Chapter 14, Problem 14.12QAP

Interpretation Introduction

(a)

Interpretation:

The concentration in samples that yielded absorbance of 0.143 should be expressed in ppm. The relative standard deviation of the result should be calculated. The calculation should be repeated assuming the absorbance data are means of three measurements.

Concept introduction:

Standard deviation of the results obtained from the calibration curve = $s_c=\frac{s_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{\left({\overline{y}}_c-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

M − number of replicates

N- number of calibration points.

Answer to Problem 14.12QAP

$x=3.66\text{ ppm}$

$s_c=0.229\text{ ppm}$

If absorbance data are means of 3 measurements, $s_c=0.170\text{ ppm}$

Explanation of Solution

$\begin{array}{l}y=0.03949x-0.00134\\ 0.143=0.03949x-0.00134\\ 0.03949x=0.14434\\ x=3.66\text{ ppm}\end{array}$

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{\left(0.143-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{1+0.167+\frac{0.4692}{1.4441}}\\ s_c=0.1874\sqrt{1+0.167+0.325}\\ s_c=0.1874\times 1.221\\ s_c=0.229\text{ ppm}\end{array}$

If the experiment was replicated three times,

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{3}+\frac{1}{6}+\frac{{\left(0.143-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{0.333+0.167+\frac{0.4692}{1.4441}}\\ s_c=0.1874\sqrt{0.333+0.167+0.325}\\ s_c=0.1874\times 0.9083\\ s_c=0.170\text{ ppm}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The concentration in samples that yielded absorbance of 0.675 should be expressed in ppm. The relative standard deviation of the result should be calculated. The calculation should be repeated assuming the absorbance data are means of three measurements.

Concept introduction:

Standard deviation of the results obtained from the calibration curve = $s_c=\frac{s_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{\left({\overline{y}}_c-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

M − number of replicates

N- number of calibration points.

Answer to Problem 14.12QAP

$x=17.13\text{ ppm}$

$s_c=0.204\text{ ppm}$

If absorbance data are means of 3 measurements, $s_c=0.135\text{ ppm}$

Explanation of Solution

$\begin{array}{l}y=0.03949x-0.00134\\ 0.675=0.03949x-0.00134\\ 0.03949x=0.67634\\ x=17.13\text{ ppm}\end{array}$

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{\left(0.675-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{1+0.167+\frac{0.0234}{1.4441}}\\ s_c=0.1874\sqrt{1+0.167+0.016}\\ s_c=0.1874\times 1.087\\ s_c=0.204\text{ ppm}\end{array}$

If the experiment was replicated three times,

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{3}+\frac{1}{6}+\frac{{\left(0.675-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{0.333+0.167+\frac{0.0234}{1.4441}}\\ s_c=0.1874\sqrt{0.333+0.167+0.016}\\ s_c=0.1874\times 0.7183\\ s_c=0.135\text{ ppm}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The concentration in samples that yielded absorbance of 0.068 should be expressed in ppm. The relative standard deviation of the result should be calculated. The calculation should be repeated assuming the absorbance data are means of three measurements.

Concept introduction:

Standard deviation of the results obtained from the calibration curve = $s_c=\frac{s_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{\left({\overline{y}}_c-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

M − number of replicates

N- number of calibration points.

Answer to Problem 14.12QAP

$x=1.76\text{ ppm}$

$s_c=0.294\text{ ppm}$

If absorbance data are means of 3 measurements, $s_c=0.178\text{ ppm}$

Explanation of Solution

$\begin{array}{l}y=0.03949x-0.00134\\ 0.068=0.03949x-0.00134\\ 0.03949x=0.06934\\ x=1.76\text{ ppm}\end{array}$

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{\left(0.068-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{1+0.167+\frac{0.5776}{1.4441}}\\ s_c=0.1874\sqrt{1+0.167+0.400}\\ s_c=0.1874\times 1.567\\ s_c=0.294\text{ ppm}\end{array}$

If the experiment was replicated three times,

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{3}+\frac{1}{6}+\frac{{\left(0.068-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{0.333+0.167+\frac{0.5776}{1.4441}}\\ s_c=0.1874\sqrt{0.333+0.167+0.400}\\ s_c=0.1874\times 0.949\\ s_c=0.178\text{ ppm}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The concentration in samples that yielded absorbance of 1.009 should be expressed in ppm. The relative standard deviation of the result should be calculated. The calculation should be repeated assuming the absorbance data are means of three measurements.

