/*! 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 25 Zinc is used as an internal stan... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Zinc is used as an internal standard in an analysis of thallium by differential pulse polarography. A standard solution of \(5.00 \times 10^{-5} \mathrm{M} \mathrm{Zn}^{2+}\) and \(2.50 \times 10^{-5} \mathrm{M} \mathrm{Tl}^{+}\) has peak currents of \(5.71 \mu \mathrm{A}\) and \(3.19 \mu \mathrm{A}\), respectively. An 8.713 -g sample of a zinc-free alloy is dissolved in acid, transferred to a \(500-\mathrm{mL}\) volumetric flask, and diluted to volume. A \(25.0-\) \(\mathrm{mL}\) portion of this solution is mixed with \(25.0 \mathrm{~mL}\) of \(5.00 \times 10^{-4} \mathrm{M}\) \(\mathrm{Zn}^{2+}\). Analysis of this solution gives peak currents of \(12.3 \mu \mathrm{A}\) and of \(20.2 \mu \mathrm{A}\) for \(\mathrm{Zn}^{2+}\) and \(\mathrm{Tl}^{+}\), respectively. Report the \(\% \mathrm{w} / \mathrm{w} \mathrm{Tl}\) in the alloy.

Short Answer

Expert verified
The alloy contains 1.72% w/w Tl.

Step by step solution

01

Calculate the ratio of currents in the standard solution

Using the given peak currents for zinc and thallium in the standard solution, calculate the ratio:\[ \text{Ratio}_{\text{standard}} = \frac{I_{\text{Tl}}}{I_{\text{Zn}}} = \frac{3.19\, \mu A}{5.71\, \mu A} = 0.559 \]
02

Calculate the ratio of currents in the sample solution

Using the given peak currents for zinc and thallium in the sample solution, calculate the ratio:\[ \text{Ratio}_{\text{sample}} = \frac{I_{\text{Tl}}}{I_{\text{Zn}}} = \frac{20.2\, \mu A}{12.3\, \mu A} = 1.642 \]
03

Determine the concentration of thallium in the sample solution

Using the ratios from both the standard and sample solutions, calculate the concentration of thallium in the sample. The concentration of zinc in the mixed solution is given by: \[ \text{Concentration}_{\text{Zn, mixed}} = \frac{5.00 \times 10^{-4} \ M \times 25.0 \ mL}{50.0 \ mL} = 2.50 \times 10^{-4} \ M \] With concentration of Zn known, the concentration of Tl can be determined using the ratio:\[ \text{Concentration}_{\text{Tl, sample}} = \text{Concentration}_{\text{Zn, mixed}} \times \frac{\text{Ratio}_{\text{sample}}}{\text{Ratio}_{\text{standard}}} \] Plugging in values:\[ \text{Concentration}_{\text{Tl, sample}} = 2.50 \times 10^{-4} \times \frac{1.642}{0.559} = 7.34 \times 10^{-4} \ M \]
04

Calculate the amount of thallium in the full solution

The concentration of thallium in the sample solution refers to the mixed solution after the alloy sample was added. Calculate the moles of thallium in the full 50 mL solution:\[ \text{Moles}_{\text{Tl, mixed}} = \text{Concentration}_{\text{Tl, sample}} \times 0.050 \ L = 7.34 \times 10^{-4} \times 0.050 = 3.67 \times 10^{-5} \ mol \]
05

Calculate the amount of thallium in the original alloy solution

Determine the original amount of thallium from the 500 mL total volume knowing only 25 mL of it was used:\[ \text{Moles}_{\text{Tl, original}} = \text{Moles}_{\text{Tl, mixed}} \times \frac{500\ mL}{25\ mL} = 3.67 \times 10^{-5} \times 20 = 7.34 \times 10^{-4} \ mol \]
06

Convert moles of thallium to mass and calculate percentage

Convert moles of thallium to grams using its molar mass (204.38 g/mol):\[ \text{Mass}_{\text{Tl}} = 7.34 \times 10^{-4} \ mol \times 204.38 \ g/mol = 0.150 \ g \]Now, compute the \(\%w/w\) of thallium in the alloy:\[ \%\ w/w = \frac{\text{Mass}_{\text{Tl}}}{\text{Mass}_{\text{alloy}}} \times 100\% = \frac{0.150}{8.713} \times 100\% = 1.72\% \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Internal Standard
Using an internal standard is a clever technique for enhancing accuracy in chemical analysis. It involves adding a known quantity of a substance to a sample. This substance, called the internal standard, serves as a reference point.

