Analyte Estimation using Spectrophotometric Methods and ... · 5 Analyte Estimation using Spectrophotometric Methods and Excel © Neeraj Kumar BioChem Calculations Ans. Given- = 1.8011

0 Analyte Estimation using Spectrophotometric Methods and Excel © Neeraj Kumar BioChem Calculations



Contents

1. Beer-Lambert’s Law

1A. Plotting and Using Concentration vs Absorbance Graph

1.B. Determining unknown [Analyte] using one Standard

1C. Two Analyte Forms, One Wavelength

1D. Two Analytes, Two Wavelength

2. (External) Standard Addition Methods

2A. Single Aliquot Standard Addition

2B. Multiple Aliquots Standard Addition


1. Beer-Lambert’s Law

Many molecules including organic molecules with conjugated bonds system and

transition metals give distinct absorption spectra in the UV-VIS region. The wavelength at

which an analyte absorbs maximum light, and subsequently the highest absorption peak

of the spectrum is observed, is denoted by its max. Among the hundreds of wavelengths

() absorbed by the analyte, the spectrophotometry readings are taken at max to

maximize the absorption intensity and minimize other absorption interferences by the

solution matrix. The sample solution is often read against a suitable blank (distilled water,

solvent or reagent blank) to minimize matrix interferences, too.

Three factors, namely concentration, pathlength and extinction coefficient proportionally

affect the absorbance of an analyte. At a specified temperature, the absorption of light by

the analyte generally increases with its concentration in a certain range. Greater is the

concentration, greater would be the number of analyte molecules in a fixed volume of

solution to absorb the light- thus, greater would be absorption. Absorption also increases

with pathlength- the distance that the incident light travels through the solution.

Increasing the pathlength for a solution, say from 1 to 2 cm while keeping the cross-

section constant, increases the volume of solution, and in turn, the number of analyte

molecules. Increased number of analyte molecules further yield increased absorption. The

third factor, extinction coefficient or absorptivity, also affects absorption proportionally

but remains unaffected of the previous two factors. It is the temperature-dependent

intrinsic property of the analyte and is the proportionality constant of the Beer-Lambert’s

law equation.

When light from a monochromatic light source of Incident intensity I0 passes through a

solution, some of it is absorbed, scattered and/or reflected by the analyte of interest as

well as the all other components of the solution. The remaining light with a transmitted

intensity I comes out of the solution as transmitted light. The ratio of I to I0 is called

transmittance of the solution. The detector of a UV-VIS spectrophotometer reads %

transmittance of a solution. However, transmittance does not exhibit linearity with the

concentration of the analyte in solution. Instead, the analyte concentration in solution

exhibits linearity with its absorbance.

Transmittance (T) and absorbance (A) are related to the following equations-

Transmittance, T = % Transmittance = x 100

Absorbance, A = -log10 Absorbance, A = log10 = -log10 T

I

I0 I0

I

I

I0

1

T


Absorbance, A = log10 Absorbance, A = 2 – log10 %T

Beer law or Beer-Lambert’s law states that the absorbance of a solution is proportional to

its concentration and pathlength. Mathematically,

Absorbance, A C L

And, A = C L - Equation 1

where,

= absorptivity at specified wavelength and temperature (M-1cm-1)

L = path length (in cm)

C = molar concentration of the solute

Though the standard unit of is M-1 cm-1, it can be presented in any unit depending on

the combinations of units of concentration and pathlength. Similarly, different units of

concentration can be used as required. When the unit of concertation is molarity, the

absorptivity or absorption coefficient is termed molar absorptivity or molar absorption

coefficient.

Example 1: A solution of 2.1043 x 10-4 M indophenol gives absorbance of 0.379 in a 1.00

cm cuvette. Calculate the molar absorptivity, of indophenol.

Ans. Given-

C = 2.1043 x 10-4 M ; A = 0.379

L = 1.0 cm

Now, putting the values in Beer-Lambert’s Law equation, A = C L

= A / CL = 0.379 / (2.1043 x 10-4 M x 1.0 cm) = 1.8011 x 103 M-1 cm-1

Example 2: The molar absorptivity, of indophenol under these experimental conditions

is 1.8011 x 103 M-1 cm-1. What is the expected absorbance of a 2.1043 x 10-4 M indophenol

solution through a 10 mm cuvette?

Ans. Given-

C = 2.1043 x 10-4 M ; = 1.8011 x 103 M-1 cm-1

L = 10 mm = 1.0 cm


A = 1.8011 x 103 M-1 cm-1 x 2.1043 x 10-4 M x 1.0 cm = 0.379


is 1.8011 x 103 M-1 cm-1. What is the expected concentration of indophenol solution if its

absorbance is 0.379 through a 10 mm cuvette?

% T

100


Ans. Given-

= 1.8011 x 103 M-1 cm-1 ; A = 0.379

L = 10 mm = 1.0 cm


C = A / L = 0.379 / (1.8011 x 103 M-1 cm-1 x 1.0 cm) = 2.1043 x 10-4 M


is 1.8011 x 103 M-1 cm-1. What is the pathlength for a 2.1043 x 10-4 M indophenol solution

that gives an absorbance of 0.379.

Ans. Given-

C = 2.1043 x 10-4 M ; = 1.8011 x 103 M-1 cm-1

A = 0.379


L = A / C = 0.379 / (1.8011 x 103 M x 2.1043 x 10-4 M) = 1.00 cm

Example 5: A student wishes to measure the iron content in well water. She prepares a

7.64 x 10-4 M standard Fe3+ solution. An 11.0 mL aliquot of this standard solution is treated

with an excess of HNO3 and KSCN to form a red complex and subsequently diluted to

55.0 mL. The absorbance of this diluted reference or standard solution is read in a cell of

1.00 cm pathlength.

A 20.0 mL sample of the well water is also treated with an excess of HNO3 and KSCN to

form a red complex and subsequently diluted to 100.0 mL. This diluted sample is placed

in a variable pathlength cell. The absorbance of the sample solution at a pathlength of

2.15 cm exactly matches that of the reference solution at a pathlength of 1.00 cm. What

is the concentration of iron in the well water?

Ans. Beer-Lambert’s Law, A = C L

where,

A = Absorbance

= molar absorptivity at the specified wavelength (M-1cm-1)

L = path length (in cm)

C = Molar concentration of the solute

Note that the extinction coefficient is a constant for the specified (red) complex under

specified experimental conditions. Since both the standard and well water is treated in

the same way (reaction) and remains constant.

Step 1: Preparation of working reference aliquot: 11.0 mL standard solution with 7.64 x

10-4 M Fe3+ is diluted up to a final volume of 55.0 mL.


Using C1V1 (standard solution) = C2V2 (diluted reference solution)

Or, 7.64 x 10-4 M x 11.0 mL = C2 x 55.0 mL

Or, C2 = (7.64 x 10-4 M x 11.0 mL) / 55.0 mL = 1.5820 x 10-4 M

Hence, [Fe3+] in the reference solution = 1.5820 x 10-4 M

Step 2: For the simplicity of expression, we label the standard/ reference solution as

solution 1. The well water is labelled 2. The final working solution whose absorbances are

taken are referred as aliquot of the respective solutions.

Now,

Abs of standard aliquot, A1 = 1 x (1.5820 x 10-4 M x 1.00 cm) - equation 1

Abs of well water aliquot, A2 = (C2 x 2.15 cm) - equation 2

Where, C2 = concentration of Fe in the well water

Given, A1 = A2. So, equating equations 1 and 2-

1 x (1.5820 x 10-4 M x 1.00 cm) = (C2 x 2.15 cm)

Also note that remains constant. So, 1 = 2

Or, 1.5820 x 10-4 M x 1.00 cm = C2 x 2.15 cm

Hence, C2 = (1.5820 x 10-4 M x 1.00 cm) / 2.15 cm = 7.1070 x 10-5 M

Hence, [Fe3+] in final working well water aliquot = 7.1070 x 10-5 M

Step 3: Calculating [Fe3+] in original well water: Given, 20.0 mL well water is diluted up to

100.0 mL.

