Calculate the equilibrium constant at 25°C for the reaction: 4 Fe(CN)3(s) + 3 Cd^2+(aq) ⇄ 3 Cd(CN)4^2¯ (aq) + 4 Fe^3+(aq)

Question

Using the following information: Fe(CN)3(s) ⇄ Fe³⁺(aq) + 3 CN¯(aq)

K = 4.8×10-39 (at 25°C) Cd²⁺(aq) + 4 CN¯(aq) ⇄ Cd(CN)₄^2¯ (aq) Kc = 3.0 1018 (at 25°C)

Answer 1

Are you familiar with Hess' Law? If not, you should read about it. We will use this to answer the question.

We have a target equation of interest which is...

4Fe(CN)₃(s) + 3Cd²⁺(aq) <==> 3Cd(CN)₄^2-(aq) + 4Fe³⁺(aq)

We are given 2 other equation, which are...

(1) Fe(CN)₃(s) <==> Fe³⁺(aq) + 3CN-(aq) K = 4.8x10^-39

(2) Cd2+(aq) + 4CN-(aq) <==> Cd(CN)₄^2- K = 3.0x10¹⁸

We can use these two reactions to obtain our target equation as follows (Hess Law):

multiply (1) by 4 to get: 4Fe(CN)₃(s) <==> 4Fe³⁺(aq) + 12CN-(aq) K = (4.8x10-39)⁴ = 5.31x10^-154

multiply (2) by 3 to get: 3Cd2+(aq) + 12CN-(aq) <==> 3Cd(CN)₄^2- K = (3.0x10¹⁸)³ = 2.7x10⁵⁵

Add these together and cancel like terms on opposite sides to obtain the following target equation:

4Fe(CN)₃(s) + 3Cd²⁺(aq) <==> 4Fe³⁺(aq) + 3Cd(CN)₄^2- K = 5.31x10^-154 x 2.7x10⁵⁵ = 1.4x10^-98

(PLEASE CHECK THE MATH)