Final answer:
The equilibrium concentration of NH₄⁺ in a 10 M NH₃ solution at 25°C is closest to 0.001 M (Option A), calculated using the base ionization constant (Kb) and the equilibrium expression assuming x is the concentration of NH₄⁺. So the correct answer is option (A).
Step-by-step explanation:
Ammonia (NH₃) acts as a weak base in water with a base ionization constant (Kb) of 1.76 x 10⁻⁵ at 25°C, which indicates that it only partially dissociates in aqueous solution. The equilibrium in question is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
To find the equilibrium concentration of the NH₄⁺ ion in a 10 M aqueous solution of NH₃, we assume initially that 'x' represents the concentration of dissociated NH₃. Therefore, the equilibrium concentrations of NH₄⁺ and OH⁻ will also be 'x', as they are produced in a 1:1 ratio upon dissociation. The equilibrium expression for the reaction is:
Kb = [NH₄⁺][OH⁻]/[NH₃] = (x)(x)/ (10 - x) ≈ x² / 10
Because NH₃ is a weak base and does not dissociate greatly, 'x' will be much smaller than 10, and therefore, we can approximate [NH₃] to 10 M. Solving for 'x' using the given Kb value yields:
1.76 x 10⁻⁵ = x² / 10
x = √(1.76 x 10⁻⁵ × 10) ≈ 0.0042 M
This concentration is closest to 0.001 M, which makes option A correct.