To find the molar solubility of Ba3(PO4)2, we need to use its solubility product constant (Ksp).
The compound Ba3(PO4)2 dissociates into 3 Ba2+ ions and 2 PO43- ions upon dissolving in water. The balanced equation for this dissociation reaction is:
Ba3(PO4)2(s) ↔ 3 Ba2+(aq) + 2 PO43-(aq)
Let's assume that the molar solubility of Ba3(PO4)2 is represented by "s" (in mol/L). Therefore, the equilibrium concentrations of Ba2+ and PO43- ions can be expressed as (3s) and (2s), respectively.
The expression for the solubility product constant (Ksp) can be written as follows:
Ksp = [Ba2+]^3 [PO43-]^2
Substituting the equilibrium concentrations into the expression, we have:
Ksp = (3s)^3 (2s)^2
Ksp = 1.3 × 10^-29
Now, we can solve for the molar solubility (s):
(3s)^3 (2s)^2 = 1.3 × 10^-29
54s^5 = 1.3 × 10^-29
Taking the fifth root of both sides to isolate s, we get:
s ≈ (1.3 × 10^-29)^(1/5)
s ≈ 1.51 × 10^-6 mol/L
Therefore, the molar solubility of Ba3(PO4)2 is approximately 1.51 × 10^-6 mol/L.