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The analysis of \(\mathrm{NO}_{3}^{-}\) in aquarium water is carried out by CZE using \(\mathrm{IO}_{4}^{-}\) as an internal standard. A standard solution of \(15.0 \mathrm{ppm} \mathrm{NO}_{3}^{-}\) and 10.0 ppm \(\mathrm{IO}_{4}^{-}\) gives peak heights (arbitrary units) of 95.0 and 100.1 , respectively. A sample of water from an aquarium is diluted 1: 100 and sufficient internal standard added to make its concentration \(10.0 \mathrm{ppm}\) in \(\mathrm{IO}_{4}^{-}\). Analysis gives signals of 29.2 and 105.8 for \(\mathrm{NO}_{3}^{-}\) and \(\mathrm{IO}_{4}^{-},\) respectively. Report the \(\mathrm{ppm} \mathrm{NO}_{3}^{-}\) in the sample of aquarium water.

Short Answer

Expert verified
The concentration of \(\mathrm{NO}_{3}^{-}\) in the aquarium water sample is 440 ppm.

Step by step solution

01

Calculate the Response Factor

The response factor, R, for the standard solution is found using the given peak heights for both the \(\mathrm{NO}_{3}^{-}\) and \(\mathrm{IO}_{4}^{-}\). The formula to calculate R is: \[ R = \frac{\text{Peak Height of } \mathrm{NO}_{3}^{-}}{\text{Concentration of } \mathrm{NO}_{3}^{-}} \div \frac{\text{Peak Height of } \mathrm{IO}_{4}^{-}}{\text{Concentration of } \mathrm{IO}_{4}^{-}} \]Substitute the given values: \[ R = \frac{95.0}{15.0} \div \frac{100.1}{10.0} \]Calculate R: \[ R = 0.0633 \]
02

Determine the Concentration in the Diluted Sample

The concentration of \(\mathrm{NO}_{3}^{-}\) in the diluted sample can be calculated with the response factor. Use the signals from the sample to find the concentration of \(\mathrm{NO}_{3}^{-}\) in the diluted solution. The formula is given by: \[ C_{\text{diluted}}(\mathrm{NO}_{3}^{-}) = \frac{\text{Signal of } \mathrm{NO}_{3}^{-}}{\text{Signal of } \mathrm{IO}_{4}^{-}} \cdot \frac{\text{Concentration of } \mathrm{IO}_{4}^{-}}{R} \]Substitute in the known values: \[ C_{\text{diluted}}(\mathrm{NO}_{3}^{-}) = \frac{29.2}{105.8} \cdot \frac{10.0}{0.0633} \]Calculate the concentration: \[ C_{\text{diluted}}(\mathrm{NO}_{3}^{-}) = 4.4 \, \mathrm{ppm} \]
03

Calculate the Original Concentration in the Sample

Since the sample was diluted 1:100 before analysis, the concentration calculated in Step 2 is not the original concentration. To find the original concentration of \(\mathrm{NO}_{3}^{-}\) in the aquarium water, multiply the diluted concentration by the dilution factor: \[ C_{\text{original}}(\mathrm{NO}_{3}^{-}) = C_{\text{diluted}}(\mathrm{NO}_{3}^{-}) \times 100 \]Insert the value from the previous step: \[ C_{\text{original}}(\mathrm{NO}_{3}^{-}) = 4.4 \times 100 = 440 \, \mathrm{ppm} \]

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Key Concepts

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

Capillary Zone Electrophoresis
Capillary Zone Electrophoresis (CZE) is a versatile analytical technique employed for the separation of ionic species. It involves the use of a narrow-bore capillary tube, through which an electric field is applied. Within this setup, ions migrate towards the electrode of opposite charge. This movement allows for the separation based on the ions' charge-to-size ratio.
CZE is particularly useful for
  • environmental analysis
  • pharmaceuticals
  • biochemical applications
In this exercise, CZE is used to analyze nitrate (\(\mathrm{NO}_{3}^{-}\)) in aquarium water. The technique is effective because it offers high-resolution separation and requires minimal sample preparation. It's valuable for students to understand why CZE is chosen in many real-world applications and how it contributes to precise measurements.
Internal Standards
An internal standard is a crucial element in quantitative analysis. It is a compound, similar yet different from the target analyte, added in a constant amount to all samples, standards, and blanks. In this case, iodate (\(\mathrm{IO}_{4}^{-}\)) serves as the internal standard.
The purpose of using internal standards includes:
  • Compensating for extraction errors
  • Improving precision and accuracy of the analysis
  • Normalizing variations in instrument response
The concentration consistency of the internal standard allows for corrections of signal variations, leading to more reliable results. Implementing internal standards in CZE helps to enhance data consistency, especially when dealing with sample complexities.
Dilution Factors
The concept of dilution factors is essential when preparing samples for analysis. A dilution factor represents the ratio of solute to solvent, indicating how much a solution has been diluted. In the example given, aquarium water is diluted 1:100.
This practice is necessary when the concentration of an analyte, like \(\mathrm{NO}_{3}^{-}\) in this instance, is too high to measure directly. Diluting the sample ensures it falls within the proper range for accurate analysis. Calculating the original concentration involves multiplying by the dilution factor used.
Understanding dilution factors is crucial for:
  • Preparing samples for accurate quantification
  • Avoiding saturation of detection systems
  • Aiding in reliable analytical results
Students should master this concept to effectively handle sample preparation in various scientific fields.
Response Factor Calculation
Calculating the response factor is vital for accurately determining the concentration of an analyte in a sample. It provides a scale relating the signal from the detector to the concentration of the analyte. The response factor (\( R \)) considers both the analyte and internal standard.
Here's how it's calculated:
  • Take the ratio of the peak height of the analyte to its concentration.
  • Then divide by the ratio of the internal standard's peak height to its concentration.
In practical terms, response factors dignify how different substances respond during analysis. For example, in the problem, \( R \) is used to equate the signals from the diluted sample to the concentrations present. Accurately calculating \( R \) ensures effective conversion, essential for precise quantification in straightforward analytical tasks.

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Most popular questions from this chapter

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.

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.

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.

Haddad and associates report the following retention factors for the reversed- phase separation of salicylamide and caffeine. \({ }^{25}\) \(\begin{array}{ccccccc}\% \text { methanol } & 30 \% & 35 \% & 40 \% & 45 \% & 50 \% & 55 \% \\ k_{\text {sal }} & 2.4 & 1.6 & 1.6 & 1.0 & 0.7 & 0.7 \\\ k_{\text {caff }} & 4.3 & 2.8 & 2.3 & 1.4 & 1.1 & 0.9\end{array}\) (a) Explain the trends in the retention factors for these compounds. (b) What is the advantage of using a mobile phase with a smaller \(\% \mathrm{v} / \mathrm{v}\) methanol? Are there any disadvantages?

A mixture of \(n\) -heptane, tetrahydrofuran, 2 -butanone, and \(n\) -propanol elutes in this order when using a polar stationary phase such as Carbowax. The elution order is exactly the opposite when using a nonpolar stationary phase such as polydimethyl siloxane. Explain the order of elution in each case.

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