Chapter 15, Problem 15.BE

Interpretation Introduction

Interpretation:

The cell voltage at each volumes of added $\text{EDTA}$ has to be calculated.

Concept Introduction:-

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

The unreacted concentration in the solution is determined by

$\left[\text{X}\right]\text{=}\left(\text{fraction}\text{of}\left[\text{X}\right]\text{remaining}\right)\left(\text{initial}\text{concentration}\text{of}\text{X}\right)\left(\text{dilution}\text{factor}\right)$

Ion-selective electrode:

An ion-selective electrode (ISE) best known as a specific ion electrode (SIE), is a sensor (or transducer) that transforms the activity of a specific ion dissolved in a solution into an electrical potential.

Ion-selective electrode obeys following equation.

$\text{E=constant-0}\text{.05916log}\left[{\text{CN}}^{\text{-}}\right]$

Explanation of Solution

Given:

The given reactions are the small amount of ${\text{HgY}}^{\text{2-}}$ added to analyte equilibrates with a tiny amount of ${\text{Hg}}^{\text{2+}}$

The titration reaction is a reduction reaction.

$\begin{array}{l}{\text{Hg}}^{\text{2+}}{\text{+Y}}^{\text{4-}}\stackrel{}{\rightleftharpoons }{\text{HgY}}^{\text{2-}}\\ {\text{K}}_{\text{f}}\text{=}\frac{\left[{\text{HgY}}^{\text{2-}}\right]}{\left[{\text{Hg}}^{\text{2+}}\right]\left[{\text{Y}}^{\text{4-}}\right]}{\text{=10}}^{\text{21}\text{.5}}\mathrm{...}\left(\text{1}\right)\end{array}$

The Nernst equation for the cell is

${\text{E=E}}_{\text{+}}{\text{-E}}_{\text{-}}\text{=}\left(\text{0}\text{.852-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\frac{\text{1}}{\left[{\text{Hg}}^{\text{2+}}\right]}\right)\right){\text{-E}}_{\text{-}}\mathrm{...}\left(2\right)$

Where $\left[{\text{Hg}}^{\text{2+}}\right]$ can be substituted with equation 1 that is $\left[{\text{HgY}}^{\text{2-}}\right]{\text{/K}}_{\text{f}}\left[{\text{Y}}^{\text{4-}}\right]$

$\begin{array}{l}{\text{E=E}}_{\text{+}}{\text{-E}}_{\text{-}}\text{=}\left(\text{0}\text{.852-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\frac{\left[{\text{Y}}^{\text{4-}}\right]{\text{K}}_{\text{f}}}{\left[{\text{HgY}}^{\text{2-}}\right]}\right)\right){\text{-E}}_{\text{-}}\\ \text{=0}{\text{.852-E}}_{\text{-}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\frac{{\text{K}}_{\text{f}}}{\left[{\text{HgY}}^{\text{2-}}\right]}\right)\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left[{\text{Y}}^{\text{4-}}\right]\mathrm{...}\left(3\right)\end{array}$

To calculate the voltage we have to calculate the concentration of $\left[{\text{HgY}}^{\text{2-}}\right]$ and $\left[{\text{Y}}^{\text{4-}}\right]$ at every point.

The concentration of $\left[{\text{HgY}}^{\text{2-}}\right]\text{=1}{\text{.0×10}}^{\text{4-}}\text{M}$ .

If $\text{V=0}$ the dilution will be effected ${\text{K}}_{\text{f}}\left({\text{HgY}}^{\text{2-}}\right)\gg {\text{K}}_{\text{f}}\left({\text{MgY}}^{\text{2-}}\right)$

$\left[\text{X}\right]\text{=}\left(\text{fraction}\text{of}\left[\text{X}\right]\text{remaining}\right)\left(\text{initial}\text{concentration}\text{of}\text{X}\right)\left(\text{dilution}\text{factor}\right)$

At $\text{0}\text{mL:}$ $\frac{\left[{\text{Hg}}^{\text{+}}\right]}{\left[{\text{Hg}}^{\text{2+}}\right]\left[\text{EDTA}\right]}{\text{=α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{(for}{\text{HgY}}^{\text{4-}}\text{)}$

