Chapter 12, Problem 12.6P

(a)

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

Answer to Problem 12.6P

The volume of solution at equivalent point is $\text{100}\text{.0}\text{mL}$

Explanation of Solution

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

Given,

Volume of known solution is $\text{100}\text{.0 mL}$

Concentration of known solution is $\text{0}\text{.0500 M}$

Concentration of $\text{EDTA}$ solution $\text{0}\text{.0500 M}$

$\text{pH}\text{=}\text{9}\text{.0}$

According to the volumetric principle,

$\begin{array}{l}{\text{V}}_{\text{e}}\text{×0}\text{.0500}\text{M}\text{=}\text{100}\text{.0 mL×}\text{0}\text{.0500}\text{M}\\ \text{=}\frac{\text{100}\text{.0 mL×}\text{0}\text{.0500}\text{M}}{\text{0}\text{.500}\text{M}}\\ \text{=}\text{100}\text{.0}\text{mL}\end{array}$

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

Conclusion

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

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\text{M}}^{\text{n+}}$ given solution should be calculated.

Concept introduction:

Molarity:

The number of moles of solute present in liter of volume of solution is given by molarity.

$\text{Molarity}\text{=}\frac{\text{moles}}{\text{Volume}\text{(L)}}$

Answer to Problem 12.6P

The concentration of ${\text{M}}^{\text{n+}}$ is given solution is $\text{0}\text{.0167}\text{M}$

Explanation of Solution

To determine the concentration of ${\text{M}}^{\text{n+}}$ is given solution.

Given,

Equivalent point in $\text{EDTA}$ titration is $\text{100}\text{.0 mL}$

Volume of solution is $\text{V}\text{=}\frac{\text{1}}{\text{2}}{\text{V}}_{\text{e}}$

Where,

$\frac{\text{1}}{\text{2}}$ is a fraction remaining

Total volume of solution is,

$\begin{array}{l}\text{V}\text{=}\frac{\text{1}}{\text{2}}{\text{V}}_{\text{e}}\\ =\frac{100}{2}=50\\ =100+50\\ =150\end{array}$

The concentration of ${\text{M}}^{\text{n+}}$ ion is,

$\begin{array}{l}{\text{[Mn}}^{\text{n+}}\text{]}\text{=}\left(\frac{\text{1}}{\text{2}}\right)\text{×(0}\text{.0500M)×}\left(\frac{\text{100}}{\text{150}}\right)\\ \text{=}\text{0}\text{.0167}\text{M}\end{array}$

The fraction remaining and volume of solution are plugged in above equation to give the concentration of ${\text{M}}^{\text{n+}}$ ion.

Conclusion

The concentration of ${\text{M}}^{\text{n+}}$ was calculated.

(c)

Interpretation Introduction

Interpretation:

At $\text{9}\text{pH}$, the degree of dissociation ${\text{α}}_{{\text{Y}}^{\text{4-}}}$ of $\text{EDTA}$ should be given.

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 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 $\text{EDTA}$ at $\text{9}\text{pH}$ is $\text{0}\text{.041}$

Conclusion

At $\text{9}\text{pH}$, the degree of dissociation ${\text{α}}_{{\text{Y}}^{\text{4-}}}$ of $\text{EDTA}$ was given.

(d)

Interpretation Introduction

Interpretation:

The conditional formation constant 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 conditional formation constant

Given,

The formation constant ${\text{K}}_{\text{f}}$ is ${\text{10}}^{12.00}$

From the standard data table the ${\text{α}}_{{\text{Y}}^{\text{4-}}}$ of $\text{EDTA}$ at $\text{9}\text{pH}$ is $\text{0}\text{.041}$

$\begin{array}{l}\text{The formation constant}{\text{K}}_{\text{f}}\text{'}\text{=}{\text{α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\\ ={10}^{12.00}\times 0.041\\ =4.1\times {10}^{10}\end{array}$

The degree of dissociation and formation constant are plugged ion above equation to give conditional formation constant.

Conclusion

The conditional formation constant was calculated.