By comparing the signals from the internal standard and the target analyte, fluctuations in measurement conditions can be corrected. In our exercise, zinc is used as the internal standard. It is important to choose an internal standard that behaves similarly under analysis but does not interfere. Zinc meets these criteria when analyzing thallium because it produces a distinct, non-overlapping signal. Thus, any deviations in measurement are detected and corrected by monitoring the zinc's constant behavior.

When conducting differential pulse polarography, calibrate your instrument by comparing the zinc signal in both your sample and the known standard solutions, ensuring reliable results. This careful calibration allows for accurate quantification of the analyte, in this case, thallium.
Thallium Analysis
In the realm of analytical chemistry, thallium analysis is a task that utilizes advanced techniques for accuracy. Differential pulse polarography, as used in this exercise, involves measuring the current that passes through an electrochemical cell as the voltage is varied. Different elements exhibit distinct electrochemical behavior.

Thallium's electrochemical properties allow it to be distinguished and quantified based on these behaviors. In our given problem, we have a known concentration of thallium in a standard solution. By analyzing the current resulting from thallium during polarography, we gain essential quantitative data.

Calibration and Measurements

The calibration of the polarographic instrument using the standard zinc solution allows thallium's accurate measurement. By comparing thallium's peak current deviations in sample and standard solutions, analysts can calculate the unknown thallium concentration.

Ultimately, this method provides highly specific detection of thallium, even in complex matrices.
Zinc Concentration
Zinc concentration plays a critical role in our analysis as it acts as the internal standard. This role ensures that comparative analysis can be conducted reliably. The initial standard solution presents a known zinc concentration of \(5.00 \times 10^{-5} \ M\).

For accurate readings, this concentration is essential when constructing initial calibration curves. Post-sample dilution, the new zinc concentration in the mixed solution becomes \(2.50 \times 10^{-4} \ M\). This value aids calculation by providing a precise reference against which the thallium concentration can be measured.

The constant zinc concentration helps determine the voltage ratio shifts. These shifts assess the presence and quantity of thallium, based on deviations from expected values. Using consistent zinc concentrations in samples is paramount to achieving reliable results and determining unknown quantities in analysis applications.
Peak Current Ratio
The peak current ratio is a significant element in this analysis technique. It pinpoints the proportion of one analyte's current relative to the internal standard's. By calculating this ratio, quantitative results become possible even when the actual concentrations are unknown.

For example, the initial standard solution gives a ratio \(\text{Ratio}_{\text{standard}} = 0.559\), calculated by dividing thallium's peak current by zinc's. Similarly, in the sample solution, the ratio is \(\text{Ratio}_{\text{sample}} = 1.642\). A change in ratio indicates a shift in analyte concentration.

Importance of Ratios

These changes facilitate the computation of thallium concentration within the sample. This is achieved by creating a relationship between the known zinc concentration and the observed peak current ratio deviations.

