I am stumped on this chemistry

Question

Methylamine, CH3NH2, is a weak base that is the building block of many commercially used organic substances.

1.Write the equation for the dissociation of methylamine in water. Label the conjugate acid of methylamine.

2.Determine the pH of a solution of methylamine with a concentration of 0.250 M. Methylamine has a known Kb of 4.17 x 10-14.

3.What is the value of the acid dissociation constant, Ka, for the conjugate acid of methylamine? Which is stronger, methylamine or its conjugate acid? Why?

4.Methanaminium (CH3NH3Cl) is an acidic salt made from methylamine. What concentration of methanaminium is needed to produce a solution with a pH of 2.00?

Answer 1

1). CH3NH2 + H2O ==> CH3NH3⁺ + OH^- (OH^- is the conjugate base)

2). Kb = [CH3NH3⁺][OH^-] / [CH3NH2] = (x)(x) / 0.250-x and assuming x is small we can ignore it in 0.25-x

4.17x10^-4 = x²/0.250 (NOTE: the Kb given in the problem is incorrect as 4.17x10^-14. That can't be)

x² = 1.04x10^-4

x = 1.02x10^{-2 =} [OH-] and it is small compared to 0.25 (less than 5%) so assumption was valid

pOH = -log [OH-] = -log 1.02x10^-2 = 1.99

pH = 14 - pOH = 12.0

3). KaKb = Kw

4.17x10^-4 Ka = 1x10^-14 and Kb = 2.4x10^-9

Methylamine is a stronger base than the conjugate acid is as an acid because of the values of Ka and Kb

4). Look at hydrolysis of CH3NH3Cl and find pH

CH3NH3⁺ + H2O ==> H3O⁺ + CH3NH2

Use Ka for CH3NH3⁺ from #3 above and solve for [CH3NH3⁺] using 1x10^-2 for [H3O⁺] and [CH3NH2]