To determine the pH of a 0.1 M solution of anilinium chloride (C₆H₅NH₃⁺), we need to consider its acidic nature due to the presence of the anilinium ion. This ion is the conjugate acid of aniline, and its dissociation in water is described by the equation: C₆H₅NH₃⁺ ⇌ C₆H₅NH₂ + H⁺. The equilibrium constant for this dissociation is the acid dissociation constant (Ka), given as 10⁻⁶.
We start by writing the expression for Ka: Ka = [C₆H₅NH₂][H⁺]/[C₆H₅NH₃⁺]. Assuming the initial concentration of C₆H₅NH₃⁺ is 0.1 M and the change in concentration at equilibrium is 'x', the expression becomes 10⁻⁶ = (x)(x)/(0.1-x). For small x, 0.1-x can be approximated as 0.1, simplifying the equation: 10⁻⁶ = x²/0.1.
Solve for x: x² = (10⁻⁶)(0.1) = 10⁻⁷, thus x = √(10⁻⁷) = 10⁻³.⁵.
The concentration of H⁺ ions [H⁺] = x = 10⁻³.⁵ M. Therefore, the pH = -log₁₀(10⁻³.⁵) = 3.5.
Verify the solution range: The calculated pH value of 3.5 is within the provided range of 3.5 to 3.5, confirming correctness.
