What is the equilibrium constant for the solubility of FeCO₃ (Ksp = 2.1 × 10⁻¹¹) in NaCN? (Kf of Fe(CN)₆⁴⁻ is 1.5 × 10³⁵)

Answer 1

FeCO₃(s) <==> Fe²⁺(aq) + CO₃^2-(aq) ... Ksp = 2.1x10^-11

Fe²⁺ + 6CN^- ==> Fe(CN)₆^4- ... Kf = 1.5x10³⁵

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FeCO₃(s) + 6CN^-(aq) <==> CO₃^2-(aq) + Fe(CN)₆^4-(aq) ... Keq = Ksp x Kf

Keq = (2.1x10^-11)(1.5x10³⁵)

Keq = 3.2x10²⁴