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VLD [36.1K]
2 years ago
9

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 f

ollowing 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?

Someone Help Me
Chemistry
2 answers:
LenKa [72]2 years ago
8 0

<u>Answer:</u> The concentration of ML_2^{2+} at equilibrium is 0.045 M.

<u>Explanation:</u>

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.

Mrac [35]2 years ago
7 0
The chemical reaction is
<span>                                 M2+(aq) + 2L(aq) <==> ML22+(aq)
</span>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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