Again, using C1V1 (original well water) = C2V2 (well water aliquot)

Or, C1 x 20.0 mL = 7.1070 x 10-5 M X 100.0 mL

Or, C1 = (7.1070 x 10-5 M X 100.0 mL) / 20.0 mL = 3.5535 x 10-4 M

Therefore, [Fe3+] in original well water sample = 3.5535 x 10-4 M

Example 6: The iron content of a well water sample is determined by UV-VIS

spectroscopy. A 10.0 mL aliquot of well water is transferred to a 100.0 mL volumetric flask

and 20 mL of 0.10 M nitric acid is added to oxidized Fe2+ to Fe3+. Next 30 mL of a 0.1 M

sodium thiocyanate solution is added to form the deep red complex of Fe(III) thiocyanate.

The solution is diluted to the mark with distilled water. A reagent blank is similarly

prepared. If the absorbance of the Fe(III) thiocyanate solution is 0.5483 and the reagent

blank is 0.0282, calculate the concentration of iron for the water sample. The molar

absorptivity for Fe(III) thiocyanate is 3200 M-1 cm-1. Assume a pathlength of 1 cm and the

complete reaction with the thiocyanate is-

Fe3+(aq) + SCN-(aq) ----------> FeSCN2+(aq)

Ans. Step 1: Calculate [Fe] in the 100.0 mL solution


Actual absorbance of the solution = Observed absorbance – Absorbance of the blank

= 0.5483 – 0.0282 = 0.5201

Putting the values in the Beer-Lambert’s Law, A = C L

0.5201 = 3200 M-1 cm-1 x C x 1.00 cm

Or, C = 0.5201 / (3200 M-1 cm-1 x 1.00 cm) = 1.6253 x 10-4 M

Hence, [FeSCN2+] in the solution, C = 1.6253 x 10-4 M

Following the stoichiometry of the balanced reaction, 1 mol Fe3+ forms 1 mol FeSCN2+.

So, [Fe] in the solution must be equal to [FeSCN2+].

Hence, [Fe] in the solution = 1.6253 x 10-4 M

Step 2: Calculate [Fe] in the well water sample

Given- 10.0 mL aliquot of the well water sample is diluted to a final volume of 100.0 mL.

From step 1, [Fe] in the 100.0 mL solution is 1.6253 x 10-4 M.

Now, using C1V1 (well water sample) = C2V2 (100 mL solution)

Or, C1 x 10.0 mL = 1.6253 x 10-4 M x 100.0 mL

Or, C1 = (1.6253 x 10-4 M x 100.0 mL) / 10.0 mL = 1.6253 x 10-3 M

Hence, [Fe] in the well water sample = 1.6253 x 10-3 M


1.A. Plotting and Using Concentration vs Absorbance Graph

Example 5: The absorbances of different concentrations of an unknown dye are tabulated

below. Calculate the molar absorptivity, of the dye.

Concentration (M): 5.94 x 10-6 4.36 x 10-6 2.97 x 10-6 1.58 x 10-6 3.92 x 10-7

Absorbance 0.842 0.600 0.410 0.200 0.030

Ans.

Step 1: Tabulate the concentration and respective absorbance data in an Excel sheet. By

default, Excel puts the first row on X-axis and the second row on Y-axis. Remember to put

the independent variables (here, concentration) in the first row, and the dependent

variable (absorbance) in the second row.

Step 2: Once the graph is generated, we need to enter Chart Title and label the axes.

Follow the steps-

I. Simply click (Left click of the mouse) on the graph.

II. Click on the Design Tab

III. Click on “Add Chart Element”

IV. Under the “Axis Title” select “Primary Horizontal”. Enter the X-axis unit in the

textbox appearing at the bottom of the graph. In this case, the X-axis represents


the concentration of the dye. So, type “[Dye], M” in the textbox. Or, you can type

“Molar Concentration of Dye” or similar phrases to represent the X-axis.

V. Similarly, select “Primary Vertical” from the “Axis Title” tab and enter a suitable

Y-axis label. In this case, the Y-axis represents the absorbance values. So, a Y-axis

label can be “Absorbance”.

VI. Double click on the “Chart Title” textbox at the top of the graph. Then enter the

chart title. In this case, “Absorbance vs Concentration Graph” seems to be a good

chart title.

VII. The background of the chart as well as Chart Styles can also be changed as

required to fit the need or for beautification by simply selecting the graph designs

from VII. However, it is not required at this moment.

Step 3: Deriving the Slope of the Graph: For a given data table for a linear graph, the

three methods can be used to determine the slope and y-intercept.

A. From Data Table (without requiring the Graph): The slope can be manually

calculated from the data table without requiring to plot a graph. Take the difference

between the first and last absorbance (dY) and concentration (dX) values. Any two data


points can be taken, but accuracy increases with the inclusion of more data points. So,

take the first and last data columns for slope calculation as shown below-

Slope =

Note that the manually calculated value of the slope is not equal to that of the actual or

Excel-generated value because it does not account regression like Excel. Also, note that

the unit of the slope is equal to the unit of the Y-axis divided by the unit of the X-axis. In

this case, the Y-axis represents the dimensionless absorbance values, so it does not have

a unit. The X-axis represents concentration in terms of molarity, so has a unit of molarity,

M. So, the slope = (dY / dX) yields a unit of M-1 for this graph.

B. From the best-fit line of the Graph: This method also takes the difference

between the first and last absorbance (dY) and concentration (dX) values but requires a

graph and its best-fit line drawn. Since this method derives the slope from the best-fit line

but not directly from the data table, the resultant value of the slope is better than the

above method A.

To derive the slope, we need to take any two points- the farther they’re on the best-fit

line, the better would be the outcome because of the inclusion of more data points.

However, this method may not always seem feasible without compromising accuracy.

Consider this case, the y-axis values can be approximated equal or very close to the actual

value on the best-fit line because of measurable subdivisions on the respective y-axis

scale. However, the subdivisions on the X-axis are not clearly measurable because of the

large magnitude of values between two adjacent subdivisions. So, the resultant slope is

most likely to exhibit deviation from the actual or Excel-generated value.

To minimize the extent of error, we need to take the two points (need not the data points,

these can be any two points on the best-fit line) with an exact intercept on both the axes.

If this is not possible, then choose two points with an exact intercept on the axis with

lesser clear subdivisions, in this case, the X-axis. So, we proceed with graph keeping in

mind-

- choose two data points with an exact intercept on both the axes

- if above criterion not met, then choose two points with an exact intercept on the

axis with lesser clear subdivisions (in this case the X-axis), and

- the two points on the best-fit line shall be as farther as possible to include more

data points, which in turn, would increase the accuracy of the slope.

dY

dX =

(5.94 x 10-6 – 3.92 x 10-7) M

0.842 – 0.030 = 146359 M-1


A possible scenario of taking two points on the best-fit line is shown in the graph below.

We’ll use the respective X- and Y-values at the two points to calculate the slope.

So,

Slope =

C. From the trendline equation in Excel: Follow the steps-

I. Right-click on the calibration curve of the graph. Select “Format Trendline” at the

bottom of the popup screen.

II. By default, the options are presented for the “Linear” graph (shown by BLUE box).

If the “Linear” option is not selected for some reason, re-select it before

proceeding.

III. Select the box “Display Equation on Chart”. It shows the linear regression

equation or the trendline equation for the graph.

IV. Select the box “Display R-squared value on Chart”. It shows R2 value for linear

regression. Though R2 value is generally not required in such cases, it provides

dY

dX =

(5.40 x 10-6 – 6.00 x 10-7) M

0.760 – 0.060 = 145833 M-1

YB - YA

XB - XA =


valuable information about the linearity of the graph. Closer is the value of R2 to 1,

greater is the linearity of the calibration curve and greater would be the accuracy

of calculations based on its linear regression equation.

D. Use LINEST function in Excel: Follow the steps-

I. Select any cell in the Excel sheet.

II. Enter =LINEST and select the fxLINEST from the popup. It shows the options as

shown below-

Where,

known_ys = the Y-axis values = select Y-axis (absorbance) values

[known_xs] = the X-axis values = select X-axis (concentration) values

[const] = type TRUE

[stats] = type TRUE

III. When all values are entered, press ENTER. It returns the value of slope in the

selected cell.

IV. The LINEST command also gives many more statistical data. In this chapter, the

y-intercept, and uncertainties in slope and y-intercept may also be required at

some points. To get these values, first, drag the slope-cell value to the right cell,


then drag that to one cell down. Now, enter Ctrl+shift+enter in the formula bar

at the top. It gives values in all four cells.

