Use Ksp & Kf to find equilibrium constant for the reaction

Question

Use the values of K_sp FeS_(s) = 3.72 x 10^-19 and

Kf Fe(CN)₆⁴⁻(aq) = 1.5 x 1035 and

to find the equilibrium constant for the reaction.

FeS_(s) + 6 CN⁻ (aq) ⇌ Fe(CN)₆ ⁴⁻ (aq) + S²⁻

Please solve using Hess's Law or another way.

Answer 1

We can use Hess's law as follows:

Write the individual reactions for the dissolution of FeS and for the complex formation of Fe(CN)₆^4-

FeS(s) ==> Fe²⁺(aq) + S^2-(aq) ... Ksp = 3.72x10^-19 ... dissolution of FeS

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

Add these two reactions together

FeS + 6CN^- ==> Fe(CN)₆^4- + S^2-

The equilibrium constant for this will be the product of the Ksp x Kf

K = (3.72x10^-19) (1.5x10³⁵)

K = 5.58x10¹⁶