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

The freezing point depression can be calculated using the pure solvent freezing point and the molality of the solution.

At the freezing point, the vapor pressure of both the solid and liquid form of a compound must be equal.

The freezing point of a substance is the temperature at which the solid and liquid forms are in equilibrium.

To reattain equilibrium, the freezing point of the solute and solvent mixture is lowered relative to the original pure solvent.