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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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

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RomeliaThurston RomeliaThurston · Answer 2 · 2019-08-20T01:53:59-04:00

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.