Such ratios provide insight into the sample's chemical composition, transforming peak currents into actionable concentration data, which allows scientists to report, as in our task, the weight percentage of thallium in an alloy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Abass and co-workers developed an amperometric biosensor for \(\mathrm{NH}_{i}^{+}\) that uses the enzyme glutamate dehydrogenase to catalyze the following reaction $$ \begin{array}{l} \text { 2- oxyglutarate }(a q)+\mathrm{NH}_{4}^{+}(a q) \\ \quad+\operatorname{NADH}(a q)=\text { glutamate }(a q)+\mathrm{NAD}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array} $$ where \(\mathrm{NADH}\) is the reduced form of nicotinamide adenine dinucleotide. \({ }^{28}\) The biosensor actually responds to the concentration of \(\mathrm{NADH}\), however, the rate of the reaction depends on the concentration of \(\mathrm{NH}_{4}^{+}\). If the initial concentrations of 2 -oxyglutarate and NADH are the same for all samples and standards, then the signal is proportional to the concentration of \(\mathrm{NH}_{4}^{+}\). As shown in the following table, the sensitivity of the method is dependent on \(\mathrm{pH}\). \begin{tabular}{cc} \(\mathrm{pH}\) & sensitivity \(\left(\mathrm{nA} \mathrm{s}^{-1} \mathrm{M}^{-1}\right)\) \\ \hline 6.2 & \(1.67 \times 10^{3}\) \\ 6.75 & \(5.00 \times 10^{3}\) \\ 7.3 & \(9.33 \times 10^{3}\) \\ 7.7 & \(1.04 \times 10^{4}\) \\ 8.3 & \(1.27 \times 10^{4}\) \\ 9.3 & \(2.67 \times 10^{3}\) \end{tabular} Two possible explanations for the effect of \(\mathrm{pH}\) on the sensitivity of this analysis are the acid-base chemistry of \(\mathrm{NH}_{4}^{+}\) and the acid-base chemistry of the enzyme. Given that the \(\mathrm{p} K_{\mathrm{a}}\) for \(\mathrm{NH}_{4}^{+}\) is 9.244 , explain the source of this pH-dependent sensitivity.

Wang and Taha described an interesting application of potentiometry, which they call batch injection. \({ }^{24}\) As shown in Figure \(11.56,\) an ionselective electrode is placed in an inverted position in a large volume tank, and a fixed volume of a sample or a standard solution is injected toward the electrode's surface using a micropipet. The response of the electrode is a spike in potential that is proportional to the analyte's concentration. The following data were collected using a \(\mathrm{pH}\) electrode and a set of \(\mathrm{pH}\) standards. \begin{tabular}{cc} \(\mathrm{pH}\) & potential (mV) \\ \hline 2.0 & +300 \\ 3.0 & +240 \\ 4.0 & +168 \\ 5.0 & +81 \\ 6.0 & +35 \\ 8.0 & -92 \\ 9.0 & -168 \\ 10.0 & -235 \\ 11.0 & -279 \end{tabular} Determine the \(\mathrm{pH}\) of the following samples given the recorded peak potentials: tomato juice, \(167 \mathrm{mV} ;\) tap water, \(-27 \mathrm{mV} ;\) coffee, \(122 \mathrm{mV}\).

Differential pulse polarography is used to determine the concentrations of lead, thallium, and indium in a mixture. Because the peaks for lead and thallium, and for thallium and indium overlap, a simultaneous analysis is necessary. Peak currents (in arbitrary units) at \(-0.385 \mathrm{~V}\), \(-0.455 \mathrm{~V}\), and \(-0.557 \mathrm{~V}\) are measured for a single standard solution, and for a sample, giving the results shown in the following table. Report the \(\mu \mathrm{g} / \mathrm{mL}\) of \(\mathrm{Pb}^{2+}, \mathrm{Tl}^{+}\) and \(\mathrm{In}^{3+}\) in the sample. standards \(\quad\) peak currents (arb. units) at \begin{tabular}{lcccc} analyte & \(\mu \mathrm{g} / \mathrm{mL}\) & \(-0.385 \mathrm{~V}\) & \(-0.455 \mathrm{~V}\) & \(-0.557 \mathrm{~V}\) \\ \hline \(\mathrm{Pb}^{2+}\) & 1.0 & 26.1 & 2.9 & 0 \\ \(\mathrm{Tl}^{+}\) & 2.0 & 7.8 & 23.5 & 3.2 \\ \(\mathrm{In}^{3+}\) & 0.4 & 0 & 0 & 22.9 \\ \hline \multicolumn{2}{c} { sample } & 60.6 & 28.8 & 54.1 \end{tabular}

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.

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 \(\mathrm{mg} / 100 \mathrm{~mL}\) ?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.