$\left[{\text{Y}}^{\text{4-}}\right]{\text{=α}}_{{\text{Y}}^{\text{4-}}}\left[\text{EDTA}\right]\text{=9}{\text{.7×10}}^{\text{-14}}\text{M}$

Using equation 3,

$\text{E}\text{=0}\text{.852-0}\text{.241-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{10}}^{\text{21}\text{.5}}}{\text{1}{\text{.0×10}}^{\text{-4}}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\text{9}{\text{.7×10}}^{\text{-14}}\right)\text{=0}\text{.242}\text{V}$

$\text{E}\text{=0}\text{.242}\text{V}$

At $\text{10}\text{.0}\text{mL:}$

${\text{V}}_{\text{e}}\text{=25}\text{.0}\text{mL}$ Thus $\frac{\text{10}}{\text{25}}\text{of}{\text{Mg}}^{\text{2+}}$ is in the form ${\text{MgY}}^{\text{2-}}$

And $\frac{\text{15}}{\text{25}}\text{of}\text{Mg}$

$\left[{\text{Y}}^{\text{4-}}\right]\text{=}\frac{\left[{\text{MgY}}^{\text{2-}}\right]}{\left[{\text{Mg}}^{\text{2+}}\right]}{\text{/K}}_{\text{f}}$

$\left[{\text{Y}}^{\text{4-}}\right]\text{=}\frac{\left[\text{10}\right]}{\left[\text{15}\right]}\text{/6}{\text{.2×10}}^{\text{8}}\text{=1}{\text{.08×10}}^{\text{-9}}\text{M}$

$\left[{\text{HgY}}^{\text{2-}}\right]\text{=}\left(\frac{\text{50}\text{.0}\text{mL}}{\text{60}\text{.0}\text{mL}}\right)\left(\text{1}{\text{.0×10}}^{\text{-4}}\text{M}\right)$= $\text{8}{\text{.33×10}}^{\text{-5}}\text{M}$

$\text{E}\text{=0}\text{.852-0}\text{.241-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{10}}^{\text{21}\text{.5}}}{\text{8}{\text{.33×10}}^{\text{-5}}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\text{1}{\text{.08×10}}^{\text{-9}}\right)\text{=0}\text{.120}\text{V}$

At $\text{20}\text{.0}\text{mL:}$

$\left[{\text{Y}}^{\text{4-}}\right]\text{=}\frac{\left[{\text{MgY}}^{\text{2-}}\right]}{\left[{\text{Mg}}^{\text{2+}}\right]}{\text{/K}}_{\text{f}}\text{=}\left(\frac{\text{20}}{\text{5}}\right)\text{/6}{\text{.2×10}}^{\text{8}}\text{=6}{\text{.45×10}}^{\text{-9}}\text{M}$

$\left[{\text{HgY}}^{\text{2-}}\right]\text{=}\frac{\left[{\text{MgY}}^{\text{2-}}\right]}{\left[{\text{Mg}}^{\text{2+}}\right]}{\text{/K}}_{\text{f}}\text{=}\left(\frac{\text{50}\text{.0}}{\text{70}\text{.0}}\right)\text{/1}{\text{.0×10}}^{\text{-4}}\text{=7}{\text{.14×10}}^{\text{-5}}\text{M}$

$\text{E}\text{=0}\text{.852-0}\text{.241-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{10}}^{\text{21}\text{.5}}}{\text{7}{\text{.14×10}}^{\text{-5}}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\text{6}{\text{.45×10}}^{\text{-9}}\right)\text{=0}\text{.095}\text{V}$

At $\text{24}\text{.9}\text{mL:}$

$\left[{\text{Y}}^{\text{4-}}\right]\text{=}\frac{\left[{\text{MgY}}^{\text{2-}}\right]}{\left[{\text{Mg}}^{\text{2+}}\right]}{\text{/K}}_{\text{f}}\text{=}\left(\frac{\text{24}\text{.9}}{0.1}\right)\text{/6}{\text{.2×10}}^{\text{8}}\text{=4}\text{.0}\times {\text{10}}^{\text{-7}}\text{M}$

