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Chapter 12: Problem 30

Janusa and coworkers describe the determination of chloride by CZE. \(^{29}\) Analysis of a series of external standards gives the following calibration curve. $$ \text { area }=-883+5590 \times \text { ppm } \mathrm{Cl}^{-} $$ A standard sample of \(57.22 \% \mathrm{w} / \mathrm{w} \mathrm{Cl}^{-}\) is analyzed by placing \(0.1011-\mathrm{g}\) portions in separate \(100-\mathrm{mL}\) volumetric flasks and diluting to volume. Three unknowns are prepared by pipeting \(0.250 \mathrm{~mL}, 0.500 \mathrm{~mL},\) and \(0.750 \mathrm{~mL}\) of the bulk unknown in separate \(50-\mathrm{mL}\) volumetric flasks and diluting to volume. Analysis of the three unknowns gives areas of \(15310,31546,\) and \(47582,\) respectively. Evaluate the accuracy of this analysis.

Short Answer

Expert verified
The analysis shows chloride concentrations very close to 578 ppm, indicating reasonable accuracy with slight variation.

Step by step solution

01

Calculate the Chloride Concentration of the Standard

First, we need to determine how much actual chloride is in the standard solution. We start with 0.1011 g of the substance, which is 57.22% chloride by weight. To find the weight of chloride:\[(0.1011 \text{ g}) \times 0.5722 = 0.0578 \text{ g of } Cl^-\]Since the solution is diluted to 100 mL:\[\text{ppm of } Cl^- = \frac{0.0578 \text{ g} \times 1000}{100 \text{ mL}}= 578 \text{ ppm}\]
02

Determine Chloride Concentrations for Each Unknown Solution

We have three samples with different amounts of the bulk unknown and varying volumes. First, calculate the chloride concentration for each unknown based on the given calibration curve.For the first unknown (area = 15310):\[15310 = -883 + 5590 \times \text{ppm }\text{ppm} = \frac{15310 + 883}{5590} = 2.88 \text{ ppm}\]For the second (area = 31546):\[31546 = -883 + 5590 \times \text{ppm}\text{ppm} = \frac{31546 + 883}{5590} = 5.77 \text{ ppm}\]For the third (area = 47582):\[47582 = -883 + 5590 \times \text{ppm}\text{ppm} = \frac{47582 + 883}{5590} = 8.69 \text{ ppm}\]
03

Back-Calculation to Determine Accuracy of Analysis

Use the prepared volumes and resulting concentrations to evaluate the accuracy. Each unknown is made from a bulk unknown, so calculate the relative concentration based on the 50 mL volumetric flask. - For 0.250 mL of bulk: \[\text{ppm}_{\text{bulk}} = 2.88 \text{ ppm} \times \frac{50 \text{ mL}}{0.250 \text{ mL}} = 576 \text{ ppm}\]- For 0.500 mL: \[\text{ppm}_{\text{bulk}} = 5.77 \text{ ppm} \times \frac{50 \text{ mL}}{0.500 \text{ mL}} = 577 \text{ ppm}\]- For 0.750 mL: \[\text{ppm}_{\text{bulk}} = 8.69 \text{ ppm} \times \frac{50 \text{ mL}}{0.750 \text{ mL}} = 579 \text{ ppm}\]Compare against the standard concentration value of 578 ppm. The results show that the method's accuracy is reasonable with slight variation around the expected value.

