Chapter 17, Problem 17.26P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ range for predominant sulfurous acid forms has to be found.

Concept Introduction:

$\text{pH}$ is calculated by Henderson-HasselBalch equation given as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Explanation of Solution

Equilibrium for dissociation of is given as follows:

  ${\text{H}}_2{\text{SO}}_3\rightleftharpoons {\text{HSO}}_3{}^{-}+{\text{H}}^{+}$

Value of $\text{pH}$ is calculated by Henderson-HasselBalch equation given as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{HSO}}_3{}^{-}\right]}{\left[{\text{H}}_2{\text{SO}}_3\right]}$        (1)

Substitute 1.857 for $\text{p}{K}_{\text{a}}$ in equation (1).

  $\text{pH}=1.857+\log \frac{\left[{\text{HSO}}_3{}^{-}\right]}{\left[{\text{H}}_2{\text{SO}}_3\right]}$        (2)

Equation (2) indicates there is large $\left[{\text{H}}_2{\text{SO}}_3\right]$ relative to $\left[{\text{HSO}}_3{}^{-}\right]$ at $\text{pH}$ of 1.857.

Value of $\text{pH}$ is calculated by Henderson-HasselBalch equation given as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{SO}}_3^{2-}\right]}{\left[{\text{HSO}}_3^{-}\right]}$        (3)

Substitute 7.172 for $\text{p}{K}_{\text{a}}$ in equation (3).

  $\text{pH}=7.172+\log \frac{\left[{\text{SO}}_3^{2-}\right]}{\left[{\text{HSO}}_3^{-}\right]}$        (4)

Equation (4) indicates there is large $\left[{\text{SO}}_3^{2-}\right]$ relative to $\left[{\text{HSO}}_3^{-}\right]$ at $\text{pH}$ of 7.172. In between the two $\text{pH}$, the intermediate form $\left[{\text{HSO}}_3^{-}\right]$ is major species.

(b)

Interpretation Introduction

Interpretation:

Balanced cathodic and anodic half-reactions have to be found.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode.

Explanation of Solution

Half cell reaction for reduction of water that occurs at cathode is written as follows:

  ${\text{H}}_2\text{O}\left(s\right)+{\text{e}}^{-}\to \frac{1}{2}{\text{H}}_2\left(g\right)+{\text{OH}}^{-}$

Half cell reaction for anodic compartment when iodide gets oxidized to triodide ion is written as follows:

  $3{\text{I}}^{-}\to 2{\text{I}}_3^{-}+2{\text{e}}^{-}$

(c)

Interpretation Introduction

Interpretation:

Balanced reactions amongst ${\text{I}}_3^{-}$ -${\text{HSO}}_3^{-}$ and  ${\text{I}}_3^{-}$ -${\text{S}}_2{\text{O}}_3^{2-}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Unbalanced reaction is formulated as follows:

  ${\text{I}}_3^{-}+{\text{HSO}}_3^{-}\to {\text{I}}^{-}+{\text{SO}}_4^{2-}$

Add 3 as coefficient of ${\text{I}}^{-}$ to get equal ${\text{I}}^{-}$ on both sides.

  ${\text{I}}_3^{-}+{\text{HSO}}_3^{-}\to 3{\text{I}}^{-}+{\text{SO}}_4^{2-}$

Since there are 4 $\text{O}$ atoms on right, add one water molecule to left to have equal $\text{O}$ atoms.

  ${\text{I}}_3^{-}+{\text{HSO}}_3^{-}+{\mathrm{H}}_{2}O\to 3{\text{I}}^{-}+{\text{SO}}_4^{2-}$

Since there are 3 $\text{H}$ atoms on right, add three ${\text{H}}^{+}$ ions to right to have equal $\text{H}$ atoms.

Finally, verify each side has equal charge. Thus balanced reaction amongst ${\text{I}}_3^{-}$ and ${\text{HSO}}_3^{-}$ is written as follows:

  ${\text{I}}_3^{-}+{\text{HSO}}_3^{-}+{\mathrm{H}}_{2}O\to 3{\text{I}}^{-}+{\text{SO}}_4^{2-}+3{\text{H}}^{+}$

Unbalanced reaction between ${\text{I}}_3^{-}$ and ${\text{S}}_2{\text{O}}_3^{2-}$ is formulated as follows:

  ${\text{I}}_3^{-}+{\text{S}}_2{\text{O}}_3^{2-}\rightleftharpoons {\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$

Add 3 as coefficient of ${\text{I}}^{-}$ and 2 as coefficient of ${\text{S}}_2{\text{O}}_3^{2-}$ to get equal  ${\text{I}}^{-}$ , $\text{S}$ and ${\text{S}}_4\text{O}$ on both sides.

