Question

Sulfites are used worldwide in the wine industry as antioxidant and antimicrobial agents. However, sulfites have also been identified as causing certain allergic reactions suffered by asthmatics, and the FDA mandates that sulfites be identified on the label if they are present at levels of 10 ppm (parts per million) or higher. The analysis of sulfites in wine uses the "Ripper method" in which a standard iodine solution, prepared by the reaction of iodate and iodide ions, is used to titrate a sample of the wine. The iodine is formed in the reaction $$ \mathrm{IO}_{3}^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O} $$ The iodine is held in solution by adding an excess of \(\mathrm{I}^{-}\), which combines with \(\mathrm{I}_{2}\) to give \(\mathrm{I}_{3}^{-}\). In the titration, the \(\mathrm{SO}_{3}^{2-}\) is converted to \(\mathrm{SO}_{2}\) by acidification, and the reaction during the titration is $$ \mathrm{SO}_{2}+\mathrm{I}_{3}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{SO}_{4}^{2-}+3 \mathrm{I}^{-}+4 \mathrm{H}^{+} $$ Starch is added to the wine sample to detect the end point, which is signaled by the formation of a dark blue color when excess iodine binds to the starch molecules. In a certain analysis, \(0.0421 \mathrm{~g}\) of \(\mathrm{NaIO}_{3}\) was dissolved in dilute acid and excess NaI was added to the solution, which was then diluted to a total volume of \(100.0 \mathrm{~mL}\) A \(50.0 \mathrm{~mL}\) sample of wine was then acidified and titrated with the iodine- containing solution. The volume of iodine solution required was \(2.47 \mathrm{~mL}\). (a) What was the molarity of the iodine (actually, \(\left.\mathrm{I}_{3}^{-}\right)\) in the standard solution? (b) How many grams of \(\mathrm{SO}_{2}\) were in the wine sample? (c) If the density of the wine was \(0.96 \mathrm{~g} / \mathrm{mL}\), what was the percentage of \(\mathrm{SO}_{2}\) in the wine? (d) Parts per million (ppm) is calculated in a manner similar to percent (which is equivalent to parts per hundred). $$ \mathrm{ppm}=\frac{\text { grams of component }}{\text { grams of sample }} \times 10^{6} \mathrm{ppm} $$ What was the concentration of sulfite in the wine, expressed as parts per million \(\mathrm{SO}_{2} ?\)

Short answer

The concentration of sulfite in the wine, expressed as \text{SO}_2, is approximately 23.79 ppm.

Step-by-step solution

Calculate the Molarity of the Iodine Solution

Using the balanced equation for the formation of iodine (\text{IO}_3^- + 5 \text{I}^- + 6 \text{H}^+ \longrightarrow 3 \text{I}_2 + 3 \text{H}_2 \text{O}), find the moles of \text{NaIO}_3 (which provides the \text{IO}_3^-). The molar mass of NaIO_3 (22.989 + 126.9 + 3*15.999) is 174.888 g/mol. Using the mass of NaIO_3 (0.0421 g), calculate the moles: \[ \frac{0.0421\, g}{174.888\, g/mol} = 2.406 \times 10^{-4} \, mol \, \text{NaIO}_3 \] Since the ratio of moles of NaIO_3 to I_2 (actually I_3^- in solution) is 1:3 from the balanced equation, calculate the moles of I_2: \[ 2.406 \times 10^{-4} \, mol \, \text{NaIO}_3 \times \frac{3}{1} = 7.218 \times 10^{-4} \, mol \, I_2 \] Now, with the total solution volume of 100.0 mL, the molarity of the I_2 (as I_3^-) is: \[ \frac{7.218 \times 10^{-4} \, mol \, I_2}{0.1000\, L} = 0.007218\, M \, I_3^- \]

