Final answer:
The equilibrium concentration of ML₂²⁺ is calculated by finding the change in concentration of L (0.090 M), determining the corresponding change of M²⁺ (0.045 M), and hence, the concentration of ML₂²⁺ formed (0.045 M).
Step-by-step explanation:
To calculate the equilibrium concentration of ML₂²⁺, we start by considering the initial concentrations of both M²⁺ and L, which are 0.100 M each, due to the equal volumes mixed. At equilibrium, we know that the concentration of L is 0.0100 M. Since the reaction consumes 2 moles of L for every mole of M²⁺ that reacts, we can determine the change in concentration for L and use that to find the final concentration of ML₂²⁺.
The change in concentration of L at equilibrium will be equal to 0.100 M - 0.0100 M = 0.090 M. Given the stoichiometry of 2:1, this means that half the change in L will be the change in M²⁺, or 0.090 M / 2 = 0.045 M of M²⁺ reacted to form ML₂²⁺. The final concentration of ML₂²⁺ is therefore 0.045 M, because for every 1 mole of M²⁺, 1 mole of ML₂²⁺ forms.