/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The concentration of \(\mathrm{S... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 19

The concentration of \(\mathrm{SO}_{2}\) in air is determined by bubbling a sample of air through a trap that contains \(\mathrm{H}_{2} \mathrm{O}_{2} .\) Oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\) results in the formation of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is then determined by titrating with \(\mathrm{NaOH}\). In a typical analysis, a sample of air is passed through the peroxide trap at a rate of \(12.5 \mathrm{~L} / \mathrm{min}\) for \(60 \mathrm{~min}\) and required \(10.08 \mathrm{~mL}\) of \(0.0244 \mathrm{M} \mathrm{NaOH}\) to reach the phenolphthalein end point. Calculate the \(\mu \mathrm{L} / \mathrm{L} \mathrm{SO}_{2}\) in the sample of air. The density of \(\mathrm{SO}_{2}\) at the temperature of the air sample is \(2.86 \mathrm{mg} / \mathrm{mL}\).

Short Answer

Expert verified
3.67 \(\mu\text{L/L}\) of \(\text{SO}_2\) in the air.

Step by step solution

01

Calculate Total Volume of Air Sampled

The air is bubbled through the trap for 60 minutes at a rate of \(12.5 \text{ L/min}\). Calculate the total volume of the air sample: \[ 12.5 \text{ L/min} \times 60 \text{ min} = 750 \text{ L} \] Thus, the total volume of the air sample is 750 L.
02

Determine Moles of NaOH Used in Titration

Calculate the moles of \(\text{NaOH}\) used in the titration. The volume of \(\text{NaOH}\) used is \(10.08 \text{ mL}\) or \(0.01008 \text{ L}\). The concentration of \(\text{NaOH}\) is \(0.0244 \text{ M}\). Convert to moles: \[ \text{moles NaOH} = 0.01008 \text{ L} \times 0.0244 \text{ mol/L} = 2.46 \times 10^{-4} \text{ mol} \] Thus, 2.46 \(\times 10^{-4}\) mol of NaOH was used.
03

Calculate Moles of H2SO4 Formed

The balanced chemical equation for the titration is \(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\). From this reaction, 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{H}_2\text{SO}_4\). Thus, moles of \(\text{H}_2\text{SO}_4\) formed: \[ \text{moles H}_2\text{SO}_4 = \frac{2.46 \times 10^{-4} \text{ mol NaOH}}{2} = 1.23 \times 10^{-4}\] It indicates 1.23 \(\times 10^{-4}\) mol of \(\text{H}_2\text{SO}_4\) was formed.
04

Calculate Grams of SO2 in Air Sample

From the reaction, the moles of \(\text{SO}_2\) are equal to the moles of \(\text{H}_2\text{SO}_4\) since each mole of \(\text{SO}_2\) produces one mole of \(\text{H}_2\text{SO}_4\). Molar mass of \(\text{SO}_2\) is \(64.07 \text{ g/mol}\). Calculate grams of \(\text{SO}_2\): \[ \text{mass SO}_2 = 1.23 \times 10^{-4} \text{ mol} \times 64.07 \text{ g/mol} = 7.88 \times 10^{-3} \text{ g} \] Therefore, there are 0.00788 grams of \(\text{SO}_2\) in the air sample.
05

Convert Grams of SO2 to Microliters

Convert grams of \(\text{SO}_2\) to microliters using the given density (2.86 mg/mL). First, convert grams to milligrams: \[ 0.00788 \text{ g} = 7.88 \text{ mg} \] Given the density, \(2.86 \text{ mg/mL}\), calculate volume: \[ \text{volume in mL} = \frac{7.88 \text{ mg}}{2.86 \text{ mg/mL}} = 2.755 \text{ mL} \] Finally, convert mL to \(\mu\text{L}\): \[ 2.755 \text{ mL} = 2755 \mu\text{L} \] Therefore, there are 2755 microliters of \(\text{SO}_2\) in the total air sample.
06

Calculate Microliters per Liter of SO2 in Air

Finally, calculate \(\mu\text{L/L}\) of \(\text{SO}_2\) in the air by dividing the microliters of \(\text{SO}_2\) by the total air volume in liters: \[ \frac{2755 \mu\text{L}}{750 \text{ L}} = 3.67 \mu\text{L/L} \] Thus, there are approximately 3.67 \(\mu\text{L/L}\) of \(\text{SO}_2\) in the air sample.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

