Chapter U6.121, Problem 2E

Interpretation Introduction

Interpretation :

The reason behind the solutions of ${\text{Fe}}^{3+}$ , ${\text{SCN}}^{-}$ ${\text{and FeSCN}}^{2+}$ with different colors to be at equilibrium must be explained.

Concept Introduction :

Equilibrium is a state for a reversible reaction when the rate of a forward and backward reaction is the same.

Answer to Problem 2E

${\text{Fe}}^{3+}$ reacts with ${\text{SCN}}^{-}$ to produce ${\text{FeSCN}}^{2+}$ which has blood-red coloration.

Explanation of Solution

The formation of ${\text{FeSCN}}^{2+}$ (blood red) from component ions ${\text{Fe}}^{3+}$ (green) and ${\text{SCN}}^{-}$ (colorless) is a reversible reaction. For this reaction, there is one equilibrium constant which is the ratio of the concentration of products to that of the reactants. The equilibrium reaction is shown below.

  $\begin{array}{l}{\text{Fe}}^{\text{3+}}{\text{(aq)+SCN}}^{\text{-}}\text{(aq)}\rightleftharpoons {\text{FeSCN}}^{\text{2+}}\text{(aq)}\end{array}$

  $\begin{array}{c}\text{K=}\frac{{\text{[FeSCN}}^{\text{2+}}\text{(aq)]}}{{\text{[Fe}}^{\text{3+}}{\text{(aq)][SCN}}^{\text{-}}\text{(aq)]}}\text{.}\end{array}$

Conclusion

Thus, ${\text{Fe}}^{3+}$ , ${\text{SCN}}^{-}$ and ${\text{FeSCN}}^{2+}$ all can be in equilibrium though these different colors.