Problem 107 - Chemistry and Chemical Reactivity

Question

Sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) is used as a "fixer" in black-and-white photography. Suppose you have a botule of sodium thiosulfate and want to determine its purity. The thiosulfate ion can be oxidized with \(\mathrm{I}_{2}\) according to the balanced, net ionic equation $$ \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq}) $$ 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 Na2S2O3 is 96.79%.

Step 1: Calculate Moles of Iodine

Using the molarity and volume of iodine (I_2), calculate the moles of iodine used in the reaction.\[\text{Moles of } I_2 = 0.246 \text{ M} \times \frac{40.21 \text{ mL}}{1000 \text{ mL/L}} = 0.00989 \text{ moles of } I_2\]

Step 2: Calculate Moles of Thiosulfate

Using the balanced equation, determine the moles of thiosulfate ions (S_2O_3^{2-}) that react with iodine. The stoichiometry is a 1:2 ratio.\[\text{Moles of } S_2O_3^{2-} = 2 \times 0.00989 = 0.01978 \text{ moles}\]

Step 3: Calculate Mass of Sodium Thiosulfate

Convert moles of thiosulfate ions to grams of sodium thiosulfate using its molar mass. The molar mass of Na_2S_2O_3 is approximately 158.11 g/mol.\[\text{Mass of } Na_2S_2O_3 = 0.01978 \text{ moles} \times 158.11 \text{ g/mol} = 3.128 \text{ g}\]

Step 4: Calculate Weight Percent

Find the weight percent of sodium thiosulfate in the impure sample by comparing the calculated mass by the total mass of the sample.\[\text{Weight percent} = \left( \frac{3.128 \text{ g}}{3.232 \text{ g}} \right) \times 100\%\]\[= 96.79\%\]

Key Concepts

Chemical Stoichiometry

Chemical stoichiometry is like the recipe of chemistry, detailing how reactants combine to form products in fixed ratios. It uses balanced chemical equations to determine the quantitative relationships between substances in a reaction. In the problem, we look at the balanced reaction between iodine \[\mathrm{I}_2\] and thiosulfate ions \[\mathrm{S}_2\mathrm{O}_3^{2-}\]. This reaction is crucial for determining the purity of sodium thiosulfate.

• A balanced chemical equation is critical because it ensures that the law of conservation of mass is fulfilled. This means that the number of each type of atom remains constant throughout the reaction.
• In the given equation, one mole of iodine reacts with two moles of thiosulfate ions. This 1:2 ratio is essential for calculating how much thiosulfate ions react with the iodine used.
• Stoichiometry allows us to use this ratio to find the number of moles of thiosulfate ions that have reacted, based on the amount of iodine used in the titration.

Understanding stoichiometry helps you scale chemical reactions up or down and is fundamental for calculations in titration analysis.

Molarity and Volume

Molarity is a measure of concentration that lets you know how much solute is present in a given volume of solution. It's a critical part of titration analysis, as it provides a way to calculate the exact amount of reactant used. In this problem, the iodine solution has a molarity of 0.246 M.
Volume, in the context of a titration, is the precise measurement of how much of a solution you use to reach a reaction's completion. For our iodine solution, this was measured at 40.21 mL.

• Converting the volume from milliliters to liters (by dividing by 1000) lets us calculate the moles of iodine used through the formula: \[\text{Moles} = \text{Molarity} \times \text{Volume in Liters}\].
• In the given exercise, we calculated that 0.00989 moles of iodine were present in our reaction flask.

The molarity and volume are essential tools that enable us to determine the number of moles involved in a chemical reaction, thereby advancing our stoichiometric calculations.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species and often play a vital role in titration analysis. In the case of our exercise, iodine \[\mathrm{I}_2\] is reduced to iodide ions \[\mathrm{I}^-\], and thiosulfate ions \[\mathrm{S}_2\mathrm{O}_3^{2-}\] are oxidized to tetrathionate ions \[\mathrm{S}_4\mathrm{O}_6^{2-}\].
Understanding the concept of oxidation and reduction helps make sense of how substances change during a reaction:

• Oxidation is the process of losing electrons, while reduction is the gaining of electrons.
• In the reaction provided, each molecule of iodine gains two electrons and is reduced to two iodide ions. Concurrently, two thiosulfate ions lose electrons and form one tetrathionate ion.
• These changes in electron states are usually accompanied by color changes, which are critical for identifying the endpoint of a titration.

Redox reactions are foundational for many chemical processes, from industrial applications to biological systems, and allow us to precisely quantify materials in titration procedures.