Question

Sodium nitrite, \(\mathrm{NaNO}_{2}\), is frequently added to processed meats as a preservative. The amount of nitrite ion in a sample can be determined by acidifying to form nitrous acid (HNO \(_{2}\) ), letting the nitrous acid react with an excess of iodide ion, and then titrating the \(\mathrm{I}_{3}^{-}\) ion that results with thiosulfate solution in the presence of a starch indicator. The unbalanced equations are (1) \(\mathrm{HNO}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{NO}+\mathrm{I}_{3}^{-}\) (in acidic solution) (2) \(\mathrm{I}_{3}^{-}+\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}\) (a) Balance the two redox equations. (b) When a nitrite-containing sample with a mass of \(2.935 \mathrm{~g}\) was analyzed, \(18.77 \mathrm{~mL}\) of \(0.1500 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution was needed for the reaction. What is the mass percent of \(\mathrm{NO}_{2}^{-}\) ion in the sample?

Short answer

The mass percent of \( \mathrm{NO}_2^- \) in the sample is 2.94\%.

Step-by-step solution

Balance the first equation

To balance the equation \( \text{HNO}_2 + \text{I}^- \rightarrow \text{NO} + \text{I}_3^- \), let's first balance nitrogen and iodine. Since both sides have one nitrogen atom, the nitrogen is balanced. Now look at iodine: 1 iodide ion yields \( \text{I}_3^- \), so we start with three iodide ions: \( 3\text{I}^- \).To balance the equation: \[\text{HNO}_2 + 3\text{I}^- + 2\text{H}^+ \rightarrow \text{NO} + \text{I}_3^- + 2\text{H}_2\text{O}\]

Balance the second equation

Balance the equation \( \text{I}_3^- + \text{S}_2\text{O}_3^{2-} \rightarrow \text{I}^- + \text{S}_4\text{O}_6^{2-} \). Start with iodide electrons: \( \text{I}_3^- \) breaks down into 3 iodide ions.Hence, the balanced equation is: \[\text{I}_3^- + 2\text{S}_2\text{O}_3^{2-} \rightarrow 3\text{I}^- + \text{S}_4\text{O}_6^{2-}\]

Determine moles of \( \mathrm{S}_2\mathrm{O}_3^{2-} \) used

Calculate the moles of sodium thiosulfate used in the titration: \[moles = M \times V = 0.1500 \, \text{mol/L} \times 0.01877 \, \text{L} = 0.0028155 \, \text{mol}\]

Calculate moles of \( \mathrm{NO}_2^- \) initially present

Using the stoichiometric relationship from balanced equations, 2 moles of \( \text{S}_2\text{O}_3^{2-} \) react with 1 mole of \( \text{I}_3^- \), which originates from 1 mole of \( \text{NO}_2^- \).Thus, moles of \( \text{NO}_2^- \) = \( 0.0028155 \, \text{mol} \div 1.5 = 0.001877 \, \text{mol} \).

Calculate mass of \( \mathrm{NO}_2^- \) and percentage

Calculate molar mass of \( \text{NO}_2^- \): 14.01 (N) + 2 \times 16.00 (O) = 46.01 \, \text{g/mol}Mass of \( \text{NO}_2^- \)= \( 0.001877 \, \text{mol} \times 46.01 \, \text{g/mol} = 0.08636 \, \text{g} \).Mass percent = \( \left( \frac{0.08636}{2.935} \right) \times 100 = 2.94\% \).

Key concepts

Sodium Nitrite

Sodium nitrite, known by its chemical formula \( \mathrm{NaNO}_{2} \), plays a crucial role in the preservation of processed meats. It acts by inhibiting the growth of bacteria, particularly those responsible for botulism, a severe form of food poisoning.
Upon the acidification of sodium nitrite, nitrous acid (\( \mathrm{HNO}_{2} \)) is formed. This is an important step because \( \mathrm{HNO}_{2} \) undergoes a redox reaction that is integral to the subsequent analysis and titration process.
Understanding this process requires knowledge of redox chemistry, where electron transfer takes place. In this particular reaction, the iodide ion serves as a reducing agent, donating electrons, while the nitrous acid acts as an oxidizing agent, accepting those electrons.
The formula is given as:
\( \mathrm{HNO}_{2} + 3\mathrm{I}^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{NO} + \mathrm{I}_3^{-} + 2\mathrm{H}_2\mathrm{O} \).
This balanced equation ensures that all the atoms and charge are balanced across the reaction, which is a fundamental aspect of redox reactions.

Iodide Ion

The iodide ion, symbolized as \( \mathrm{I}^{-} \), is an essential participant in the redox titration process involving sodium nitrite. When the iodide ions are introduced to nitrous acid under acidic conditions, an intricate redox reaction occurs.
In this reaction, iodide ions reduce the nitrous acid, resulting in the formation of \( \mathrm{I}_3^{-} \) ions and nitric oxide (\( \mathrm{NO} \)). The role of iodide is to provide the necessary electrons to facilitate the reduction of \( \mathrm{HNO}_{2} \).
It's important to understand how iodide ions transform during the reaction. Initially, they exist as separate ions. They combine to form \( \mathrm{I}_3^{-} \) through the process:
\( \mathrm{HNO}_{2} + 3\mathrm{I}^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{NO} + \mathrm{I}_3^{-} + 2\mathrm{H}_2\mathrm{O} \).
By achieving this transformation, iodide ions play a pivotal role in the progression of the redox titration process.

Thiosulfate Solution

The use of a thiosulfate solution is a key component in the final stage of this redox titration process. In the presence of a starch indicator, the thiosulfate ions (\( \mathrm{S}_2\mathrm{O}_3^{2-} \)) play a decisive role in determining the endpoint of the titration.
The reaction between thiosulfate ions and \( \mathrm{I}_3^{-} \) ions results in the reconversion of \( \mathrm{I}_3^{-} \) back to iodide ions, while forming tetrathionate ions (\( \mathrm{S}_4\mathrm{O}_6^{2-} \)). This is represented in the balanced equation:
\( \mathrm{I}_3^{-} + 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 3\mathrm{I}^{-} + \mathrm{S}_4\mathrm{O}_6^{2-} \).
During this reaction, the deep blue-black complex formed with starch and \( \mathrm{I}_3^{-} \) disappears, signaling the endpoint of the titration. This visual cue allows for the precise determination of the amount of thiosulfate solution used, which is crucial for calculating the concentration of nitrite ions in the original sample. Understanding the interplay between these substances enriches one's grasp of titration processes.