V. The number of decimal places can be increased or decreased as required.

Step 4: Finally, calculate the molar absorptivity,

Molar absorptivity, =

We have, Slope = 145912 M-1 ; [from the linear regression equation]

Pathlength = 1.00 cm ; [given]

Now, putting the values in above expression-

Molar absorptivity, = = 145912 M-1 cm-1 = 1.46 x 105 M-1 cm-1

Slope

Pathlength

1.0 cm

145912 M-1


Example 6: The absorbances of different concentrations of the standard solutions of Ca2+

are tabulated below. A 0.1000-gram sample, when diluted to 1:50 dilution, gives an

absorbance of 0.400. Calculate % (w/w) calcium in the original sample.

Concentration (ppm): 1 3 6 10 14 16

Absorbance 0.340 0.420 0.560 0.750 0.940 1.050

Ans. Step 1: Construct the standard calibration curve and generate the linear regression

equation or the trendline equation in excel for the graph.

Step 2: Calculate [Ca2+] in the diluted unknown aliquot using the absorbance of the

unknown and the linear regression equation of the standard graph.

We have, linear regression equation is y = 0.0474x + 0.2821.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is,

according to the trendline (linear regression) equation, 1 absorbance unit (= 1Y) is equal

to 0.0474 units on the X-axis plus 0.2821. Note that the unit of the slope is inverse of the

unit of the X-axis.

Given, the absorbance of the unknown aliquot = 0.400


Now, putting the value of y = 0.400 in the above linear regression equation-

0.400 = 0.0474x + 0.2821

Hence, x =

Hence, [Ca2+] in the diluted unknown solution = 2.487 ppm = 2.487 mg L-1

Step 3: Calculate the amount of Ca2+ in the original sample

Given, the original sample is diluted to a 1:50 dilution. That is, 1.0 g of the original

sample is diluted to a final volume of 50.0 mL.

So,

Total volume of the diluted unknown solution = (50 mL / 1 g) x 0.1000 g = 5.0 mL

Now,

Mass of Ca2+ in the diluted unknown solution = [Ca2+] x Vol. of soln.

= 2.487 mg L-1 x 5.0 mL

= 2.487 mg L-1 x 0.005 L = 0.012435 mg

Step 4: Calculate % Ca2+ in original sample:

We have-

Mass of original sample = 0.1000 g = 100.0 mg

Mass of Ca2+ in the original sample = 0.012435 mg

Now,

% Ca2+ (w/w) = (Mass of Ca2+ / Mass of original sample) x 100

= (0.012435 mg / 100.0 mg) x 100 = 0.0124%

Example 7: A student dissolved 0.500 grams of a Kool-Aid powder sample into a final

volume of 250.0 mL. The resultant solution gives an absorbance of 0.200. The standard

graph for FD&C Red 40 dye yielded the linear regression equation of y = 2500x + 0.0037,

where the unit of concentration on the X-axis is the terms of molarity. Calculate % (w/w)

of the dye in the original sample.

Ans. Step 1: Calculate [dye] in the 250.0 mL of the solution using the absorbance of the

unknown and the linear regression equation of the standard graph.

We have, linear regression equation is y = 2500x - 0.0037.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is,

according to the trendline (linear regression) equation, 1 absorbance unit (= 1Y) is equal

0.400 – 0.2821

0.0474 ppm-1 = 2.487 ppm


to 2500 units on the X-axis minus 0.0037. Note that the unit of the slope is inverse of the

unit of the X-axis.

Given, the absorbance of the unknown aliquot = 0.200

Now, putting the value of y = 0.360 in the above linear regression equation-

0.200 = 2500x – 0.0037

Hence, x =

Hence, [dye] in the diluted solution = 8.148 x 10-5 M

Step 2: Determine mass of FD&C Red 40 in 250.0 mL of solution

Total volume of solution prepared = 250.0 mL = 0.250 L

Now,

Moles of FD&C Red 40 in total vol. of solution = [FD&C Red 40] x Vol. of soln. in L

= 8.148 x 10-5 M x 0.250 L = 2.0370 x 10-5 mol

And,

Mass of FD&C Red 40 in total vol. of solution = Moles x molar mass

= 2.0370 x 10-5 mol x 496.42 g mol-1 = 0.0101121 g

Step 3: Determine % mass of FD&C Red 40 in 0.500 g of Kool-Aid powder

0.500 g of Kool-Aid powder was used to prepare 250.0 mL solution. So, the mass of the

dye in both these samples must be the same.

That is, the mass of FD&C Red 40 in 0.500 g Kool-Aid powder = 0.0101121 g

Now,

% dye in sample = (mass of dye / mass of sample) x 100 = 2.022 %

Example 8: The signal intensity (arbitrary units) of various standard solutions of vitamin

B2 is tabulated below. Determine the concentration and uncertainty of an unknown

vitamin B2 solution with the signal intensity of 15.4. Also, calculate the 95% confidence

interval for this measured value.

[Vit B2], g mL-1 0.000 0.100 0.200 0.400 0.800 Unknown

Absorbance 0.0 5.8 12.2 22.3 43.3 15.4

Ans. Plot the graph, label the axes and generate the trendline the equation as shown

below. Use the LINEST function of Excel to calculate the slope (m), uncertainty is slope (x

slope), y-intercept and uncertainty in y-intercept (y-intercept) as explained previously.

The solution is presented in the following steps-

0.200 + 0.0037

2500 M-1 = 8.1480 x 10-5 M




1.B. Determining unknown [Analyte] using one Standard

On many occasions, where the concentration of an analyte in an unknown sample is

expected to be in a certain range of a certain concentration range, this method may be

adopted.

Example 9: A 3.96 x 10-4 M standard solution of analyte A gives an absorbance of 0.624

at 238 nm in a 10 mm cuvette. The absorbance of an unknown solution of the same

analyte gives an absorbance of 0.400 under the same conditions. Calculate the

concentration of the analyte in the unknown solution.

Ans.

Method 1: Through the calculation of Molar absorptivity,

Step 1: Calculate molar absorptivity, using absorbance of the standard solution

Using Beer’s law,

Molar absorptivity, =

Step 2: Since the values of molar absorptivity, remains constant for an analyte at

specified experimental conditions, it can be used to calculate the unknown [analyte] from

its absorbance.

Now,

Unknown [Analyte], C =

Method 2: Using,

Let the suffixes std and unk represent respective values for absorbance and concentration.

Step 1: Calculate molar absorptivity, using absorbance of the standard solution using

Beer’s law,

Molar absorptivity, = - equation E8.1

Step 2: Similarly, for the unknown solution-

Molar absorptivity, = - equation E8.2

Step 3: Since the values of molar absorptivity, remains constant for an analyte at

specified experimental conditions, 1 must be equal to 2.

So,

3.96 x 10-4 M x 1.00 cm

0.624 = 1575.758 M-1 cm-1 =

A

C L

= 2.54 x 10-4 M 0.400

= A

L 1575.758 M-1 cm-1 x 1.00 cm

Astd

Cstd L

Cunk L

Aunk


1 = 2 =

Hence,

- Equation E8.3

And,

Cunk = - Equation E8.4

Now, putting the values in equation E8.4,

Unknown [Analyte], Cunk =

Example 10: A 3.96 x 10-4 M standard solution of analyte A gives an absorbance of 0.624

at 238 nm in a 10 mm cuvette. The blank gives an absorbance of 0.029. The absorbance

of an unknown solution of the same analyte gives an absorbance of 0.400 under the same

conditions. Calculate the concentration of the analyte in the unknown solution.

Ans.

Actual absorbance of standard solution = Abs of standard solution – Abs of blank

= 0.624 – 0.029 = 0.595

Actual absorbance of unknown solution = Abs of unknown solution – Abs of blank

= 0.400 – 0.029 = 0.371

Now, putting the values in equation E8.4,

Unknown [Analyte], Cunk =

Cstd L

Astd Aunk =

Cunk L

Cunk

Astd

Aunk =

Cstd

Astd x Aunk

Cstd

0.624

3.96 x 10-4 M x 0.400 = 2.54 x 10-4 M

0.595

3.96 x 10-4 M x 0.371 = 2.50 x 10-4 M


1.C. Two Analyte Forms, One Wavelength

This method includes the quantitative estimation of two analytes with different max

(wavelength for maximum absorption is different for the two analytes under

consideration) simultaneously present in a solution. Though each of the two analytes

exhibits absorption maxima at their own respective max, they also yield some lower

absorption value at the max of the other analyte. So, the total absorption of the solution

at any wavelength is the sum of the individual absorption of each analyte.

