/*! 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 18 The concentration of \(\mathrm{H... [FREE SOLUTION] | 91Ó°ÊÓ

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The concentration of \(\mathrm{H}_{2} \mathrm{~S}\) in the drainage from an abandoned mine is determined by a coulometric titration using KI as a mediator and \(I_{3}^{-}\) as the titrant. $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{I}_{5}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)=2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+3 \mathrm{I}^{-}(a q)+\mathrm{S}_{(s)} $$ A \(50.00-\mathrm{mL}\) sample of water is placed in a coulometric cell, along with an excess of KI and a small amount of starch as an indicator. Electrolysis is carried out at a constant current of \(84.6 \mathrm{~mA}\), requiring \(386 \mathrm{~s}\) to reach the starch end point. Report the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) in the sample in \(\mu \mathrm{g} / \mathrm{mL}\)

Short Answer

Expert verified
The concentration of \( \mathrm{H}_{2} \mathrm{S} \) is 46.1 \( \mu \mathrm{g/mL} \).

Step by step solution

01

Calculate Total Charge Passed

The first step is to calculate the total charge in coulombs that passed through the electrolysis setup. This is done using the formula for charge, which is: \[ Q = I \times t \]where \( I \) is the current in amperes and \( t \) is the time in seconds.Substituting the values: \[ Q = 84.6 \times 10^{-3} \times 386 = 32.6356 \text{ C} \]
02

Determine Moles of Electrons

Next, we calculate the moles of electrons passed using the Faraday's constant \( F = 96485 \text{ C/mol} \):\[ ext{Moles of electrons} = \frac{Q}{F} = \frac{32.6356}{96485} \approx 3.382 \times 10^{-4} \text{ mol} \]
03

Relate Moles of Electrons to Moles of \( \mathrm{H}_{2} \mathrm{S} \)

In the reaction provided:\[ \mathrm{H}_{2} \mathrm{~S} + \mathrm{I}_{5}^{-} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}_{3} \mathrm{O}^{+} + 3 \mathrm{I}^{-} + \mathrm{S} \]The net reaction shows that each mole of \( \mathrm{H}_{2} \mathrm{S} \) is oxidized by five moles of electrons (given that \( \mathrm{I}_{5}^{-} \) being the overall form of iodine that accepts 2 electrons to form \( \mathrm{I}^{-} \)). Hence,\[ \text{Moles of } \mathrm{H}_{2} \mathrm{S} = \frac{3.382 \times 10^{-4}}{5} = 6.764 \times 10^{-5} \text{ mol} \]
04

Convert Moles of \( \mathrm{H}_{2} \mathrm{S} \) to Mass

Calculate the mass of \( \mathrm{H}_{2} \mathrm{S} \) using its molar mass (34.08 g/mol):\[ \text{Mass of } \mathrm{H}_{2} \mathrm{S} = 6.764 \times 10^{-5} \times 34.08 = 2.305 \times 10^{-3} \text{ g} \]
05

Determine Concentration in \( \mu \mathrm{g}/ \mathrm{mL} \)

Convert the mass of \( \mathrm{H}_{2} \mathrm{S} \) to micrograms and divide by the sample volume to find the concentration:\[ 2.305 \times 10^{-3} \text{ g} \times 10^6 = 2305 \text{ } \mu\text{g} \]For a 50 mL sample:\[ \frac{2305 \text{ } \mu\text{g}}{50.00 \text{ mL}} = 46.1 \text{ } \mu\text{g/mL} \]

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Key Concepts

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

Hydrogen Sulfide Concentration
Hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)) is a toxic gas often found in industrial byproducts such as in water running from abandoned mines. Determining its concentration accurately is crucial for environmental safety. In this specific exercise, coulometric titration is employed to measure the concentration of \(\mathrm{H}_2\mathrm{S}\) by monitoring the electrolysis of water in the presence of potassium iodide (\(\mathrm{KI}\)) and an iodine-based mediator. Here, \(\mathrm{I}_3^-\) serves as the titrant. The reaction’s endpoint is detected using starch, which turns blue in the presence of iodine, signaling the end of \(\mathrm{H}_2\mathrm{S}\) conversion. Understanding this process helps accurately determine levels of toxic substances and ensures appropriate remediation steps can be taken.
Faraday's Constant
Faraday's constant (\( F \)) is a crucial element in electrochemistry and equates to approximately \( 96485 \, \text{C/mol} \). It represents the charge of one mole of electrons. In the context of this problem, Faraday's constant is used to find the moles of electrons that passed through during electrolysis by dividing the total charge (\( Q \)) by \( F \). Calculating the exact number of electrons involved in the reaction helps in determining the amount of \(\mathrm{H}_2\mathrm{S}\) converted, which directly correlates to its concentration. This fundamental constant connects the observable electrolysis with theoretical chemical changes, providing essential insights to quantify substances in a titration.
Electrolysis Calculation
Electrolysis involves using an electric current to drive a non-spontaneous chemical reaction. Understanding it involves calculating the total charge passed during the process, which is determined through the formula \( Q = I \times t \), where \( I \) is current in amperes, and \( t \) is time in seconds. In this exercise, the calculation provides the total charge passed, which is then used to find the moles of electrons that facilitated the reaction. Since each mole of \(\mathrm{H}_2\mathrm{S}\) requires five moles of electrons, knowing the number of electrons helps determine how much \(\mathrm{H}_2\mathrm{S}\) has reacted. This calculation is essential in the coulometric titration method used in detecting concentrations of chemical substances.
Iodometric Reaction
An iodometric reaction is a type of redox titration where iodine species participate in the reaction. In this exercise, the reaction between \(\mathrm{H}_2\mathrm{S}\) and \(\mathrm{I}_5^-\) forms \(\mathrm{I}^-\), indicating the reaction progress. The iodine solution changes color in the presence of starch at the endpoint, visually signifying the completion of the titration. These reactions are highly valued for their specificity and simplicity, making them well-suited for detecting and quantifying \(\mathrm{H}_2\mathrm{S}\). The application of such titrations helps students and professionals to measure concentrations of specific substances accurately, benefiting diverse sectors such as environmental monitoring and chemical manufacturing.
Molar Mass Conversion
Molar mass is the weight of one mole of a substance. Converting moles of a substance to its mass involves multiplying by its molar mass. For \(\mathrm{H}_2\mathrm{S}\), which has a molar mass of 34.08 \, \mathrm{g/mol}, this conversion allows us to find the weight of hydrogen sulfide in a sample. This calculated mass can then be converted to micrograms for easier interpretation and reporting, especially in environmental contexts where regulations might dictate such units (\(\mu \mathrm{g/mL}\)). Understanding and applying molar mass conversions is a foundational skill in chemistry, facilitating the quantity analysis of chemical reactions and substances.

