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. pH=

Answer 1

At the equivalence point, the moles of HCl = moles CH₃NH₂ and all HCl and CH₃NH₂ has been converted to CH₃NH₃⁺. [CH₃NH₃⁺] will be 0.120 M since you will use equal volumes of 0.240 M HCl and CH₃NH₂. So, you need to look at the hydrolysis of CH₃NH₃⁺

CH₃NH₃⁺ + H₂O ==> H₃O⁺ + CH₃NH₂ where CH₃NH₃⁺ is acting as an acid. So, we need the Ka.

KaKb = 1x10^-14

Ka = 1x10^-14 / 5.0x10^-4 = 2x10^-11

Ka = 2x10^-11 = [H₃O⁺][CH₃NH₂] / [CH₃NH₃⁺]

2x10^-11 = (x)(x) / 0.120

x² = 2.4x10^-12

x = 1.55x10^-6 M = [H₃O⁺]

pH = -log [H₃O⁺] = -log 1.55x10^-6

pH = 5.81