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.

Step by step solution

01

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.

02

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.

03

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.

04

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 = \( 0.00356 / 2.17 = 0.00164 \) .

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