Answer:
C₆H₅NH₂ + H₂O = C₆H₅NH₃⁺ + OH⁻
[OH⁻] = 1.1 × 10⁻⁵ M
pOH = 5.0
pH = 9.0
[H₃O⁺] = 1.0 × 10⁻⁹ M
Explanation:
Let's consider when aniline behaves as a Bronsted-Lowry base.
C₆H₅NH₂ + H₂O = C₆H₅NH₃⁺ + OH⁻
We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(3.9 × 10⁻¹⁰ × 0.30) = 1.1 × 10⁻⁵ M
where,
Kb: base dissociation constant
Cb: concentration of the base
The pOH is:
pOH = -log [OH⁻]
pOH = -log 1.1 × 10⁻⁵ = 5.0
The pH is:
pH + pOH = 14
pH = 14 - pOH = 14 - 5.0 = 9.0
The [H₃O⁺] is:
pH = -log [H₃O⁺]
[H₃O⁺] = antilog -pH = antilog -9.0 = 1.0 × 10⁻⁹ M
