Question

One way to analyze a sample for nitrite ion is to acidify a solution containing \(\mathrm{NO}_{2}^{-}\) and then allow the \(\mathrm{HNO}_{2}\) that is formed to react with iodide ion in the presence of excess \(\mathrm{I}^{-}\). The reaction is $$ 2 \mathrm{HNO}_{2}+2 \mathrm{H}^{+}+3 \mathrm{I}^{-} \longrightarrow 2 \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{3}- $$ Then the \(\mathrm{I}_{3}^{-}\) is titrated with \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution using starch as an indicator. $$ \mathrm{I}_{3}^{-}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 3 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-} $$ In a typical analysis, a \(1.104 \mathrm{~g}\) sample that was known to contain \(\mathrm{NaNO}_{2}\) was treated as described above. The titration required \(29.25 \mathrm{~mL}\) of \(0.3000 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solu- tion to reach the end point. (a) How many moles of \(\mathrm{I}_{3}^{-}\) had been produced in the first reaction? (b) How many moles of \(\mathrm{NO}_{2}^{-}\) had been in the original \(1.104 \mathrm{~g}\) sample? (c) What was the percentage by mass of \(\mathrm{NaNO}_{2}\) in the original sample?

Short answer

The moles of I3- produced are \frac{29.25 mL \times 0.3000 M}{1000 mL/L \times \frac{1}{2} moles I3- moles Na2S2O3^{-1}}. The moles of NO2- in the original sample are the moles of I3- times the ratio 3 moles NO2- per 2 moles I3-. The percentage by mass of NaNO2 in the sample is \frac{moles of NO2- \times molar mass of NaNO2 \times 100%}{1.104 g}.

Step-by-step solution

- Determine moles of Na2S2O3

Calculate the moles of sodium thiosulfate (\text{Na}\(_2\)S\(_2\)O\(_3\)) used in the titration by using the molarity equation: moles = molarity x volume (in Liters).

- Equate moles of Na2S2O3 to moles of I3-

Using the stoichiometry of the titration reaction, determine the moles of triiodide ion (I\(_{3}^{-}\)) by relating it to the moles of sodium thiosulfate (Na\(_{2}\)S\(_{2}\)O\(_{3}\)) using the mole ratio from the balanced equation.

- Determine moles of NO2- produced

Calculate the moles of nitrite ion (NO\(_{2}^{-}\)) that reacted to form triiodide ion (I\(_{3}^{-}\)) in the first reaction by using the stoichiometry between I\(_{3}^{-}\) and NO\(_{2}^{-}\).

- Calculate the mass of NaNO2

Using the molar mass of sodium nitrite (NaNO\(_{2}\)), convert the moles of NO\(_2^{-}\) to grams.

- Calculate the percentage by mass of NaNO2

Divide the mass of NaNO\(_{2}\) calculated in step 4 by the initial mass of the sample and multiply by 100% to find the percentage by mass of NaNO\(_{2}\) in the sample.

Key concepts

Nitrite Ion Quantification

Understanding the quantification of nitrite ions (o_2^-) is crucial when analyzing chemical samples. In the described process, acidified conditions convert nitrite ions to nitrous acid (o_2), which subsequently reacts with iodide ions (o^-) to form triiodide (o_3^-). This forms the basis for quantification.

Through titration, thiosulfate (o_2g_3^{2-}) is added until it fully reacts with the triiodide, indicating the amount of nitrite originally present. Titration serves as a reliable method to determine concentrations due to its precise and measurable nature. The endpoint is detected with the help of starch, which forms a complex with triiodide, resulting in a color change that signals completion.

For students, it's important to visualize this step-by-step reaction and comprehend the concepts of acid-base chemistry and redox reactions that underpin this analytical technique. Clearly noting the color changes and understanding the role of the indicator expedites mastery of practical titration and analytical skills.

Stoichiometry Calculations

Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It is the grounding theory for many quantitative analyses in chemistry, including titration.

To properly calculate the stoichiometry in our problem, it's essential to grasp the concept of the mole ratio, which comes from the balanced chemical equations. For instance, the balanced equation signifies that for every mole of triiodide (o_3^-) produced, two moles of thiosulfate (o_2g_3^{2-}) are required. When this ratio is applied to the known concentration and volume of thiosulfate used, it allows for the back-calculation to find the moles of triiodide and, ultimately, the moles of nitrite ions.

Simple ratio and proportion exercises can enhance a student's familiarity with these calculations, ensuring they understand not only how to perform them but also the reasons why these relationships exist within chemical equations.

Percentage by Mass Calculation

The percentage by mass (also called mass percent or weight percent) is a critical calculation in analytical chemistry, providing a dimensionless measure of the concentration of a component in a mixture. In the given exercise, after finding the moles of nitrite ion (o_2^-), we converted it to mass using the molar mass of sodium nitrite (ano_2) and then calculated its percentage by mass in the sample.

To calculate this percentage, divide the mass of the component (sodium nitrite in this instance) by the total mass of the sample and multiply by 100%. This gives us a clear picture of the purity or concentration of the substance within the mixture. It is essential for learners to practice these calculations across various examples to become adept at quickly and accurately determining the composition of different chemical samples.

Remember, accuracy in such calculations is paramount, as they are fundamental in determining the quality and usability of chemical materials in various industries, including food, pharmaceuticals, and environmental monitoring.