\(12.97\) In an attempt to determine the concentration of sulfur dioxide in the air near a power plant, two students set up a bubbler that passes air through \(50.00 \mathrm{~mL}\) of \(1.0 \times 10^{-4} \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq})\). The temperature is \(22^{\circ} \mathrm{C}\), and the atmospheric pressure 753 Torr. The air is pumped for \(2.5 \mathrm{~h}\) at a flow rate of \(3.0 \mathrm{~L} \cdot \mathrm{h}^{-1}\). The students then returned to the laboratory and titrated the solution with \(1.5 \times 10^{-4} \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) with phenolphthalein indicator to see how much \(\mathrm{NaOH}\) was left unreacted. They found that \(30.2 \mathrm{~mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) was required to reach the stoichiometric point. (a) Write the balanced chemical equation for the reaction of \(\mathrm{SO}_{2}\) and water. (b) What amount of \(\mathrm{NaOH}\) (in mol) had reacted with the \(\mathrm{SO}_{2}\) ? (c) What was the concentration of sulfur dioxide in the air, in parts per million?

To understand the reaction, we start by writing the balanced chemical equation for the reaction of sulfur dioxide (\text{SO}_2) with water (\text{H}_2\text{O}). The reaction produces sulfurous acid (\text{H}_2\text{SO}_3).\[ \text{SO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_3(aq) \]

To find the amount of \text{NaOH} that reacted with \text{SO}_2, first calculate the moles of \text{HCl} used in the titration.\[ \text{Moles of HCl} = \text{Concentration of HCl} \times \text{Volume of HCl} \]\[ \text{Moles of HCl} = (1.5 \times 10^{-4} \text{ M}) \times (30.2 \times 10^{-3} \text{ L}) \]\[ \text{Moles of HCl} = 4.53 \times 10^{-6} \text{ mol} \]Since \text{HCl} reacts with \text{NaOH} 1:1, the same amount of \text{NaOH} was left unreacted. Initially, there were \text{50.0}\text{ mL of } 1.0 \times 10^{-4} \text{ M NaOH}, which is\[ \text{Initial moles of NaOH} = 1.0 \times 10^{-4} \text{ M} \times 50.0 \times 10^{-3} \text{ L} = 5.0 \times 10^{-6} \text{ mol} \]The amount of \text{NaOH} that reacted with \text{SO}_2 is the difference between the initial moles and the moles that were left unreacted.\[ \text{Moles of NaOH that reacted} = \text{Initial moles of NaOH} - \text{Moles of HCl} \]\[ \text{Moles of NaOH that reacted} = 5.0 \times 10^{-6} \text{ mol} - 4.53 \times 10^{-6} \text{ mol} \]\[ \text{Moles of NaOH that reacted} = 0.47 \times 10^{-6} \text{ mol} \]

To find the concentration of \text{SO}_2 in air, you first need to calculate the volume of air that has reacted. Since the flow rate of the air is \text{3.0 L/h}, over \text{2.5 h}, the total volume is\[ \text{Total volume of air} = 3.0 \text{ L/h} \times 2.5 \text{ h} = 7.5 \text{ L} \]Next, we convert this volume from liters at the given conditions to liters at standard temperature and pressure (STP) using the combined gas law because concentrations are typically reported at STP.\[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \]Where \text{V_2} is the volume at STP conditions, and \text{n_1} and \text{n_2} represent the same amount of substance in different conditions. Since \text{n_1} = \text{n_2}, the expression simplifies to:\[ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} \]Given that \text{P_1} = 753 \text{ Torr}, \text{P_2} = 760 \text{ Torr} (1 atm), \text{T_1} = 295 \text{ K} (22°C + 273), and \text{T_2} = 273 \text{ K} (0°C),\[ V_2 = 7.5 \text{ L} \times \frac{753 \text{ Torr}}{760 \text{ Torr}} \times \frac{273 \text{ K}}{295 \text{ K}} \]\[ V_2 = 7.171 \text{ L} \]Finally, the concentration of \text{SO}_2 in ppm can be calculated using the number of moles of \text{SO}_2 that reacted and the corrected volume of air at STP conditions.\[ \text{Concentration of SO}_2 = \frac{\text{Moles of SO}_2}{\text{Volume of air at STP}} \times 10^6 \]\[ \text{Concentration of SO}_2 = \frac{0.47 \times 10^{-6} \text{ mol}}{7.171 \text{ L}} \times 10^6 \]\[ \text{Concentration of SO}_2 \text{ in ppm} = 65.5 \]