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The following lifetimes were measured for the chloride quenching of quinine sulfate given in Example 15-1. The fluorescence intensities are given in the example.

(a) Plot fluorescence intensity versus [Cl-].

(b) Plot the ratio of intensity to lifetime, F-t versus [Cl-].

(c) Develop a normalization factor to correct the measured fluorescence intensity to that of the solution without quencher.

(d) Plot on the same graph F versus [Cl-] and Fcorr versus [Cl-] .

Short Answer

Expert verified

a)

b)

c) Normalization factor = τ0τ

d)

Step by step solution

01

Part (a) Step 1: Given Information

Fluorescence intensity versus [Cl-] should be plotted.

02

Part (a) Step 2: Explanation

Fluorescence intensity is the dependent variable and the concentration of Cl- is the independent variable.

Fluorescence intensity

[Cl-], M

180.0

0.000

87.5

0.005

58.0

0.010

43.2

0.015

35.0

0.020

28.5

0.025

24.5

0.030

19.1

0.040

15.7

0.050

03

Part (b) Step 1: Given Information

The ratio of intensity to lifetime, F-τ versus [Cl-] should be plotted.

04

Part (b) Step 2: Explanation

F/τ

[Cl-], M

9.9

0.000

9.8

0.005

10.2

0.010

9.6

0.015

9.7

0.020

10.2

0.025

9.8

0.030

10.1

0.040

9.8

0.050

05

Part (c) Step 1: Given Information

A normalization factor should be developed to correct the measured fluorescence intensity to that of the solution without quencher.

06

Part (c) Step 2: Explanation

According to the graph of ratio of fluorescence to the lifetime versus chloride concentration, the ration remains constant even though the concentration of the quencher increases. ThereforeF0F=τ0τF0=F.τ0τ

F0 – corrected fluorescence (fluorescence without quencher)

F- observed fluorescence (with quencher)

τ0 –fluorescence life time without the quencher

τ – fluorescence lifetime with the quencher

so the normalization factor isτ0τ

07

Part (d) Step 1: Given Information

F versus [Cl-] and Fcorr versus [Cl-] should be plotted on the same graph.

08

Part (d) Step 2: Explanation

F

Fcorr

[Cl-], M

180.0

180.0

0.000

87.5

177.9

0.005

58.0

184.2

0.010

43.2

173.8

0.015

35.0

176.0

0.020

28.5

184.2

0.025

24.5

177.4

0.030

19.1

182.0

0.040

15.7

177.6

0.050

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Most popular questions from this chapter

Discuss the major reasons why molecular phosphorescence spectrometry has not been as widely used as molecular fluorescence spectrometry.

The determination in Problem 15-9 was modified to use the standard-addition method. In this case, a 3.925-g tablet was dissolved in sufficient 0.10 M HCl to give 1.000 L. Dilution of a 20.00-mL aliquot to 100 mL yielded a solution that gave a reading of 415 at 347.5 nm. A second 20.00-mL aliquot was mixed with 10.0 mL of 50-ppm quinine solution before dilution to 100 mL. The fluorescence intensity of this solution was 503. Calculate the percentage of quinine in the tablet.

The following volumes of a standard 10.0 ppb F- solution were added to four 10.00-mL aliquots of a water sample: 0.00, 1.00, 2.00, and 3.00 mL. Precisely 5.00 mL of a solution containing an excess of the strongly absorbing Al-acid alizarin garnet R complex was added to each of the four solutions, and they were each diluted to 50.0 mL. The fluorescence intensities of the four solutions were as follows:

(a) Explain the chemistry of the analytical method.

(b) Construct a plot of the data.

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plot, similar to Equation 1-4.

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(e) Calculate the standard deviation of the slope and intercept.

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(g) Calculate the standard deviation of the result in (e).

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