Question

Enter your answer in the provided box. Calculate the pH at the equivalence point for the following titration: 0.40 M HCl versus 0.40 M methylamine (CH3NH2). The Ka of methylammonium is 2.3 × 10−11

Answer 1

CH₃NH₂ + HCl ---> CH₃NH₃⁺ + Cl^-

Suppose 1 L of CH₃NH₂

since the concentrations of HCl and CH₃NH₂ are the same we need 1 L of HCl to reach the equivalence point
Total volume = 2
At the equilvalence point 0.40 mol CH₃NH₂ + 0.40 mole H+ ----> 0.40mole CH₃NH₃⁺

[CH₃NH₃⁺] = 0.40 / 2 L = 0.20 M

CH₃NH₃⁺ + H₂O <-----> CH₃NH₂ + H₃O⁺

Ka = 2.3 x 10^-11

Ka = [H₃O⁺][CH₃NH₂] / [CH₃NH₃⁺]

at equilibrium [CH₃NH₂] = [H₃O⁺] = x, [CH₃NH₃⁺] = 0.20 - x

Therefore
2.3 x 10^-11 = x² / 0.20 - x

since 0.20 >>> x

2.3 x 10^-11 = x² / 0.20

x = [H+]= 2.14 x 10^-6 M

pH = - log (2.14 x 10^-6)

pH = 5.66