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?

, show all calculation with comments.

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

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