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Key Concepts

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

Chloride Determination
Chloride determination in capillary zone electrophoresis (CZE) relies on understanding the chloride content within a given sample. In this exercise, the sample begins with 0.1011 g containing 57.22% chloride by weight.
By multiplying, we find the actual weight of chloride to be 0.0578 g. Diluting this into 100 mL gives us a chloride concentration of 578 ppm. Chloride determination is integral to various chemical analyses because chloride ions are prevalent in many chemical and biological systems. The goal here is to accurately gauge the presence of these ions within a solution. CZE plays a role by separating ions in a sample, making it easier for analysts to pinpoint the concentration of chloride present. This is crucial for ensuring consistency in operations like food safety testing, water quality assessments, or any process where chloride concentration is monitored.
Calibration Curve
A calibration curve in the context of CZE is a valuable tool for translating experimental data into meaningful analysis outcomes. In this exercise, the calibration curve provided is: \[ \text{area} = -883 + 5590 \times \text{ppm} \ Cl^{-} \]This equation symbolizes the relationship between the measured area in the electrophoresis and the concentration of chloride ions present.
When you know the area, you can calculate ppm, which tells you the parts per million of chloride in your sample.Creating an accurate calibration curve is vital. It requires analysis of standards with known concentrations and plotting the response, such as the area under a peak from an electropherogram. When unknown samples are assessed, their response can be matched to this curve to gauge concentration levels.
  • The calibration curve must be linear over the concentration range of interest for accuracy.
  • Regular checks and recalibrations ensure validity since equipment may drift over time.
Calibration curves are a staple in quantitative analysis, converting data into a form that speaks to real-world concentrations.
Quantitative Analysis
In quantitative analysis, the aim is to determine the quantity or concentration of a substance in a sample. In this case, we're quantitatively analyzing chloride ions using CZE and calibration curve-derived equations.
Using the equation from our calibration curve, we compute the concentrations for unknowns by solving for ppm based on the area values obtained from CZE. Calculations for three samples gave ppm values of 2.88, 5.77, and 8.69 respectively. Such results help identify the bulk unknown's chloride concentration when volumes are back-calculated. Quantitative analysis facilitates the understanding of sample content that is necessary when scrutinizing substances for compliance with regulatory standards.
  • Exactness is pivotal; even small inaccuracies can lead to significant deviations in critical fields like pharmaceuticals or environmental monitoring.
  • It links experimental data to tangible quantities much needed for reliable and usable results.
It's about transforming qualitative observations into numbers we can compare, evaluate, and rely upon.

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Most popular questions from this chapter

The amount of caffeine in an analgesic tablet was determined by HPLC using a normal calibration curve. Standard solutions of caffeine were prepared and analyzed using a \(10-\mu L\) fixed-volume injection loop. Results for the standards are summarized in the following table. \begin{tabular}{cc} concentration \((\mathrm{ppm})\) & signal (arb. units) \\ \hline 50.0 & 8354 \\ 100.0 & 16925 \\ 150.0 & 25218 \\ 200.0 & 33584 \\ 250.0 & 42002 \end{tabular} The sample is prepared by placing a single analgesic tablet in a small beaker and adding \(10 \mathrm{~mL}\) of methanol. After allowing the sample to dissolve, the contents of the beaker, including the insoluble binder, are quantitatively transferred to a \(25-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. The sample is then filtered, and a \(1.00-\mathrm{mL}\) aliquot transferred to a \(10-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. When analyzed by HPLC, the signal for caffeine is found to be \(21469 .\) Report the milligrams of caffeine in the analgesic tablet.

The amount of camphor in an analgesic ointment is determined by GC using the method of internal standards. \({ }^{21}\) A standard sample is prepared by placing \(45.2 \mathrm{mg}\) of camphor and \(2.00 \mathrm{~mL}\) of a \(6.00 \mathrm{mg} / \mathrm{mL}\) internal standard solution of terpene hydrate in a \(25-\mathrm{mL}\) volumetric flask and diluting to volume with \(\mathrm{CCl}_{4}\). When approximately \(2-\mu \mathrm{L}\) sample of the standard is injected, the FID signals for the two components are measured (in arbitrary units) as 67.3 for camphor and 19.8 for terpene hydrate. A 53.6-mg sample of an analgesic ointment is prepared for analysis by placing it in a \(50-\mathrm{mL}\) Erlenmeyer flask along with \(10 \mathrm{~mL}\) of \(\mathrm{CCl}_{4}\). After heating to \(50^{\circ} \mathrm{C}\) in a water bath, the sample is cooled to below room temperature and filtered. The residue is washed with two \(5-\mathrm{mL}\) portions of \(\mathrm{CCl}_{4}\) and the combined filtrates are collected in a \(25-\mathrm{mL}\) volumetric flask. After adding \(2.00 \mathrm{~mL}\) of the internal standard solution, the contents of the flask are diluted to volume with \(\mathrm{CCl}_{4}\). Analysis of an approximately \(2-\mu \mathrm{L}\) sample gives FID signals of 13.5 for the terpene hydrate and 24.9 for the camphor. Report the \(\% \mathrm{w} / \mathrm{w}\) camphor in the analgesic ointment.

