Calculate the molar concentration of uncomplexed Zn²⁺ in a solution that contains 0.20 mol of Zn(NH₃)₄²⁺ per liter. The overall K_f for Zn(NH₃)₄²⁺ is 3.8 × 10⁹. A) 6.7 × 10^-4 M B) 2.9 × 10^-3 M C) 2.0 × 10^-13 M D) 8.8 × 10^-3 M E) none of these

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To solve this problem, we need to use the equilibrium constant expression: K_f = [Zn²⁺] / [Zn(NH₃)₄²⁺] We know the value of K_f and we are given the concentration of Zn(NH₃)₄²⁺, so our task is to find [Zn²⁺]. To do this, we can use the fact that the total concentration of zinc in the solution will be: [zinc]total = [Zn²⁺] + [Zn(NH₃)₄²⁺] We know that [Zn(NH₃)₄²⁺] = 0.20 mol/L, so we just need to solve for [Zn²⁺] in the equation above. Rearranging, we get: [Zn²⁺] = [zinc]total - [Zn(NH₃)₄²⁺] [Zn²⁺] = (0.20 mol/L) - [Zn(NH₃)₄²⁺] Now we can substitute this into our equilibrium constant expression and solve for [zinc]total: 3.8 × 10⁹ = [(0.20 mol/L) - [Zn(NH₃)₄²⁺]] / [Zn(NH₃)₄²⁺] Simplifying: [Zn(NH₃)₄²⁺] = 0.20 mol/L [Zn²⁺] = 0.152 mol/L [zinc]total = 0.352 mol/L Finally, we can calculate the molar concentration of uncomplexed Zn²⁺: [Zn²⁺] = 0.152 mol/L Total volume of solution = 1 L Molarity of Zn²⁺ = [Zn²⁺] / total volume = 0.152 mol/L Therefore, the answer is B) 2.9 × 10^-3 M.

Calculate the Molar Concentration of Uncomplexed Zn2+ in a Solution

Calculate the Molar Concentration of Uncomplexed Zn²⁺ in a Solution