Answer:
Kps = 3.07 x 10⁻⁸
pKsp= 7.51
Step-by-step explanation:
First, we calculate the molar solubility of silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):
2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s
Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:
Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻
I 0 0
C +3s +s
E 3s s
Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴
Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:
Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸
The pKsp value is:
pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51