Calculate the pH at the equivalence point for the titration of 0.200 M 0.200 M methylamine ( CH 3 NH 2 ) (CH3NH2) with 0.200 M HCl . 0.200 M HCl. The K b Kb of methylamine is 5.0 × 10 − 4 . 5.0×10−4. pH =

pH = 5.85

Explanation:

The reaction of methylamine with HCl is:

CH₃NH₂ + HCl → CH₃NH₃⁺ + Cl⁻

As the titration was made with two solutions with the same molarity, the final concentration of the CH₃NH₃⁺ ion is 0.100M ([CH₃NH₃⁺] = 0.100M)

Then, the conjugate acid will be in equilibrium with water as follows:

CH₃NH₃⁺(aq) + H₂O(l) ⇄ CH₃NH₂(aq) + H₃O⁺(aq)

And the equilibrium constant, Ka, of this reaction is:

Kw/Kb = Ka =  1.0x10⁻¹⁴ / 5.0x10⁻⁴= 2x10⁻¹¹

2x10⁻¹¹ =  [CH₃NH₂] [H₃O⁺] / [CH₃NH₃⁺]

As both CH₃NH₂ and H₃O⁺ comes from the same equilibrium we can say:

[CH₃NH₂] = [H₃O⁺] = X

And as we know [CH₃NH₃⁺] = 0.100M:

2x10⁻¹¹ =  [X] [X] / [0.100M]

2x10⁻¹² = X²

X = 1.414x10⁻⁶M = [H₃O⁺]

And as pH= -log [H₃O⁺]

b.

the accuracy of the data.