(e)

Interpretation Introduction

Interpretation:

At $\text{V}\text{=}{\text{V}}_{\text{e}}$, the concentration of ${\text{M}}^{\text{n+}}$ 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 12.6P

At $\text{V}\text{=}{\text{V}}_{\text{e}}$, the concentration of ${\text{M}}^{\text{n+}}$ ion is $\text{7}{\text{.8×10}}^{\text{-7}}$

Explanation of Solution

To calculate the concentration of ${\text{M}}^{\text{n+}}$ ion at $\text{V}\text{=}{\text{V}}_{\text{e}}$

Given,

Total volume of solution is $\text{200}\text{mL}$

Concentration of titrants is $\text{0}\text{.0500}\text{M}$

${\text{K}}_{\text{f}}=4.1\times {10}^{10}$

The concentration of ${\text{MY}}^{\text{n-4}}$ is,

$\begin{array}{l}{\text{[MY}}^{\text{n-4}}\text{]}\text{=}\text{0}\text{.0500}\text{M}\left(\frac{\text{100}}{\text{200}}\right)\\ \text{=}\text{0}\text{.0250}\text{M}\end{array}$

We know,

$\begin{array}{l}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}\\ \text{4}{\text{.1×10}}^{\text{10}}\text{=}\frac{\text{0}\text{.0250-x}}{{\text{x}}^{\text{2}}}\\ \text{x}\text{=}\text{7}{\text{.8×10}}^{\text{-7}}\end{array}$

The calculated concentration of ${\text{MY}}^{\text{n-4}}$ and ${\text{K}}_{\text{f}}$ are plugged in above equation to give concentration of ${\text{M}}^{\text{n+}}$ ion at $\text{V}\text{=}{\text{V}}_{\text{e}}$.

Conclusion

At $\text{V}\text{=}{\text{V}}_{\text{e}}$, the concentration of ${\text{M}}^{\text{n+}}$ ion was calculated.

(f)

Interpretation Introduction

Interpretation:

At $\text{V = 1}{\text{.100 V}}_{\text{e}}$, the concentration of ${\text{M}}^{\text{n+}}$ 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 12.6P

At $\text{V = 1}{\text{.100 V}}_{\text{e}}$, the concentration of ${\text{M}}^{\text{n+}}$ ion is $\text{2}{\text{.4×10}}^{\text{-10}}$

Explanation of Solution

To calculate the concentration of ${\text{M}}^{\text{n+}}$ ion at $\text{V = 1}{\text{.100 V}}_{\text{e}}$

Given,

Total volume of $\text{EDTA}$ is $\text{100}\text{mL}$

Concentration of $\text{EDTA}$ is $\text{0}\text{.0500}\text{M}$

${\text{K}}_{\text{f}}=4.1\times {10}^{10}$

The concentration of $\text{EDTA}$ is,

$\begin{array}{l}\text{[EDTA]}\text{=}\text{0}\text{.0500}\text{M}\left(\frac{\text{10}\text{.0}}{\text{210}}\right)\\ \text{=}2.\text{38}\times {\text{10}}^{-3}\text{M}\end{array}$

The concentration of ${\text{[MY}}^{\text{n-4}}\text{]}$ is,

$\begin{array}{l}{\text{[MY}}^{\text{n-4}}\text{]}\text{=}\text{0}\text{.0500}\text{M}\left(\frac{\text{100}}{\text{210}}\right)\\ \text{=}3.\text{38}\times {\text{10}}^{-2}\text{M}\end{array}$

We know,

$\begin{array}{l}{\text{K}}_{\text{f}}\text{=}\frac{{\text{[MY}}^{\text{n-4}}\text{]}}{{\text{[M}}^{\text{n+}}{\text{][Y}}^{\text{4-}}\text{]}}\\ \text{4}{\text{.1×10}}^{\text{10}}\text{=}\frac{\text{[2}{\text{.38×10}}^{\text{-2}}\text{]}}{{\text{[m}}^{\text{n+}}\text{]}\text{[2}{\text{.38×10}}^{\text{-3}}\text{]}}\\ {\text{[m}}^{\text{n+}}\text{]}\text{=}\text{2}{\text{.4×10}}^{\text{-10}}\end{array}$

The calculated concentrations of ${\text{MY}}^{\text{n-4}}$, ${\text{[Y}}^{\text{4-}}\text{]}$ and ${\text{K}}_{\text{f}}$ are plugged in above equation to give concentration of ${\text{M}}^{\text{n+}}$ ion at $\text{V = 1}{\text{.100 V}}_{\text{e}}$.

Conclusion

At $\text{V = 1}{\text{.100 V}}_{\text{e}}$, the concentration of ${\text{M}}^{\text{n+}}$ ion was calculated.