Question

Calculate the pH at the equivalence point for the titration of 0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of methylamine is 5.0× 10–4.

Answer 1

Since the concentrations of CH₃NH₂ and HCl are equal (both 0.240M), you need equal volumes of acid and base to react at the equivalence point. I'll assume 1 L for both of them, so the total volume would be 2 L.

Therefore the final concentrations would be 0.240 * (1/2) = 0.120 M for both reactants.

Now at the equivalence point, almost all the base has reacted with the HCl, and we have the following products:

CH₃NH₃⁺ and Cl⁻ and concentrations are already calculated above. so, we need to know the Ka:

Ka = 1x10^-14 / 5x10^-4 = 2x10^-11

Now the overall reaction and a ICE chart:

r: CH₃NH₃⁺ <--------> CH₃NH₂ + H⁺

i: 0.120 0 0

e: 0.120-x x x

2x10^-11 = x² / 0.120-x Kb is small so 0.120-x can be neglected and stay as 0.120

2x10^-11 * 0.120 = x²

x = 1.5491x10^-6 M = [H⁺]

pH = -log(1.5491x10^-6)

pH = 5.81

Hope this helps