Chapter 12, Problem 12.16P

(a)

Interpretation Introduction

Interpretation:

The ${\text{pCu}}^{\text{2+}}$ should be calculated at $\text{0}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ .

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

${\text{MY}}^{\text{n4-}}\underset{}{\rightleftharpoons }{\text{MY}}^{\text{n+}}+{\text{Y}}^{\text{4-}}$

The formation constant for above complex reaction is,

${\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}$

Where,

${\text{K}}_{\text{f}}$ is formation constant

${\text{[M}}^{\text{+}}\text{]}$ is concentration of metal ion

${\text{[Y}}^{\text{4-}}\text{]}$ is concentration of ligand

${\text{[MY}}^{\text{n4-}}\text{]}$ is concentration of the acid

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

$\begin{array}{l}\text{α}{}_{{\text{Y}}^{\text{4-}}}\text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{{\text{[H}}_{\text{6}}{\text{Y}}^{\text{2+}}{\text{]+[H}}_{\text{5}}{\text{Y}}^{\text{+}}{\text{]+[H}}_{\text{4}}{\text{Y]+[H}}_{\text{3}}{\text{Y}}^{\text{-}}{\text{]+[H}}_{\text{2}}{\text{Y}}^{\text{2-}}{\text{]+[H}}_{\text{3}}{\text{Y}}^{\text{3-}}{\text{]+[Y}}^{\text{4-}}\text{]}}\\ \text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{\text{[EDTA]}}\end{array}$

If pH is fixed, the degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ is constant and ${\text{K}}_{\text{f}}$ will be ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}\text{][EDTA]}}$

Where,

${\text{K}}_{\text{f}}\text{' is the conditional formation constant}$

Fraction of free metal ion:

In the metal ligand equilibria the number of moles of metal and ligand are equilibrium with number of mole of complex formed.

$\begin{array}{l}\text{M+}\text{L}\underset{}{\rightleftharpoons }\text{ML}{\text{β}}_1\text{=}\frac{\text{[ML]}}{\text{[M][L]}}\\ \text{M+}2\text{L}\underset{}{\rightleftharpoons }{\text{ML}}_2{\text{β}}_2\text{=}\frac{{\text{[ML}}_2\text{]}}{{\text{[M][L]}}^2}\\ \text{M+}\text{nL}\underset{}{\rightleftharpoons }{\text{ML}}_{\text{n}}{\text{β}}_{\text{n}}\text{=}\frac{{\text{[ML}}_{\text{n}}\text{]}}{{\text{[M][L]}}^{\text{n}}}\end{array}$

Where,

${\text{β}}_{\text{n}}$ is cumulative formation constant

Fraction of free metal ion ${\text{α}}_{\text{M}}$ is,

${\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M}}_{\text{tot}}\text{]}}$

$\begin{array}{l}{\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M]{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}\text{}}}\\ =\frac{1}{{\text{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}}\end{array}$

Where,

$\text{M}\text{is free metal ion}$

Answer to Problem 12.16P

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{0}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ is $\text{15}\text{.03}$

Explanation of Solution

To calculate the ${\text{pCu}}^{\text{2+}}$ at given volume of $\text{EDTA}$

Given,

$\text{50 }\text{.0 mL of 0}{\text{.001 M Cu}}^{\text{2+}}$

$\text{0}\text{.0100 M EDTA}$

$\text{pH}\text{=11}\text{.00}$

End point is $\text{50}\text{.0}\text{mL}$

Volume of ${\text{NH}}_{\text{3}}$ is $\text{1}\text{.00}\text{mL}$

The complex formation reaction is,

${\text{Cu}}^{\text{2+}}\text{+EDTA}\underset{}{\overset{}{\rightleftharpoons }}{\text{CuY}}^{\text{2-}}$

At the give $\text{pH}\text{=11}\text{.0}$ the degree of dissociation is fixed and it is,

