Question

What is the pH of a 0.29 *M* solution of methylammonium
chloride, CH_{3}NH_{3}Cl?

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)

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