Chapter 15, Problem 15.16QAP

Interpretation Introduction

(a)

Interpretation:

The chemistry of the analytical method should be explained.

Concept introduction:

Quenching is the process of nonradiative energy transfer from an excited species to another molecule. This requires a contact between the excited species and a quenching agent. The rate of quenching is higher when the quencher concentration is high, because of higher number of collisions during the lifetime of the excited state.

Answer to Problem 15.16QAP

Al-acid alizarin garnet R acts as a quencher. The decrease in fluorescence is due to absorbing emission energy from F- by Al-acid alizarin garnet R complex.

Explanation of Solution

In this experiment standard addition technique has been used. Four water samples with same volume are taken and they have been spiked with standard F^-. Different volumes of standard F- solution are added to the water samples prepared. Then excess, but same volumes of Al-acid alizarin garnet R are added to the four solutions. This Al-acid alizarin garnet R acts as a quencher. The complex absorbs emission energy from F^-. So, instead of producing fluorescence from emission energy, it is quenched by the complex. So, the fluorescence intensity decreases.

Interpretation Introduction

(b)

Interpretation:

A plot of data should be constructed.

Concept introduction:

Standard addition method is used to analyze complex samples which matrix effect is significant. One common approach is standard addition technique is spiking of sample. In this method, one or more increments of a standard solution is added to the several aliquots of sample containing same volume. Each solution is then diluted to a fixed volume and experiment is proceeded.

Answer to Problem 15.16QAP

Explanation of Solution

Meter reading is the dependent variable and the volume of standard added is the independent variable.

Interpretation Introduction

(c)

Interpretation:

A relationship for multiple standard additions should be derived. An equation for the unknown concentration should be obtained in terms of the slope and the intercept of the standard addition plot.

Concept introduction:

Several aliquots with volume V_x of an unknown solution which have a concentration C_x are measured into volumetric flasks with volume V_t and then a standard solution with concentration Csis added in variable volumes of V_s. Then other reagents that aid for the detection are added and each volumetric flask is topped up to the maximum volume. Then instrumental measurements are made on each solution. If the instrumental response is proportional to the concentration of analyte, following equation can be derived.

$S=\frac{k{V}_s{c}_s}{V_t}+\frac{k{V}_x{c}_x}{V_t}\to \left(1\right)$

k is a proportionality constant. According to above equation the instrumental signal versus V_s is a plot in the form of $S=m{V}_s+b$, where,

$m=\frac{k{c}_s}{V_t}$

$b=\frac{k{V}_x{c}_x}{V_t}$

Concentration of unknown then can be obtained by following equation.

$c_x=\frac{b{c}_s}{m{V}_x}\to \left(2\right)$

Answer to Problem 15.16QAP

$S=\frac{k{V}_x{c}_x}{V_t}-\frac{k{V}_s{c}_s}{V_t}$

$c_x=-\frac{b{c}_s}{m{V}_x}$

Explanation of Solution

The fluorescence decreases with the increasing amounts of F- standard. The control sample has the highest fluorescence meter reading. So the equation 1 can be rewrite as

The meter reading versus the volume of standard added has a linear relationship. Above equation can be written in the form of $S=-m{V}_s+b$

$S=-\frac{k{V}_s{c}_s}{V_t}+\frac{k{V}_x{c}_x}{V_t}$ where,

$\begin{array}{l}m=-\frac{k{c}_s}{V_t}\\ b=\frac{k{V}_x{c}_x}{V_t}\end{array}$

An equation for the concentration of unknown can be derived by taking the ratio of intercept and the slope of fluorescence Vs Vs graph.

$\begin{array}{l}\frac{b}{m}=\frac{k{V}_x{c}_x/V_t}{-k{c}_s/V_t}=-\frac{V_x{c}_x}{c_s}\\ c_x=-\frac{b{c}_s}{m{V}_x}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The equation for the line representing the decrease in fluorescence relative to volume of standard should be determined.

Concept introduction:

In the least squares method, in order to find the slope and the intercept the three quantities S_yy, S_xx and S_xy should be determined.

$\begin{array}{l}{S}_{yy}={\displaystyle \sum {\left(y_i-\overline{y}\right)}^2}\\ S_{xx}={\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}\\ S_{xy}={\displaystyle \sum \left(x_i-\overline{x}\right)}\left(y_i-\overline{y}\right)\end{array}$

The slope of the line, $m=\frac{S_{xy}}{S_{xx}}$

The intercept, $b=\overline{y}-m\overline{x}$

$\overline{y}$ - mean of y values

$\overline{x}$ - mean of x values

Answer to Problem 15.16QAP

$y=-13.22x+68.23$

Explanation of Solution

The three quantities S_yy, S_xx and S_xy can be easily calculated using a spreadsheet and then the slope and the intercept can be calculated using the given equations.

