Calculate the pH at the equivalence point for the titration

Calculate the pH at the equivalence point for the titration of 0.160 M methylamine (CH3NH2) with 0.160 M HCl. The Kb of methylamine is 5.0× 10–4.

Please explain when you do this. The answers aren\'t making sense to me as so many assumptions are hapenning.

Solution

CH₃NH₂ + HCl <----> CH₃NH₃⁺ + Cl^-

At the equivalence point , moles of HCl = moles of CH₃NH₂. In 1 L, there will be 0.160 moles of each, which will react to produce 1 mole of CH₃NH₃⁺, which is now in 2 L of solution, so you have 0.08 M CH₃NH₃⁺

The dissociation of CH₃NH₃⁺ is

CH₃NH₃⁺ <-------> CH₃NH2 + H⁺ Ka = 10^-14/Kb = 2.0*10^-11

Ka = [H⁺]*[CH₃NH2]/[CH₃NH₃⁺]

Let x = [H+]; then [CH3NH2] = x and [CH3NH3+] = 0.08 - x

since Ka is small, x will be small compared to 0.080 and we have

Ka = x²/0.080

x = 1.26*10^-6
[H+] = 1.26*10^-6

or, pH = 5.89