Question

Coulometric titration of sulfite in wine. Sulfur dioxide is added to many foods as a preservative. In aqueous solution, the following species are in equilibrium:

Bisulfite reacts with aldehydes in food near neutral pH:

Sulfite is released from the adduct in 2MNaOH and can be analyzed by its reaction with ${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$ to give ${\mathbf{I}}^{\mathbf{-}}$and sulfate. Excess${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$ must be present for quantitative reaction.

Here is a coulometric procedure for analysis of total sulfite in white wine. Total sulfite means all species in Reaction and the adduct in Reaction . We use white wine so that we can see the color of a starch-iodine end point.

1. Mix 9.00 mL of wine plus 0.8gNaOH and dilute to 10.00mL. The releases sulfite from its organic adducts.

2. Generate ${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$at the working electrode (the anode) by passing a known current for a known time through the cell in Figure 17 - 10. The cell containsofacetate buffer () plus. In the cathode compartment, is reduced to ${\mathbf{H}}_{\mathbf{2}}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}$. The frit retards diffusion of into the main compartment, where it would react with ${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$ to give${\mathbf{IO}}^{\mathbf{-}}$.

3. Generate ${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$ at the anode with a current of for.

4. Inject 2.000mL of the wine/ solution into the cell, where the sulfite reacts with leaving excess.

5. Add of thiosulfate to consume by Reaction and leave excess thiosulfate.

6. Add starch indicator to the cell and generate fresh${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$ with a constant current of 10.0mA. A time of 131s was required to consume excess thiosulfate and reach the starch end point.

(a) In what pH range is each form of sulfurous acid predominant?

(b) Write balanced half-reactions for the anode and cathode.

(c) At pH 3.7, the dominant form of sulfurous acid isand the dominant form of sulfuric acid is ${\mathbf{HSO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$. Write balanced reactions between andand between ${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$and ${\mathbf{HSO}}_{\mathbf{3}}^{\mathbf{-}}$thiosulfate.

(d) Find the concentration of total sulfite in undiluted wine.

Short answer

1. When pH <1.86 the${\mathrm{H}}_2{\mathrm{SO}}_3$ is predominant.
2. The balanced half-reactions are:${\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to 1/2{\mathrm{H}}_2+{\mathrm{OH}}^{-}$ .
3. The balanced reactions between and :${\mathrm{I}}_3^{-}\mathrm{and}{\mathrm{HSO}}_3^{-}$

${\mathrm{I}}_3^{-}+{\mathrm{HSO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\to 3{\mathrm{I}}^{-}+{\mathrm{SO}}_4^{2-}+3{\mathrm{H}}^{+}$

The balanced reactions between and thiosulfate

${\mathrm{I}}_3^{-}+2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\to {\mathrm{S}}_4{\mathrm{O}}_6^{2-}+3{\mathrm{I}}^{-}$.

1. The concentration of total sulphite in undiluted wine is .

Step-by-step solution

Deifinition of titration.

Titration is a method or procedure for estimating the concentration of a dissolved material by calculating the least quantity of known-concentration reagent necessary to produce a particular effect in interaction with a known volume of the test solution.

Step 2: The range in which sulphurous acid is predominant.

a)

Equilibrium, ${\mathrm{H}}_2{\mathrm{SO}}_3\stackrel{p{K}_1}{\to }{\mathrm{HSO}}_3^{-}\stackrel{p{K}_2}{\to }{\mathrm{SO}}_3^{2-}$

The ${\mathrm{pK}}_1$is 1.86 , while the ${\mathrm{pK}}_2$is 7.17 .

So, when pH < 1.86 the ${\mathrm{H}}_2{\mathrm{SO}}_3$is predominant.

When pH range is between and , the dominant form is .

When pH > 7.17, the dominant form is${\mathrm{SO}}_3^{2-}$ .

Step 3: The balanced half-reactions for the anode and cathode.

b)

The anode's balanced half-reactions are:${31}^{-}\to I_3^{-}+2e^{-}$

For the cathode, the balanced half-reactions are: ${\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to 1/2{\mathrm{H}}_2+{\mathrm{OH}}^{-}$.

Step 4: The balanced reactions between  and  and between  and thiosulfate.

c)

The balanced reactions between$\mathrm{I}\stackrel{-}{{}_3}$ and${\mathrm{HSO}}_3^{-}$ :

${\mathrm{I}}_3^{-}+{\mathrm{HSO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\to 3{\mathrm{I}}^{-}+{\mathrm{SO}}_4^{2-}+3{\mathrm{H}}^{+}$

The balanced reactions between and thiosulfate

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

Step 5: The concentration of total sulfite in undiluted wine.

d)

The charge is:

$\begin{array}{r}q=l\cdot t\\ =10\cdot {10}^{-3}\mathrm{A}\cdot 240\mathrm{s}\\ q=2.4\mathrm{C}\end{array}$

The mole of electron is:

