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

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)

10.0 mL of a 0.100 mol L<sup>-1</sup> solution of a metal ion M<sup>2+</sup> is mixed with 10.0 mL of a 0.100 mol L<sup>-1</sup> solution of a substance L. The following equilibrium is established: M<sup>2+</sup>(aq) + 2L(aq) ML<sub>2</sub><sup>2+</sup>(aq) At equilibrium the concentration of L is found to be 0.0100 mol L<sup>-1</sup>. What is the equilibrium concentration of ML<sub>2</sub><sup>2+</sup>, in mol L<sup>-1</sup>? A) 0.100 mol L<sup>-1</sup> B) 0.050 mol L<sup>-1</sup> C) 0.025 mol L<sup>-1</sup> D) 0.0200 mol L<sup>-1</sup> E) 0.0100 mol L<sup>-1</sup>

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?

Quizplus · Accepted Answer

Since the solution contains a 1:2 stoichiometric ratio of M²⁺ to L, we can use the initial concentration of L (0.100 mol L^-1) to find the initial concentration of M²⁺ (0.050 mol L^-1). At equilibrium, all of the M²⁺ will react with L to form ML₂²⁺, so the equilibrium concentration of M²⁺ will be zero. Therefore, the equilibrium concentration of ML₂²⁺ will also be 0.050 mol L^-1, which is equal to the initial concentration of M²⁺. Finally, we use the volume of the solution (20.0 mL) to calculate the equilibrium concentration of ML₂²⁺, which is 0.0200 mol L^-1.

100 ML of a 0