Spreadsheet equation for auxiliary complexing agent. Consider the titration of metal M (initial concentration = C_M, initial volume = V_M) with EDTA (concentration = C_EDTA, volume added = V_EDTA) in the presence of an auxiliary complexing ligand (such as ammonia). Follow the derivation in Section 12-4 to show that the master equation for the titration is

$\mathbf{\varphi }\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{EDTA}}{\mathbf{V}}_{\mathbf{EDTA}\mathbf{}}}{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{\prime \prime }}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{-}\frac{\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{\prime \prime }}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}}{{\mathbf{C}}_{\mathbf{M}}}}{{\mathbf{K}}_{\mathbf{f}}^{\mathbf{\prime \prime }}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{+}\frac{\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{\prime \prime }}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}^{\mathbf{2}}}{{\mathbf{C}}_{\mathbf{EDTA}\mathbf{}}}}$

where ${\mathbf{K}}_{\mathbf{f}}^{\mathbf{\text{'}\text{'}}}$is the conditional formation constant in the presence of auxiliary complexing agent at the fixed pH of the titration (Equation 12-18) and [M]_free is the total concentration of metal not bound to EDTA. [M]_free is the same as [M] in Equation 12-15. The result is equivalent to Equation 12-11, with [M] replaced by [M]_free and${\mathbf{K}}_{\mathbf{f}}$replaced by ${\mathbf{K}}_{\mathbf{f}}^{\mathbf{\text{'}\text{'}}}$.

The following equation can be written in place of 12-8

${\mathrm{M}}_{\mathrm{free}}+\mathrm{EDTA}\rightleftharpoons \mathrm{M}\left(\mathrm{EDTA}\right)$

The equilibrium constant can be written as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{f}}^{\text{'}\text{'}}=\frac{\left[\mathrm{M}\left(\mathrm{EDTA}\right)\right]}{{\left[\mathrm{M}\right]}_{\mathrm{free}}\left[\mathrm{EDTA}\right]}----------------\left(1\right) \\ {\left[\mathrm{M}\right]}_{\mathrm{free}}=\mathrm{Concentration}\mathrm{of}\mathrm{all}\mathrm{metals}\mathrm{not}\mathrm{bound}\mathrm{to}\mathrm{EDTA} \\ \left[\mathrm{EDTA}\right]=\mathrm{Concentration}\mathrm{of}\mathrm{all}\mathrm{metals}\mathrm{not}\mathrm{bound}\mathrm{to}\mathrm{metals} \\ \left[\mathrm{M}\left(\mathrm{EDTA}\right)\right]={\mathrm{K}}_{\mathrm{f}}^{\text{'}\text{'}}{\left[\mathrm{M}\right]}_{\mathrm{free}}\left[\mathrm{EDTA}\right]--------------\left(2\right) \end{gathered}$$

Mass balance for metal:$[\mathrm{M}{]}_{\mathrm{Free}}+[\mathrm{M}\left(\mathrm{EDTA}\right)]=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}}--------(3)$

Mass balance for EDTA: $\left[EDTA\right]+\left[M\right(EDTA\left)\right]=\frac{C_{EDTA}{V}_{EDTA}}{V_M+V_{EDTA}}-------\left(4\right)$

Substituting Eq (2) into Eq (3) we get

$$\begin{gathered} {\left[\mathrm{M}\right]}_{\mathrm{Free}}+{\mathrm{K}}_{\mathrm{f}}^{\text{'}\text{'}}{\left[\mathrm{M}\right]}_{\mathrm{Free}}\left[\mathrm{EDTA}\right]=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}} \\ {\left[\mathrm{M}\right]}_{\mathrm{Free}}\left(1+{\mathrm{K}}_{\mathrm{f}}^{\text{'}\text{'}}\left[\mathrm{EDTA}\right]\right)=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}}--------------------\left(5\right) \end{gathered}$$

Substituting Eq (2) into Eq (4) we get

$$\begin{gathered} [\mathrm{EDTA}]+{\mathrm{K}}_{\mathrm{f}}\left[\mathrm{M}{]}_{\mathrm{free}}\right[\mathrm{EDTA}]=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}} \\ [\mathrm{EDTA}]\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{fes}}\right)=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}} \\ [\mathrm{EDTA}]=\frac{\left(\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}}\right)}{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}\right)}-----------------\left(6\right) \end{gathered}$$

Substituting Eq(6) into Eq (5)

$$\begin{gathered} [\mathrm{M}{]}_{\mathrm{free}}(1+{\mathrm{K}}_{\mathrm{f}}\frac{\left(\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EOTA}}}\right)}{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{fee}}\right)})=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}} \\ [\mathrm{M}{]}_{\mathrm{Fiee}}\left(\frac{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{tee}}\right)\left({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}\right)+{\mathrm{K}}_{\mathrm{f}}{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{\left(1+{\mathrm{K}}_{\mathrm{f}}^{\ast }[\mathrm{M}{]}_{\mathrm{tree}}\right)\left({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}\right)}\right)=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}})} \\ \frac{1}{[\mathrm{M}{]}_{\mathrm{free}}}\left(\frac{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}\right){\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{fee}}\right)\left({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}\right)+{\mathrm{K}}_{\mathrm{f}}{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}\right)=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}} \end{gathered}$$

$$\begin{gathered} \frac{1+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{tee}}-\frac{[\mathrm{M}{]}_{\mathrm{tee}}+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{tee}}}{{\mathrm{C}}_{\mathrm{M}}}}{{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}+\frac{[\mathrm{M}{]}_{\mathrm{tee}}+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{tee}}^2}{{\mathrm{C}}_{\mathrm{EDTA}}}}=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EOTA}}}{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}} \\ \frac{1+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{tee}}-\frac{[\mathrm{M}{]}_{\mathrm{tree}}+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{tee}}}{{\mathrm{C}}_{\mathrm{M}}}}{{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}+\frac{[\mathrm{M}{]}_{\mathrm{tee}}+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}^2}{{\mathrm{C}}_{\mathrm{EDTA}}}}=\mathrm{\varphi } \end{gathered}$$