Question

A way to analyze a sample for nitrite ion is to acidify a solution containing NO, and then allow the HNO, that is formed to react with iodide ion in the presence of excess F. The reaction is: 2HNO,+2H+32NO+2H0+ Then the ly is titrated with Na,5,0, salution using starch as an indicator +2503+S,0 hatupica anahpis, 0.958 g sample that was known to contain NaNo, was treated as described above. The titration require 0300 M Na$,0,solution to reach the end point was treated as described above. The titration required 28.75 ml How may moles of l, had been produced in the first reaction? How many moles of NO, had been in the original 0.958 g sample? hat was the percentage by mass of NaNO, in the original sample

Answer 1

The moles of I3− produced are 0.0043125, the same as the original moles of NO2− in the sample. To find the percentage by mass of NaNO2, calculate the mass of NaNO2 using its molar mass, divide by the sample mass, and multiply by 100, yielding 31.06%.

Step-by-step explanation:

Understanding the Titration of Nitrite Ions

To analyze a sample for nitrite ions, the substance is acidified, leading to the production of iodine (I3−) from iodide ions, which is then titrated with sodium thiosulfate (Na2S2O3). In the scenario provided, 0.958 g of the sample containing sodium nitrite (NaNO2) requires 28.75 mL of 0.300 M Na2S2O3 to reach the endpoint.

First, calculate the moles of sodium thiosulfate: (0.02875 L) × (0.300 M) = 0.008625 mol Na2S2O3.

Using the stoichiometry of the reaction, which implies that 2 moles of thiosulfate reacts with 1 mole of iodine (I3−), the moles of iodine produced in the reaction are: 0.0043125 mol I3−.

The original moles of nitrite (NO2−) would also be 0.0043125 mol, due to one-to-one stoichiometry in the formation of iodine from nitrite. To determine the percentage by mass of NaNO2 in the original sample, we use the moles of NaNO2 and its molar mass: (0.0043125 mol) × (69.00 g/mol) = 0.29756 g of NaNO2. Divide this by the mass of the sample and multiply by 100 to get the percentage: (0.29756 g / 0.958 g) × 100% = 31.06%.