Question

calculate the solubility of silver phosphate in 0.1 M AgNo3. Ksp of Ag3PO4-1.1*10^-16​

Answer 1

The solubility of silver phosphate in a 0.1 M solution of silver nitrate is equal to 1.1*10^-13 M.

• Each mole of silver phosphate dissociates to give three moles of silver ions and one mole of phosphate ions.
• Silver nitrate being soluble in water dissociates completely.
• Let solubility be equal to S. So, concentration of silver ions is equal to (3S+0.1) and the concentration of phosphate ions is equal to S.
• Ksp = [Ag⁺]³.[PO₄³⁻] ⇒ Ksp = (S+0.1)³.S ⇒ Ksp ≈ S.10⁻³
• Therefore the value of S is equal to 1.1*10⁻¹³M