The formation constant ${\text{K}}_{\text{f}}$ for ${\text{Cu}}^{\text{2+}}$ is ${10}^{18.78}$

Degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}\text{=}\text{0}\text{.81}$

The Degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ of ${\text{Cu}}^{\text{2+}}$ at $\text{00}\text{.0}\text{mL}$ of $\text{EDTA}$ with $\text{1}\text{.00}\text{M}{\text{NH}}_{\text{3}}$

For $\text{00}\text{.0}\text{mL}$

From the standard data,

$\begin{array}{l}\text{log}{\text{β}}_{\text{1}}\text{of}{\text{Cu}}^{\text{2+}}\text{=3}\text{.99}\\ \text{log}{\text{β}}_{\text{2}}\text{of}{\text{Cu}}^{\text{2+}}\text{=7}\text{.33}\\ \text{log }{\text{β}}_{\text{3}}\text{of}{\text{Cu}}^{\text{2+}}\text{= 10}\text{.06}\\ \text{log}{\text{β}}_{\text{4}}\text{of}{\text{Cu}}^{\text{2+}}\text{=12}\text{.03}\end{array}$

To taking a log of above values,

$\begin{array}{l}{\text{β}}_{\text{1}}\text{of}{\text{Cu}}^{\text{2+}}\text{=}\text{9}{\text{.8×10}}^{\text{3}}\\ {\text{β}}_{\text{2}}\text{of}{\text{Cu}}^{\text{2+}}\text{=}\text{2}{\text{.1×10}}^{\text{7}}\\ {\text{β}}_{\text{3}}\text{of}{\text{Cu}}^{\text{2+}}\text{=1}{\text{.15×10}}^{\text{10}}\\ {\text{β}}_{\text{4}}\text{of}{\text{Cu}}^{\text{2+}}\text{=}\text{1}{\text{.07×10}}^{\text{12}}\end{array}$

Therefore,

The Degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ of ${\text{Cu}}^{\text{2+}}$ at $\text{00}\text{.0}\text{mL}$ of $\text{EDTA}$ with $\text{1}\text{.00}\text{M}{\text{NH}}_{\text{3}}$ is,

$\begin{array}{l}{\text{αCu}}^{\text{2+}}\text{=}\frac{\text{1}}{\text{1+β(1}{\text{.00)+β}}_{\text{2}}{\text{(1}\text{.00)}}^{\text{2}}{\text{+β}}_{\text{3}}{\text{(1}\text{.00)}}^{\text{3}}{\text{+β}}_{\text{4}}{\text{(1}\text{.00)}}^{\text{4}}}\\ \text{=}\text{9}{\text{.2}}_{\text{3}}{\text{×10}}^{\text{-13}}\end{array}$

We know,

$\begin{array}{l}\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\\ ={4.8}_8\times {10}^{18}\end{array}$

$\begin{array}{l}\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\\ {\text{K}}_{\text{f}}\text{=}\frac{{\text{K}}_{\text{f}}\text{'}}{{\text{α}}_{{\text{Y}}^{\text{4-}}}}\\ \text{=}\frac{\text{4}{\text{.8}}_{\text{8}}{\text{×10}}^{\text{18}}}{\text{9}{\text{.2}}_{\text{3}}{\text{×10}}^{\text{-13}}}\\ ={4.5}_1\times {10}^6\end{array}$

The calculated $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ and given ${\text{K}}_{\text{f}}$ are plugged and rearrange in the above equation to give a ${\text{K}}_{\text{f}}\text{'}$

At $\text{0}\text{.00}\text{mL}$ $\text{EDTA}$, the concentration of ${\text{Cu}}^{\text{2+}}$ is $\text{0}\text{.001}\text{M}$

