Problem 97 - Chemistry and Chemical Reactivity

Question

Sodium sulfate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) is used as a "fixer" in black-and-white photography. Suppose you have a bottle of sodium sulfate and want to determine its purity. The thiosulfate ion can be oxidized with \(\mathrm{I}_{2}\) according to the balanced, net ionic equation \(\mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})\) If you use \(40.21 \mathrm{mL}\) of \(0.246 \mathrm{M} \mathrm{I}_{2}\) in a titration, what is the weight percent of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) in a \(3.232-\mathrm{g}\) sample of impure material?

Step-by-Step Solution

Answer

The weight percent of \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \) is approximately 96.72\%.

Step 1: Find moles of iodine, \( \mathrm{I}_2 \)

To find the number of moles of iodine, use the molarity formula \( \text{Molarity} = \frac{\text{moles}}{\text{liters}} \). Convert 40.21 mL to liters: \( \frac{40.21}{1000} = 0.04021 \) L. Then, calculate moles: \[ \text{moles of } \mathrm{I}_2 = 0.246 \text{ M} \times 0.04021 \text{ L} = 0.00989646 \text{ moles} \]

Step 2: Use the stoichiometry of the reaction

According to the balanced equation \( \mathrm{I}_{2} + 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-} \), two moles of thiosulfate ions react with one mole of iodine. So, the moles of \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \) are \( 2 \times 0.00989646 = 0.01979292 \text{ moles} \).

Step 3: Calculate grams of sodium thiosulfate, \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \)

To convert moles to grams, use molar mass: \( \text{molar mass of } \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 = 2(22.99) + 2(32.07) + 3(16.00) = 158.11 \text{ g/mol} \). Thus, the mass of \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \) is \( 0.01979292 \times 158.11 = 3.126 \text{ g} \).

Step 4: Calculate the weight percent of \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \)

The weight percent is calculated using the formula: \( \text{Weight percent} = \frac{\text{mass of } \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3}{\text{mass of impure sample}} \times 100 \). Thus, \( \frac{3.126}{3.232} \times 100 \approx 96.72\% \).

Key Concepts

Titration

Titration is a common laboratory technique used to determine the concentration of a solution. In this scenario, its purpose is to find the purity of a sodium thiosulfate solution. Titration involves the gradual addition of a standard solution, here iodine (\(\mathrm{I}_2\)), to a sample until the reaction reaches completion. This point is indicated by an observable change, which is often visual, like a color shift.

• The titrant is the iodine solution with a known concentration (0.246 M).
• The analyte is the sodium thiosulfate solution in the sample.

During titration, the volume of the titrant used is crucial for calculations. Here, 40.21 mL of the iodine solution was used, indicating the amount needed to reach the reaction endpoint. This volume tells us how much thiosulfate was present and is key in further calculations.

Stoichiometry

Stoichiometry is the part of chemistry that relates the quantities of reactants and products in a chemical reaction. Based on the balanced equation provided, \[\mathrm{I}_{2} + 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\]we see that one mole of iodine reacts with two moles of thiosulfate ions.
This stoichiometric relationship allows us to use the previously calculated moles of iodine—0.00989646 moles—to find out how many moles of thiosulfate ions are in the sample. We multiply the moles of iodine by 2, the stoichiometric coefficient for thiosulfate in the reaction, resulting in 0.01979292 moles of thiosulfate ions. This conversion from iodine to thiosulfate is crucial because it helps us further determine the amount of sodium thiosulfate present in terms of mass.

Molarity

Molarity is the measurement of the concentration of a solution expressed in moles of solute per liter of solution. It's an essential concept in titration experiments because it quantifies how much solute is present in a known volume of solvent. Here, the iodine solution has a molarity of 0.246 M.
To find the number of moles of iodine present in 40.21 mL of this solution, first convert milliliters to liters - 40.21 mL is 0.04021 L. Using the formula:\[\text{Moles of } \mathrm{I}_2 = \text{Molarity} \times \text{Volume (L)}\]Thus, \[ 0.246 \text{ M} \times 0.04021 \text{ L} = 0.00989646 \text{ moles} \].
Understanding molarity allows us to quantitatively assess how much iodine is involved in the titration process, a critical factor that informs the succeeding stoichiometry and purity analysis steps.

Purity Analysis

Purity analysis is the process used to ascertain the proportion of the desired compound in a sample that may contain impurities. Here, it involves determining the percentage of sodium thiosulfate in the entire sample weight. The sample's weight is known, 3.232 g, and we have calculated the amount of pure sodium thiosulfate in the sample, which is 3.126 g.
The weight percent formula is:\[\text{Weight percent} = \left( \frac{\text{mass of pure compound}}{\text{total sample mass}} \right) \times 100\%\]Plugging in the values:\[\frac{3.126}{3.232} \times 100 \approx 96.72\%\]This calculation reveals that the sodium thiosulfate in the impure sample is 96.72%, indicating excellent purity, with 3.28% potentially being other substances. This method of purity analysis is crucial in many fields, such as quality control in pharmaceuticals and materials chemistry.