Chapter 12, Problem 12.7P

Interpretation Introduction

Interpretation:

The equivalence volume of $\text{EDTA}$ titration should be calculated.

Concept introduction:

Volumetric principle:

The volume and concentration of unknown solution is determined by it is titrate with known volume and concentration solution.

The volume and concentration of unknown solution is required equivalent volume and concentration of known solution in the volumetric titration.

${\text{V}}_{\text{1}}{\text{M}}_{\text{1}}\text{=}{\text{V}}_{\text{2}}{\text{M}}_{\text{2}}$

Where,

${\text{V}}_{\text{1}}$ is volume of known solution

${\text{N}}_{\text{1}}$ is concentration of known solution

${\text{V}}_{\text{1}}$ is volume of unknown solution

${\text{V}}_{\text{1}}$ is concentration of unknown solution

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 ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}$

Where,

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

${\text{α}}_{{\text{Y}}^{\text{4-}}}$ is degree of dissociation

Answer to Problem 12.7P

The volume of solution at equivalent point is $13.14\text{mL}$

The conditional formation value is $5.1\times {10}^{11}$

Explanation of Solution

To determine the volume of solution at equivalent point in $\text{EDTA}$ titration.

Given,

Volume of known solution is $\text{25}\text{.00 mL}$

Concentration of known solution is $\text{0}\text{.02026M}$

Concentration of $\text{EDTA}$ solution $\text{0}\text{.03855 M}$

$\text{pH}\text{=}6.\text{0}$

According to the volumetric principle,

$\begin{array}{l}{\text{V}}_{\text{e}}\text{×0}\text{.03855}\text{M}\text{=}25.0\text{ mL×}\text{0}\text{.020}\text{M}\\ \text{=}\frac{\text{25}\text{.0 mL×}\text{0}\text{.020}\text{M}}{\text{0}\text{.03855}\text{M}}\\ \text{=}13.14\text{mL}\end{array}$

The given values are plugged in above equation to give a volume of solution at equivalent point in $\text{EDTA}$ titration.

To give the degree of dissociation ${\text{α}}_{{\text{Y}}^{\text{4-}}}$ at $\text{9}\text{pH}$

Given,

From the standard data table the ${\text{α}}_{{\text{Y}}^{\text{4-}}}$ of ${\text{Co}}^{\text{2+}}$ at $\text{9}\text{pH}$ is,

$\begin{array}{l}\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\\ =(1.8\times {10}^{-5})({10}^{16.45})\\ =5.1\times {10}^{11}\end{array}$

The conditional formation value is $5.1\times {10}^{11}$

Conclusion

The equivalence volume of $\text{EDTA}$ titration was calculated.

(a)

Interpretation Introduction

Interpretation:

The ${\text{pCo}}^{\text{2+}}$ at $\text{12}\text{mL}$ is should be calculated.

Conditional Formation Constant:

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

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

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

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}$

Where,

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

${\text{α}}_{{\text{Y}}^{\text{4-}}}$ is degree of dissociation

Answer to Problem 12.7P

The ${\text{pCo}}^{\text{2+}}$ at $\text{12}\text{mL}$ is $\text{2}\text{.93}$

Explanation of Solution

To determine the ${\text{pCo}}^{\text{2+}}$ at $\text{12}\text{mL}$

Given,

Equivalent point in $\text{EDTA}$ titration is $\text{13}\text{.14}\text{mL}$

Volume of different is $\text{13}\text{.14-12=}\text{1}\text{.14}$

$\begin{array}{l}{\text{[Ca}}^{\text{2+}}\text{]}\text{=}\left(\frac{\text{13}\text{.14-12}\text{M}}{\text{0}\text{.03855}\text{M}}\right)\left(\text{0}\text{.02026}\text{M}\right)\left(\frac{\text{25}\text{.00}}{\text{37}\text{.00}}\right)\\ \text{=1}{\text{.19×10}}^{\text{-3}}\\ {\text{pCa}}^{\text{2+}}\text{=}{\text{10}}^{\text{1}{\text{.19×10}}^{\text{-3}}}\\ \text{=}\text{2}\text{.93}\end{array}$

The volume difference and concentration are plugged in above equation to give ${\text{pCo}}^{\text{2+}}$ at $\text{12}\text{mL}$.

Conclusion

The ${\text{pCo}}^{\text{2+}}$ was calculated.

(b)

Interpretation Introduction

Interpretation:

At the equivalent point, the ${\text{pCo}}^{\text{2+}}$ should be calculated.

