Question

The amount of ozone, \(\mathrm{O}_{3}\), in a mixture of gases can be determined by passing the mixture through a solution of excess potassium iodide, KI. Ozone reacts with the iodide ion as follows: $$\begin{aligned} \mathrm{O}_{3}(\mathrm{g})+3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) & \longrightarrow \\ \mathrm{O}_{2}(\mathrm{g})+\mathrm{I}_{3}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$ The amount of \(I_{3}^{-}\) produced is determined by titrating with thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}:\) $$\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \longrightarrow 3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})$$ A mixture of gases occupies a volume of \(53.2 \mathrm{L}\) at \(18^{\circ} \mathrm{C}\) and \(0.993 \mathrm{atm} .\) The mixture is passed slowly through a solution containing an excess of KI to ensure that all the ozone reacts. The resulting solution requires \(26.2 \mathrm{mL}\) of \(0.1359 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) to titrate to the end point. Calculate the mole fraction of ozone in the original mixture.

Short answer

The mole fraction of ozone in the original mixture is 0.00164.

Step-by-step solution

Calculate the moles of thiosulfate ions

The moles of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \) is calculated using the formula \( moles = Volume(L) \times Molarity(M) \). The Volume is \( 26.2mL = 0.0262L \) and the Molarity is \( 0.1359M \). Therefore, \( moles = 0.0262 \times 0.1359 = 0.00356 \) moles.

Determine the moles of ozone

From the second reaction equation it can be concluded that one \( \mathrm{I}_{3}^{-} \) reacts with two \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\). In the first reaction equation one \( \mathrm{O}_{3} \) reacts with one \( \mathrm{I}_{3}^{-} \). Combining both reactions, it can be inferred that one \( \mathrm{O}_{3} \) reacts with two \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\). Therefore the moles of \( \mathrm{O}_{3} \) in the mixture are the same as the moles of \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \). Hence, the moles of ozone are also \( 0.00356 \) moles.

Calculate the moles of total gas mixture

Using the ideal gas law \( PV = nRT \), the total moles of the gas can be calculated where P=pressure, V=volume, n=number of moles, R=gas constant, T=temperature (in Kelvin). Given, P = 0.993 atm, V = 53.2 L, R=0.0821 atm.L/mol.K and T = 18 C = 291 K. By rearranging the formula to find n, number of moles = PV/RT \( = 0.993 \times 53.2 / (0.0821 \times 291) = 2.17 \) moles.

Calculate the mole fraction of ozone

The mole fraction of a component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components in the mixture. The mole fraction of ozone = moles of ozone / total moles of gas = \( 0.00356 / 2.17 = 0.00164 \) .

Key concepts

Ozone Reaction

In the provided exercise, we deal with the fascinating world of ozone chemistry. Ozone, or \( \mathrm{O}_3 \), is a gas known for its reactivity. In the reaction mentioned, ozone reacts with iodide ions (\( \mathrm{I}^- \)) in the presence of water to produce iodine triiodide ions (\( \mathrm{I}_3^- \)), oxygen, and hydroxide ions (\( \mathrm{OH}^- \)). This chemical reaction is pivotal in gauging the amount of ozone in a mixture.

When you pass the gaseous mixture through a solution containing excess potassium iodide (KI), the ozone present reacts completely. This ensures that all ozone is accounted for as it transforms iodide ions into iodine triiodide ions. The reaction is such that it gives us a stoichiometric relationship: one mole of ozone produces one mole of \( \mathrm{I}_3^- \).

• This transformation is essential because it allows us to indirectly measure the quantity of ozone present in the gaseous mixture by tracking iodine triiodide.
• Understanding these stoichiometric relationships is crucial for solving the problem, by linking the moles of ozone to the moles in the subsequent reaction.

Thiosulfate Ion Titration

Once we have the \( \mathrm{I}_3^- \) ions from the ozone reaction, they can be titrated using thiosulfate ions \( \mathrm{S}_2 \mathrm{O}_3^{2-} \). Titration is an analytical technique that helps identify the concentration of a dissolved substance through a controlled chemical reaction.

Here, \( \mathrm{I}_3^- \) ions react with thiosulfate ions in a 1:2 ratio. This means for every mole of \( \mathrm{I}_3^- \), two moles of thiosulfate are required to complete the reaction. When the titration is complete, the \( \mathrm{I}_3^- \) ions are reduced back to \( \mathrm{I}^- \) ions.

• The conversion back to iodide ions effectively marks the end of the reaction, and the amount of thiosulfate used in this process is a direct measurement of the \( \mathrm{I}_3^- \) ions originally present.
• By calculating the moles of thiosulfate consumed, one can deduce the amount of iodine triiodide, and by extension, the original quantity of ozone.

Ideal Gas Law

The Ideal Gas Law is an essential tool in this calculation, linking the physical properties of gases. It is formulated as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature.

It's worth noting in this exercise that conversion must occur for the temperature to Kelvin (from degrees Celsius), a necessary step due to the absolute scale required in gas law calculations: \( T = 18\, ^{\circ} C + 273 = 291\, \text{K} \).

• Using the given values in the problem, you can determine the total moles of the original gas mixture. This forms the basis for the further calculation of mole fractions.
• The knowledge of the total moles allows us to then determine the mole fraction of ozone by comparing the moles of ozone (from titration-derived calculations) to the total moles of gas.

Understanding these principles allows you to calculate the mole fraction effectively, completing the quantitative analysis of ozone in the gas mixture.