$\begin{array}{l}{\text{[Cu}}^{\text{2+}}\text{]}\text{=}{\text{αCu}}^{\text{2+}}{\text{+C}}_{{\text{Cu}}^{\text{2+}}}\\ \text{=}\text{9}{\text{.2}}_{\text{3}}{\text{×10}}^{\text{-16}}\text{M}\\ {\text{pCu}}^{\text{2+}}\text{=}{\text{-log}}_{\text{10}}{\text{[Cu}}^{\text{2+}}\text{]}\\ \text{=}\text{15}\text{.03}\end{array}$

The calculated ${\text{αCu}}^{\text{2+}}$ and given concentration of ${\text{Cu}}^{\text{2+}}$ are plugged in above equation to gives a ${\text{pCu}}^{\text{2+}}$.

The ${\text{pCu}}^{\text{2+}}$ at $\text{0}\text{.00}\text{mL}$ volume of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{15}\text{.03}$

Conclusion

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{1}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{pCu}}^{\text{2+}}$ should be calculated at $\text{1}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$.

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

${\text{MY}}^{\text{n4-}}\underset{}{\rightleftharpoons }{\text{MY}}^{\text{n+}}+{\text{Y}}^{\text{4-}}$

The formation constant for above complex reaction is,

${\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}$

Where,

${\text{K}}_{\text{f}}$ is formation constant

${\text{[M}}^{\text{+}}\text{]}$ is concentration of metal ion

${\text{[Y}}^{\text{4-}}\text{]}$ is concentration of ligand

${\text{[MY}}^{\text{n4-}}\text{]}$ is concentration of the acid

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

$\begin{array}{l}\text{α}{}_{{\text{Y}}^{\text{4-}}}\text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{{\text{[H}}_{\text{6}}{\text{Y}}^{\text{2+}}{\text{]+[H}}_{\text{5}}{\text{Y}}^{\text{+}}{\text{]+[H}}_{\text{4}}{\text{Y]+[H}}_{\text{3}}{\text{Y}}^{\text{-}}{\text{]+[H}}_{\text{2}}{\text{Y}}^{\text{2-}}{\text{]+[H}}_{\text{3}}{\text{Y}}^{\text{3-}}{\text{]+[Y}}^{\text{4-}}\text{]}}\\ \text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{\text{[EDTA]}}\end{array}$

If pH is fixed, the degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ is constant and ${\text{K}}_{\text{f}}$ will be ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}\text{][EDTA]}}$

Where,

${\text{K}}_{\text{f}}\text{' is the conditional formation constant}$

Fraction of free metal ion:

In the metal ligand equilibria the number of moles of metal and ligand are equilibrium with number of mole of complex formed.

$\begin{array}{l}\text{M+}\text{L}\underset{}{\rightleftharpoons }\text{ML}{\text{β}}_1\text{=}\frac{\text{[ML]}}{\text{[M][L]}}\\ \text{M+}2\text{L}\underset{}{\rightleftharpoons }{\text{ML}}_2{\text{β}}_2\text{=}\frac{{\text{[ML}}_2\text{]}}{{\text{[M][L]}}^2}\\ \text{M+}\text{nL}\underset{}{\rightleftharpoons }{\text{ML}}_{\text{n}}{\text{β}}_{\text{n}}\text{=}\frac{{\text{[ML}}_{\text{n}}\text{]}}{{\text{[M][L]}}^{\text{n}}}\end{array}$

Where,

${\text{β}}_{\text{n}}$ is cumulative formation constant

Fraction of free metal ion ${\text{α}}_{\text{M}}$ is,

${\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M}}_{\text{tot}}\text{]}}$

$\begin{array}{l}{\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M]{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}\text{}}}\\ =\frac{1}{{\text{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}}\end{array}$

Where,

$\text{M}\text{is free metal ion}$

Answer to Problem 12.16P

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{1}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ is $\text{15}\text{.05}$