Concept Information:

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 ${\text{K}}_{\text{f}}\text{'}$

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}$

Where,

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

${\text{α}}_{{\text{Y}}^{\text{4-}}}$ is degree of dissociation

Explanation of Solution

To calculate the ${\text{pCo}}^{\text{2+}}$ at the equivalent point

The equilibrium reaction is,

${\text{Co}}^{\text{2+}}\text{+EDTA}\underset{}{\rightleftharpoons }{\text{CoY}}^{\text{2-}}$

The concentration of ${\text{CoY}}^{\text{2-}}$ is,

$\begin{array}{l}{\text{[CoY}}^{\text{2-}}\text{]}\text{=}\left(\frac{\text{25}\text{.00}}{\text{38}\text{.00}}\right)\left(\text{0}\text{.02026}\text{M}\right)\\ \text{=}\text{1}{\text{.33×10}}^{\text{-2}}\text{M}\end{array}$

From the above calculation the ${\text{CoY}}^{\text{2-}}$ and the concentration of ${\text{Co}}^{\text{2+}}$ is,

$\begin{array}{l}{\text{Co}}^{\text{2+}}\text{+}\text{EDTA}\underset{}{\rightleftharpoons }{\text{CoY}}^{\text{2-}}\\ \text{x}\text{x}\text{1}{\text{.33×10}}^{\text{-2}}\text{-x}\end{array}$

Hence, the ${\text{Co}}^{\text{2+}}$ concentration is, taken x

$\begin{array}{l}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{=}\frac{\text{1}{\text{.33×10}}^{\text{-2}}\text{-x}}{{\text{x}}^{\text{2}}}\\ \text{5}{\text{.1×10}}^{\text{11}}\text{=}\frac{\text{1}{\text{.33×10}}^{\text{-2}}\text{-x}}{{\text{x}}^{\text{2}}}\\ \text{[x]}\text{=}\text{1}{\text{.6×10}}^{\text{-7}}\text{M}\\ \text{[x]=}{\text{[Co}}^{\text{2+}}\text{]}\text{=}{\text{-log}}_{\text{10}}\text{[1}{\text{.6×10}}^{\text{-7}}\text{]}\\ \text{=}\text{6}\text{.79}\end{array}$

The ${\text{K}}_{\text{f}}\text{'}$ and ${\text{CoY}}^{\text{2-}}$ concentrations are plugged in above equation and do some rearrangement to give the ${\text{pCo}}^{\text{2+}}$ at equivalent point.

The ${\text{pCo}}^{\text{2+}}$ at equivalent point is $\text{6}\text{.79}$.

Conclusion

At the equivalent point, the ${\text{pCo}}^{\text{2+}}$ was calculated.

(c)

Interpretation Introduction

Interpretation:

At $\text{14 mL}$, the ${\text{pCo}}^{\text{2+}}$ ion should be calculated.

Concept Information:

Conditional Formation Constant:

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

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

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

$\text{The formation constant}{\text{K}}_{\text{f}}\text{'=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}$

Where,

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

${\text{α}}_{{\text{Y}}^{\text{4-}}}$ is degree of dissociation

Answer to Problem 12.7P

At $\text{14 mL}$, the ${\text{pCo}}^{\text{2+}}$ ion is $\text{10}\text{.52}$

Explanation of Solution

To calculate the ${\text{pCo}}^{\text{2+}}$ ion at $\text{14 mL}$

Given,

To calculate the ${\text{pCo}}^{\text{2+}}$ at $\text{14 mL}$

The equilibrium reaction is,

${\text{Co}}^{\text{2+}}\text{+EDTA}\underset{}{\rightleftharpoons }{\text{CoY}}^{\text{2-}}$

The concentration of ${\text{CoY}}^{\text{2-}}$ is,

$\begin{array}{l}{\text{[CoY}}^{\text{2-}}\text{]}\text{=}\left(\frac{\text{25}\text{.00}}{\text{39}\text{.00}}\right)\left(\text{0}\text{.02026}\text{M}\right)\\ \text{=}\text{1}{\text{.30×10}}^{\text{-2}}\text{M}\end{array}$

Formal concentration of EDTA is,

$\begin{array}{l}\text{[EDTA]}\text{=}\left(\frac{\text{14}\text{.0-13}\text{.14}}{\text{39}}\right)\left(\text{0}\text{.03855}\text{M}\right)\\ \text{=}\text{8}{\text{.50×10}}^{\text{-4}}\text{M}\end{array}$

The ${\text{pCo}}^{\text{2+}}$ ion at $\text{14 mL}$ is,

$\begin{array}{l}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[CoY}}^{\text{n-4}}\text{]}}{{\text{[Co}}^{\text{2+}}{\text{][Y}}^{\text{4-}}\text{]}}\\ {\text{K}}_{\text{f}}\text{=}\text{5}{\text{.1×10}}^{\text{11}}\text{M}\\ {\text{[Co}}^{\text{2+}}\text{]=}\frac{\text{1}{\text{.30×10}}^{\text{2}}\text{M}}{\text{8}{\text{.50×10}}^{\text{-4}}\text{M(5}{\text{.1×10}}^{\text{11}}\text{)}}\\ {\text{pCo}}^{\text{2+}}\text{=}{\text{-log}}_{\text{10}}{\text{[Co}}^{\text{2+}}\text{]}\\ \text{=}\text{10}\text{.52}\end{array}$

The calculated EDTA concentration and ${\text{K}}_{\text{f}}$ are plugged in above equation to give ${\text{pCo}}^{\text{2+}}$ ion at $\text{14 mL}$ is $\text{10}\text{.52}$.

Conclusion

At $\text{14 mL}$, the ${\text{pCo}}^{\text{2+}}$ ion was calculated.