Example 11: Acid-base indicators are themselves acids or bases. Consider an indicator,

HIn, which dissociates according to the equation-

HIn H- + In-

The molar absorptivity, ε, is 2080 M-1 cm-1 for HIn and 14200 M-1 cm-1 for In-, at a

wavelength of 440 nm.

a. Write an expression for the absorbance of a solution containing HIn at a concentration

[HIn] and In- at a concentration [In-] in a cell of pathlength 1.00 cm. The total absorbance

is the sum of absorbances of each component.

b. A solution containing an indicator at a formal concentration of 1.84 x 10-4 M is adjusted

to pH 6.23 and found to exhibit an absorbance of 0.868 at 440 nm. Calculate pKa for this

indicator.

Ans. a. I. For HIn, given-

C = [HIn] , e = 2080 M-1 cm-1 , L = 1.0 cm


AHIn = 2080 M-1 cm-1 x [HIn] x 1.0 cm = 2080 [HIn] M-1

II. For In-, given-

C = [In-] , = 14200 M-1 cm-1 , L = 1.0 cm


AIn- = 14200 M-1 cm-1 x [In-] x 1.0 cm = 14200 [In-] M-1

b. Step 1: Given- the total concentration of HIn and In- in the solution is 1.84 x 10-4 M.

So, [HIn] + [In-1] = 1.84 x 10-4 M - Equation E10.1

Also given- total absorption of the solution at 440 nm is 0.868.

So, AHIn + AIn- = 0.868

Or, 2080 [HIn] M-1 + 14200 [In-] M-1 = 0.868

Hence, 2080 [HIn] + 14200 [In-] = 0.868 M - Equation E10.2

Now, (Equation E10.1 x 2080) – Equation E10.2

Ka


Step 2: Given- pH of the solution = 6.23

Or, [H+] = 10-pH = 10-6.23 = 5.888 x 10-7 M

The acid dissociation constant, Ka of the acid-base indicator HIn can be given as follow-

Note that the pKa values calculated by method 1 and 2 differ slightly. We would have

expected to get the same pKa value from both the methods. What did go wrong?

Note that the pH in the Henderson-Hasselbalch equation is the actual pH of a weak acid

(or weak base, etc. as specified), in this case, the indicated HIn. However, the pH

mentioned in the question is the adjusted (but not the actual) pH. So, the use of the

adjusted pH instead of the actual pH in method 2 is the reason for deviation from the

actual (expected) pKa value. Also, keep in mind that the deviation of the calculated pKa

value from the expected pKa value in method 2 would be proportional to the extent of

adjustment of the actual pH of the solution.

Method 1 calculates the acid-dissociation constant, Ka and subsequently pKa using the

actual value of [H+] in the solution from its pH. For any specified pH, [H+] is always a

constant. Since all the values used in method 1 are exact, the result value of pKa is also

exact and actual. So, method 1 shall be preferred over method 2 in cases with adjustment

of pH. In cases where pH adjustment of the indicator solution has not been done, both

the method shall yield the same value of pKa.


Example 12: The oxidized form of a flavoprotein (FOx) that functions as a one-electron

reducing agent has a molar absorptivity of 1.12 x 104 M-1 cm-1 at 457 nm at pH 7.00. For

the reduced form (FRed), = 3.82 x 103 M-1 cm-1 at 457 nm at pH 7.00.

FOx + e- FRed E0’ = -0.128 V

The substrate (S) is the molecule reduced by the protein-

FRed + S FOx + S-

Both S and S- are colorless. A solution at pH 7.00 was prepared by mixing enough of the

reduced protein plus substrate (FRed + S) to produce initial concentrations [FRed] = [S] =

5.70 x 10-5 M. The absorbance at 457 nm was 0.500 in a 1.00 cm cell.

a. Calculate the concentrations of FOx and FRed from the absorbance data.

b. Calculate the concentrations of S and S-.

c. Calculate the value of E0’ for the reaction S + e- S-

Ans. a. Step 1: Let the equilibrium concentrations of the FRed and FOx be [FRed] and [FOx],

respectively. Since the FOx is derived from FRed, and both the forms interconvert into each

other to establish the equilibrium, the sum of these forms at equilibrium must be equal

to the initial [FRed]i = 5.70 x 10-5 M. The absorbance of the solution at equilibrium (or, after

completion of reaction) is equal to the sum of the absorbances of the FRed and FOx species.

So, [FRed] + [FOx] = 5.70 x 10-5 M - Equation E11.1

Using Beer-Lambert’s Law equation, A = C L, the absorbances of the two forms of the

flavoprotein can be given as-

Abs of FRed (ARed) = 3.82 x 103 M-1 cm-1 x [FR-E] x 1.0 cm = 3.82 x 103 [FR-E] M-1

Abs of FOx (AOx) = 1.12 x 104 M-1 cm-1 x [FO-E] x 1.0 cm = 1.12 x 104 [FO-E] M-1

And, the sum of absorbance is equal to 0.500 as given-

ARed + AOx = 3.82 x 103 [FRed] M-1 + 1.12 x 104 [FOx] M-1 = 0.500

So, 3.82 x 103 [FRed] + 1.12 x 104 [FOx] = 0.500 M - Equation E11.2

Step 2: (Equation E11.1 x 3.82 x 103) – Equation E11.2


b. Following the stoichiometry of the balanced reaction, 1 mol of the reduced form (FRed)

reacts with 1 mol substrate (S) to form 1 mol reduced substrate (S-). So, at equilibrium,

the concentration of [S-] is equal to the [FOx] formed. And, the equilibrium [S] is equal to

the remaining [S] in the solution.

Hence, Equilibrium [S-] = [FOx] = 3.825 x 10-5 M

And, Equilibrium [S] = Initial [S] – [S] consumed = Initial [S] – Equilibrium [S-]

= 5.70 x 10-5 M - 3.825 x 10-5 M = 1.875 x 10-5 M

c. Since the question does not mention the reaction temperature, it’s assumed to be

25.00C or 298.15 K (standard temperature).

When a half-reaction is reversed, the numeric sign of the resultant reaction is also

reversed. Hence, of E0’ for the given reaction S + e- S- = - 0.154 V


1.D. Two Analyte, Two Wavelengths

Example 13: Cobalt and nickel ions form colored complexes with 2,3-quinoxalinedithiol.

These complexes have molar absorptivity of Co = 36400 M-1 cm-1 and Ni = 5520 M-1 cm-

1 at 510 nm, and Co = 1240 M-1 cm-1 and Ni = 17500 M-1 cm-1 at 656 nm. A 0.635-gram

sample containing Ni and Co ions was dissolved and diluted to 100.0 mL. A 50.0 mL

aliquot was treated to eliminate interferences; an excess 2,3-quinoxalinedithiol was added,

and the volume of the solution was adjusted to 100.0 mL. This solution had an absorbance

of 0.347 at 510 nm and 0.228 at 656 nm in a 1-cm cell. Given the following atomic masses:

Co = 58.933 g mol-1 and Ni = 58.69 g mol-1.

I. Calculate the number of moles of Ni and Co ions contained in the sample.

II. Determine the parts per million of cobalt and nickel in the sample.

Ans. Step 1: Beer-Lambert’s Law: A = C L

Where, A = absorbance , = molar absorptivity (M-1 cm-1)

L = Path length (cm) , C = Concentration

Given-

The total absorbance of the solution at any wavelength equals the sum of absorbances of

the individual analytes in it. Let the concentrations of Co and Ni in the final solution

(whose absorbance is taken) be X and Y molar, respectively.

At 510 nm: Total abs = Abs of Co + Abs of Ni

Or, 0.347 = (36400 M-1 cm-1 x X M x 1.0 cm) + (5520 M-1 cm-1 x Y M x 1.0 cm)

Hence, 36400X + 5520Y = 0.347 - Equation E12.1




Now, (Comparing equation 1 x 1240) – (Equation 2 x 36400)

45136000 X + 6844800 Y = 430.28

-45136000 X - 637000000 Y = -8299.2

-630155200 Y = -7868.92

510 nm 656 nm

Co 36400 1240

Ni 5520 17500

Total Abs 0.347 0.228


Or, Y = -7868.92 / -630155200 = 1.249 x 10-5

Hence, [Ni] in the final aliquot = Y M = 1.249 x 10-5 M

And, Putting the values of Y in equation E12.1-

X = [0.347 – (5520 x 1.249 x 10-5)] / 36400 = 7.639 x 10-6

Hence, [Co] in the final aliquot = X M = 7.639 x 10-6 M

Step 2: I. Calculate the number of moles of Ni and Co ions contained in the sample.

