Final answer:
The molar solubility of Li₃PO₄ in water is approximately 8.8 × 10⁻⁵ M.
Step-by-step explanation:
The molar solubility of Li₃PO₄ can be calculated using the solubility product constant (Ksp). The balanced equation for the dissolution of Li₃PO₄ is:
Li₃PO₄(s) → 3Li⁺(aq) + PO₄³⁻(aq)
In this equation, the stoichiometry shows that for every 1 mol of Li₃PO₄ that dissolves, 3 mol of Li⁺ ions and 1 mol of PO₄³⁻ ions are produced.
Considering that the solubility product constant (Ksp) for Li₃PO₄ is 3.2 × 10⁻⁹, we can set up the following expression:
Ksp = [Li⁺]³[PO₄³⁻]
Since the stoichiometry tells us that the concentration of PO₄³⁻ ions is equal to the molar solubility of Li₃PO₄ (let's call it x), we have:
Ksp = (3x)³(x) = 27x⁴
Plugging in the value for Ksp, we get:
3.2 × 10⁻⁹ = 27x⁴
Solving for x, the molar solubility of Li₃PO₄, we find:
x = (3.2 × 10⁻⁹)^(1/4)
The molar solubility of Li₃PO₄ in water is approximately 8.8 × 10⁻⁵ M.