Explanation of Solution

To calculate the ${\text{pCu}}^{\text{2+}}$ at given volume of $\text{EDTA}$

Given,

$\text{50 }\text{.0 mL of 0}{\text{.001 M Cu}}^{\text{2+}}$

$\text{0}\text{.0100 M EDTA}$

$\text{pH}\text{=11}\text{.00}$

End point is $\text{50}\text{.0}\text{mL}$

Volume of ${\text{NH}}_{\text{3}}$ is $\text{1}\text{.00}\text{mL}$

The complex formation reaction is,

${\text{Cu}}^{\text{2+}}\text{+EDTA}\underset{}{\overset{}{\rightleftharpoons }}{\text{CuY}}^{\text{2-}}$

At the give $\text{pH}\text{=11}\text{.0}$ the degree of dissociation is fixed and it is,

The formation constant ${\text{K}}_{\text{f}}$ for ${\text{Cu}}^{\text{2+}}$ is ${10}^{18.78}$

At $\text{1}\text{.00}\text{mL}$ $\text{EDTA}$, the concentration of ${\text{C}}_{{\text{Cu}}^{\text{2+}}}$ is,

The volume difference is $\text{50+1}\text{.00}\text{=}\text{51}$

$\begin{array}{l}{\text{C}}_{{\text{Cu}}^{\text{2+}}}\text{=}\left(\frac{\text{49}\text{.00}}{\text{50}\text{.0}}\right)\left(\text{0}\text{.001}\text{M}\right)\left(\frac{\text{50}\text{.00}}{\text{51}\text{.00}}\right)\\ \text{=}\text{9}{\text{.6×10}}^{\text{-4}}\text{M}\end{array}$

$\begin{array}{l}{\text{[Cu}}^{\text{2+}}\text{]}\text{=}{\text{αCu}}^{\text{2+}}{\text{+C}}_{{\text{Cu}}^{\text{2+}}}\\ \text{=}{8.8}_7{\text{×10}}^{\text{-16}}\text{M}\\ {\text{pCu}}^{\text{2+}}\text{=}{\text{-log}}_{\text{10}}{\text{[Cu}}^{\text{2+}}\text{]}\\ \text{=}\text{15}\text{.05}\end{array}$

The calculated concentration of ${\text{Cu}}^{\text{2+}}$ is plugged in above equation to gives a ${\text{pCu}}^{\text{2+}}$.

The ${\text{pCu}}^{\text{2+}}$ at $\text{1}\text{.00}\text{mL}$ of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{15}\text{.05}$

Conclusion

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{1}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ .

(c)

Interpretation Introduction

Interpretation:

The ${\text{pCu}}^{\text{2+}}$ should be calculated at $\text{45}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$.

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

${\text{MY}}^{\text{n4-}}\underset{}{\rightleftharpoons }{\text{MY}}^{\text{n+}}+{\text{Y}}^{\text{4-}}$

The formation constant for above complex reaction is,

${\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}$

Where,

${\text{K}}_{\text{f}}$ is formation constant

${\text{[M}}^{\text{+}}\text{]}$ is concentration of metal ion

${\text{[Y}}^{\text{4-}}\text{]}$ is concentration of ligand

${\text{[MY}}^{\text{n4-}}\text{]}$ is concentration of the acid

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

$\begin{array}{l}\text{α}{}_{{\text{Y}}^{\text{4-}}}\text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{{\text{[H}}_{\text{6}}{\text{Y}}^{\text{2+}}{\text{]+[H}}_{\text{5}}{\text{Y}}^{\text{+}}{\text{]+[H}}_{\text{4}}{\text{Y]+[H}}_{\text{3}}{\text{Y}}^{\text{-}}{\text{]+[H}}_{\text{2}}{\text{Y}}^{\text{2-}}{\text{]+[H}}_{\text{3}}{\text{Y}}^{\text{3-}}{\text{]+[Y}}^{\text{4-}}\text{]}}\\ \text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{\text{[EDTA]}}\end{array}$

If pH is fixed, the degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ is constant and ${\text{K}}_{\text{f}}$ will be ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}\text{][EDTA]}}$

Where,

${\text{K}}_{\text{f}}\text{' is the conditional formation constant}$

Fraction of free metal ion:

In the metal ligand equilibrium, the number of moles of metal and ligand are equilibrium with number of mole of complex formed.

