Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

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 minimum detectable concentration ${\text{y}}_{\text{dl}}$ can be defined as,

$\text{signal}\text{detection}\text{limit}$ ${\text{y}}_{\text{dl}}\text{=}{\text{y}}_{\text{blank}}\text{+}\text{3s}$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$\text{0}\text{.0006,0}\text{.0012,0}\text{.0022,0}\text{.0005,0}\text{.0016,0}\text{.0008,0}\text{.00017,0}\text{.0010,0}\text{.0011}$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

${\text{y}}_{\text{dl}}\text{=}{\text{y}}_{\text{blank}}\text{+}\text{3s}$

${\text{y}}_{\text{dl}}\text{=}\text{0}\text{.00118+}\text{3}\left(\text{0}\text{.00688}\right)\text{=0}\text{.003112}$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$.

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

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

$\text{minimum detectable concentration}\text{=}\frac{\text{3s}}{\text{m}}$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $\text{8}{\text{.6×10}}^{\text{-8}}\text{M}$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$\text{0}\text{.0006,0}\text{.0012,0}\text{.0022,0}\text{.0005,0}\text{.0016,0}\text{.0008,0}\text{.00017,0}\text{.0010,0}\text{.0011}$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24\times {10}^4{M}^{-1}$

The calculated standard deviation of $9$ samples is0.000644.

$\begin{array}{l}\text{minimum}\text{detectable}\text{concentration}\text{=}\frac{\text{3s}}{\text{m}}\\ \text{=}\frac{\text{3}\left(\text{0}\text{.000644}\right)}{\text{2}{\text{.24×10}}^{\text{4}}{\text{M}}^{\text{-1}}}\text{=8}{\text{.6×10}}^{\text{-8}}\text{M}\end{array}$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $\text{8}{\text{.6×10}}^{\text{-8}}\text{M}$

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

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

$\text{lower limit of quantitation}\text{=}\frac{\text{10s}}{\text{m}}$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$\text{lower limit of quantitation = }\frac{\text{10s}}{\text{m}}$

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $\text{2}{\text{.9×10}}^{\text{-7}}\text{M}$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$\text{0}\text{.0006,0}\text{.0012,0}\text{.0022,0}\text{.0005,0}\text{.0016,0}\text{.0008,0}\text{.00017,0}\text{.0010,0}\text{.0011}$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24\times {10}^4{M}^{-1}$

The calculated standard deviation of $9$ samples is0.000644.

$\begin{array}{l}\text{lower limit of quantitation}\text{=}\frac{\text{10s}}{\text{m}}\\ \text{=}\frac{10\left(\text{0}\text{.000644}\right)}{\text{2}{\text{.24×10}}^{\text{4}}{\text{M}}^{\text{-1}}}\text{=2}{\text{.9×10}}^{\text{-7}}\text{M}\end{array}$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $\text{2}{\text{.9×10}}^{\text{-7}}\text{M}$