魅影直播

Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

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

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

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 ${\mathbf{K}}_{\mathbf{f}}^{\mathbf{"}}$.

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

$\mathbf{蠒}\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{EDTA}}{\mathbf{V}}_{\mathbf{EDTA}\mathbf{}}}{{\mathbf{C}}_{\mathbf{M}}{\mathbf{V}}_{\mathbf{M}}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{鈥测赌�}}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{鈭�}\frac{\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{鈥测赌�}}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}}{{\mathbf{C}}_{\mathbf{M}}}}{{\mathbf{K}}_{\mathbf{f}}^{\mathbf{鈥测赌�}}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{+}\frac{\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}\mathbf{+}{\mathbf{K}}_{\mathbf{f}}^{\mathbf{鈥测赌�}}\mathbf{[}\mathbf{M}{\mathbf{]}}_{\mathbf{free}\mathbf{}}^{\mathbf{2}}}{{\mathbf{C}}_{\mathbf{EDTA}\mathbf{}}}}$

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 ${\mathbf{K}}_{\mathbf{f}}^{\mathbf{\text{'}\text{'}}}$.

At what pH does ${\mathit{伪}}_{{\mathbf{y}}^{\mathbf{4}\mathbf{-}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$?

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:

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

Let 伪M be the fraction of metal in the form M, 伪_ML be the fraction in the form ML, and ${\mathbf{伪}}_{{\mathbf{ML}}_{\mathbf{2}}}$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" $\begin{array}{r}{\mathbf{伪}}_{\mathbf{M}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{尾}}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{尾}}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}\\ {\mathbf{伪}}_{\mathbf{ML}}\mathbf{=}\frac{{\mathbf{尾}}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}}{\mathbf{1}\mathbf{+}{\mathbf{尾}}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{尾}}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}\\ {\mathbf{伪}}_{{\mathbf{ML}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{尾}}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{尾}}_{\mathbf{1}}\mathbf{\left[}\mathbf{L}\mathbf{\right]}\mathbf{+}{\mathbf{尾}}_{\mathbf{2}}\mathbf{[}\mathbf{L}{\mathbf{]}}^{\mathbf{2}}}\end{array}$

The concentrations of ML and ML2are

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

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

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

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:

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

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

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

The concentrations of ML and ML₂ are

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

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

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

Titration of M with L to form ML and ML₂. Use theequation from Problem 12-21, where M is Cu²⁺ and L is acetate.Consider adding 0.500 M acetate to 10.00 mL of 0.050 0 M Cu²⁺ atpH 7.00 (so that all ligand is present as ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CO}}_{\mathbf{2}}^{\mathbf{-}}$, not CH₃CO₂H).Formation constants forCu(CH₃CO₂)⁺ and Cu(CH₃CO₂)₂ aregiven in Appendix I. Construct a spreadsheet in which the inputis pL and the output is [L], V_L, [M], [ML], and [ML₂]. Prepare agraph showing concentrations of L, M, ML, and ML₂as V_L rangesfrom 0 to 3 mL

Explain why the change from red to blue in Reaction 12-19 occurs suddenly at the equivalence point instead of gradually throughout the entire titration.

List four methods for detecting the end point of an EDTA Titration