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 bottle 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 sodium thiosulfate is 96.78%.

Step 1: Calculate Moles of Iodine

To find the moles of iodine (\(\mathrm{I}_{2}\)) used in the reaction, multiply the volume of (\(\mathrm{I}_{2}\)) solution by its molarity. Convert the volume from milliliters to liters: 40.21 \, \mathrm{mL} = 0.04021 \, \mathrm{L}. Then use the equation \(\text{moles} = \text{molarity} \times \text{volume (L)}\): \[\text{moles of } \mathrm{I}_{2} = 0.246 \, \mathrm{mol/L} \times 0.04021 \, \mathrm{L} = 0.009897 \text{ moles of } \mathrm{I}_{2}\].

Step 2: Use Stoichiometry to Find Moles of Sodium Thiosulfate

From the balanced equation, \(1 \text { mole of } \mathrm{I}_{2}\) reacts with \(2 \text { moles of } \mathrm{S}_{2}\mathrm{O}_{3}^{2-}\). Therefore, \(0.009897 \text{ moles of } \mathrm{I}_{2}\) will react with \(2 \times 0.009897 = 0.019794 \text{ moles of } \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\).

Step 3: Calculate Mass of Pure Sodium Thiosulfate

Find the molar mass of \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\) (\(= 2 \times (22.99) + 2 \times (32.07) + 3 \times (16.00) = 158.11 \, \mathrm{g/mol}\)). Then calculate the mass: \[\text{mass of } \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} = 0.019794 \, \text{moles} \times 158.11 \, \mathrm{g/mol} = 3.128 \, \mathrm{g}\].

Step 4: Determine Weight Percent of Sodium Thiosulfate

The weight percent is given by \(\frac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100\). Plug in the numbers: \[\text{Weight percent of } \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} = \frac{3.128 \, \mathrm{g}}{3.232 \, \mathrm{g}} \times 100 = 96.78\%\].

Key Concepts

Sodium Thiosulfate

Sodium thiosulfate, a compound with the formula \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\), plays a key role in various applications, especially in the field of photography, where it acts as a fixer in black-and-white film processing. Its ability to dissolve sliver halides is what makes it effective in this role. Beyond photography, it is also used in chemical laboratories as a component in titration experiments because it participates in clear, measurable reactions.

This compound consists of two sodium atoms, two sulfur atoms, and three oxygen atoms. Typically, it appears as a colorless or white crystalline solid that dissolves well in water. This solubility is crucial for its application in solutions needed for titration reactions. Sodium thiosulfate also has antiseptic and antifungal properties, which makes it useful in various medical applications as well.

Chemical Purity

Chemical purity is a crucial aspect when working with any chemical sample whether for industrial, laboratory, or pharmaceutical purposes. The purity of a chemical such as sodium thiosulfate directly affects the accuracy of any experimental results or reactions in which it is involved. In the context of titration, determining the purity involves quantifying the actual concentration of the active substance versus any impurities or inert materials present.

The weight percent calculation—comparing the mass of pure sodium thiosulfate against the total mass of the sample—helps in assessing the purity. For a precise determination, you need to account for all possible variables like the purity level of every reagent used, consistent temperature control, and proper calibration of measurement equipment. High purity means the chemical will behave predictably in reactions, ensuring reliable and reproducible experimental outcomes.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It allows chemists to use balanced chemical equations to predict reactants or products needed in sufficient quantities for a desired result.

In this example, the stoichiometric relation comes from the balanced equation: \[ \mathrm{I}_2(\mathrm{aq}) + 2 \mathrm{S}_2\mathrm{O}_3^{2-}(\mathrm{aq}) \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) + \mathrm{S}_4\mathrm{O}_6^{2-}(\mathrm{aq}) \] This shows us that 1 mole of iodine \(\mathrm{I}_2\) reacts with 2 moles of thiosulfate ions \(\mathrm{S}_2\mathrm{O}_3^{2-}\). By understanding this relationship, you can deduce how much sodium thiosulfate would react completely with a given amount of iodine in the solution, which is critical for calculating purity in titration.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between substances, influencing their oxidation states. In this titration example, iodine \(\mathrm{I}_2\) is reduced to iodide ions \(\mathrm{I}^{-}\) while the thiosulfate ion \(\mathrm{S}_2\mathrm{O}_3^{2-}\) is oxidized to tetrathionate \(\mathrm{S}_4\mathrm{O}_6^{2-}\).

In a redox titration, you can observe color changes that help indicate the end point of the titration. The initial brown color of the iodine solution fades as it is consumed in the reaction with thiosulfate, reaching completion when all iodine is reduced. Such characteristics make redox reactions particular useful in titration processes, allowing us to determine the concentration of unknown substances using known reactant quantities.