$\begin{array}{r}\mathrm{n}\left({\mathrm{e}}^{-}\right)=\frac{\mathrm{q}}{\mathrm{F}}\\ =\frac{2.4\mathrm{C}}{96485\mathrm{C}/\mathrm{mol}}\\ \mathrm{n}\left({\mathrm{e}}^{-}\right)=2.49\cdot {10}^{-5}\mathrm{mol}\end{array}$

The n of${\mathrm{I}}_3^{-}$is:

$\begin{array}{r}\frac{n\left(I_3^{-}\right)}{n\left(e^{-}\right)}=\frac{1}{2}n\left(I_3^{-}\right)\\ =\frac{n\left(e^{-}\right)}{2}\\ =\frac{2.49\cdot {10}^{-5}\mathrm{mol}}{2}\\ n\left(I_3^{-}\right)=1.24\cdot {10}^{-5}\mathrm{mol}\end{array}$

Now calculate of :

$\begin{array}{r}n\left({\mathrm{S}}_2{\mathrm{O}}_3^{2-}\right)=c\left({\mathrm{S}}_2{\mathrm{O}}_3^{2-}\right)\cdot \mathrm{V}\left({\mathrm{S}}_2{\mathrm{O}}_3^{2-}\right)n\left({\mathrm{S}}_2{\mathrm{O}}_3^{2-}\right)\\ =0.0507\mathrm{M}\cdot 0.5\cdot {10}^{-3}\mathrm{Ln}\left({\mathrm{S}}_2{\mathrm{O}}_3^{2-}\right)\\ =2.54\cdot {10}^{-5}\mathrm{mol}\end{array}$

The n of${\mathrm{I}}_3^{-}$ is:

$\begin{array}{r}\frac{n\left(I_3^{-}\right)}{n\left(S_2{O}_3^{2-}\right)}=\frac{1}{2}n\left(I_3^{-}\right)\\ =\frac{n\left(S_2{O}_3^{2-}\right)}{2}\\ =\frac{2.54\cdot {10}^{-5}\mathrm{mol}}{2}\\ n\left(I_3^{-}\right)=1.27\cdot {10}^{-5}\mathrm{mol}\end{array}$

The charge is:

$\begin{array}{r}\mathrm{q}=\mathrm{l}\cdot \mathrm{t}\\ =10\cdot {10}^{-3}\mathrm{A}\cdot 131\mathrm{s}\\ \mathrm{q}=1.31\mathrm{C}\end{array}$

The mole of electron is:

$\begin{array}{r}n\left(e^{-}\right)=\frac{q}{F}\\ =\frac{1.31\mathrm{C}}{96485\mathrm{C}/\mathrm{mol}}\\ n\left(e^{-}\right)=1.36\cdot {10}^{-5}\mathrm{mol}\end{array}$

The n of$I_3^{-}$ is:

$\begin{array}{r}\frac{n\left(I_3^{-}\right)}{n\left(e^{-}\right)}=\frac{1}{2}n\left(I_3^{-}\right)\\ =\frac{n\left(e^{-}\right)}{2}\\ =\frac{1.36\cdot {10}^{-5}\mathrm{mol}}{2}\\ n\left(I_3^{-}\right)=6.79\cdot {10}^{-6}\mathrm{mol}\end{array}$

Now calculate how much$I_3^{-}$ is left,

$\begin{array}{r}n{\left(I_3^{-}\right)}_{\text{left}}=\left(1.27\cdot {10}^{-5}-6.79\cdot {10}^{-6}\right)\mathrm{mol}\\ n{\left(I_3^{-}\right)}_{\text{left}}=5.91\cdot {10}^{-6}\mathrm{mol}\end{array}$

The sulfite in wine consumed:

$\begin{array}{r}n=1.24\cdot {10}^{-5}-5.91\cdot {10}^{-6}\mathrm{mol}\\ n=6.49\cdot {10}^{-6}\mathrm{mol}\end{array}$

Because$1{\mathrm{moll}}_3^{-}$ reacts with mol sulfite, the wine contained$6.49.{10}^{-6}$ mol sulfite in 2.00 mL the injected for analysis.

9.000 mLwine was diluted to 10.00mL .

So, the n of sulfite is:

$\begin{array}{r}n\left(\text{sulfite}\right)=\frac{10}{9}\cdot 6.49\cdot {10}^{-6}\mathrm{mol}\\ n\left(\text{sulfite}\right)=7.21\cdot {10}^{-6}\mathrm{mol}\end{array}$

So, the concentration of sulfite is:

$\begin{array}{r}c\left(\text{sulfite}\right)=\frac{n\left(\text{sulfite}\right)}{V_{\text{injected}}}\\ n\left(\text{sulfite}\right)=\frac{7.21\cdot {10}^{-6}\mathrm{mol}}{2\cdot {10}^{-3}\mathrm{L}}\\ n\left(\text{sulfite}\right)=3.61\cdot {10}^{-3}\mathrm{mol}\\ n\left(\text{sulfite}\right)=3.61\mathrm{mM}\end{array}$