- 0.29 y M, 0.29 y M

Ka = 2.4x10^-11 = [CH3NH2]x[H+(aq)] / [CH3NH3Cl(aq)]

=>  2.4x10^-11 = (0.29y x 0.2y) / 0.29(1 - y)

y <<1, (1 - y) is nearly equals to 1.

=> y = underroot(2.4x10^-11 / 0.29) = 9.10x10^-6 M

Hence [H+(aq)] = 0.29y = 0.29 x 9.10x10^-6 M = 2.64x10^-6 M

=> pH = - log[H+(aq)] = - log(2.64x10^-6 M) = 5.58 (answer)