10.0 mL of a 0.100 mol L⁻¹ solution of a metal ion M²⁺ is mixed with 10.0 mL of a 0.100 mol L⁻¹ solution of a substance L. The following equilibrium is established: M²⁺(aq) + 2L(aq) ⇄ ML₂²⁺(aq)
At equilibrium the concentration of L is found to be 0.0100 mol L⁻¹. What is the equilibrium concentration of ML₂²⁺, in mol L⁻¹?

Question

Quizplus · Accepted Answer

Using the equilibrium equation: M²⁺(aq) + 2L(aq) ⇄ ML₂²⁺(aq) The initial concentration of M²⁺ and L is 0.100 mol L⁻¹, and the concentration of L at equilibrium is 0.0100 mol L⁻¹. Let x be the equilibrium concentration of ML₂²⁺ in mol L⁻¹. Then from the balanced equation, we know that at equilibrium: [ML₂²⁺] = x mol L⁻¹ [L] = (0.100 - x) mol L⁻¹ [M²⁺] = (0.100 - x) mol L⁻¹ Substitute into the equilibrium equation and solve for x: (0.100 - x)(0.100 - x)/(x) = 1/2 x = 0.0200 mol L⁻¹ Therefore, the equilibrium concentration of ML₂²⁺ is 0.0200 mol L⁻¹.

100 ML of a 0