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Most popular questions from this chapter

Thioamide drugs are determined by cathodic stripping analysis. \(^{30}\) Deposition occurs at \(+0.05 \mathrm{~V}\) versus an \(\mathrm{SCE}\). During the stripping step the potential is scanned cathodically and a stripping peak is observed at \(-0.52 \mathrm{~V}\). In a typical application a 2.00 -mL sample of urine is mixed with \(2.00 \mathrm{~mL}\) of a \(\mathrm{pH} 4.78\) buffer. Following a \(2.00 \mathrm{~min}\) deposition, a peak current of \(0.562 \mu \mathrm{A}\) is measured. A \(0.10-\mathrm{mL}\) addition of \(a 5.00 \mu \mathrm{M}\) solution of the drug is added to the same solution. A peak current of \(0.837 \mu \mathrm{A}\) is recorded using the same deposition and stripping conditions. Report the drug's molar concentration in the urine sample.

Differential pulse voltammetry at a carbon working electrode is used to determine the concentrations of ascorbic acid and caffeine in drug formulations \({ }^{25}\) In a typical analysis a \(0.9183-\mathrm{g}\) tablet is crushed and ground into a fine powder. A \(0.5630-\mathrm{g}\) sample of this powder is transferred to a \(100-\mathrm{mL}\) volumetric flask, brought into solution, and diluted to volume. A \(0.500-\mathrm{mL}\) portion of this solution is then transferred to a voltammetric cell that contains \(20.00 \mathrm{~mL}\) of a suitable supporting electrolyte. The resulting voltammogram gives peak currents of \(1.40 \mu \mathrm{A}\) and \(3.88 \mu \mathrm{A}\) for ascorbic acid and for caffeine, respectively. A 0.500 \(\mathrm{mL}\) aliquot of a standard solution that contains \(250.0 \mathrm{ppm}\) ascorbic acid and 200.0 ppm caffeine is then added. A voltammogram of this solution gives peak currents of \(2.80 \mu \mathrm{A}\) and \(8.02 \mu \mathrm{A}\) for ascorbic acid and caffeine, respectively. Report the milligrams of ascorbic acid and milligrams of caffeine in the tablet.

Calculate the activity of \(\mathrm{KI}, x\), in the following electrochemical cell if the potential is \(+0.294 \mathrm{~V}\) $$ \mathrm{Ag}\left|\mathrm{AgCl}(s), \mathrm{NaCl}(a q, 0.1) \| \mathrm{KI}(a q, x), \mathrm{I}_{2}(s)\right| \mathrm{Pt} $$

A potentiometric electrode for HCN uses a gas-permeable membrane, a buffered internal solution of \(0.01 \mathrm{M} \mathrm{KAg}(\mathrm{CN})_{2},\) and a \(\mathrm{Ag}_{2} \mathrm{~S}\) ISE electrode that is immersed in the internal solution. Consider the equilibrium reactions that take place within the internal solution and derive an equation that relates the electrode's potential to the concentration of \(\mathrm{HCN}\) in the sample.

Sittampalam and Wilson described the preparation and use of an amperometric sensor for glucose. \(^{27}\) The sensor is calibrated by measuring the steady- state current when it is immersed in standard solutions of glucose. A typical set of calibration data is shown here. \begin{tabular}{cc} [glucose] \((\mathrm{mg} / 100 \mathrm{~mL})\) & current (arb. units) \\ \hline 2.0 & 17.2 \\ 4.0 & 32.9 \\ 6.0 & 52.1 \\ 8.0 & 68.0 \\ 10.0 & 85.8 \end{tabular} A \(2.00-\mathrm{mL}\) sample is diluted to \(10 \mathrm{~mL}\) in a volumetric flask and a steady-state current of 23.6 (arbitrary units) is measured. What is the concentration of glucose in the sample in \(\mathrm{mg} / 100 \mathrm{~mL}\) ?

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