The original sample diluted to prepare the final aliquot (whose OD or absorbance is taken)

as follow-

I. 0.635 g of the original sample is dissolved and diluted to a final volume of 100.0

mL. Let us label it as solution A.

II. 50.0 mL of solution A is treated and diluted to a final volume of 100 mL. This

solution (the final aliquot), gives the specified absorbances at two wavelengths as

mentioned in the question. Let us label this 100 mL solution as solution B.

Note that 50.0 mL of solution A is used to prepare 100.0 mL of solution B.

Now, using C1V1 (solution B) = C2V2 (solution A)

[Co] in solution A, C2 = (C1V1) / V2 = (7.639 x 10-6 M x 100 mL) / 50 mL = 1.5278 x 10-5 M

[Ni] in solution A, C2 = (C1V1) / V2 = (1.249 x 10-5 M x 100 mL) / 50 mL = 2.498 x 10-5 M

Now,

Moles of Co in solution A = [Co] in solution A x Vol. in liters

= 1.5278 x 10-5 M x 0.100 L = 1.5278 x 10-6 mol

Moles of Ni in solution A = [Ni] in solution A x Vol. in liters

= 2.498 x 10-5 M x 0.100 L = 2.498 x 10-6 mol

Since 100.0 mL of solution A is prepared from 0.635 g of the original sample, the number

of moles of analytes in the original sample must be equal to that of 100.0 mL of solution

A.

Hence,

Moles of Co in 0.635 g original sample = 1.5278 x 10-6 mol

Moles of Ni in 0.635 g original sample = 2.498 x 10-6 mol

II. Determine the parts per million of cobalt and nickel in the sample.

[Co],ppm in original sample = Mass of Co in g / Mass of sample in grams

= [(Mol x atomic mass) x 106 g g-1] / Mass of sample in grams

= [(1.5278 x 10-6 mol x 58.933 g mol-1) x 106 g g-1] / 0.635 g

= [(9.0038 x 10-5 g) x 106 g g-1] / 0.635 g

= 90.038 g / 0.635 g = 141.79 g g-1 = 141.79 ppm

And, [Ni],ppm in original sample = Mass of Ni in g / Mass of sample in grams



= [(2.498 x 10-6 mol x 58.69 g mol-1) x 106 g g-1] / 0.635 g

= [(9.0038 x 10-5 g) x 106 g g-1] / 0.635 g

= 146.61 g / 0.635 g = 141.79 g g-1 = 230.88 ppm

Example 14: Molar absorptivity data for the cobalt and nickel complexes with 2,3-

quinoxalinedithiol are Co = 36400 M-1 cm-1 and Ni = 5520 M-1 cm-1 at 510 nm, and Co =

1240 M-1 cm-1 and Ni = 17500 M-1 cm-1 at 656 nm. A 0.425-gram sample was dissolved

and diluted to 50.0 mL. A 25.0.0 mL aliquot was treated to eliminate interferences; after

addition of an excess 2,3-quinoxalinedithiol, the volume of the solution was adjusted to

50.0 mL. This solution had an absorbance of 0.446 at 510 nm and 0.326 at 656 nm in a

1.00-cm cell. Given the following atomic masses: Co = 58.933 g mol-1 and Ni = 58.69 g

mol-1. Calculate the concentrations in parts per million of cobalt and nickel in the sample.

Ans. Step 1: Beer-Lambert’s Law: A = C L

Where, A = absorbance , = molar absorptivity (M-1 cm-1)

L = Path length (cm) , C = Concentration

Given-

The total absorbance of the solution at any wavelength equals the sum of absorbances of

the individual analytes in it. Let the concentrations of Co and Ni in the final solution

(whose absorbance is taken) be X and Y molar, respectively.







Now, (Comparing equation 1 x 1240) – (Equation 2 x 36400)

45136000 X + 6844800 Y = 553.04

-45136000 X - 637000000 Y = -11866.4

-630155200 Y = -11313.36

510 nm 656 nm

Co 36400 1240

Ni 5520 17500

Total Abs 0.446 0.326


Or, Y = -11313.36 / -630155200 = 1.795 x 10-5

Hence, [Ni] in the final aliquot = Y M = 1.795 x 10-5 M

And, Putting the values of Y in equation E13.1-

X = [0.446 – (5520 x 1.795 x 10-5)] / 36400 = 9.530 x 10-6

Hence, [Co] in the final aliquot = X M = 9.530 x 10-6 M

Step 2: Calculate the number of moles of Ni and Co ions contained in the sample.

The original sample diluted to prepare the final aliquot (whose OD or absorbance is taken)

as follow-

I. 0.425 g of the original sample is dissolved and diluted to a final volume of 50.0

mL. Let us label it as solution A.

II. 25.0 mL of solution A is treated and diluted to a final volume of 50 mL. This

solution (the final aliquot), gives the specified absorbances at two wavelengths as

mentioned in the question. Let us label this 50 mL solution as solution B.

Note that 25.0 mL of solution A is used to prepare 50.0 mL of solution B.

Now, using C1V1 (solution B) = C2V2 (solution A)

[Co] in solution A, C2 = (C1V1) / V2 = (9.530 x 10-6 M x 50 mL) / 25 mL = 1.906 x 10-5 M

[Ni] in solution A, C2 = (C1V1) / V2 = (1.795 x 10-5 M x 50 mL) / 25 mL = 3.590 x 10-5 M

Now,

Moles of Co in solution A = [Co] in solution A x Vol. in liters

= 1.906 x 10-5 M x 0.050 L = 9.530 x 10-7 mol

Moles of Ni in solution A = [Ni] in solution A x Vol. in liters

= 3.590 x 10-5 M x 0.050 L = 1.795 x 10-6 mol

Since 50.0 mL of solution A is prepared from 0.435 g of the original sample, the number

of moles of analytes in the original sample must be equal to that of 50.0 mL of solution

A.

Hence,

Moles of Co in 0.435 g original sample = 9.530 x 10-7 mol

Moles of Ni in 0.635 g original sample = 1.795 x 10-6 mol

Determine the parts per million of cobalt and nickel in the sample.

[Co],ppm in original sample = Mass of Co in g / Mass of sample in grams


= [(9.530 x 10-6 mol x 58.933 g mol-1) x 106 g g-1] / 0.435 g

= [(5.616 x 10-5 g) x 106 g g-1] / 0.435 g

= 56.16 g / 0.435 g = 129.11 g g-1 = 129.11 ppm

And, [Ni],ppm in original sample = Mass of Ni in g / Mass of sample in grams


= [(Moles x atomic mass) x 106] g / Mass of sample in grams

= [(1.795 x 10-6 mol x 58.69 g mol-1) x 106 g g-1] / 0.435 g

= [(1.054 x 10-4 g) x 106 g g-1] / 0.435 g

= 105.37 g / 0.435 g = 305.41 g g-1 = 305.44 ppm


2. (External) Standard Addition Methods

2.A. Single Aliquot Standard Addition

Standard addition numerical may sometimes appear a bit difficult at first. Keeping in mind

that these problems are also based on Beer-Lambert’s law would be useful to solve such

problems. The approach can be molded in different ways by different authors. Two

approaches or methods shall be used to illustrate the solution strategies for single aliquot

external addition methods.

Example 15: An unknown sample of Cu2+ gave an absorbance of 0.262 in an atomic

absorption analysis. Then 1.0 mL of the solution containing 100.0 ppm Cu2+ was mixed

with 95.0 ml of the unknown, and the mixture was diluted to 100.0 mL in a volumetric

flask. The absorbance of the new solution was 0.500. Find the [Cu2+] in the unknown.

Ans. Method 1: Using Beer-Lambert’s law

Step 1: For the original unknown solution.

Using Beer-Lambert’s Law A = C L

A1 = C1 L

Where, A1 = Absorbance of the original unknown solution = 0.262

CX = [Cu2+] in the original unknown solution

In the above expression, A1 is a known quantity, C1 is the unknown to be determined,

and L are unknown but remain constant for a specified analyte under the same set of

experimental conditions. Since and L remain the same for the given set of experimental

conditions, their product would be a constant for both the original unknown solution as

well as the spiked (standard-addition) aliquot. So, we need to derive both the equations

(one for each case) in terms of the product of and L.

