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

$\mathbf{\varphi }\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{ETDA}}{\mathbf{V}}_{\mathbf{ETDA}}}{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{"}}{\left[M\right]}_{\mathbf{free}}\mathbf{-}{\displaystyle \frac{{\mathbf{\left[}\mathbf{M}\mathbf{\right]}}_{\mathbf{free}}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{"}}{\mathbf{\left[}\mathbf{M}\mathbf{\right]}}_{\mathbf{free}}}{{\mathbf{C}}_{\mathbf{M}}}}}{{\mathbf{K}}_{\mathbf{f}}^{\mathbf{"}}{\mathbf{\left[}\mathbf{M}\mathbf{\right]}}_{\mathbf{free}}\mathbf{+}{\displaystyle \frac{{\mathbf{\left[}\mathbf{M}\mathbf{\right]}}_{\mathbf{free}}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{"}}{\mathbf{\left[}\mathbf{M}\mathbf{\right]}}_{\mathbf{free}}^{\mathbf{2}}}{{\mathbf{C}}_{\mathbf{ETDA}}}}}$

where role="math"> 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 K_f replaced by ${\mathbf{K}}_{\mathbf{f}}^{\mathbf{"}}$.

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

${\mathrm{M}}_{\mathrm{free}}+\mathrm{ETDA}𝆏\mathrm{M}\left(\mathrm{ETDA}\right)$

The equilibrium constant can be written as

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

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

Mass balance for EDTA: $\left[\mathrm{ETDA}\right]+\left[\mathrm{M}\left(\mathrm{EDTA}\right)\right]=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{V}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{EDTA}}}----------\left(4\right)$

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

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

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

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

Substituting Eq(6) into Eq (5)

$$\begin{gathered} [\mathrm{M}{]}_{\mathrm{Free}}\left(1+{\mathrm{K}}_{\mathrm{f}}\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{tee}}\right)}\right)=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}} \\ [\mathrm{M}{]}_{\mathrm{Free}}\left(\frac{\left(1+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{fee}}\right)\left({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}\right)+{\mathrm{K}}_{\mathrm{i}}{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}\right)\left({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}\right)}\right)=\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{\left({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{EDTA}}\right)} \\ \frac{1}{[\mathrm{M}{]}_{\mathrm{Free}}}\left(\frac{\left(1+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{tree}}\right){\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{\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}}}\right)=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}} \\ \frac{1+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{ree}}-\frac{[\mathrm{M}{]}_{\mathrm{tree}}+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{ree}}}{{\mathrm{C}}_{\mathrm{M}}}}{{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}+\frac{[\mathrm{M}{]}_{\mathrm{tee}}+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{free}}^2}{{\mathrm{C}}_{\mathrm{EDTA}}}}=\frac{{\mathrm{C}}_{\mathrm{EDTA}}{\mathrm{V}}_{\mathrm{EDTA}}}{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}} \\ \frac{1+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{tee}}-\frac{[\mathrm{M}{]}_{\mathrm{free}}+{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{fee}}}{{\mathrm{C}}_{\mathrm{M}}}}{{\mathrm{K}}_{\mathrm{f}}[\mathrm{M}{]}_{\mathrm{free}}+\frac{[\mathrm{M}{]}_{\mathrm{tee}}+{\mathrm{K}}_{\mathrm{f}}^{\mathrm{\prime }}[\mathrm{M}{]}_{\mathrm{fee}}^2}{{\mathrm{C}}_{\mathrm{EDTA}}}}=\mathrm{\varphi } \end{gathered}$$

Therefore, the expression given in the question is proved.