Spreadsheet equation for formation of the complexes ML and ML₂.Consider the titration of metal M (initial concentration = C_M, initial volume = V_M) with ligand L (concentration = C_L, volume added = V_L), which can form 1:1 and 2 : 1 complexes:

$$\begin{gathered} \mathbf{M}\mathbf{+}\mathbf{L}\mathbf{\leftrightharpoons }\mathbf{}\mathbf{}\mathbf{}\mathbf{ML}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\frac{\left[\mathrm{ML}\right]}{\left[M\right]\left[L\right]} \\ \mathbf{M}\mathbf{+}\mathbf{2}\mathbf{L}\mathbf{\leftrightharpoons }\mathbf{}\mathbf{}\mathbf{}{\mathbf{ML}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}\frac{\left[{\mathrm{ML}}_2\right]}{\left[M\right]{\left[L\right]}^{\mathbf{2}}} \end{gathered}$$

Let α_M be the fraction of metal in the form M, α_ML be the fraction in the form ML, and ${\mathit{\alpha }}_{\mathbf{M}{\mathbf{L}}_{\mathbf{2}}}$ be the fraction in the form ML₂. Following the derivation in Section 12-5, you could show that these fractions are given by

$$\begin{gathered} {\mathit{\alpha }}_{\mathbf{M}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}} \\ {\mathit{\alpha }}_{\mathbf{M}}\mathit{L}\mathbf{=}\frac{\mathbf{\left(}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\right[}\mathbf{L}\mathbf{\left]}\mathbf{\right)}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}} \\ {\mathit{\alpha }}_{\mathbf{M}{\mathbf{L}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}} \end{gathered}$$

The concentrations of ML and ML₂ are

$$\begin{gathered} \mathbf{\left[}\mathbf{ML}\mathbf{\right]}\mathbf{=}{\mathbf{\alpha }}_{\mathbf{ML}}\frac{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}} \\ \mathbf{\left[}\mathbf{ML}\mathbf{\right]}\mathbf{=}{{\mathbf{\alpha }}_{\mathbf{ML}}}_{\mathbf{2}}\frac{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}} \end{gathered}$$

because$\frac{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}}$ is the total concentration of all metal in the solution. The mass balance for ligand is

$\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}\mathbf{\left[}\mathbf{ML}\mathbf{\right]}\mathbf{+}\mathbf{2}\mathbf{\left[}{\mathbf{ML}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{L}}{\mathbf{V}}_{\mathbf{L}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}}$

By substituting expressions for [ML] and [ML₂] into the mass balance, show that the master equation for a titration of metal by ligand is

$\mathbf{\varphi }\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{L}}{\mathbf{V}}_{\mathbf{L}}}{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}\mathbf{=}\frac{{\mathbf{\alpha }}_{\mathbf{ML}}\mathbf{+}\mathbf{2}{\mathbf{\alpha }}_{{\mathbf{ML}}_{\mathbf{2}}}\mathbf{+}{\displaystyle \frac{\mathbf{L}}{{\mathbf{C}}_{\mathbf{M}}}}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{L}}{{\mathbf{C}}_{\mathbf{L}}}}}$

Titration of metal M having initial concentration C_Mand initial volume V_M with ligand L having concentration C_L and added volume V_L.This can form 1:1 and 2:1 complex:

The reactions involved are as follows

$$\begin{gathered} \mathrm{M}+\mathrm{L}\leftrightharpoons \mathrm{ML}{\mathrm{\beta }}_1=\frac{\left[\mathrm{ML}\right]}{\left[\mathrm{M}\right]\left[\mathrm{L}\right]} \\ \mathrm{M}+2\mathrm{L}\leftrightharpoons {\mathrm{ML}}_2{\mathrm{\beta }}_2=\frac{\left[{\mathrm{ML}}_2\right]}{\left[\mathrm{M}\right]{\left[\mathrm{L}\right]}^2} \end{gathered}$$

Let α_M be the fraction of metal in the form M, α_ML be the fraction in the form ML, and be the fraction in the form ML₂.

$$\begin{gathered} {\alpha }_{\mathrm{M}}=\frac{1}{1+{\mathrm{\beta }}_1\left[\mathrm{L}\right]+{\mathrm{\beta }}_2[\mathrm{L}{]}^2} \\ {\alpha }_{\mathrm{M}}L=\frac{\left({\mathrm{\beta }}_1\right[\mathrm{L}\left]\right)}{1+{\mathrm{\beta }}_1\left[\mathrm{L}\right]+{\mathrm{\beta }}_2[\mathrm{L}{]}^2} \\ {\alpha }_{{\mathrm{ML}}_2}=\frac{{\mathrm{\beta }}_2[\mathrm{L}{]}^2}{1+{\mathrm{\beta }}_1\left[\mathrm{L}\right]+{\mathrm{\beta }}_2[\mathrm{L}{]}^2} \end{gathered}$$

The concentrations of ML and ML₂ are

$$\begin{gathered} \left[\mathrm{ML}\right]={\mathrm{\alpha }}_{\mathrm{ML}}\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}} \\ \left[\mathrm{ML}\right]={{\mathrm{\alpha }}_{\mathrm{ML}}}_2\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}} \end{gathered}$$

$\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}}$= total concentration of all metal in the solution.

Mass balance for ligand:$\left[\mathrm{L}\right]+\left[\mathrm{ML}\right]+2\left[{\mathrm{ML}}_2\right]=\frac{{\mathrm{C}}_{\mathrm{L}}{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}}$

Substituting the expression for [ML] and [ML₂] in the mass balance equation we get

$\left[L\right]+{\alpha }_{ML}\frac{C_M{V}_M}{V_M+V_L}+2{\alpha }_{M{L}_2}\frac{C_M{V}_M}{V_M+V_L}=\frac{C_L{V}_L}{V_M+V_L}$

Multiplying both sides by V_M+V_L

$$\begin{gathered} \left[L\right]V_M+\left[L\right]V_L+{\alpha }_{M}L{C}_M{V}_M+2{\alpha }_{M{L}_2}{C}_M{V}_M=C_L{V}_L \\ {V}_L(L-C_L)=V_M\left(-\left[L\right]-{\alpha }_{M}L{C}_M-2{\alpha }_{M{L}_2}{C}_M\right) \\ \frac{V_L}{V_M}=\frac{\left(-\left[L\right]-{\alpha }_{M}L{C}_M-2{\alpha }_{M{L}_2}{C}_M\right)}{\left(L-C_L\right)} \\ \frac{{\displaystyle \frac{1}{C_M}}\times V_L)/)}{{\displaystyle \frac{1}{C_L}}\times V_M}=\frac{{\displaystyle \frac{1}{C_M}}\left(-\left[L\right]-{\alpha }_{ML}{C}_M-2{\alpha }_{M{L}_2}{C}_M\right)}{{\displaystyle \frac{1}{C_L\left(L-C_L\right)}}} \\ \frac{C_L\times V_L}{C_M\times V_M}=\frac{{\displaystyle \frac{-\left[L\right]}{C_M}}-{\alpha }_{ML}-2{\alpha }_{M{L}_2}}{{\displaystyle \frac{L}{C_L}}-1} \\ \frac{C_L\times V_L}{C_M\times V_M}=\frac{{\displaystyle \frac{\left[L\right]}{C_M}}+{\alpha }_{M}L+2{\alpha }_{M{L}_2}}{1-{\displaystyle \frac{L}{C_L}}} \end{gathered}$$

Therefore, the final expression given is proved.