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) Picture 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?

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Answer 1

The chemical reaction is
                                 M2+(aq) + 2L(aq) <==> ML22+(aq)
Intial concentration   0.10           0.10        
Change                       -x              -2x               +x
Equilibrium                0.10 - x    0.01 = 0.10 - 2x  x
Solving for x
0.01 = 0.10 - 2x
x = 0.045
The equilibrium concentration of ML22+ is 0.045 mol L-1

Answer 2

Answer: The concentration of $ML_2^{2+}$ at equilibrium is 0.045 M.

Explanation:

We are given:

$[M^{2+}]_{initial}=0.100M$

$[L]_{initial}=0.100M$

For the given chemical equation:

                        $M^{2+}(aq.)+2L(aq.)\rightleftharpoons ML_2^{2+}(aq.)$

Initially:             0.100M        0.100M

At eqllm:          0.100 - x      0.100 - 2x          x

We are also given:

$[L]_{eqllm}=0.0100M$

Equating the two values:

$0.0100=0.100-2x\\\\x=0.045M$

Hence, the concentration of $ML_2^{2+}$ at equilibrium is 0.045 M.