10.0 mL of a 0.100 mol/L solution of a metal ion M²⁺ is mixed with 10.0 mL of a 0.100 mol/L solution of a ligand L. A reaction occurs in which the product is ML₃. Approximately, what is the maximum concentration of ML₃²⁺, in mol/L, which could result from this reaction? A)0.100 B)0.050 C)0.033 D)0.025 E)0.017

Question

Quizplus · Accepted Answer

The reaction requires 3 moles of L for every mole of M²⁺. Since both solutions are mixed in equal volumes, the concentration of each is halved. Thus, M²⁺ is 0.050 mol/L and L is 0.050 mol/L. L is the limiting reagent, and only 1/3 of its concentration can form ML₃²⁺, resulting in 0.017 mol/L.

100 ML of a 0