Final answer:
The molar solubility of calcium phosphate is calculated using the Ksp value of 1.3 × 10⁻²⁶ and the relationships between ion concentrations at equilibrium. The final molar solubility is mathematically derived to be lower than 1.0 × 10⁻⁶ M, reflecting its status as a sparingly soluble salt.
Step-by-step explanation:
The molar solubility of calcium phosphate is determined by the solubility product constant (Ksp). For calcium phosphate, the dissolution equation in water can be written as:
Ca3(PO4)2 (s) ⇌ 3Ca²⁺ (aq) + 2PO4³⁻ (aq)
From this, we can express the Ksp as Ksp = [Ca²⁺]³[PO4³⁻]². If we assume that molar solubility of Ca3(PO4)2 is 's', then we have the following relationships at equilibrium:
[Ca²⁺] = 3s and [PO4³⁻] = 2s.
Substituting these expressions into the Ksp equation gives us:
Ksp = (3s)³(2s)² = 108s⁵
Solving for s we find that the molar solubility of calcium phosphate is:
s = (Ksp/108)^(1/5)
Now, substituting the provided Ksp value of 1.3 × 10⁻²⁶ into the equation:
s = (1.3 × 10⁻²⁶ / 108)^(1/5)
Calculating this yields a molar solubility for calcium phosphate, typically lower than 1.0 × 10⁻⁶ M, indicating that it is a sparingly soluble salt.