$\begin{array}{l}\text{M+}\text{L}\underset{}{\rightleftharpoons }\text{ML}{\text{β}}_1\text{=}\frac{\text{[ML]}}{\text{[M][L]}}\\ \text{M+}2\text{L}\underset{}{\rightleftharpoons }{\text{ML}}_2{\text{β}}_2\text{=}\frac{{\text{[ML}}_2\text{]}}{{\text{[M][L]}}^2}\\ \text{M+}\text{nL}\underset{}{\rightleftharpoons }{\text{ML}}_{\text{n}}{\text{β}}_{\text{n}}\text{=}\frac{{\text{[ML}}_{\text{n}}\text{]}}{{\text{[M][L]}}^{\text{n}}}\end{array}$

Where,

${\text{β}}_{\text{n}}$ is cumulative formation constant

Fraction of free metal ion ${\text{α}}_{\text{M}}$ is,

${\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M}}_{\text{tot}}\text{]}}$

$\begin{array}{l}{\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M]{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}\text{}}}\\ =\frac{1}{{\text{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}}\end{array}$

Where,

$\text{M}\text{is free metal ion}$

Answer to Problem 12.16P

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{45}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ is $\text{16}\text{.30}$

Explanation of Solution

To calculate the ${\text{pCu}}^{\text{2+}}$ at given volume of $\text{EDTA}$

Given,

$\text{50 }\text{.0 mL of 0}{\text{.001 M Cu}}^{\text{2+}}$

$\text{0}\text{.0100 M EDTA}$

$\text{pH}\text{=11}\text{.00}$

End point is $\text{50}\text{.0}\text{mL}$

Volume of ${\text{NH}}_{\text{3}}$ is $\text{1}\text{.00}\text{mL}$

The complex formation reaction is,

${\text{Cu}}^{\text{2+}}\text{+EDTA}\underset{}{\overset{}{\rightleftharpoons }}{\text{CuY}}^{\text{2-}}$

At the give $\text{pH}\text{=11}\text{.0}$ the degree of dissociation is fixed and it is,

The formation constant ${\text{K}}_{\text{f}}$ for ${\text{Cu}}^{\text{2+}}$ is ${10}^{18.78}$

At $\text{45}\text{.00}\text{mL}$ $\text{EDTA}$, the concentration of ${\text{C}}_{{\text{Cu}}^{\text{2+}}}$ is,

The volume difference is $\text{50+ 45=}95$

$\begin{array}{l}{\text{C}}_{{\text{Cu}}^{\text{2+}}}\text{=}\left(\frac{\text{5}\text{.00}}{\text{50}\text{.0}}\right)\left(\text{0}\text{.001}\text{M}\right)\left(\frac{\text{50}\text{.00}}{\text{95}\text{.00}}\right)\\ \text{=}5.26{\text{×10}}^{\text{-5}}\text{M}\end{array}$

$\begin{array}{l}{\text{[Cu}}^{\text{2+}}\text{]}\text{=}{\text{αCu}}^{\text{2+}}{\text{+C}}_{{\text{Cu}}^{\text{2+}}}\\ \text{=}{5.0}_7{\text{×10}}^{\text{-17}}\text{M}\\ {\text{pCu}}^{\text{2+}}\text{=}{\text{-log}}_{\text{10}}{\text{[Cu}}^{\text{2+}}\text{]}\\ \text{=}\text{16}\text{.30}\end{array}$

The calculated concentration of ${\text{Cu}}^{\text{2+}}$ is plugged in above equation to gives a ${\text{pCu}}^{\text{2+}}$.

