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: 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

Let x be the equilibrium concentration of MS1U1B12+S1U1P0 produced. Then, using the stoichiometry of the balanced equation, the equilibrium concentration of L is (0.100 - x) mol L¯S1U1P11S1S1P0.

Using the equilibrium constant expression:

Kc = [MS1U1B12+S1U1P0][L] 
     ____________________ 
        [ML][H+]

we can plug in the given values and solve for x:

3.6 x 10¯S12U1PPS1S1S1S1S1S1S1S1S1S1S1S1S1S1S10¯S11 = (x)(0.0100) 
                             ____________________ 
                                 (0.100 - x)(0.100)

Simplifying,

3.6 x 10¯S12U1PPS1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S10¯S11 = 0.1x 
                                                        _________ 
                                                  1 - 0.01x

x = 0.0200 mol L¯S1U1P11S1S1P0

Therefore, the equilibrium concentration of MS1U1B12+S1U1P0 is 0.0200 mol L¯S1U1P11S1S1P0, which corresponds to choice (D).

100 ML of a 0