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) 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? multiple choice 0.100 mol l–1 0.050 mol l–1 0.025 mol l–1 0.0200 mol l–1 0.0100 mol l–1

0.0200 mol L⁻¹

Explanation:

M²⁺(aq) + 2L(aq) → ML₂²⁺(aq)

keq = [ML₂²⁺] / [M²⁺]

Because the mixture volume is (10+10) 20 mL, the concentrations of the species at the beginning of the reaction are:

M²⁺ ⇒ 0.100 M * 10mL/20mL = 0.05 M M²⁺

L ⇒ 0.100 M * 10mL/20mL = 0.05 M L

ML₂²⁺ ⇒ 0.100 M * 10mL/20mL = 0.05 M ML₂²⁺

If at equilibrium the concentration of L is 0.01 M, it means that the moles of L that reacted are:

0.05 M * 0.02 L - 0.01 M * 0.02 L = 8x10⁻⁴ mol L

Now we convert the moles of L into moles of ML₂²⁺

8x10⁻⁴ mol L * (1mol ML₂²⁺ / 2mol L) = 4x10⁻⁴ mol ML₂²⁺

Finally we divide the moles by the volume in order to calculate the concentration:

4x10⁻⁴ mol ML₂²⁺ / 0.020 L = 0.02 M = 0.02 mol L⁻¹

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