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Detection limit. In spectrophotometry, we measure the concentration of an analyte by its absorbance of light. A low-concentration sample was prepared and nine replicate measurements gave absorbances of 0.0047,0.0054,0.0062,0.0060,0.0046,0.0056,0.0052,0.0044, and 0.0058. Nine reagent blanks gave values of 0.0006,0.0012, 0.0022,0.0005,0.0016,0.0008,0.0017,0.0010, and 0.0011.

a) Find the absorbance detection limit with equation 5-3.

b) The calibration curve is a graph of absorbance versus concentration. Absorbance is a dimensionless quantity. The slope of the calibration curve is m=2.24x104M-1Find the concentration detection limit with Equation 5-5.

(c) Find the lower limit of quantitation with Equation 5-6.

Short Answer

Expert verified

a) The absorbance detection limit of the sample was calculated as

0.003112.

b) The minimum detectable concentration of the sample was calculated as

8.6x10-8M.

c) The lower limit of quantitation of the sample was calculated as

2.9x10-7M

Step by step solution

01

Concept used

Detection limit:

This is the concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The absorbance detectable concentration ydlcan be defined as,:

signal detection limitydl=yblank+3s

Minimum detectable concentration=3sm

Lower limit of quantitation=10sm

Where,

s=Standard deviation.

m=Slope of linear calibration curve.

02

Calculate the absorbance detection limit of the sample

a)The low concentrations samples near the detection limit are.

0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011.

The mean reading of the nine blanks is 0.001189.

The calculated standard deviation of the nine samples is 0.000644.

ydl=yblank+3sydl=0.00118+30.00688=0.022

03

Calculate the minimum detectable concentration of the sample

b)The low concentrations samples near the detection limit are. 0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011.

The mean reading of nine blanks is 0.001189.

The slope of the calibration curve is 2.24x104M1

The calculated standard deviation of the nine samples is 0.000644.minimumdetectableconcentration=3sm

=30.0006442.24x104M-1=8.6x10-8M

04

Calculate the lower limit quantitation of the sample

c)The low concentrations samples near the detection limit are.

0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011.

The mean reading of nine blanks is 0.001189.

The slope of the calibration curve is 2.24x104M-1.

The calculated standard deviation of the nine samples is 0.000644.

lower limit of quantitation

Lowerlimitofquantitation=10sm=100.0006442.24x104M=2.9x107M

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Most popular questions from this chapter

What are the three parts of quality assurance? What questions are asked in each part and what actions are taken in each part?

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