That is, L = A1 / Cx - Equation 1E.1

Step 2: For the spiked and finally diluted solution

To derive the expression in terms of L, we first need to determine the total [Cu2+] in the

spiked aliquot. There are two sources of [Cu2+] in the spiked aliquot- one [Cu2+] from the

original unknown solution, and the other [Cu2+] from the standard solution. We need to

calculate [Cu2+] in the final diluted solution from both the sources separately, and then

add them to get the total [Cu2+] in the spiked solution.

In the finally diluted spiked aliquot,

using C1V1 (original solution) = C2V2 (spiked solution)

[Cu2+] from standard solution = C1V1 / V2 = (100.0 ppm x 1.0 mL) / 100.0 mL = 1.0 ppm


[Cu2+] from unknown solution = (Cx x 95.0 mL) / 100.0 mL = 0.95 Cx

And,

Total [Cu2+] in the spiked solution, [Cu2+]T = [Cu2+] from (standard +unknown) solution

Or, [Cu2+]T = 1.0 ppm + 0.95 Cx

Now, using Beer-Lambert’s Law for the spiked solution

A2 = CT L

Where, A2 = Absorbance of the spiked and diluted solution = 0.500

CT = [Cu2+] in the spiked and diluted solution = 1.0 ppm + 0.95 Cx

Or, L = A2 / CT - Equation 1E.2

As explained above, and L remain constant for the given set of experimental conditions.

Now, comparing equations 1 and 2-

L = A1 / C1 = A2 / CT

Or, A1 / C1 = A2 / CT - Equation 1E.3

Hence,

(Abs / [Analyte]) of unknown = (Abs / [Analyte]) of spiked soln.

-equation 1E.4

Equation 1E.4 shall serve as the principal equation to solve the numerical of single

standard addition methods.

Step 3: Putting the values in equation 1E.3 or 1E.4-

0.262 / Cx = 0.500 / (1.0 ppm + 0.95 Cx)

Or, 0.262 x (1.0 ppm + 0.95 Cx) = 0.500 Cx

Or, 0.262 ppm + 0.2489 Cx = 0.500 Cx

Or, 0.500 Cx - 0.2489 Cx = 0.262 ppm

Or, 0.2511 Cx = 0.262 ppm

So, Cx = 0.262 ppm / 0.2511 = 1.043 ppm

Hence, [Cu2+] in the original unknown solution, Cx = 1.043 ppm

Method 2: Accounting increase in absorbance of the spiked solution

Following Beer-Lambert’s law, the absorbance of a solution is proportional to the

concertation of the analyte. It also means that an increase in concentration shall reflect a

proportional increase in the absorbance.

That is,

([Analyte] / Abs) of original soln. = Increase in ([Analyte] / Abs) of spiked soln.

- Equation 1E.5

Step 1: Let [Cu2+] in the original unknown solution be Cx.


Given-

Absorbance of the original unknown solution = 0.262

Absorbance of the spiked solution = 0.500

So, increase in absorbance due to spiking = 0.500 – 0.262 = 0.238

In the finally diluted spiked aliquot,

using C1V1 (original solution) = C2V2 (spiked solution)

[Cu2+] from standard solution = C1V1 / V2 = (100.0 ppm x 1.0 mL) / 100.0 mL = 1.0 ppm

[Cu2+] from unknown solution = (Cx x 95.0 mL) / 100.0 mL = 0.95 Cx

And,

Total [Cu2+] in the spiked solution, [Cu2+]T = [Cu2+] from (standard +unknown) solution

Or, [Cu2+]T = 1.0 ppm + 0.95 Cx

Now, increase in [Cu2+] due to spiking = [Cu2+]T – [Cu2+] of unknown soln.

= (1.0 ppm + 0.95 Cx) – Cx = 1.0 ppm - 0.05 Cx

Step 2: Using Equation 1E.5-

Cx / 0.262 = (1.0 ppm - 0.05 Cx) / 0.238

Or, 0.238 Cx = 0.262 x (1.0 ppm - 0.05 Cx) = 0.262 ppm – 0.0131 Cx

Or, 0.238 Cx + 0.0131 Cx = 0.262 ppm

Or, Cx = 0.262 ppm / 0.2511 = 1.043 ppm

Hence, [Cu2+] in the original unknown solution, Cx = 1.043 ppm

Example 16: The lithium concentration in serum taken from a patient being treated with

lithium for maniac-depressive illness was analyzed using flame emission spectroscopy. A

sample of the serum gave a reading of 372 unit for the intensity of the 671 nm red

emission line. Then, 1.00 mL of a 11.3 mM lithium standard was added to 9.00 mL of

serum. This spiked serum gave an intensity reading of 767 units at the 671 nm emission

line. What is the original concentration of Li+ in the serum?

Ans. Method 1: Given-

The intensity of the original serum sample = 372 units

The intensity of the Spiked serum solution = 767 units

[Li+] in the standard solution = 11.3 mM

Vol. of Standard lithium solution taken = 1.00 mL

Vol. of the original serum taken = 9.00 mL

Now,

The spiked soln. has its [Li+] from two sources - I. from the original sample, and II. from

the addition of standard solution (standard addition). Let [Li+] in the original sample be X.


Using C1V1 (original soln.) = C2V2 (spiked soln.) for the spiked solution-

[Li+] from original serum sample = (X x 9.00 mL) / 10.0 mL = 0.900 X

[Li+] from Std. soln. = (11.3 mM x 1.0 mL) / 10.0 mL = 1.130 mM

So, Total [Li+] in the spiked soln., [Li+]T = 0.900 X + 1.130 mM

Now, Using Equation 1E.4-

(Intensity / [Analyte]) of Original sample = (Intensity / [Analyte]) of spiked soln.

Or, 372 / X = 767 / (0.900 X + 1.130 mM)

Or, 0.900 X + 1.130 mM = 767 X / 372 = 2.06183 X

Or, 2.06183 X - 0.900 X = 1.130 mM

So, X = 1.130 mM / 1.16183 = 0.973 mM

Hence, [Li+] in the original serum sample, X = 0.973 mM

Method 2: Given-

The intensity of the original serum sample = 372 units

The intensity of the Spiked serum solution = 767 units

[Li+] in the standard solution = 11.3 mM

Vol. of Standard lithium solution taken = 1.00 mL

Vol. of the original serum taken = 9.00 mL

The spiked soln. has its [Li+] from two sources - I. from the original sample, and II. from

the addition of standard solution (standard addition). Let [Li+] in the original sample be X.


[Li+] from original serum sample = (X x 9.00 mL) / 10.0 mL = 0.900 X

[Li+] from Std. soln. = (11.3 mM x 1.0 mL) / 10.0 mL = 1.130 mM

So, Total [Li+] in the spiked soln., [Li+]T = 0.900 X + 1.130 mM

Now,

Increase in intensity due to spiking = (767 – 372) AU = 395 AU

Increase in [Li+] due to spiking = [Li+]T – X

= (0.900 X + 1.130 mM) – X = -0.100X + 1.130 mM


([Analyte] / Intensity) of original sample = Increase in ([Analyte] / Intensity) of spiked soln.

Or, X / 372 = (-0.100X + 1.130 mM) / 395

Or, 395X / 372 = -0.100X + 1.130 mM

Or, 1.06183 X + 0.100 X = 1.130 mM

Or, 1.16183 X = 1.130 mM

So, X = 1.130 mM / 1.16183 = 0.973 mM


Hence, [Li+] in the original serum sample, X = 0.973 mM

Example 17: Copper was determined in river water by atomic absorption

spectrophotometer and the method of standard additions. For the addition, 250.0 mL of

a 1000.0 g mL-1 Cu standard was added to 150.0 ml of the unknown solution. The

following data were obtained-

Absorbance of reagent blank = 0.021

The absorbance of the sample (original water sample) = 0.472

The absorbance of sample plus addition – blank = 1.027

1. Calculate the copper concentration in the sample

2. Later studies showed that the reagent blank used to obtain the above data was

inadequate and that the actual blank absorbance was 0.100. Find the copper

concentration with the appropriate blank, and determine the % error caused by using an

improper blank.

