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Chapter 9: Problem 53

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.

Short Answer

Expert verified
The concentration of CO is approximately 112 ppm.

Step by step solution

01

Determine moles of Na2S2O3 used

First, we calculate the moles of sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)) used in the titration. We use the formula: \[\text{Moles} = \text{Volume} \times \text{Molarity}.\] Hence, \[\text{Moles of Na}_2\text{S}_2\text{O}_3 = 7.17\,\text{mL} \times \frac{0.00329\,\text{mol}}{1000\,\text{mL}} = 2.36 \times 10^{-5} \text{mol}.\]
02

Determine moles of CO

The reaction between \(\text{I}_3^-\) and \(\text{Na}_2\text{S}_2\text{O}_3\) indicates a 1:1 molar ratio between iodine triiodide (\(\text{I}_3^-\)) and \(\text{CO}\). Therefore, the moles of \(\text{CO}\) are the same as the moles of \(\text{Na}_2\text{S}_2\text{O}_3\), which is \(2.36 \times 10^{-5} \text{mol}.\)
03

Calculate mass of CO

To find the mass of \(\text{CO}\), we multiply the moles of \(\text{CO}\) by its molar mass. The molar mass of \(\text{CO}\) is \(28.01\,\text{g/mol}\). Thus,\[\text{Mass of CO} = 2.36 \times 10^{-5} \text{mol} \times 28.01\,\text{g/mol} = 6.61 \times 10^{-4} \text{g}.\]
04

Determine total mass of air sample

To find the mass of the air sample, we multiply the volume by the density. The sample volume is \(4.79\,\text{L}\), which is \(4790\,\text{mL}\). The density is \(1.23 \times 10^{-3}\,\text{g/mL}\). Thus,\[\text{Mass of air} = 4790\,\text{mL} \times 1.23 \times 10^{-3}\,\text{g/mL} = 5.90\,\text{g}.\]
05

Calculate parts per million (ppm) of CO

Parts per million (ppm) is calculated as \(\text{ppm} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6\). Substituting in the values, we get\[\text{ppm CO} = \left(\frac{6.61 \times 10^{-4}\,\text{g}}{5.90\,\text{g}}\right) \times 10^6 \approx 112 \,\text{ppm}.\]

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Key Concepts

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

Gas Analysis
Gas analysis is an essential practice in analytical chemistry, helping us understand the composition of gaseous mixtures. When we want to determine the concentration of a specific gas, like carbon monoxide (CO) in air, we typically use a chemical reaction to convert the gas into a more easily measured form.
In the problem you looked at, carbon monoxide is passed through a tube containing iodine pentoxide (\(\mathrm{I}_2\mathrm{O}_5\)). The reaction of CO with \(\mathrm{I}_2\mathrm{O}_5\) results in carbon dioxide (CO2) and iodine (\(\mathrm{I}_2\)). This reaction allows us to isolate iodine, which will then be analyzed to find the concentration of CO.
By analyzing gases with precision, we are able to measure even tiny concentrations of pollutants and determine air quality efficiently. Understanding the different components in gas analysis helps scientists and engineers control and improve environmental conditions.
Titration Methods
Titration is a fundamental technique in analytical chemistry. It involves adding a titrant of known concentration to a solution until a chemical reaction occurs to completion. In this exercise, the iodine (\(\mathrm{I}_2\)) formed from the CO reaction is distilled and converted into \(\mathrm{I}_3^-\) ions using potassium iodide (KI).
The next step involves a redox titration, where \(\mathrm{I}_3^-\) is titrated with a sodium thiosulfate solution (\(\mathrm{Na}_2\mathrm{~S}_2\mathrm{O}_3\)).- **End point detection:** The endpoint of the titration is typically detected when the iodine color changes or disappears, indicating that all \(\mathrm{I}_3^-\) has reacted.- **Precision in measurement:** The precise measurement of volumes and concentrations is essential to ensure accuracy in calculating the concentration of the analyte, in this case, carbon monoxide.
Titration methods provide a way to determine unknown concentrations based on well-established stoichiometric relationships.
Chemical Reactions
Chemical reactions drive the conversion processes crucial for gas and solution analysis. In this specific problem, it begins with the reaction between carbon monoxide and iodine pentoxide:- \(5\mathrm{CO} + \mathrm{I}_2\mathrm{O}_5 \rightarrow 5\mathrm{CO}_2 + \mathrm{I}_2\)This step is vital for converting the initial target gas into a measurable form, iodine (\(\mathrm{I}_2\)).
Next, the \(\mathrm{I}_2\) reacts with surplus KI to form tri-iodide ions (\(\mathrm{I}_3^-\)):- \(\mathrm{I}_2 + \mathrm{I}^- \rightarrow \mathrm{I}_3^-\)
These reactions are monitored using stoichiometry, which is a key aspect of determining the amounts of reactants and products involved. Understanding these chemical interactions allows chemists to draw relevant conclusions from seemingly small environmental samples.
Parts Per Million Calculation
The parts per million (ppm) calculation is important to express the concentration of a solute in a solution, especially when dealing with gases like CO in air. In this exercise, you determined the mass of CO relative to the total mass of the air sample.To calculate ppm, use the formula:- \(\text{ppm} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6\)
The mass of the solute (CO) was obtained from the moles calculated through titration and its molar mass. The mass of the solution was the total mass of the air sample based on its volume and density.
Understanding ppm provides a standardized way to compare concentrations, crucial for scientific studies involving environmental gases and pollutants, ensuring assessments remain clear and universally interpretable.