The concentrations of \(\mathrm{Cl}^{-}, \mathrm{NO}_{3}^{-},\) and \(\mathrm{SO}_{4}^{2-}\) are determined by ion chromatography. A \(50-\mu \mathrm{L}\) standard sample of \(10.0 \mathrm{ppm} \mathrm{Cl}^{-}, 2.00 \mathrm{ppm}\) \(\mathrm{NO}_{3}^{-},\) and \(5.00 \mathrm{ppm} \mathrm{SO}_{4}^{2-}\) gave signals (in arbitrary units) of \(59.3,\) \(16.1,\) and 6.08 respectively. A sample of effluent from a wastewater treatment plant is diluted tenfold and a \(50-\mu \mathrm{L}\) portion gives signals of 44.2 for \(\mathrm{Cl}^{-}, 2.73\) for \(\mathrm{NO}_{3}^{-},\) and 5.04 for \(\mathrm{SO}_{4}^{2-} .\) Report the parts per million for each anion in the effluent sample.

Loconto and co-workers describe a method for determining trace levels of water in soil. \(^{19}\) The method takes advantage of the reaction of water with calcium carbide, \(\mathrm{CaC}_{2}\), to produce acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\). By carrying out the reaction in a sealed vial, the amount of acetylene produced is determined by sampling the headspace. In a typical analysis a sample of soil is placed in a sealed vial with \(\mathrm{CaC}_{2}\). Analysis of the headspace gives a blank corrected signal of \(2.70 \times 10^{5} .\) A second sample is prepared in the same manner except that a standard addition of \(5.0 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} / \mathrm{g}\) soil is added, giving a blank-corrected signal of \(1.06 \times 10^{6} .\) Determine the milligrams \(\mathrm{H}_{2} \mathrm{O} / \mathrm{g}\) soil in the soil sample.

Bohman and colleagues described a reversed-phase HPLC method for the quantitative analysis of vitamin \(\mathrm{A}\) in food using the method of standard additions. \({ }^{27}\) In a typical example, a \(10.067-\mathrm{g}\) sample of cereal is placed in a 250 -mL Erlenmeyer flask along with \(1 \mathrm{~g}\) of sodium ascorbate, \(40 \mathrm{~mL}\) of ethanol, and \(10 \mathrm{~mL}\) of \(50 \% \mathrm{w} / \mathrm{v} \mathrm{KOH}\). After refluxing for \(30 \mathrm{~min}, 60 \mathrm{~mL}\) of ethanol is added and the solution cooled to room temperature. Vitamin \(\mathrm{A}\) is extracted using three \(100-\mathrm{mL}\) portions of hexane. The combined portions of hexane are evaporated and the residue containing vitamin A transferred to a \(5-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. A standard addition is prepared in a similar manner using a \(10.093-\mathrm{g}\) sample of the cereal and spiking with \(0.0200 \mathrm{mg}\) of vitamin \(\mathrm{A}\). Injecting the sample and standard addition into the HPLC gives peak areas of, respectively, \(6.77 \times 10^{3}\)and \(1.32 \times 10^{4}\). Report the vitamin \(\mathrm{A}\) content of the sample in milligrams/100 g cereal.
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