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:

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

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

role="math" localid="1667801924683" $\begin{array}{r}{\mathbf{\alpha }}_{\mathbf{M}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}\\ {\mathbf{\alpha }}_{\mathbf{ML}}\mathbf{=}\frac{{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}\\ {\mathbf{\alpha }}_{{\mathbf{ML}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}\end{array}$

The concentrations of ML and ML2are

$$\begin{gathered} \left[\mathrm{ML}\right]\mathbf{=}{\mathbf{\alpha }}_{\mathbf{ML}}\frac{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}} \\ \left[{\mathrm{ML}}_2\right]\mathbf{=}{\mathbf{\alpha }}_{{\mathbf{ML}}_{\mathbf{2}}}\frac{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}} \end{gathered}$$

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

$\left[L\right]\mathbf{+}\left[\mathrm{ML}\right]\mathbf{+}\mathbf{2}\left[{\mathrm{ML}}_2\right]\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}{{\mathbf{V}}_{\mathbf{M}}\mathbf{+}{\mathbf{V}}_{\mathbf{L}}}$

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

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

Step by step solution

01

Information Given

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

$\begin{array}{clc}\mathrm{M}+\mathrm{L}𝆏& \mathrm{ML}& {\mathrm{\beta }}_1=\frac{\left[\mathrm{ML}\right]}{\left[\mathrm{M}\right]\left[\mathrm{L}\right]}\\ \mathrm{M}+2\mathrm{L}𝆏& {\mathrm{ML}}_2& {\mathrm{\beta }}_2=\frac{\left[{\mathrm{ML}}_2\right]}{\left[\mathrm{M}\right][\mathrm{L}{]}^2}\end{array}$

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

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

The concentrations of ML and ML₂ are

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

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

02

Mass Balance

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

03

Derivation

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

$\left[\mathrm{L}\right]+{\mathrm{\alpha }}_{\mathrm{ML}}\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}}+2{\mathrm{\alpha }}_{{\mathrm{ML}}_2}\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}}=\frac{{\mathrm{C}}_{\mathrm{L}}{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{L}}}$

Multiplying both sides by V_M+V_L

$$\begin{gathered} \left[\mathrm{L}\right]{\mathrm{V}}_{\mathrm{M}}+\left[\mathrm{L}\right]{\mathrm{V}}_{\mathrm{L}}+{\mathrm{\alpha }}_{\mathrm{ML}}{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}+2{\mathrm{\alpha }}_{\mathrm{ML}}{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}={\mathrm{C}}_{\mathrm{L}}{\mathrm{V}}_{\mathrm{L}} \\ {\mathrm{V}}_{\mathrm{L}}\left(\mathrm{L}-{\mathrm{C}}_{\mathrm{L}}\right)={\mathrm{V}}_{\mathrm{M}}\left(-\left[\mathrm{L}\right]-{\mathrm{\alpha }}_{\mathrm{ML}}{\mathrm{C}}_{\mathrm{M}}-2{\mathrm{\alpha }}_{{\mathrm{ML}}_2}{\mathrm{C}}_{\mathrm{M}}\right) \\ \frac{{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{M}}}=\frac{\left(-\left[\mathrm{L}\right]-{\mathrm{\alpha }}_{\mathrm{ML}}{\mathrm{C}}_{\mathrm{M}}-2{\mathrm{\alpha }}_{\mathrm{ML}}{\mathrm{C}}_{\mathrm{M}}\right)}{\left(\mathrm{L}-{\mathrm{C}}_{\mathrm{L}}\right)} \\ \frac{\frac{1}{{\mathrm{C}}_{\mathrm{M}}}\times {\mathrm{V}}_{\mathrm{L}}}{\frac{1}{{\mathrm{C}}_{\mathrm{L}}}\times {\mathrm{V}}_{\mathrm{M}}}=\frac{\frac{1}{{\mathrm{C}}_{\mathrm{M}}}\left(-\left[\mathrm{L}\right]-{\mathrm{\alpha }}_{\mathrm{ML}}{\mathrm{C}}_{\mathrm{M}}-2{\mathrm{\alpha }}_{{\mathrm{ML}}_2}{\mathrm{C}}_{\mathrm{M}}\right)}{\frac{1}{{\mathrm{C}}_{\mathrm{L}}}\left(\mathrm{L}-{\mathrm{C}}_{\mathrm{L}}\right)} \\ \frac{{\mathrm{C}}_{\mathrm{L}}\times {\mathrm{V}}_{\mathrm{L}}}{{\mathrm{C}}_{\mathrm{M}}\times {\mathrm{V}}_{\mathrm{M}}}=\frac{-\frac{\left[\mathrm{L}\right]}{{\mathrm{C}}_{\mathrm{M}}}-{\mathrm{\alpha }}_{\mathrm{ML}}-2{\mathrm{\alpha }}_{{\mathrm{ML}}_2}}{\frac{\mathrm{L}}{{\mathrm{C}}_{\mathrm{L}}}-1} \\ \frac{{\mathrm{C}}_{\mathrm{L}}{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}=\frac{\frac{\left[\mathrm{L}\right]}{{\mathrm{C}}_{\mathrm{M}}}+{\mathrm{\alpha }}_{\mathrm{ML}}+2{\mathrm{\alpha }}_{{\mathrm{ML}}_2}}{1-\frac{\mathrm{L}}{{\mathrm{C}}_{\mathrm{L}}}} \end{gathered}$$

Therefore, the final expression given is proved.

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