Write a balanced equation that shows the products formed when aniline dissolves in water (use the Bronsted-Lowry definition to show proton transfer from water to aniline). Calculate the pH, pOH, [H3O+], and [OH-] of a 0.30 M solution of aniline. The Kb value for aniline is 3.9 x 10-10.

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

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.