SO2 concentration
The concentration of sulfur dioxide (\(\mathrm{SO}_2\)) in the air is a crucial measurement in analytical chemistry due to its impact on environmental and human health. To determine this concentration, a sample of air containing \(\mathrm{SO}_2\) is typically captured using a chemical method. In the given exercise, the air sample is passed through a trap containing hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_2\)), which reacts with \(\mathrm{SO}_2\) to form sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)). This process effectively converts \(\mathrm{SO}_2\) from a gaseous state into a measurable form that can be further analyzed.
  • Understanding how \(\mathrm{SO}_2\) concentration impacts air quality helps regulate industrial emissions.
  • This analytical approach ensures precise measurement, even in cases of dilute \(\mathrm{SO}_2\) presence in large volumes of air.
The concentration is calculated as resin milliliters of \(\mathrm{SO}_2\) per liter of air (\(\mu\text{L/L}\)), providing a direct way to express air quality in terms of known pollutants.
Titration calculation
Titration is a fundamental method used in analytical chemistry to determine the quantity of a specific substance within a solution. Here, titration measures the amount of sulfuric acid, indirectly quantifying the original \(\mathrm{SO}_2\). In the exercise, sodium hydroxide (\(\mathrm{NaOH}\)) is used in a titration process to neutralize the sulfuric acid formed, reaching an endpoint indicated by phenolphthalein.
The task involves calculating the number of moles of \(\mathrm{NaOH}\) used:
  • Calculate the \(\mathrm{NaOH}\) volume in liters (\(0.01008\text{ L}\)
  • Use the concentration (\(0.0244\text{ M}\)
  • The result is the moles of \(\mathrm{NaOH}\) utilized, equaling \(2.46 \times 10^{-4}\)
This step is complemented by the chemical equation \(\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}\), leveraging stoichiometry to determine moles of \(\mathrm{H}_2\mathrm{SO}_4\) produced. This balanced equation is critical for the precise transformation of titration data into actionable information about the initial \(\mathrm{SO}_2\).
Gas analysis
Gas analysis is an essential component of understanding the composition of gaseous mixtures in environmental and industrial contexts. In this exercise, air containing sulfur dioxide (\(\mathrm{SO}_2\)) is analyzed through a series of chemical reactions and calculations to determine its concentration.
Analyzing gases often involves converting gaseous components into a measurable form. Here, \(\mathrm{SO}_2\) in air interacts with \(\mathrm{H}_2\mathrm{O}_2\) to form sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)), which can then be quantified through titration with \(\mathrm{NaOH}\). The total volume of air sampled can be calculated by multiplying the flow rate by the sampling time:
  • Flow rate: \(12.5\text{ L/min}\)
  • Time: \(60\text{ min}\)
  • Total air volume: \(750\text{ L}\)
This process highlights the importance of carefully controlled sampling and conversion processes in gas analysis, emphasizing accuracy and consistency in analytical measurements.
Chemical reactions
Chemical reactions underpin the entire process of determining the concentration of \(\mathrm{SO}_2\) in air. In this exercise, the key reaction involves the conversion of sulfur dioxide to sulfuric acid by hydrogen peroxide:\[\mathrm{SO}_2 + \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{SO}_4}\]This oxidation reaction is step one in converting a gaseous pollutant into a form that can be further processed and quantified. The formation of \(\mathrm{H}_2\mathrm{SO}_4\) completes the series of chemical events needed to titrate and ultimately measure the \(\mathrm{SO}_2\) concentration in air. Furthermore, titration itself relies on a neutralization reaction:\[\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}\]The relationship between the reactants and products in these reactions emphasizes the importance of stoichiometry in analytical chemistry, connecting the chemical transformations to measurable outcomes. Understanding these reactions enables chemists to manipulate variables accurately in laboratory settings, ensuring reliable data collection and analysis in environmental chemistry applications.