$\left[{\text{HgY}}^{\text{2-}}\right]\text{=}\frac{\left[{\text{MgY}}^{\text{2-}}\right]}{\left[{\text{Mg}}^{\text{2+}}\right]}{\text{/K}}_{\text{f}}\text{=}\left(\frac{\text{50}\text{.0}}{\text{74}\text{.9}}\right)\text{/1}{\text{.0×10}}^{\text{-4}}\text{=6}{\text{.48×10}}^{\text{-5}}\text{M}$

$\text{E}\text{=0}\text{.852-0}\text{.241-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{10}}^{\text{21}\text{.5}}}{\text{6}{\text{.48×10}}^{\text{-5}}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\text{6}{\text{.68×10}}^{\text{-9}}\right)\text{=0}\text{.041V}$

After $\text{25}\text{mL}$, this is the equivalence point where $\left[{\text{Mg}}^{\text{2+}}\right]\text{=}\left[\text{EDTA}\right]$

$\frac{\left[{\text{Mg}}^{\text{+}}\right]}{\left[{\text{Mg}}^{\text{2+}}\right]\left[\text{EDTA}\right]}{\text{=α}}_{{\text{Y}}^{\text{4-}}}{\text{K}}_{\text{f}}\text{(for}{\text{MgY}}^{\text{4-}}\text{)}$

$\begin{array}{l}\frac{\left(\frac{\text{50}\text{.0}}{\text{75}\text{.0}}\right)\left(\text{0}\text{.0100}\right)\text{-x}}{{\text{x}}^{\text{2}}}\text{=1}{\text{.85×10}}^{\text{8}}\\ \text{x=6}{\text{.0×10}}^{\text{-6}}\text{M}\end{array}$

$\left[{\text{Y}}^{\text{4-}}\right]\text{=}{\alpha }_{{\text{Y}}^{\text{4-}}}\left(\text{6}{\text{.0×10}}^{-6}\right)\text{=1}\text{.80}\times {\text{10}}^{\text{-6}}\text{M}$

$\left[{\text{HgY}}^{\text{2-}}\right]\text{=}\left(\frac{50.0}{75.0}\right)\left(\text{1}{\text{.0×10}}^{-4}\right)\text{=6}\text{.67}\times {\text{10}}^{\text{-5}}\text{M}$

$\text{E}\text{=0}\text{.852-0}\text{.241-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{10}}^{\text{21}\text{.5}}}{\text{6}{\text{.67×10}}^{\text{-5}}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\text{1}{\text{.0×10}}^{\text{-6}}\text{M}\right)\text{=0}\text{.021V}$

$\text{E=0}\text{.021V}$

At $\text{26}\text{.0}\text{mL}$ the excess EDTA in the solution has to be calculated.

$\left[{\text{Y}}^{\text{4-}}\right]{\text{=α}}_{{\text{Y}}^{\text{4-}}}\left(\text{EDTA}\right)\text{=}\left(\text{0}\text{.30}\right)\left[\left(\frac{\text{1}\text{.0}\text{mL}}{\text{76}\text{.0}\text{mL}}\right)\left(\text{0}\text{.0200}\text{M}\right)\right]\text{=7}{\text{.89×10}}^{\text{-6}}\text{M}$

$\left[{\text{HgY}}^{\text{2-}}\right]\text{=}\left(\frac{50.0}{76.0}\right)\left(\text{1}{\text{.0×10}}^{-4}\right)\text{=6}\text{.58}\times {\text{10}}^{\text{-5}}\text{M}$

$\text{E}\text{=0}\text{.852-0}\text{.241-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{10}}^{\text{21}\text{.5}}}{\text{6}{\text{.58×10}}^{\text{-5}}}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\left(\text{7}\text{.89}\times {\text{10}}^{\text{-6}}\text{M}\right)\text{=0}\text{.027V}$

$\text{E=-0}\text{.027}\text{V}$

Conclusion

The cell voltage at each volumes of added $\text{EDTA}$ was calculated.