Chapter 14, Problem 14.43P

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{f}}$ value for the metal –EDTA complex has to be calculated.

Concept introduction:

Nernst equation can be used to determine the cell potential at any instant, the difference from the standard state.

${\text{E = E}}^{\text{o }}\text{- }\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\frac{{\left[\text{B}\right]}^{\text{b}}{\left[{\text{H}}^{\text{+}}\right]}^{\text{m}}}{{\left[\text{A}\right]}^{\text{a}}}$

Where:

$\text{E}$ = cell potential under nonstandard conditions
$\text{E}{}_{}^{\text{o}}$ = cell potential under standard conditions
n = number of moles of electrons exchanged in the electrochemical reaction

Finding $\text{K}$ from ${\text{E}}^{\text{o}}$

$\text{K}\text{=}\frac{{\text{10}}^{{\text{nE}}^{\text{o}}}}{\text{0}\text{.05916}}\text{at}{\text{25}}^{\text{o}}\text{C}$

Where $\text{K}$ is equilibrium constant

Answer to Problem 14.43P

${\text{K}}_{\text{f}}$ value for the metal –EDTA complex is $\text{7}\text{.1×}{\text{10}}^{\text{14}}$.

Explanation of Solution

$\begin{array}{l}{\text{M}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{M}}_{\text{(s)}}{\text{E}}_{\text{+}}^{\text{o}}\text{=}\text{-0}\text{.266}\text{V}\\ \underset{\_}{{}^{\text{-}}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}\text{(0}\text{.50}\text{bar)}{\text{E}}_{\text{-}}^{\text{o}}\text{=}\text{0}\text{V}}\\ {\text{H}}_{\text{2(g)}}\text{+}{\text{M}}^{\text{2+}}\underset{}{\rightleftharpoons }{\text{2H}}^{\text{+}}\text{+}{\text{M}}_{\text{(s)}}{\text{E}}_{}^{\text{o}}\text{=}\text{-0}\text{.266}\text{V}\\ \text{E=}\text{-0}\text{.266-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{{\text{P}}_{{\text{H}}_{\text{2}}}\left[{\text{M}}^{\text{2+}}\right]}\end{array}$

Concentration of acid the left half-cell is found by titration of 28 mL of the tetraprotic pyrophosphoric acid with $\text{72 mL of KOH}$

$\begin{array}{l}\text{28}\text{mL}\text{of0}\text{.0100}\text{M}{\text{H}}_{\text{4}}\text{P}\text{=}\text{0}\text{.280}\text{mmol}\\ \text{72}\text{mL}\text{of0}\text{.0100}\text{M}\text{KOH}\text{=}\text{0}\text{.720}\text{mmol}\end{array}$

$\text{0}\text{.280}\text{mmol}{\text{OH}}^{\text{-}}$ consumed $\text{0}{\text{.280 mmol of H}}_{\text{4}}\text{P}$ giving $\text{0}{\text{.280 mmol of H}}_3{\text{P}}^{\text{-}}$ and

$\text{(0}\text{.720}\text{-}\text{0}\text{.280)}\text{=}\text{0}{\text{.440 mmol of OH}}^{\text{-}}$

$\begin{array}{l}\text{0}{\text{.280 mmol of OH}}^{\text{-}}\text{is}\text{consumes}\text{0}\text{.280}\text{mmol}\text{of}{\text{H}}_{\text{3}}{\text{P}}^{\text{-}}\text{giving0}\text{.280}\text{mmol}\text{of}{\text{H}}_{\text{2}}{\text{P}}^{\text{2-}}\\ \text{and}\text{(0}\text{.440-0280}\text{)}\text{=}\text{0}\text{.160}\text{mmol}\text{of}{\text{OH}}^{\text{-}}\text{finally}\text{0}\text{.160}\text{mmol}\text{of}{\text{OH}}^{\text{-}}\text{reacts}\text{with}\text{0}\text{.280}\text{mmol}{\text{ofH}}_{\text{2}}{\text{P}}^{\text{2-}}\\ \text{to}\text{create}\text{0}\text{.160}\text{mmol}\text{of}{\text{HP}}^{\text{3-}}\text{leaving}\text{0}\text{.120}\text{mmol}\text{of}\text{unreacted}{\text{H}}_{\text{2}}{\text{P}}^{\text{2-}}\end{array}$

$\begin{array}{l}{\text{H}}_{\text{4}}\text{P}\text{+}{\text{OH}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{P}}^{\text{2-}}\text{+}{\text{HP}}^{\text{3-}}\\ \text{0}\text{.280}\text{mmol}\text{0}\text{.720}\text{mmol}\text{0}\text{.120}\text{mmol}\text{0}\text{.160}\text{mmol}\\ \text{pH=}{\text{pK}}_{\text{3}}\text{+}\text{log}\frac{\left[{\text{HP}}^{\text{3-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{P}}^{\text{2-}}\right]}\\ \text{=}\text{6}\text{.70}\text{+}\text{log}\frac{\text{0}\text{.160}}{\text{0}\text{.120}}\\ \text{=}\text{6}\text{.82}\\ \left[{\text{H}}^{\text{+}}\right]\text{=1}\text{.50}\text{×}{\text{10}}^{\text{-7}}\text{M}\end{array}$

The known values is substituted into the Nernst equation gives

$\begin{array}{l}\text{-0}\text{.246}\text{=}\text{-0}\text{.266}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{{\text{P}}_{{\text{H}}_{\text{2}}}\left[{\text{M}}^{\text{2+}}\right]}\\ \text{=}\text{-0}\text{.266}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\left[\text{1}{\text{.5×10}}^{\text{-7}}\right]}^{\text{2}}}{\text{0}\text{.50}\left[{\text{M}}^{\text{2+}}\right]}\\ \text{=}\left[{\text{M}}^{\text{2+}}\right]\\ \text{=}\text{2}\text{.13}\text{×}{\text{10}}^{\text{-13}}\text{M}\end{array}$

The right half-cell we have the equilibrium

$\begin{array}{l}{\text{M}}^{\text{2+}}\text{+}{\text{F}}_{\text{EDTA}}\underset{}{\rightleftharpoons }{\text{MY}}^{\text{2-}}\\ \text{initial}\text{mmol/}\text{mL}\frac{\text{0}\text{.280}}{\text{100}}\frac{\text{0}\text{.720}}{\text{100}}\text{-}\\ \text{final}\text{mmol/mL}\text{small}\frac{\text{0}\text{.440}}{\text{100}}\frac{\text{0}\text{.280}}{\text{100}}\\ {\text{K}}_{\text{f}}\text{=}\frac{\left[{\text{MY}}^{\text{2-}}\right]}{\left[{\text{M}}^{\text{2+}}\right]\text{α}{}_{{\text{Y}}^{\text{4-}}}{\text{F}}_{\text{EDTA}}}\\ \text{=}\frac{\text{0}\text{.280/100}}{\text{(2}\text{.13}\text{×}{\text{10}}^{\text{-13}}\text{)(0}\text{.00042)(0}\text{.440/100)}}\text{=}\text{7}\text{.1×}{\text{10}}^{\text{14}}\end{array}$

Conclusion

${\text{K}}_{\text{f}}$ value for the metal –EDTA complex was found to be $\text{7}\text{.1×}{\text{10}}^{\text{14}}$