The number of moles of NH , that must be added to 2 L of 0.80 M AgNO in order to reduce the concentr — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Calculate initial moles of Ag⁺** Moles of AgNO₃ = Molarity × Volume = 0.80 M × 2 L = 1.6 mol Ag⁺ **Step 2: Set up the complex formation equilibrium** Ag⁺ + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ Formation constant: K_f = [Ag(NH₃)₂]⁺/([Ag⁺][NH₃]²) = 1.0 × 10⁸ **Step 3: Use the equilibrium expression with final [Ag⁺]** Given: [Ag⁺]_final = 5.0 × 10⁻⁸ M 1.0 × 10⁸ = [Ag(NH₃)₂]⁺/((5.0 × 10⁻⁸)[NH₃]²) **Step 4: Determine [Ag(NH₃)₂]⁺ formed** By mass balance: [Ag(NH₃)₂]⁺ = initial [Ag⁺] − final [Ag⁺] [Ag(NH₃)₂]⁺ = (1.6 mol/2 L) − (5.0 × 10⁻⁸ M) ≈ 0.80 M **Step 5: Solve for [NH₃]** 1.0 × 10⁸ = 0.80/((5.0 × 10⁻⁸)[NH₃]²) [NH₃]² = 0.80/(1.0 × 10⁸ × 5.0 × 10⁻⁸) = 0.80/5 = 0.16 [NH₃] = 0.40 M **Step 6: Calculate moles of NH₃ needed** From stoichiometry