Concept introduction:

Standard deviation of the results obtained from the calibration curve = $s_c=\frac{s_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{\left({\overline{y}}_c-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

M − number of replicates

N- number of calibration points.

Answer to Problem 14.12QAP

$x=25.58\text{ ppm}$

$s_c=0.204\text{ ppm}$

If absorbance data are means of 3 measurements, $s_c=0.135\text{ ppm}$

Explanation of Solution

$\begin{array}{l}y=0.03949x-0.00134\\ 1.009=0.03949x-0.00134\\ 0.03949x=1.01034\\ x=25.58\text{ ppm}\end{array}$

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{\left(1.009-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{1+0.167+\frac{0.0327}{1.4441}}\\ s_c=0.1874\sqrt{1+0.167+0.0227}\\ s_c=0.1874\times 1.091\\ s_c=0.204\text{ ppm}\end{array}$

If the experiment was replicated three times,

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{3}+\frac{1}{6}+\frac{{\left(1.009-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{0.333+0.167+\frac{0.0327}{1.4441}}\\ s_c=0.1874\sqrt{0.333+0.167+0.0227}\\ s_c=0.1874\times 0.7229\\ s_c=0.135\text{ ppm}\end{array}$

Interpretation Introduction

(e)

Interpretation:

The concentration in samples that yielded absorbance of 1.512 should be expressed in ppm. The relative standard deviation of the result should be calculated. The calculation should be repeated assuming the absorbance data are means of three measurements.

Concept introduction:

Standard deviation of the results obtained from the calibration curve = $s_c=\frac{s_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{\left({\overline{y}}_c-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

M − number of replicates

N- number of calibration points.

Answer to Problem 14.12QAP

$x=38.32\text{ ppm}$

$s_c=0.229\text{ ppm}$

If absorbance data are means of 3 measurements, $s_c=0.170\text{ ppm}$

Explanation of Solution

$\begin{array}{l}y=0.03949x-0.00134\\ 1.512=0.03949x-0.00134\\ 0.03949x=1.51334\\ x=38.32\text{ ppm}\end{array}$

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{\left(1.512-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{1+0.167+\frac{0.4678}{1.4441}}\\ s_c=0.1874\sqrt{1+0.167+0.324}\\ s_c=0.1874\times 1.221\\ s_c=0.229\text{ ppm}\end{array}$

If the experiment was replicated three times,

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{3}+\frac{1}{6}+\frac{{\left(1.512-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{0.333+0.167+\frac{0.4678}{1.4441}}\\ s_c=0.1874\sqrt{0.333+0.167+0.324}\\ s_c=0.1874\times 0.908\\ s_c=0.170\text{ ppm}\end{array}$

Interpretation Introduction

(f)

Interpretation:

The concentration in samples that yielded absorbance of 0.546 should be expressed in ppm. The relative standard deviation of the result should be calculated. The calculation should be repeated assuming the absorbance data are means of three measurements.

Concept introduction:

Standard deviation of the results obtained from the calibration curve = $s_c=\frac{s_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{\left({\overline{y}}_c-\overline{y}\right)}^2}{m^2{S}_{xx}}}$

M − number of replicates

N- number of calibration points.

Answer to Problem 14.12QAP

$x=13.86\text{ ppm}$

$s_c=0.207\text{ ppm}$

If absorbance data are means of 3 measurements, $s_c=0.139\text{ ppm}$

Explanation of Solution

$\begin{array}{l}y=0.03949x-0.00134\\ 0.546=0.03949x-0.00134\\ 0.03949x=0.54734\\ x=13.86\text{ ppm}\end{array}$

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{\left(0.546-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{1+0.167+\frac{0.0795}{1.4441}}\\ s_c=0.1874\sqrt{1+0.167+0.0551}\\ s_c=0.1874\times 1.105\\ s_c=0.207\text{ ppm}\end{array}$

If the experiment was replicated three times,

$\begin{array}{l}{s}_c=\frac{0.0074}{0.03949}\sqrt{\frac{1}{3}+\frac{1}{6}+\frac{{\left(0.546-0.828\right)}^2}{{\left(0.03949\right)}^2\times 926}}\\ s_c=0.1874\sqrt{0.333+0.167+\frac{0.0795}{1.4441}}\\ s_c=0.1874\sqrt{0.333+0.167+0.0551}\\ s_c=0.1874\times 0.7451\\ s_c=0.139\text{ ppm}\end{array}$