Question

Write equations to show the stepwise reaction of ${\mathbf{\text{Cd(H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O)}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2 +}}}$ in an aqueous solution of $\mathbf{\text{KI}}$to form role="math" localid="1663616318558" ${\mathbf{\text{CdI}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2 -}}}$. Show that${\mathbf{K}}_{\mathbf{f}\mathbf{\left(}\mathbf{overall}\mathbf{\right)}}\mathbf{=}{\mathbf{K}}_{\mathbf{f}\mathbf{1}}\mathbf{\times }{\mathbf{K}}_{\mathbf{f}\mathbf{2}}\mathbf{\times }{\mathbf{K}}_{\mathbf{f}\mathbf{3}}\mathbf{\times }{\mathbf{K}}_{\mathbf{f}\mathbf{4}}$ .

Short answer

It is proved that the stepwise reactions of $\text{Cd}{\left({\text{H}}_{\text{2}}\text{O}\right)}_4$result in –

$$\begin{gathered} {\mathrm{Kf}}_{\mathrm{overall}}=\frac{\left[{\mathrm{CdI}}_4^{2-}\right]}{\left[\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}\right]\times {\left[{\mathrm{I}}^{-}\right]}^4} \\ ={\mathrm{Kf}}_1\times {\mathrm{Kf}}_2\times {\mathrm{Kf}}_3\times {\mathrm{Kf}}_4 \end{gathered}$$

Step-by-step solution

Concept Introduction

The production of a complex ion from its core ion and associated ligands is described by this formation constant, K_f. This constant is also known as stability constant or an association constant, and its units vary depending on the reaction it describes.

Stepwise Equation

When complex ion is being formed, they often bond one ligand at a time. The appropriate K_f value for one ligand at a time bonded to central metal ion is present.

First, write the equation of $\text{Cd}{\left({\text{H}}_{\text{2}}\text{O}\right)}_4$ reacting with one I^-:

role="math" localid="1663617331149" $\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}+{\mathrm{I}}^{-}\rightleftharpoons \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{I}}^{+}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

${\mathrm{K}}_{{\mathrm{f}}_1}=\frac{{\displaystyle \left[\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{I}}^{+}\right]}}{{\displaystyle \left[\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}\right]\times \left[{\mathrm{I}}^{-}\right]}}$

Now add another I^- ion :

$\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{I}}^{+}+{\mathrm{I}}^{-}\rightleftharpoons \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

${\mathrm{K}}_{{\mathrm{f}}_2}=\frac{{\displaystyle \left[\text{Cd}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{I}}_{\text{2}}\right]}}{{\displaystyle \left[\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{I}}^{-}\right]\times \left[{\mathrm{I}}^{-}\right]}}$

Now add another I^- ion:

role="math" localid="1663618412559" $$\begin{gathered} \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\mathrm{I}}_2+{\mathrm{I}}^{-}\rightleftharpoons \mathrm{Cd}\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{I}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ {\mathrm{K}}_{{\mathrm{f}}_3}=\frac{\left[\mathrm{Cd}\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{I}}_3^{-}\right]}{\left.\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\mathrm{I}}_2\right]\times \left[{\mathrm{I}}^{-}\right]} \end{gathered}$$

Now add the last I^- ion:

$$\begin{gathered} \mathrm{Cd}\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{I}}_3^{-}+{\mathrm{I}}^{-}\rightleftharpoons {\mathrm{CdI}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ {\mathrm{K}}_{{\mathrm{f}}_4}=\frac{\left[{\mathrm{CdI}}_4^{2-}\right]}{\left[\mathrm{Cd}\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{I}}_3^{-}\right]\times \left[{\mathrm{I}}^{-}\right]} \end{gathered}$$

Overall formation constant

The overall reaction is given below.

$$\begin{gathered} \frac{\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}+{\mathrm{I}}^{-}\rightleftharpoons \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{I}}^{+}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{I}}^{-}+{\mathrm{I}}^{-}\rightleftharpoons \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\mathrm{I}}_2+{\mathrm{I}}^{-}\rightleftharpoons \mathrm{Cd}\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{I}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Cd}\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{I}}_3^{-}+{\mathrm{I}}^{-}\rightleftharpoons {\mathrm{CdI}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}{\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}+4{\mathrm{I}}^{-}\rightleftharpoons {\mathrm{CdI}}_4^{2-}+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)} \end{gathered}$$

$\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}+4{\mathrm{I}}^{-}\rightleftharpoons {\mathrm{CdI}}_4^{2-}+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The overall reaction given above is the sum of all of the reactions, and it can be seen that some of the intermediate products cancel each other and we get the overall reaction. The overall K_f constant would be thus be the product of the equilibrium constants of the individual equations.

Therefore, the overall K_f value is obtained as:

$$\begin{gathered} {\mathrm{K}}_{{\mathrm{f}}_{\mathrm{overall}}}=\frac{\left[{\mathrm{CdI}}_4^{2-}\right]}{\left[\mathrm{Cd}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}\right]\times {\left[{\mathrm{I}}^{-}\right]}^4} \\ {\mathrm{K}}_{{\mathrm{f}}_{\mathrm{overall}}}={\mathrm{K}}_{{\mathrm{f}}_1}\times {\mathrm{K}}_{{\mathrm{f}}_2}\times {\mathrm{K}}_{{\mathrm{f}}_3}\times {\mathrm{K}}_{{\mathrm{f}}_4} \end{gathered}$$.