The ${\text{pCu}}^{\text{2+}}$ at $\text{45}\text{.00}\text{mL}$ of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{16}\text{.30}$

Conclusion

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{45}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ .

(d)

Interpretation Introduction

Interpretation:

The ${\text{pCu}}^{\text{2+}}$ should be calculated at $\text{50}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$.

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

${\text{MY}}^{\text{n4-}}\underset{}{\rightleftharpoons }{\text{MY}}^{\text{n+}}+{\text{Y}}^{\text{4-}}$

The formation constant for above complex reaction is,

${\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}$

Where,

${\text{K}}_{\text{f}}$ is formation constant

${\text{[M}}^{\text{+}}\text{]}$ is concentration of metal ion

${\text{[Y}}^{\text{4-}}\text{]}$ is concentration of ligand

${\text{[MY}}^{\text{n4-}}\text{]}$ is concentration of the acid

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

$\begin{array}{l}\text{α}{}_{{\text{Y}}^{\text{4-}}}\text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{{\text{[H}}_{\text{6}}{\text{Y}}^{\text{2+}}{\text{]+[H}}_{\text{5}}{\text{Y}}^{\text{+}}{\text{]+[H}}_{\text{4}}{\text{Y]+[H}}_{\text{3}}{\text{Y}}^{\text{-}}{\text{]+[H}}_{\text{2}}{\text{Y}}^{\text{2-}}{\text{]+[H}}_{\text{3}}{\text{Y}}^{\text{3-}}{\text{]+[Y}}^{\text{4-}}\text{]}}\\ \text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{\text{[EDTA]}}\end{array}$

If pH is fixed, the degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ is constant and ${\text{K}}_{\text{f}}$ will be ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}\text{][EDTA]}}$

Where,

${\text{K}}_{\text{f}}\text{' is the conditional formation constant}$

Fraction of free metal ion:

In the metal ligand equilibria the number of moles of metal and ligand are equilibrium with number of mole of complex formed.

$\begin{array}{l}\text{M+}\text{L}\underset{}{\rightleftharpoons }\text{ML}{\text{β}}_1\text{=}\frac{\text{[ML]}}{\text{[M][L]}}\\ \text{M+}2\text{L}\underset{}{\rightleftharpoons }{\text{ML}}_2{\text{β}}_2\text{=}\frac{{\text{[ML}}_2\text{]}}{{\text{[M][L]}}^2}\\ \text{M+}\text{nL}\underset{}{\rightleftharpoons }{\text{ML}}_{\text{n}}{\text{β}}_{\text{n}}\text{=}\frac{{\text{[ML}}_{\text{n}}\text{]}}{{\text{[M][L]}}^{\text{n}}}\end{array}$

Where,

${\text{β}}_{\text{n}}$ is cumulative formation constant

Fraction of free metal ion ${\text{α}}_{\text{M}}$ is,

${\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M}}_{\text{tot}}\text{]}}$

$\begin{array}{l}{\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M]{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}\text{}}}\\ =\frac{1}{{\text{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}}\end{array}$

Where,

$\text{M}\text{is free metal ion}$

Answer to Problem 12.16P

The ${\text{pCu}}^{\text{2+}}$ at $\text{50}\text{.00}\text{mL}$ of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{17}\text{.02}$