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Most popular questions from this chapter

The amount of \(\mathrm{Cr}^{3+}\) in an inorganic salt is determined by a redox titration. A portion of sample that contains approximately \(0.25 \mathrm{~g}\) of \(\mathrm{Cr}^{3+}\) is accurately weighed and dissolved in \(50 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O} .\) The \(\mathrm{Cr}^{3+}\) is oxidized to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) by adding \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\), which serves as a catalyst, and \(50 \mathrm{~mL}\) of \(10 \% \mathrm{w} / \mathrm{v}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\), which serves as the oxidizing agent. After the reaction is complete, the resulting solution is boiled for 20 minutes to destroy the excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) cooled to room temperature, and diluted to \(250 \mathrm{~mL}\) in a volumetric flask. A \(50-m L\) portion of the resulting solution is transferred to an Erlenmeyer flask, treated with \(50 \mathrm{~mL}\) of a standard solution of \(\mathrm{Fe}^{2+}\), and acidified with \(200 \mathrm{~mL}\) of \(1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), reducing the \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\). The excess \(\mathrm{Fe}^{2+}\) is then determined by a back titration with a standard solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) using an appropriate indicator. The results are reported as \(\% \mathrm{w} / \mathrm{w} \mathrm{Cr}^{3+}\). (a) There are several places in the procedure where a reagent's volume is specified (see italicized text). Which of these measurements must be made using a volumetric pipet. (b) Excess peroxydisulfate, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) is destroyed by boiling the solution. What is the effect on the reported \(\% \mathrm{w} / \mathrm{w} \mathrm{Cr}^{3+}\) if some of the \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) is not destroyed during this step? (c) Solutions of \(\mathrm{Fe}^{2+}\) undergo slow air oxidation to \(\mathrm{Fe}^{3+}\). What is the effect on the reported \(\% \mathrm{w} / \mathrm{w} \mathrm{Cr}^{3+}\) if the standard solution of \(\mathrm{Fe}^{2+}\) is inadvertently allowed to be partially oxidized?

The amount of uranium in an ore is determined by an indirect redox titration. The analysis is accomplished by dissolving the ore in sulfuric acid and reducing \(\mathrm{UO}_{2}^{2+}\) to \(\mathrm{U}^{4+}\) with a Walden reductor. The solution is treated with an excess of \(\mathrm{Fe}^{3+}\), forming \(\mathrm{Fe}^{2+}\) and \(\mathrm{U}^{6+}\). The \(\mathrm{Fe}^{2+}\) is titrated with a standard solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) In a typical analysis a 0.315 -g sample of ore is passed through the Walden reductor and treated with \(50.00 \mathrm{~mL}\) of \(0.0125 \mathrm{M} \mathrm{Fe}^{3+}\). Back titrating with 0.00987 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) requires \(10.52 \mathrm{~mL}\). What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{U}\) in the sample?

Calculate or sketch the titration curve for a \(50.0 \mathrm{~mL}\) solution of a \(0.100 \mathrm{M}\) monoprotic weak acid \(\left(\mathrm{p} K_{\mathrm{a}}=8.0\right)\) with \(0.1 \mathrm{M}\) strong base in a nonaqueous solvent with \(K_{\mathrm{s}}=10^{-20}\). You may assume that the change in solvent does not affect the weak acid's \(\mathrm{p} K_{\mathrm{a}}\). Compare your titration curve to the titration curve when water is the solvent.

The concentration of \(\mathrm{CO}\) in air is determined by passing a known volume of air through a tube that contains \(\mathrm{I}_{2} \mathrm{O}_{5}\), forming \(\mathrm{CO}_{2}\) and \(\mathrm{I}_{2}\). The \(\mathrm{I}_{2}\) is removed from the tube by distilling it into a solution that contains an excess of \(\mathrm{KI}\), producing \(\mathrm{I}_{3}^{-}\). The \(\mathrm{I}_{3}^{-}\) is titrated with a standard solution of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\). In a typical analysis a 4.79 - \(\mathrm{L}\) sample of air is sampled as described here, requiring \(7.17 \mathrm{~mL}\) of \(0.00329 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) to reach the end point. If the air has a density of \(1.23 \times 10^{-3} \mathrm{~g} / \mathrm{mL},\) determine the parts per million CO in the air.

A 0.1036 -g sample that contains only \(\mathrm{BaCl}_{2}\) and \(\mathrm{NaCl}\) is dissolved in \(50 \mathrm{~mL}\) of distilled water. Titrating with \(0.07916 \mathrm{M} \mathrm{AgNO}_{3}\) requires \(19.46 \mathrm{~mL}\) to reach the Fajans end point. Report the \(\% \mathrm{w} / \mathrm{w} \mathrm{BaCl}_{2}\) in the sample.
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