Ans. 1. Calculate Initial [Cu] in the Sample without Correction in Abs

Given-

Abs of the reagent blank = 0.021

Abs of the original sample = 0.472

Abs of (Sample + Addition – Blank) = 1.027

Corrected absorbance of blank = 0.100

[Cu] in the standard solution = 1000.0 g mL-1

Vol. of standard [Cu] soln. taken = 250.0 mL

Vol. of original sample taken = 150.0 mL

Total volume of the spiked soln. = 250.0 mL + 150.0 mL = 400.0 mL

Now,

Actual abs sample = Abs of (Sample – Blank) = 0.472 – 0.021 = 0.451

The spiked soln. has its [Cu] from two sources - I. from the original sample, and II. from

the addition of standard solution (standard addition). Let [Cu] in the original sample be X.


[Cu] from original soln. = (X x 150.0 mL) / 400.0 mL = 0.375 X

[Cu] from Std. soln. = (1000.0 g mL-1 x 250.0 mL) / 400.0 mL = 625.0 g mL-1

So, Total [Cu] in the spiked soln., [Cu]T = 0.375 X + 625.0 g mL-1

And,

Increase in Abs due to spiking = Actual Abs of (spiked soln. - Sample)

= 1.027 – 0.451 = 0.576


Increase in [Cu] due to spiking = Total [Cu] in spiked soln. - [Cu] in sample

= [Cu]T – X = (0.375 X + 625.0 g mL-1) – X

= -0.625X + 625.0 g mL-1

Using Equation 1E.5 ([Analyte] / Abs) of original soln. = Increase in ([Analyte] / Abs) of

spiked soln.-

X / 0.451 = (-0.625X + 625.0 g mL-1) / 0.576

Or, 0.576 X = 0.451 x (-0.625X + 625.0 g mL-1) = -0.2819 X + 281.875 g mL-1

Or, 0.576 X + 0.2891 X = 281.875 g mL-1

So, X = 281.875 g mL-1 / 0.8579 = 328.57 g mL-1

Hence, Initial or Uncorrected [Cu] in the original sample water = 328.57 g mL-1

2. Calculate [Cu] in Sample after Correction in Absorbance

Corrected Abs of Sample = Abs of (sample- Corrected blank) = 0.472 – 0.100 = 0.372

Corrected Abs of Spiked soln. = Abs of Spiked Soln. - (Corrected - Actual) abs of blank

= 1.027 – (0.100 – 0.021) = 0.948

Corrected Increase in Abs due to spiking = Corrected Abs of (Spiked soln. – Sample)

= 0.948 – 0.372 = 0.576

Again, Using Equation 1E.5-

X / 0.372 = (-0.625X + 625.0 g mL-1) / 0.576

Or, 0.576 X = 0.372 x (-0.625X + 625.0 g mL-1) = -0.2325 X + 232.500 g mL-1

Or, 0.576 X + 0.2325 X = 232.500 g mL-1

So, X = 232.500 g mL-1 / 0.8085 = 287.57 g mL-1

Hence, the Corrected [Cu] in the original sample water = 287.57 g mL-1

Now, Account Error in [Cu]:

Error in [Cu] = Corrected [Cu] - Initial [Cu] = (287.57 – 328.57) g mL-1 = -41.00 g mL-1

Now,

% Error in [Cu] = (Error in [Cu] / Corrected [Cu]) x 100

= (-41.00 g mL-1 / 287.57 g mL-1) x 100 = -14.26%

The -ve sign indicates that the initial or un-corrected [Cu] is lesser than the corrected [Cu].

Example 18: A 5.00 mL sample containing just Riboflavin is diluted to 6.00 mL, then

analyzed and found to emit an intensity of 4518 at 520 nm. The 5.00 mL volume sample

was spiked with 0.50 mL of 5.00 ppm riboflavin standard solution, diluted to 6.00 mL. The


intensity at the same wavelength with the spiked standard solution was 7362. Calculate

the concentration of Riboflavin in the original sample.

Ans. Step 1: Given-

The intensity of the diluted sample = 4518 units

The intensity of the Spiked and diluted solution = 7362 units

[Riboflavin] in the standard solution = 5.00 ppm

Vol. of Standard Riboflavin solution taken = 0.500 mL

Vol. of the original sample taken = 5.00 mL

Total volume of spiked and diluted solution = 6.00 mL

The spiked soln. has its [Riboflavin] from two sources - I. from the original sample, and II.

from the addition of standard solution (standard addition). Let [Riboflavin] in the original

sample be X.

Using C1V1 (diluted sample) = C2V2 (spiked, diluted soln.) for the spiked solution-

[Riboflavin] from diluted sample = (X x 5.00 mL) / 6.0 mL = 0.8333 X

[Riboflavin] from Std. soln. = (5.00 ppm x 0.5 mL) / 6.0 mL = 0.4167 ppm

So, Total [Riboflavin] in the spiked soln., [Riboflavin] T = 0.8333 X + 0.4167 ppm

Now,

Increase in intensity due to spiking = (7362 – 4518) AU = 2844

Increase in [Riboflavin] due to spiking = [Riboflavin]T – X

= (0.8333 X + 0.4167 ppm) – X = -0.1667X + 0.4167 ppm


([Analyte] / Intensity) of diluted sample = Increase in ([Analyte] / Intensity) of spiked soln.

Or, X / 4518 = (-0.1667X + 0.4167 ppm) / 2844

Or, 2844X / 4518 = -0.1667X + 0.4167 ppm

Or, 0.6295X + 0.1667 X = 0.4167 ppm

Or, 0.7962 X = 0.4167 ppm

So, X = 0.4167 ppm / 0.7962 = 0.5234 ppm

Hence, [Riboflavin] in the diluted sample, X = 0.5234 ppm

Step 2: Accounting Dilutions:

Dilution 1: 5.00 mL of the unknown original sample is spiked and diluted to 6.00 mL to

make the spiked solution.

Now, using C1V1 (diluted sample) = C2V2 (Spiked, diluted solution)

So, [Riboflavin] in diluted sample = (0.5234 ppm x 6.00 mL) / 5.00 mL = 0.628 ppm


Dilution 2: 5.00 mL of the unknown original sample is diluted to 6.00 mL to make the

un-spiked aliquot.

Now, using C1V1 (original sample) = C2V2 (diluted sample)

So, [Riboflavin] in original sample = (0.628 ppm x 6.00 mL) / 5.00 mL = 0.754 ppm

Example 19: Quinine is found in tonic water and is an aromatic based compound the

fluoresces at 460 nm after it absorbs excitation at 350 nm. A solution that contains an

unknown amount of quinine was found to have a luminescent intensity of 4343 counts.

To determine the concentration of quinine in this solution 13.50 mL of this solution was

spiked with 4.50 mL of a 23.86 ppm (w/v) quinine standard addition reagent (SAR) and

diluted to volume in a 30.00 mL volumetric flask. If the luminescent intensity of the spiked

reagent was found to 6302 counts, what is the concentration of quinine in the unknown

solution, ppm (w/w)?

Ans. Given-

The intensity of the original sample = 4343

The intensity of the Spiked and diluted solution = 6302

[Quinine] in the standard solution = 23.86 ppm

Vol. of Standard Quinine solution taken = 4.50 mL

Vol. of the original sample taken = 13.30 mL

Total volume of the spiked solution = 30.00 mL

Now,

The spiked soln. has its [Quinine] from two sources - I. from the original sample, and II.

from the addition of standard solution (standard addition). Let [Quinine] in the original

sample be X.


[Quinine] from original sample = (X x 13.50 mL) / 30.00 mL = 0.450 X

[Quinine] from Std. soln. = (23.86 ppm x 4.50 mL) / 30.00 mL = 3.579 ppm

So, Total [Quinine] in the spiked soln., [Quinine]T = 0.450 X + 3.579 ppm


(Intensity / [Analyte]) of Original sample = (Intensity / [Analyte]) of spiked soln.

Or, 4343 / X = 6302 / (0.450 X + 3.579 ppm)

Or, 0.450 X + 3.579 ppm = 6302 X / 4343 = 1.451 X

Or, 1.451 X - 0.450 X = 3.579 ppm

So, X = 3.579 ppm / 1.001 = 3.575 ppm

Hence, [Quinine] in the original tonic sample, X = 3.575 ppm


2.B Multiple Aliquots Standard Addition

Example 20: An environmental chemist working for the Environment Protection Agency

(EPA) was directed to collect razor clams from a heavily-contaminated river superfund site

and analyze them for their Cd2+ content using graphite furnace atomic absorption

spectrometry (GFAAS). The chemist dried the clams at 950C overnight and ground them

in a scientific blender, resulting in approximately 50 g of homogenized dry weight. A

representative 71.64 mg sample was taken from the approximately 50 g dry material and

dissolved in 100.0 mL of 0.1 M HCl to create a sample solution. Using the method of

standard addition, the chemist prepared five standards in 100.0 mL volumetric flasks, each

containing 5.0 ml of the sample solution. Varying amounts of a 99.0 ppb (g L-1) Cd2+

standard were added to each flask, which were then brought to volume with 0.1 M HCl.

