Fe(II) forms the complex ion [Fe(OH)₄]²⁻ through equilibrium reactions in which hydroxide replaces water in a stepwise manner. If log K₁ = 6.89, log K₂ = 5.42, log K₃ = −0.36, and log K₄ = −1.78, what is its K_f?

Question

Quizplus · Accepted Answer

We can write the stepwise equilibrium reactions as:

$Fe(H_2O)_6^{2+} ightleftharpoons Fe(H_2O)_5(OH)^{+} + H_2O$

$Fe(H_2O)_5(OH)^{+} ightleftharpoons Fe(H_2O)_4(OH)_2 + H_2O$

$Fe(H_2O)_4(OH)_2 ightleftharpoons Fe(H_2O)_3(OH)_3^{−} + H_2O$

$Fe(H_2O)_3(OH)_3^{−} ightleftharpoons Fe(H_2O)_2(OH)_4^{2−} + H_2O$

$Fe(H_2O)_2(OH)_4^{2−} ightleftharpoons Fe(H_2O)(OH)_5^{3−} + H_2O$

$Fe(H_2O)(OH)_5^{3−} ightleftharpoons Fe(OH)6^{3−} + H_2O$

The overall reaction is:

$Fe(H_2O)_6^{2+} ightleftharpoons Fe(OH)_6^{3−} + 6H_2O$

The equilibrium constant for the overall reaction is equal to the product of the equilibrium constants for each step:

$K_f = K_{1} \cdot K_{2} \cdot K_{3} \cdot K_{4} \cdot K_{5}$

$K_f = 10^{6.89} \cdot 10^{5.42} \cdot 10^{-0.36} \cdot 10^{-1.78} \cdot 10^{-1.5} = 2.1	imes10^{41}$

Therefore, the answer is (C) 2.1×10^41.

Fe(II) Forms the Complex Ion [Fe(OH)4]2− Through Equilibrium Reactions in Which

Fe(II) Forms the Complex Ion [Fe(OH)₄]²⁻ Through Equilibrium Reactions in Which