Answer: The pH of a 0.036 M solution of NH4Cl at 25°C is approximately 3.2.
Explanation:
The dissociation of NH4Cl in water can be represented as follows:
NH4Cl(s) → NH4+(aq) + Cl-(aq)
The NH4+ ion can act as a weak acid by donating a proton (H+) to water to form the hydronium ion (H3O+):
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
The equilibrium constant expression for this reaction is:
Ka = [NH3][H3O+] / [NH4+]
Since NH3 is a weak base, we can assume that its concentration is negligible compared to that of NH4+. Therefore, we can simplify the expression to:
Ka ≈ [H3O+]^2 / [NH4+]
Substituting the values given in the problem, we get:
1.8 × 10^-5 = [H3O+]^2 / 0.036
Solving for [H3O+], we get:
[H3O+] = √(1.8 × 10^-5 × 0.036) ≈ 6.1 × 10^-4 M
The pH of the solution can be calculated as:
pH = -log[H3O+] ≈ 3.2