  ${\text{I}}_3^{-}+2{\text{S}}_2{\text{O}}_3^{2-}\rightleftharpoons 3{\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$

(d)

Interpretation Introduction

Interpretation:

Concentration of total sulfite in undiluted wine has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Balanced reactions amongst ${\text{I}}_3^{-}$ and ${\text{HSO}}_3^{-}$ is written as follows:

  ${\text{I}}_3^{-}+{\text{HSO}}_3^{-}+{\mathrm{H}}_{2}O\to 3{\text{I}}^{-}+{\text{SO}}_4^{2-}+3{\text{H}}^{+}$

Balanced reactions amongst ${\text{I}}_3^{-}$ and ${\text{S}}_2{\text{O}}_3^{2-}$ is written as follows:

  ${\text{I}}_3^{-}+2{\text{S}}_2{\text{O}}_3^{2-}\rightleftharpoons 3{\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$

Since $1\text{ mol}$ electrons carry a charge of $\text{96485 C}$, so amount of iodide produced in $131\text{ s}$ is calculated as follows:

  $\begin{array}{c}\text{Amount of iodide}=\left(131\text{ s}\right)\left(\frac{1\text{ mol}}{\text{96485 C}}\right)\left(\frac{10\text{ mA}}{1\text{ s}}\right)\left(\frac{1\text{ A}}{1000\text{ mA}}\right)\left(\frac{1{\text{ mol I}}_3^{-}}{{\text{2 mol e}}^{-}}\right)\\ =6.79\times {10}^{-9}\text{ mol}\end{array}$

Excess moles of thiosulfate are calculated as follows:

  $\begin{array}{c}{\text{Excess moles of S}}_2{\text{O}}_3^{2-}\text{ =}\left(6.79\times {10}^{-9}{\text{ mol I}}^{-}\right)\left(\frac{2{\text{ mol S}}_2{\text{O}}_3^{2-}}{1{\text{ mol I}}^{-}}\right)\\ =13.5\times {10}^{-6}\text{ mol}\end{array}$

Amount of ${\text{S}}_2{\text{O}}_3^{2-}$ added initially is calculated as follows:

  $\begin{array}{c}{\text{Intial moles of S}}_2{\text{O}}_3^{2-}\text{ =}\left(0.0507\text{ M}\right)\left(0.500\text{ mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.00002535\text{ mol}\end{array}$

Therefore amount of ${\text{S}}_2{\text{O}}_3^{2-}$ consumed is calculated as follows:

  $\begin{array}{c}\mathrm{Consumed}{\text{S}}_2{\text{O}}_3^{2-}\text{ =}0.00002535\text{ mol}-13.5\times {10}^{-6}\text{ mol}\\ =11.77\text{ μmol}{\text{S}}_2{\text{O}}_3^{2-}\end{array}$

Number of moles of ${\text{SO}}_3^{2-}$ after dilution is calculated as follows:

  $\begin{array}{c}{\text{Moles of SO}}_3^{2-}\text{ =}\frac{\left(\text{6}{\text{.55 μmol SO}}_3^{2-}\right)\left(\frac{10\text{ mL}}{20\text{ mL}}\right)}{9\text{ mL}}\\ =3.64{\text{ μmol/mL SO}}_3^{2-}\end{array}$

Hence, in terms of molar concentration is calculated as follows:

  $\begin{array}{c}{\text{Molarity of SO}}_3^{2-}\text{ =}\left(\frac{3.64\text{ μmol}}{1\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\left(\frac{1\text{ mol}}{{10}^6\text{ μmol}}\right)\\ =3.64\times {10}^{-3}{\text{M SO}}_3^{2-}\\ \approx 3.64\text{ mM}\end{array}$

Thus, concentration of total sulfite in undiluted wine is $3.64\text{ mM}$.