JEE MainChemistryIonic Equilibrium
A saturated aqueous solution of silver phosphate ( Ag ₃ PO ₄ ) is analyzed at 298 K and the equilibrium concentration of silver ions ( Ag ^+ ) is found to be 3.0 10⁻⁴ mol L ⁻¹ . The solubility product constant ( K_ sp ) of Ag ₃ PO ₄ at this temperature is
Options
- A1.0 10⁻¹⁶
- B3.0 10⁻⁸
- C2.2 10⁻¹³
- D2.7 10⁻¹⁵
Correct answer
D. 2.7 10⁻¹⁵
Step-by-step solution
The dissociation of silver phosphate in water is given by: Ag ₃ PO ₄(s) 3 Ag ^+(aq) + PO ₄³⁻(aq) Let the molar solubility of Ag ₃ PO ₄ be s . Then, at equilibrium: [ Ag ^+] = 3s [ PO ₄³⁻] = s Given that [ Ag ^+] = 3.0 10⁻⁴ M , we can find s : 3s = 3.0 10⁻⁴ s = 1.0 10⁻⁴ M The expression for the solubility product constant is: K_ sp = [ Ag ^+]^3[ PO ₄³⁻] K_ sp = (3s)^3(s) = 27s^4 Substituting the value of s : K_ sp = 27 (1.0 10⁻⁴)^4 K_ sp = 27 10⁻¹⁶ = 2.7 10⁻¹⁵ Answer: 2.7 10⁻¹⁵