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100 ML of a 0

Question 57

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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) 10.0 mL of a 0.100 mol L<sup>-1</sup> solution of a metal ion M<sup>2+</sup> is mixed with 10.0 mL of a 0.100 mol L<sup>-1</sup> solution of a substance L. The following equilibrium is established: M<sup>2+</sup>(aq)  + 2L(aq)    ML<sub>2</sub><sup>2+</sup>(aq)  At equilibrium the concentration of L is found to be 0.0100 mol L<sup>-1</sup>. What is the equilibrium concentration of ML<sub>2</sub><sup>2+</sup>, in mol L<sup>-1</sup>? A)  0.100 mol L<sup>-1</sup> B)  0.050 mol L<sup>-1</sup> C)  0.025 mol L<sup>-1</sup> D)  0.0200 mol L<sup>-1</sup> E)  0.0100 mol L<sup>-1</sup> 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?


A) 0.100 mol L-1
B) 0.050 mol L-1
C) 0.025 mol L-1
D) 0.0200 mol L-1
E) 0.0100 mol L-1

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