JEE MainChemistryIonic Equilibrium
The solubility product constant ( K_ sp ) of Fe(OH) ₂ at 298 K is 4 10⁻¹⁵ . Let its solubility in pure water be S₁ and its solubility in a buffer solution of pH = 11 be S₂ . The value of the ratio S₁ S₂ is ________.
Correct answer
2500
Step-by-step solution
First, calculate the solubility in pure water ( S₁ ): For Fe(OH) ₂ , the solubility product is given by: K_ sp = 4S₁^3 4S₁^3 = 4 10⁻¹⁵ S₁^3 = 10⁻¹⁵ S₁ = 10⁻⁵ mol/L Next, calculate the solubility in the buffer solution ( S₂ ): Given pH = 11 , the pOH is 14 - 11 = 3 . Therefore, the concentration of hydroxide ions is: [ OH ^-] = 10⁻³ M In the buffer solution, the common ion effect dominates. The solubility S₂ is determined by: K_ sp = [ Fe ²⁺][ OH ^-]^2 4 10⁻¹⁵ = S₂ (10⁻³)^2 4 10⁻¹⁵ = S₂ 10⁻⁶ S₂ = 4 10⁻¹⁵ 10⁻⁶ = 4 10