Question

Calculate the pH at the equivalence point in the titration of 35.0 mL of 0.125 M methylamine (Kb = 4.4 × 10−4) with 0.250 M HCl.

Answer 1

Reaction of methyl amine with HCl (H+) is:

CH3NH2 + H⁺ ↔ CH3NH3⁺

At equivalence point:

moles of CH3NH2 = moles of HCl

moles of CH3NH2 = 0.350 * 0.125 = 0.0438 moles = moles of HCl

Volume of HCl = 0.0438 / 0.250 = 0.1752 L

Therefore, total volume at eq point = 0.350 + 0.1752 = 0.5252 L

Now, at equivalence point there is 0.0438 moles of CH3NH3+. The new equilibrium is :-

CH3NH3+ ↔ CH3NH2 + H+

where the concentration of CH3NH3+ = 0.0438/0.5252 = 0.083 M

Ka = [CH3NH2][H+]/[CH3NH3+]

Kw/Kb = x²/0.083-x

10⁻¹⁴/4.4*10⁻⁴ = x²/0.083-x

therefore x = [H+] = 1.37*10⁻⁶M

pH = -log[H+] = -log[1.37*10⁻⁶] = 5.86