Explanation of Solution

To calculate the ${\text{pCu}}^{\text{2+}}$ at given volume of $\text{EDTA}$

Given,

$\text{50 }\text{.0 mL of 0}{\text{.001 M Cu}}^{\text{2+}}$

$\text{0}\text{.0100 M EDTA}$

$\text{pH}\text{=11}\text{.00}$

End point is $\text{50}\text{.0}\text{mL}$

Volume of ${\text{NH}}_{\text{3}}$ is $\text{1}\text{.00}\text{mL}$

The complex formation reaction is,

$\begin{array}{l}{\text{Cu}}^{\text{2+}}\text{+EDTA}\underset{}{\overset{}{\rightleftharpoons }}{\text{CuY}}^{\text{2-}}\\ \left(\frac{\text{50}\text{.0}}{\text{100}\text{.0}}\right)\left(\text{0}\text{.00100}\right)\text{-x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{f}}\text{''=}\frac{\text{0}\text{.000500-x}}{{\text{x}}^{\text{2}}}\\ \text{=}\text{4}{\text{.5×10}}^{\text{6}}\\ \text{=}\text{1}{\text{.04×10}}^{\text{-5}}\text{M}\\ \text{x=}{\text{C}}_{{\text{Cu}}^{\text{2+}}}\end{array}$

$\begin{array}{l}{\text{[Cu}}^{\text{2+}}\text{]}\text{=}{\text{αCu}}^{\text{2+}}{\text{+C}}_{{\text{Cu}}^{\text{2+}}}\\ \text{=}\text{9}{\text{.6}}_{\text{2}}{\text{×10}}^{\text{-18}}\text{M}\\ {\text{pCu}}^{\text{2+}}\text{=}{\text{-log}}_{\text{10}}{\text{[Cu}}^{\text{2+}}\text{]}\\ \text{=}\text{17}\text{.02}\end{array}$

The calculated concentration of ${\text{Cu}}^{\text{2+}}$ is plugged in above equation to gives a ${\text{pCu}}^{\text{2+}}$.

The ${\text{pCu}}^{\text{2+}}$ at $\text{50}\text{.00}\text{mL}$ of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{17}\text{.02}$

Conclusion

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{45}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ .

(e)

Interpretation Introduction

Interpretation:

The ${\text{pCu}}^{\text{2+}}$ should be calculated at $\text{55}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$.

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

${\text{MY}}^{\text{n4-}}\underset{}{\rightleftharpoons }{\text{MY}}^{\text{n+}}+{\text{Y}}^{\text{4-}}$

The formation constant for above complex reaction is,

${\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}$

Where,

${\text{K}}_{\text{f}}$ is formation constant

${\text{[M}}^{\text{+}}\text{]}$ is concentration of metal ion

${\text{[Y}}^{\text{4-}}\text{]}$ is concentration of ligand

${\text{[MY}}^{\text{n4-}}\text{]}$ is concentration of the acid

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

$\begin{array}{l}\text{α}{}_{{\text{Y}}^{\text{4-}}}\text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{{\text{[H}}_{\text{6}}{\text{Y}}^{\text{2+}}{\text{]+[H}}_{\text{5}}{\text{Y}}^{\text{+}}{\text{]+[H}}_{\text{4}}{\text{Y]+[H}}_{\text{3}}{\text{Y}}^{\text{-}}{\text{]+[H}}_{\text{2}}{\text{Y}}^{\text{2-}}{\text{]+[H}}_{\text{3}}{\text{Y}}^{\text{3-}}{\text{]+[Y}}^{\text{4-}}\text{]}}\\ \text{=}\frac{{\text{[Y}}^{\text{4-}}\text{]}}{\text{[EDTA]}}\end{array}$

If pH is fixed, the degree of dissociation $\text{α}{}_{{\text{Y}}^{\text{4-}}}$ is constant and ${\text{K}}_{\text{f}}$ will be ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}\text{][EDTA]}}$

Where,

${\text{K}}_{\text{f}}\text{' is the conditional formation constant}$

Fraction of free metal ion:

In the metal ligand equilibria the number of moles of metal and ligand are equilibrium with number of mole of complex formed.