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Most popular questions from this chapter

A quantitative analysis for aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, K_{\mathrm{b}}=3.94 \times 10^{-10}\right)\) is carried out by an acid-base titration using glacial acetic acid as the solvent and \(\mathrm{HClO}_{4}\) as the titrant. A known volume of sample that contains \(3-4 \mathrm{mmol}\) of aniline is transferred to a \(250-\mathrm{mL}\) Erlenmeyer flask and diluted to approximately \(75 \mathrm{~mL}\) with glacial acetic acid. Two drops of a methyl violet indicator are added, and the solution is titrated with previously standardized \(0.1000 \mathrm{M} \mathrm{HClO}_{4}\) (prepared in glacial acetic acid using anhydrous \(\mathrm{HClO}_{4}\) ) until the end point is reached. Results are reported as parts per million aniline. (a) Explain why this titration is conducted using glacial acetic acid as the solvent instead of using water. (b) One problem with using glacial acetic acid as solvent is its relatively high coefficient of thermal expansion of \(0.11 \% /{ }^{\circ} \mathrm{C}\). For example, \(100.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) occupies \(100.22 \mathrm{~mL}\) at \(27^{\circ} \mathrm{C}\). What is the effect on the reported concentration of aniline if the standardization of \(\mathrm{HClO}_{4}\) is conducted at a temperature that is lower than that for the analysis of the unknown? (c) The procedure calls for a sample that contains \(3-4\) mmoles of aniline. Why is this requirement necessary?

The amount of iron in a meteorite is determined by a redox titration using \(\mathrm{KMnO}_{4}\) as the titrant. A \(0.4185-\mathrm{g}\) sample is dissolved in acid and the liberated \(\mathrm{Fe}^{3+}\) quantitatively reduced to \(\mathrm{Fe}^{2+}\) using a Walden reductor. Titrating with \(0.02500 \mathrm{M} \mathrm{KMnO}_{4}\) requires \(41.27 \mathrm{~mL}\) to reach the end point. Determine the \(\% \mathrm{w} / \mathrm{w} \mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample of meteorite.

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.

The concentration of cyanide, \(\mathrm{CN}^{-},\) in a copper electroplating bath is determined by a complexometric titration using \(\mathrm{Ag}^{+}\) as the titrant, forming the soluble \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) complex. In a typical analysis a \(5.00-\mathrm{mL}\) sample from an electroplating bath is transferred to a 250 -mL Erlenmeyer flask, and treated with \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}, 5 \mathrm{~mL}\) of \(20 \% \mathrm{w} / \mathrm{v}\) \(\mathrm{NaOH}\) and \(5 \mathrm{~mL}\) of \(10 \% \mathrm{w} / \mathrm{v}\) KI. The sample is titrated with 0.1012 \(\mathrm{M} \mathrm{AgNO}_{3}\), requiring \(27.36 \mathrm{~mL}\) to reach the end point as signaled by the formation of a yellow precipitate of AgI. Report the concentration of cyanide as parts per million of \(\mathrm{NaCN}\).

Voncina and co-workers suggest that a precipitation titration can be monitored by measuring \(\mathrm{pH}\) as a function of the volume of titrant if the titrant is a weak base. \({ }^{14}\) For example, when titrating \(\mathrm{Pb}^{2+}\) with \(\mathrm{K}_{2} \mathrm{CrO}_{4}\), the solution that contains the analyte initially is acidified to a \(\mathrm{pH}\) of 3.50 using \(\mathrm{HNO}_{3} .\) Before the equivalence point the concentration of \(\mathrm{CrO}_{4}^{2-}\) is controlled by the solubility product of \(\mathrm{PbCrO}_{4}\). After the equivalence point the concentration of \(\mathrm{CrO}_{4}^{2-}\) is determined by the amount of excess titrant. Considering the reactions that control the concentration of \(\mathrm{CrO}_{4}^{2-}\), sketch the expected titration curve of \(\mathrm{pH}\) versus volume of titrant.
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