1.0 m aqueous solutions of the following substances are prepared

Asked by admin @ 09/11/2021 in Chemistry viewed by 318 People

Two saturated aqueous solutions are prepared at 25 ºC.

One is made by dissolving lithium carbonate (Ksp = 8.15 x 10⁻⁴) in 100.0 mL of water until excess solid is present, while the other is prepared by dissolving lithium phosphate (Ksp = 2.37 x 10⁻⁴) in 200.0 mL of water until excess solid is present.
1) What is the molar concentration of Li¹⁺ in the lithium carbonate solution?
2) What is the molar concentration of Li¹⁺ in the lithium phosphate solution?
Please, show all calculation with comments. Thanks!

Answered by admin @ 09/11/2021

The equilibrium constant of solubility product, or Ksp, is the product of the solvated ions of a compound when dissolved in water.

Let us see the dissociation of Lithium Phosphate or (Li₂PO₄):

Li₂PO₄ ⇆ 2Li⁺ + PO₄²⁻

So, the Ksp for this dissociation is

Ksp = [Li⁺]²[PO₄²⁻] = 8.15 ˣ 10⁻⁴

Since no amount of initial moles are known, let's just use the stoichiometric coefficients. The substances in '[]' are the molar concentrations (moles/liter). We let x be the moles of the substance dissociated:

[2x mol/0.1 L]²[x mol/0.1L] = 8.15 ˣ 10⁻⁴
x = 5.88ˣ10⁻3 mol
Hence, the concentration of [Li⁺] is
[Li⁺] = 2(5.88ˣ10⁻3 mol) / 0.1 L
[Li⁺] = 0.118 M

The same procedure is applied to Lithium Carbonate (Li₂CO₃):
Li₂CO₃ ⇆ 2Li⁺ + CO₃²⁻

Ksp = [Li⁺]²[CO₃²⁻] = 2.37 ˣ 10⁻⁴
[2x mol/0.2 L]²[x mol/0.2L] = 8.15 ˣ 10⁻⁴
x = 7.797 ˣ 10⁻³ moles
Hence, the concentration of [Li⁺] is
[Li⁺] = 2(7.797 ˣ 10⁻³ moles) / 0.2 L
[Li⁺] = 0.078 M

1.0 m aqueous solutions of the following substances are prepared

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