The pH of ammonium iodide depends on the concentration of the solution. When dissolved in water, ammonium iodide undergoes hydrolysis, which is the chemical reaction between a salt and water. During hydrolysis, the ammonium ion (NH4+) reacts with water to form ammonium hydroxide (NH4OH), which is a weak base. This reaction can be represented as follows:

NH4I + H2O ⇌ NH4OH + HI

The equilibrium constant for this reaction is called the hydrolysis constant (Kb). The larger the Kb value, the stronger the base and the higher the pH of the solution.

The Kb value for ammonium hydroxide is 1.8 × 10⁻⁵ at 25 °C. This means that ammonium hydroxide is a relatively weak base, and the pH of ammonium iodide solutions will be slightly acidic.

The pH of a 0.1 M ammonium iodide solution can be calculated using the Kb expression for ammonium hydroxide:

Kb = [NH4OH][H3O+] / [NH4+]

[NH4OH] = Kb × [NH4+] / [H3O+]

Assuming complete dissociation of ammonium iodide, the concentration of ammonium ions ([NH4+]) is equal to the concentration of ammonium iodide (0.1 M).

[NH4OH] = 1.8 × 10⁻⁵ × 0.1 M / [H3O+]

[H3O+] = 0.1 M / [NH4OH]

[H3O+] = 0.1 M / (1.8 × 10⁻⁵ × 0.1 M)

[H3O+] = 5.56 × 10⁻⁴ M

pH = -log[H3O+]

pH = -log(5.56 × 10⁻⁴)

pH ≈ 3.25

Therefore, the pH of a 0.1 M ammonium iodide solution is approximately 3.25, indicating a slightly acidic solution.