91Ó°ÊÓ

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

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

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.

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

The concentrations of ML and ML₂ are

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

$\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}}$= 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

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

Multiplying both sides by V_M+V_L

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

Therefore, the final expression given is proved.