Chapter 15, Problem 15.AE

Interpretation Introduction

Interpretation:

The cell voltage at each volume of $\text{NaBr}$ has to be calculated and plotted.

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 titration reaction is a reduction reaction.

${\text{Ag}}^{\text{+}}{\text{+e}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{Ag}\left(\text{s}\right)$

The cell voltage of the reaction is written as

Cell Voltage:

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

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$ ${\text{=E}}_{\text{+}}\text{-E}\left(\text{S}\text{.C}\text{.E}\right){\text{=E}}_{\text{+}}-0.241$

Where,

${\text{E}}_{\text{+}}\text{=}\left(\text{0}\text{.799-0}\text{.05916log}\frac{\text{1}}{\left[{\text{Ag}}^{\text{+}}\right]}\right)$

$E_{+}=\left(\text{0}\text{.799-0}\text{.05916log}\frac{\text{1}}{\left[{\text{Ag}}^{\text{+}}\right]}\right)-0.241$

$\text{E =0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

Another titration reaction is

${\text{Br}}^{\text{-}}{\text{+Ag}}^{\text{+}}\to \text{AgBr}\left(\text{s}\right)$

The solubility constant for $\text{AgBr}\left(\text{s}\right)$ is $\text{5}{\text{.0×10}}^{\text{-13}}$.

The given equivalence point is ${\text{V}}_{\text{e}}=25.0$ which is between $\text{0}\text{to}\text{25}\text{mL}$

The unreacted ${\text{Ag}}^{\text{+}}$ in the solution.

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

At $\text{1}\text{.0}\text{mL:}$ $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\left(\frac{\text{24}\text{.0}\text{mL}}{\text{25}\text{.0}\text{mL}}\right)\left(\text{0}\text{.100}\text{M}\right)\left(\frac{\text{50}\text{.0}\text{mL}}{\text{51}\text{.0}\text{mL}}\right)\text{=0}\text{.0941}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{0}\text{.0941}\right]\text{=0}\text{.497}\text{V}$

At $\text{12}\text{.5}\text{mL:}$ $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\left(\frac{\text{12}\text{.5}\text{mL}}{\text{25}\text{.0}\text{mL}}\right)\left(\text{0}\text{.100}\text{M}\right)\left(\frac{\text{50}\text{.0}\text{mL}}{\text{62}\text{.5}\text{mL}}\right)\text{=0}\text{.0400}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{0}\text{.0400}\right]\text{=0}\text{.475}\text{V}$

At $\text{24}\text{.0}\text{mL:}$ $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\left(\frac{\text{1}\text{.0}\text{mL}}{\text{25}\text{.0}\text{mL}}\right)\left(\text{0}\text{.100}\text{M}\right)\left(\frac{\text{50}\text{.0}\text{mL}}{\text{74}\text{.0mL}}\right)\text{=0}\text{.00270}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{0}\text{.002700}\right]\text{=0}\text{.406}\text{V}$

At $\text{24}\text{.9}\text{mL:}$ $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\left(\frac{\text{0}\text{.10}\text{mL}}{\text{25}\text{.0}\text{mL}}\right)\left(\text{0}\text{.100}\text{M}\right)\left(\frac{\text{50}\text{.0}\text{mL}}{\text{74}\text{.9mL}}\right)\text{=2}\text{.670}\times {\text{10}}^{\text{-4}}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{2}{\text{.6×10}}^{\text{-4}}\text{M}\right]\text{=0}\text{.347}\text{V}$

After $\text{25}\text{mL}$, all $\text{AgBr}$ has precipitated and there is excess ${\text{Br}}^{\text{-}}$ in solution.

At $\text{25}\text{.1}\text{mL:}$ $\left[{\text{Br}}^{\text{-}}\right]\text{=}\left(\frac{\text{0}\text{.10}\text{mL}}{\text{75}\text{.1}\text{mL}}\right)\left(\text{0}\text{.200}\text{M}\right)\text{=2}{\text{.670×10}}^{\text{-4}}\text{M}$

$\left[{\text{Ag}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{SP}}}{\left[{\text{Br}}^{\text{-}}\right]}\text{=}\frac{\left(\text{5}{\text{.0×10}}^{\text{-13}}\right)}{\left(\text{2}{\text{.67×10}}^{\text{-4}}\right)}\text{=1}{\text{.88×10}}^{\text{-9}}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{1}{\text{.88×10}}^{\text{-9}}\text{M}\right]\text{=+0}\text{.0042}\text{V}$

At $\text{26}\text{.0}\text{mL:}$ $\left[{\text{Br}}^{\text{-}}\right]\text{=}\left(\frac{\text{0}\text{.10}\text{mL}}{\text{76}\text{.0}\text{mL}}\right)\left(\text{0}\text{.200}\text{M}\right)\text{=2}\text{.66}{\text{×10}}^{\text{-4}}\text{M}$

$\left[{\text{Ag}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{SP}}}{\left[{\text{Br}}^{\text{-}}\right]}\text{=}\frac{\left(\text{5}{\text{.0×10}}^{\text{-13}}\right)}{\left(\text{2}{\text{.66×10}}^{\text{-4}}\right)}\text{=1}{\text{.90×10}}^{\text{-9}}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{1}{\text{.90×10}}^{\text{-9}}\text{M}\right]\text{=-0}\text{.0017}\text{V}$

At $35.0\text{mL:}$ $\left[{\text{Br}}^{\text{-}}\right]\text{=}\left(\frac{\text{0}\text{.10}\text{mL}}{\text{85}\text{.0}\text{mL}}\right)\left(\text{0}\text{.200}\text{M}\right)\text{=0}\text{.0235}\text{M}$

$\left[{\text{Ag}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{SP}}}{\left[{\text{Br}}^{\text{-}}\right]}\text{=}\frac{\left(\text{5}{\text{.0×10}}^{\text{-13}}\right)}{\left(\text{0}\text{.0235}\text{M}\right)}\text{=1}{\text{.90×10}}^{\text{-9}}\text{M}$

$\text{E=0}\text{.558+0}\text{.05916log}\left[{\text{Ag}}^{\text{+}}\right]$

$\text{E}\text{=0}\text{.558+0}\text{.05916log}\left[\text{0}\text{.0235}\right]\text{=-0}\text{.073}\text{V}$

All the obtained potential are plotted as,

Figure 1

Conclusion

The cell voltage at each volume of $\text{NaBr}$ has to be calculated and plotted.