The Cd2+ content of the solution was then analyzed using GFAAS, using the absorbance

data given in the table.

Sample volume (mL) 5.0 5.0 5.0 5.0 5.0

Vol. of Std. Cd2+ soln. (mL) 0.00 2.50 5.00 7.50 10.00

Absorbance 0.080 0.119 0.163 0.200 0.241

Determine the amount of Cd2+ per gram of dry clam. Express your result as milligrams of

Cd2+ per gram dry clam.

Ans. Method 1: By determining [Cd2+] in the un-spiked aliquot: Create a data table as

shown below to calculate [Cd2+] from various standard additions. Plot absorbance vs

[Cd2+] added from standard solutions of various aliquots, label the axes appropriately.

Generate the trendline equation or linear regression equation for the standard graph.

Once the trendline equation is obtained, the required calculations can be


done in following two steps-


Hence, Cd2+ content in dry clam (dry mass) = 0.137 mg Cd2+ per gram dry clam.

Method 2: By the mass of Cd2+ in the un-spiked aliquot: Create a data table as shown

below to calculate the mass (g) of Cd2+ from various standard additions. Plot absorbance

vs mass (g) of Cd2+ added from standard solutions of various aliquots, label the axes

appropriately. Generate the trendline equation or linear regression equation for the

standard graph.

Mass (g) of Cd2+ added from standard solution = C1 x (V1 / 1000)

Where, C1 = [Cd2+] of the standard solution

V1 = Vol. of std. soln. taken (mL)


V1 / 1000 = mL / 1000 = liters

ppb x L = g L-1 x L = g

Once the trendline equation is obtained, the required calculations can be done in

following two steps-

Hence, Cd2+ content in dry clam (dry mass) = 0.137 mg Cd2+ per gram dry clam.


Example 21: Allicin is a ~0.4 wt% component in garlic with antimicrobial and possibly

anticancer and antioxidant activity. It is unstable and therefore difficult to measure. An

assay was developed in which the stable precursor alliin is added to freshly crushed garlic

and converted to allicin by the enzyme

alliinase found in garlic. Components of the

garlic are extracted and measured by

chromatography. The chromatogram shows

standard additions reported as mg alliin

added per gram of garlic. The

chromatographic peak is allicin from the

conversion of alliin.

2 Alliin –(Alliinase)→ Allicin

MW of Alliin = 177.2 g mol-1

MW of Allicin = 162.3 g mol-1

a. The standard addition procedure has a

constant total volume. Measure the

responses in the figure and prepare a graph

to find how much alliin equivalent was in the

unspiked garlic. The units of your answer will

be mg alliin/g garlic. Find the 95% confidence

interval, as well.

b. Given that 2 mol of alliin is converted to 1

mol of allicin, find the allicin content of garlic

(mg allicin/g garlic) including the 95%

confidence interval.

Ans. The graph in question does not show the detector’s response to the respective alliin

concentrations. The detector’s responses need to be manually determined. The

approximate (but not exact) values of the respective detector’s response are calculated

using graph paper. Closer is the approximated detector’s response to the respective exact

value, greater would be the accuracy of the calculated result.

The detector’s responses and their respective [alliin] are tabulated below-

Spiked [Alliin], mg g-1 0.0 3.1 7.2 8.5 20.0 38.2

Detector’s response 1.766 2.394 2.922 3.403 5.516 9.434

Plot spiked [Alliin] from standard solutions of various aliquots vs respective detector’s

response, label the axes appropriately. Generate the trendline equation or linear

regression equation for the standard graph.


Once the standard graph and corresponding trendline equation are generated, the

calculations can be done in the following steps-



Example 22: Students performed an experiment like that in figure 5-7 in which each

flask contained 25.00 mL of serum, varying additions of 2.640 M NaCl standard, and a

total volume of 50.00 mL. The detector’s responses and their respective [Na+] are

tabulated below-

S. No. 1 2 3 4 5

Vol. of Standard Soln. (mL) 0 1.000 2.000 3.000 4.000

Signal (mV) 3.13 5.40 7.89 10.30 12.48

Ans. Create a data table as shown below to calculate [Na+] from various standard

additions. Plot [Na+] added from standard solutions of various aliquots vs signal, label the

axes appropriately. Generate the trendline equation or linear regression equation for the

standard graph.


Once the trendline equation is obtained, the required calculations can be done in the

following steps-


Example 23: Tooth enamel consists mainly of the mineral calcium hydroxyapatite,

Ca10(PO4)6(OH)2. Trace elements in teeth of archeological specimens provide


anthropologists with clues about diet and diseases of ancient people. Students at Hamline

University used atomic absorption spectroscopy to measure strontium from extracted

wisdom teeth. Solutions were prepared with a constant total volume of 10.0 mL containing

0.750 mg of dissolved tooth enamel plus variable concentrations of added Sr.

S. No. 1 2 3 4 5

Added Sr, ppb 0 2.50 5.00 7.50 10.00

Signal (AU) 28.0 34.3 42.8 51.5 58.6

a. Find the concentration of Sr and its uncertainty in the 10-mL sample solution in part

per billion, ppb (= ng mL-1).

b. Find the concentration Sr in tooth enamel in parts per million (g g-1).

c. If the standard addition intercept is the major source of uncertainty, find the uncertainty

in the concentration of Sr in the tooth enamel in ppm.

d. Find 95% confidence interval for Sr in tooth enamel.

Ans. Note that the final [Sr] in all the standard aliquots are already mentioned in the

question. Plot [Sr] added from standard solutions of various aliquots vs signal, label the

axes appropriately. Generate the trendline equation or linear regression equation for the

standard graph.



Example 24: A series of volumetric flasks are being prepared for the analysis of mercury

in the drinking water supply using standard addition. To prepare flasks, 10 mL of drinking

water and a varying amount of standard are added to each flask, and the flasks are

brought to the 25.0 mL mark on the volumetric flask. The concentration of mercury

standard solution used was 20.0 micrograms per mL.

Use the data tabulated below to provide the concentration of mercury in the drinking

water in units of micrograms per mL.


S. No. 1 2 3 4 5 6

Standard [Hg] 0.00 1.00 2.00 3.00 4.00 5.00

Signal (AU) 19.65 30.35 40.28 49.78 60.44 70.06

Ans. Create a data table as shown below to calculate [Hg] from various standard additions.

Plot [Hg] added from standard solutions of various aliquots vs signal, label the axes

appropriately. Generate the trendline equation or linear regression equation for the

standard graph.



Rules for Error Propagation or Uncertainty Calculations:


About the Author

Neeraj Kumar holds a master’s

degree in Biochemistry from HNB

Garhwal University, Uttarakhand,

India. He has worked as

Junior/Senior Research Fellow in

the Department of Biotechnology,

Govt. of India, sponsored project

entitled “Elucidating the

mechanisms involved in higher

feed efficiency of bovine species by

expression of the genes regulating

mitochondrial proton leak kinetics”

at ICAR-RCER, Patna, Bihar, India.

He also served as SRF in National Fisheries Development Board, Govt. of India, sponsored

project entitled “National Surveillance Program for Aquatic Animal Diseases” at the same

institute. He has research experiences and practical expertise in biochemical analytical

techniques, molecular biology techniques, microbiological techniques, proximate

analysis, etc. Currently, he works as a freelance subject matter expert for biology and

chemistry for a US-based education technology company.

Being an optimistic and incessant learner, he has started the BioChem Calculations project

to share his skills in solving numerical questions in the biology and chemistry subjects. He

is also working on a project to establish his own R&D laboratory facilities for the

cultivation of the medicinal mushroom Cordyceps militaris. This proposed lab will further

self-sponsor projects in the fields of agriculture and bioremediations.

The author welcomes all the critics and suggestions on the topics presented here. Please

feel free to send your suggestions to [email protected].

mailto:[email protected]

Documents

Analyte Estimation using Spectrophotometric Methods and ... · 5 Analyte Estimation using Spectrophotometric Methods and Excel © Neeraj Kumar BioChem Calculations Ans. Given- = 1.8011