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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)

User Cherita
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Final answer:

The equilibrium constant for the solubility of FeCO₃ in NaCN can be determined by considering the formation of the complex ion Fe(CN)₆⁴⁻ using the given Kf value. The equilibrium expression can be set up and the concentration of Fe²⁺ can be expressed in terms of [Fe(CN)₆⁴⁻] using the Kf value. Substituting the values will give the equilibrium constant.

Step-by-step explanation:

The equilibrium constant for the solubility of FeCO₃ in NaCN can be determined by considering the formation of the complex ion Fe(CN)₆⁴⁻. The equation for the formation of Fe(CN)₆⁴⁻ is:

Fe²⁺(aq) + 6CN⁻(aq) → Fe(CN)₆⁴⁻

Using the given Kf value of Fe(CN)₆⁴⁻ (1.5 × 10³⁵), we can set up the equilibrium expression:

Ksp = [Fe(CN)₆⁴⁻] = [Fe²⁺][CN⁻]⁶

Since we are looking for the equilibrium constant for the solubility of FeCO₃, we can express the concentration of Fe²⁺ in terms of [Fe(CN)₆⁴⁻] using the equation:

[Fe²⁺] = Kf / [CN⁻]⁶

Substituting the given Kf value (1.5 × 10³⁵) and the concentration of cyanide ion ([CN⁻]) into the equation will give us the equilibrium constant for the solubility of FeCO₃ in NaCN.

User Rising
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