1. Write the Dissolution Equilibrium
Iron(II) phosphate is an ionic compound that dissolves in water according to the following equilibrium:
Fe₃(PO₄)₂(s) ⇌ 3Fe²⁺(aq) + 2PO₄³⁻(aq)
2. Set up the Ksp Expression
The solubility product constant (Ksp) is the equilibrium constant for the dissolution reaction. It's defined as the product of the ion concentrations, each raised to the power of its stoichiometric coefficient:
Ksp = [Fe²⁺]³ [PO₄³⁻]²
3. Define the Molar Solubility
Molar solubility (s) is the concentration of the metal cation (Fe²⁺ in this case) in a saturated solution.
* Since 3 moles of Fe²⁺ are produced for every 1 mole of Fe₃(PO₄)₂, the concentration of Fe²⁺ is 3s.
* Similarly, the concentration of PO₄³⁻ is 2s.
4. Substitute and Solve for s
Substitute the expressions for [Fe²⁺] and [PO₄³⁻] in terms of 's' into the Ksp expression:
Ksp = (3s)³ (2s)²
Simplify and solve for 's':
Ksp = 108s⁵
s⁵ = Ksp / 108
s = (Ksp / 108)^(1/5)
5. Look up the Ksp Value
You'll need to find the Ksp value for iron(II) phosphate. This can be found in a table of solubility products or a chemistry textbook.
Important Note: The Ksp for iron(II) phosphate is very small, indicating it's practically insoluble. The solubility will be a very small number, typically expressed in units of moles per liter (mol/L).
Example (Hypothetical Ksp):
Let's say the Ksp of iron(II) phosphate is 1.0 x 10⁻³⁶. We can now calculate the molar solubility:
s = (1.0 x 10⁻³⁶ / 108)^(1/5) ≈ 1.8 x 10⁻⁸ mol/L
Therefore, the molar solubility of iron(II) phosphate is approximately 1.8 x 10⁻⁸ mol/L, assuming the Ksp is 1.0 x 10⁻³⁶.
Remember: Always use the correct Ksp value for the specific compound.
