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Chapter 28: Q25P (page 792)

The county landfill in the diagram was monitored to verify that toxic compounds were not leaching into the local water supply. Wells drilled at 21 locations were monitored over a year and pollutants were observed only at sites\(8,11,12\), and 13 . Monitoring all 21 sites each month is very expensive. Suggest a strategy to use composite samples (Box 0-1) made from more than one well at a time to reduce the cost of routine monitoring. How will your scheme affect the minimum detectable level for pollutants at a particular site?

Short Answer

Expert verified

How the scheme affects the minimum detectable level for pollutants at a particular site is explained clearly.

Step by step solution

01

Defining composite samples.

A collection of individual samples (grab samples) taken over a specific time period (e.g., 24 hours for a daily composite). The water characteristics in a composite sample represent the conditions in the sampled flow during that time period. Subsamples can be collected at regular intervals to produce time-weighted composites, or their volume and timing can be varied to produce flowweighted composites.

02

Analzing the scheme.

In this task we will suggest a strategy to use composite samples (Box 0-1) made from more than one well at a time in order to reduce the cost of routine monitoring. Also, explain how would our scheme affect the minimum detectable level for pollutants at a particular site.

One of the possible schemes would be to monitor the wells \(8,11,12\) and 13 individually, but we would pool from other sites samples

A composite sample can be made with same volumes from wells\((1,2,3,4)\), whereas other composite samples could be made from \((5,6,7);(9,10);(14,15,16,17);(18,19,20,21)\)

(1) If there is no warning level of analyte found in composites, we could say that each well in those composites contains no analyte.

(2) If we find that analyte is present in composites, then we would separately analyze each contributor to the composite sample.

The main disadvantage of pooling samples from different \(n\) wells is that the sensitivity of analysis in each well would be reduced by \(1/n\).

How the scheme affects the minimum detectable level for pollutants at a particular site is explained above clearly.

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Most popular questions from this chapter

:(a) Describe the steps in QuEChERS and explain their purpose.

(b) Why is an internal standard used in QuEChERS?

(c) What is displayed in the total ion chromatogram in Figure 28-22?

(d) What is displayed in the extracted ion chromatogram in Figure 28-22? What is the difference between an extracted ion chromatogram and a selected ion chromatogram? Which would have greater signal-to-noise ratio?

(e) What mass spectrometric method could be used to obtain even greater signal-to-noise ratio from the same QuEChERS extract?

How does solid-supported liquid-liquid extraction differ from solid-phase extraction?

When you flip a coin, the probability of its landing on each side is \(p = q = \frac{1}{2}\)in Equations 28-2 and 28-3. If you flip it \(n\)times, the expected number of heads equals the expected number of tails \( = np = nq = \frac{1}{2}n.\)The expected standard deviation for \(n\)flips is\({\sigma _n} = \sqrt {npq} \). From Table 4-1, we expect that \(68.3\% \)of the results will lie within \( \pm 1{\sigma _n}\) and \(95.5\% \)of the results will lie within\( \pm 2{\sigma _n}\).

(a) Find the expected standard deviation for the number of heads in \({\bf{1000}}\) coin flips.

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(c) If you repeat the\({\bf{1000}}\)coin flips many times, what is the expected range for the number of heads that includes\(90\% \) of the results? (For example, your answer might be, "The range \({\bf{490}}\) to \({\bf{510}}\) will be observed \(90\% \)of the time.")

The following wet-ashing procedure was used to measure arsenic in organic soil samples by atomic absorption spectroscopy: A 0.1- to \({\bf{0}}.{\bf{5}} - \)g sample was heated in a \({\bf{150}} - {\bf{mL}}\) Teflon bomb in a microwave oven for \(2.5\;{\rm{min}}\) with \(3.5\;{\rm{mL}}\)of\(70\% \,\,\,{\rm{HN}}{{\rm{O}}_3}\). After the sample cooled, a mixture containing \(3.5\;{\rm{mL}}\)of \(70\% \,\,\,{\rm{HN}}{{\rm{O}}_3},1.5\;{\rm{mL}}\) of\(70\% \,\,{\rm{HCl}}{{\rm{O}}_4}\), and \(1.0\;{\rm{mL}}\) of \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\)was added and the sample was reheated for three \({\bf{2}}.{\bf{5}} - {\bf{min}}\) intervals with 2 -min unheated periods in between. The final solution was diluted with \(0.2{\rm{M}}\,\,\,{\rm{HCl}}\)for analysis. Why was \({\rm{HCl}}{{\rm{O}}_4}\) not introduced until the second heating?

In analyzing a lot with random sample variation, you find a sampling standard deviation of \({\bf{65}}\% .\)Assuming negligible error in the analytical procedure, how many samples must be analyzed to give \(9{\bf{5}}\% \)confidence that the error in the mean is within\(64\% \)of the true value? Answer the same question for a confidence level of \(90\% \).
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