Vs, mL	Meter reading	$\left(x_i-\overline{x}\right)$	$\left(y_i-\overline{y}\right)$	${\left(y_i-\overline{y}\right)}^2$	${\left(x_i-\overline{x}\right)}^2$	$\left(x_i-\overline{x}\right)$$\left(y_i-\overline{y}\right)$
0	68.2	-1.5	19.8	392.04	2.25	-29.7
1	55.3	-0.5	6.9	47.61	0.25	-3.45
2	41.3	0.5	-7.1	50.41	0.25	-3.55
3	28.8	1.5	-19.6	384.16	2.25	-29.4
Sum				874.22	5	-66.1
$\overline{y}$	48.4
$\overline{x}$	1.5

$m=\frac{S_{xy}}{S_{xx}}=\frac{-66.1}{5}=-13.22$

$\begin{array}{l}b=\overline{y}-m\overline{x}\\ b=48.4-\left(-13.22\times 1.5\right)\\ b=68.23\end{array}$

So the equation for the line for the standard addition plot is $y=-13.22x+68.23$

Interpretation Introduction

(e)

Interpretation:

The standard deviation of the slope and the intercept should be determined.

Concept introduction:

Standard deviation about regression, $s_r=\sqrt{\frac{S_{yy}-m^2{S}_{xx}}{N-2}}$

N − number of points used.

The standard deviation of the slope, $s_m=\sqrt{\frac{s_r^2}{S_{xx}}}$

The standard deviation of the intercept, $s_b=s_r\sqrt{\frac{{\displaystyle \sum x_i^2}}{N{\displaystyle \sum x_i^2}-{\left({\displaystyle \sum x_i}\right)}^2}}$

Answer to Problem 15.16QAP

Standard deviation of the slope = 0.194

Standard deviation of the intercept = 0.837

Explanation of Solution

in order to calculate the standard deviation of the slope or intercept, first standard deviation about regression should be determined.

$\begin{array}{l}{s}_r=\sqrt{\frac{874.22-{\left(-13.22\right)}^2\times 5}{4-2}}\\ s_r=0.435\end{array}$

Standard deviation of the slope, $s_m=\sqrt{\frac{{\left(0.435\right)}^2}{5}}=0.194$

Standard deviation of the intercept, $s_b=s_r\sqrt{\frac{14}{4\times 14-{\left(6\right)}^2}}=0.837$

Interpretation Introduction

(f)

Interpretation:

The concentration of F- in the sample in parts per billion should be determine.

Concept introduction:

$c_x=-\frac{b{c}_s}{m{V}_x}$

C_x − concentration of the unknown solution

C_s − concentration of standard solution

V_x − volume of unknown solution

m − slope of the instrumental signal versus volume of standard added curve

b − intercept of the instrumental signal versus volume of standard added curve

Answer to Problem 15.16QAP

$5.16\text{ ppb}$

Explanation of Solution

Intercept = 68.23

Slope = -13.22

Concentration of standard = 10.0 ppb

Volume of unknown = 10.00 mL

$c_x=-\frac{b{c}_s}{m{V}_x}$

$\begin{array}{l}{c}_x=-\frac{68.23\times 10.0\text{ ppb}}{\left(-13.22\right)/\text{ mL }\times 10.00\text{ mL}}\\ c_x=5.16\text{ ppb}\end{array}$

Interpretation Introduction

(g)

Interpretation:

The standard deviation of the result in (f) should be determined

Concept introduction:

$s_v=\frac{s_y}{m}\sqrt{\frac{1}{N}+\frac{{\left(0-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

S_v − standard deviation in volume

S_y − standard deviation in signal

m −slope

N − number of points

$s_c=s_v\left(\frac{c_s}{V_x}\right)$

S_c − standard deviation in concentration

Answer to Problem 15.16QAP

$s_c=-3.78\text{ ppb}$

Explanation of Solution

$\begin{array}{l}{s}_v=\frac{17.071}{-13.22}\sqrt{\frac{1}{4}+\frac{{\left(0-48.4\right)}^2}{{\left(-13.22\right)}^2\times 5}}\\ s_v=-3.78\end{array}$

$\begin{array}{l}{s}_c=-3.78\text{ mL}\left(\frac{10.0\text{ ppb}}{10.00\text{ mL}}\right)\\ s_c=-3.78\text{ ppb}\end{array}$