Calculate the pH at the equivalence point for the titration of 0.200 M methylamine (CH3NH2) with 0.200 M HCl. The 𝐾b of methylamine is 5.0×10−4.

Answer 1

Write the correctly balanced equation for the reaction:

CH₃NH₂ + HCl ==> CH₃NH₃⁺ + Cl^- .. balanced equation

At equivalence point, mols CH₃NH₂ equals mols of HCl and all CH₃NH₂ and HCl is converted to CH₃NH₃⁺Cl^-

In the current problem, the [CH3NH3+] will be 0.200 M as shown below;

CH₃NH₂ + HCl ==> CH₃NH₃⁺

0.200 M....0.200M........0...........Initial

-0.200.....-0.200.........+0.200....Change

0.................0...............0.200.....Equilibrium

To find the pH, we must now consider the hydrolysis of the 0.200 M CH₃NH₃⁺

CH₃NH₃⁺ + H2O ==> CH₃NH₂ + H₃O⁺ where CH₃NH₃⁺ is acting as an acid and H₂O as a base.

Ka = [CH₃NH₂][H₃O⁺] / [CH₃NH₃⁺] and Ka = 1x10^-14 / 5.0x10^-4 = 2.0x10^-11

2.0x10^-11 = (x)(x) / 0.200 - x (assume x is small relative to 0.200 and ignore it in denominator)

2.0x10^-11 = x² / 0.200

x² = 4.0x10^-12

x = 2.0x10^-6 (this value is very small relative to 0.200 so above assumption was valid)

[H₃O⁺] = 2.0x10^-6

pH = -log [H₃O⁺] = -log 2.0x10^-6

pH = 5.70