Answer:
pH = 3.70
Explanation:
Moles of aniline in solution are:
0.0500L × (0.0350mol / 1L) = 1.750x10⁻³ mol aniline
Aniline is in equilibrium with water, thus:
C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰
HCl reacts with aniline thus:
HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻
At equivalence point, all aniline reacts producing C₆H₅NH₃⁺. C₆H₅NH₃⁺ has its own equilibrium with water thus:
C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)
Where Ka is defined as:
Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = 2.63x10⁻⁵ (1)
As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:
[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X
[C₆H₅NH₂] = X
[H₃O⁺] = X
Replacing in (1):
2.63x10⁻⁵ = [X] [X] / [1.750x10⁻³ mol - X]
4.6x10⁻⁸ - 2.63x10⁻⁵X = X²
0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸
Solving for X:
X = -2.3x10⁻⁴ → False answer, there is no negative concentrations
X = 2.0x10⁻⁴ → Right answer
As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.
Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:
pH = 3.70