$\begin{array}{l}\text{M+}\text{L}\underset{}{\rightleftharpoons }\text{ML}{\text{β}}_1\text{=}\frac{\text{[ML]}}{\text{[M][L]}}\\ \text{M+}2\text{L}\underset{}{\rightleftharpoons }{\text{ML}}_2{\text{β}}_2\text{=}\frac{{\text{[ML}}_2\text{]}}{{\text{[M][L]}}^2}\\ \text{M+}\text{nL}\underset{}{\rightleftharpoons }{\text{ML}}_{\text{n}}{\text{β}}_{\text{n}}\text{=}\frac{{\text{[ML}}_{\text{n}}\text{]}}{{\text{[M][L]}}^{\text{n}}}\end{array}$

Where,

${\text{β}}_{\text{n}}$ is cumulative formation constant

Fraction of free metal ion ${\text{α}}_{\text{M}}$ is,

${\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M}}_{\text{tot}}\text{]}}$

$\begin{array}{l}{\text{α}}_{\text{M}}\text{=}\frac{\text{[M]}}{{\text{[M]{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}\text{}}}\\ =\frac{1}{{\text{1+β1[L]+β}}_{\text{2}}{\text{[L]}}^{\text{2}}}\end{array}$

Where,

$\text{M}\text{is free metal ion}$

Answer to Problem 12.16P

The ${\text{pCu}}^{\text{2+}}$ at $\text{55}\text{.00}\text{mL}$ of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{17}\text{.69}$

Explanation of Solution

To calculate the ${\text{pCu}}^{\text{2+}}$ at given volume of $\text{EDTA}$

Given,

$\text{50 }\text{.0 mL of 0}{\text{.001 M Cu}}^{\text{2+}}$

$\text{0}\text{.0100 M EDTA}$

$\text{pH}\text{=11}\text{.00}$

End point is $\text{50}\text{.0}\text{mL}$

Volume of ${\text{NH}}_{\text{3}}$ is $\text{1}\text{.00}\text{mL}$

The complex formation reaction is,

${\text{Cu}}^{\text{2+}}\text{+EDTA}\underset{}{\overset{}{\rightleftharpoons }}{\text{CuY}}^{\text{2-}}$

$\begin{array}{l}\text{[EDTA]=}\left(\frac{\text{5}\text{.00}}{\text{105}\text{.00}}\right)\left(\text{0}\text{.00100}\text{M}\right)\\ \text{=}\text{4}{\text{.76×10}}^{\text{-5}}\text{M}\\ \\ {\text{[CuY}}^{\text{2-}}\text{]=}\left(\frac{\text{50}\text{.00}}{\text{105}\text{.00}}\right)\left(\text{0}\text{.00100}\text{M}\right)\\ \text{=}\text{4}{\text{.76×10}}^{\text{-4}}\text{M}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{f}}\text{'}\text{=}\frac{{\text{[CuY}}^{\text{2-]}}}{{\text{[Cu}}^{\text{2+}}\text{][EDTA]}}\\ \text{=}\frac{\text{[4}{\text{.76×10}}^{\text{-5}}\text{]}}{{\text{[Cu}}^{\text{2+}}\text{][4}{\text{.76×10}}^{\text{-4}}\text{]}}\\ {\text{[Cu}}^{\text{2+}}\text{]}\text{= 2}{\text{.05×10}}^{\text{-18}}\text{M}\\ {\text{pCu}}^{\text{2+}}\text{=}{\text{-log}}_{\text{10}}{\text{[Cu}}^{\text{2+}}\text{]}\\ \text{=}\text{17}\text{.69}\end{array}$

The calculated concentration of $\text{EDTA}$ and ${\text{CuY}}^{\text{2-}}$ are plugged in above equation to gives a ${\text{pCu}}^{\text{2+}}$.

The ${\text{pCu}}^{\text{2+}}$ at $\text{55}\text{.00}\text{mL}$ of $\text{EDTA}$ the presence of ${\text{NH}}_{\text{3}}$ in $\text{EDTA}$ titration is $\text{17}\text{.69}$

Conclusion

The ${\text{pCu}}^{\text{2+}}$ was calculated at $\text{55}\text{.00}\text{mL}$ of $\text{EDTA}$ in the presence of ${\text{NH}}_{\text{3}}$ .