Determine the Amount of SO2 in the Wine Sample

Using the titration equation (\text{SO}_2 + \text{I}_3^- + 2 \text{H}_2 \text{O} \longrightarrow \text{SO}_4^{2-} + 3 \text{I}^- + 4 \text{H}^+), the moles of \text{I}_3^- consumed are equal to the moles of \text{SO}_2 reacted. Calculate the moles of I_3^- used: \[ 0.007218\, M \times 0.00247\, L = 1.783 \times 10^{-5}\, mol \, I_3^- \] Therefore, the same number of moles of SO_2 are present in the wine sample. To find the mass of SO_2: \[ 1.783 \times 10^{-5}\, mol \times 64.066\, g/mol \approx 1.142 \times 10^{-3}\, g \]

Calculate the Percentage of SO2 in the Wine

The volume of the wine sample is 50.0 mL and the density is 0.96 g/mL. Calculate the mass of wine: \[ 50.0\, mL \times 0.96\, g/mL = 48\, g \] The percentage of SO_2 in the wine is: \[ \frac{1.142 \times 10^{-3}\, g}{48\, g} \times 100\% \approx 0.00238\% \]

Calculate the PPM of SO2 in the Wine

To find the concentration in parts per million (ppm): \[ \text{ppm} = \frac{1.142 \times 10^{-3}\, g}{48\, g} \times 10^{6} = 23.79\, ppm \]

Key concepts

Titration

Titration is a laboratory technique used to determine the concentration of a solution by reacting it with a standard solution of known concentration, called the titrant. During titration, the titrant is gradually added to the sample solution until a specific chemical reaction is completed, which is usually indicated by a color change called the end point. The volume of titrant used to reach the end point is measured, and this information is used to calculate the concentration of the analyte (the substance being analyzed) in the sample.

In the context of sulfite analysis in wine, titration is essential for quantifying the amount of sulfite present. The precision of the technique means that even low levels of sulfites can be detected, which is important for meeting regulatory requirements and ensuring consumer safety.

Iodometric Titration

Iodometric titration is a type of redox titration where iodine (I2) or iodine ions such as triiodide (I3-) are used as the titrant. This method is particularly useful for analyzing substances that can oxidize iodide ions to iodine, or reduce iodine to iodide ions. The reaction involved often requires the addition of an indicator, such as starch, to signal the end point, which occurs when all the iodine has reacted.

In the exercise, iodometric titration is utilized to quantify sulfites in wine. The iodine formed reacts with the sulfite ions, and once the sulfite is fully consumed, excess iodine then binds to starch, forming a dark blue complex that indicates the end point of the titration. This visible sign allows the analyst to precisely measure the amount of iodine solution used, from which they can determine the concentration of sulfites.

Ripper Method

The Ripper method is a widely used iodometric titration technique specifically for measuring the amount of free SO2 in wine and other beverages. The process involves titrating an acidified sample of wine with a standardized iodine solution. As the reaction progresses, iodine converts sulfite ions in the wine to sulfate ions. When all the sulfite has been oxidized, any additional iodine then reacts with the starch indicator, turning the solution blue.

The Ripper method is popular because it is relatively simple and can be performed with basic laboratory equipment. However, it is important for analysts to be aware of potential interferences and ensure that conditions, such as the pH of the wine sample, are appropriately controlled for accurate results.

Molarity Calculation

Molarity, symbolized as M, is a unit of concentration that measures the number of moles of a solute per liter of solution. To calculate molarity, you divide the moles of solute by the volume of solution in liters. In formulas, it is expressed as: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

In the given exercise, molarity calculation was used to determine the concentration of the iodine (as triiodide) in the standard solution. This step is crucial in the titration process, as it provides the baseline from which the amount of sulfite in the wine sample can be inferred. Molarity calculations are fundamental in chemistry and essential for any titration analysis.

Parts Per Million (ppm)

Parts per million (ppm) is a measure of concentration that is used for very dilute solutions, where it indicates the quantity of a substance in a million parts of the total solution. It's akin to percentage, but instead of parts per hundred, ppm refers to parts per million. The formula for calculating ppm of a component in a solution is:\[ \text{ppm} = \frac{\text{grams of component}}{\text{grams of sample}} \times 10^6 \]

This unit of measurement is particularly useful for measuring trace substances, such as contaminants or additives in food and beverages like wine. In the student's exercise, the ppm calculation determines the regulatory compliance of the wine by quantifying its sulfite concentration, critical information for individuals with sulfite sensitivities and for labeling requirements.