# Analyte Estimation using Spectrophotometric Methods & Excel

**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 l_{max}. Among the hundreds of wavelengths (l) absorbed by the analyte, the spectrophotometry readings are taken at l_{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 I_{0} 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 I_{0} 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.

Though the standard unit of e 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, e 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 = e C L

e = A / CL = 0.379 / (2.1043 x 10^{-4} M x 1.0 cm) = 1.8011 x 10^{3} M^{-1} cm^{-1}

**Example 2:** The molar absorptivity, e of indophenol under these experimental conditions is 1.8011 x 10^{3} 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 ; e = 1.8011 x 10^{3} M^{-1} cm^{-1}

L = 10 mm = 1.0 cm

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

A = 1.8011 x 10^{3} M^{-1} cm^{-1} x 2.1043 x 10^{-4} M x 1.0 cm = 0.379

**Example 3:** The molar absorptivity, e of indophenol under these experimental conditions is 1.8011 x 10^{3} M^{-1} cm^{-1}. What is the expected concentration of indophenol solution if its absorbance is 0.379 through a 10 mm cuvette?

Ans. Given-

e = 1.8011 x 10^{3} M^{-1} cm^{-1} ; A = 0.379

L = 10 mm = 1.0 cm

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

C = A / eL = 0.379 / (1.8011 x 10^{3} M^{-1} cm^{-1} x 1.0 cm) = 2.1043 x 10^{-4} M

**Example 4:** The molar absorptivity, e of indophenol under these experimental conditions is 1.8011 x 10^{3} 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 ; e = 1.8011 x 10^{3} M^{-1} cm^{-1}

A = 0.379

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

L = A / eC = 0.379 / (1.8011 x 10^{3} 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 Fe^{3+} solution. An 11.0 mL aliquot of this standard solution is treated with an excess of HNO_{3} 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 HNO_{3} 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 = e C L

where,

A = Absorbance

e = molar absorptivity at the specified wavelength (M^{-1}cm^{-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 **e** remains constant.

Step 1: Preparation of working reference aliquot: 11.0 mL standard solution with 7.64 x 10^{-4} M Fe^{3+} 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, [Fe^{3+}] 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__, A_{1} = e_{1} x (1.5820 x 10^{-4} M x 1.00 cm) – equation 1

Abs of well water __aliquot__, A_{2} = e_{2} (C2 x 2.15 cm) – equation 2

Where, C2 = concentration of Fe in the well water

Given, A_{1} = A_{2}. So, equating equations 1 and 2-

e_{1} x (1.5820 x 10^{-4} M x 1.00 cm) = e_{2} (C2 x 2.15 cm)

Also note that e remains constant. So, e_{1} = e_{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, [Fe^{3+}] in final working well water aliquot = 7.1070 x 10^{-5} M

**Step 3: **Calculating [Fe^{3+}] 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, [Fe^{3+}] 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 Fe^{2+} to Fe^{3+}. 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-

### Fe^{3+}(aq) + SCN^{–}(aq) ———-> FeSCN^{2+}(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 = e 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, [FeSCN^{2+}] in the solution, C = 1.6253 x 10^{-4} M

Following the stoichiometry of the balanced reaction, 1 mol Fe^{3+} forms 1 mol FeSCN^{2+}. So, [Fe] in the solution must be equal to [FeSCN^{2+}].

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, e of the dye.

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-

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.

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.

**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 R^{2} value for linear regression. Though R^{2} value is generally not required in such cases, it provides valuable information about the linearity of the graph. Closer is the value of R^{2} 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 f_{x}LINEST 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.

**Example 6:** The absorbances of different concentrations of the standard solutions of Ca^{2+} 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.

Step 2: Calculate [Ca^{2+}] 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 = (0.400 – 0.2821) / 0.0474 = 2.487

Hence, [Ca^{2+}] in the diluted unknown solution = 2.487 ppm = 2.487 mg L^{-1}

Step 3: Calculate the amount of Ca^{2+} 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 Ca^{2+} in the diluted unknown solution = [Ca^{2+}] 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 % Ca^{2+} in original sample:

We have-

Mass of original sample = 0.1000 g = 100.0 mg

Mass of Ca^{2+} in the original sample = 0.012435 mg

Now,

% Ca^{2+} (w/w) = (Mass of Ca^{2+} / 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 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

So, X = (0.200 + 0.0037) / 2500 = 8.148 x 10^{-5}

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 B_{2} 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.

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

The solution is presented in the following steps-

**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.

**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], C_{unk }= (3.96 x 10^{-4} M / 0.595) 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 l_{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 l_{max}, they also yield some lower absorption value at the l_{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 <—-Ka—-> 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 p*K*a for this indicator.

**Ans. ****a.** I. For HIn, given-

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

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

A_{HIn} = 2080 M^{-1} cm^{-1} x [HIn] x 1.0 cm = 2080 [HIn] M^{-1}

- For In
^{–}, given-

C = [In^{–}] , e = 14200 M^{-1} cm^{-1} , L = 1.0 cm

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

A_{In}– = 14200 M^{-1} cm^{-1} x [In^{–}] x 1.0 cm = 14200 [In^{–}] M^{-1}

- 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, A_{HIn} + A_{In}– = 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

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 (F_{Ox}) that functions as a one-electron reducing agent has a molar absorptivity e of 1.12 x 10^{4} M^{-1} cm^{-1} at 457 nm at pH 7.00. For the reduced form (F_{Red}), e = 3.82 x 10^{3} M^{-1} cm^{-1} at 457 nm at pH 7.00.

#### F_{Ox} + e^{–} <———-> F_{Red} E^{0’} = -0.128 V

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

#### F_{Red} + S <———-> F_{Ox} + S^{–}

#### Both S and S^{–} are colorless. A solution at pH 7.00 was prepared by mixing enough of the reduced protein plus substrate (F_{Red} + S) to produce initial concentrations [F_{Red}] = [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 F_{Ox} and F_{Red} from the absorbance data.

#### b. Calculate the concentrations of S and S^{–}.

#### c. Calculate the value of E^{0’} for the reaction S + e^{–} S^{–}

**Ans. ****a. Step 1: **Let the equilibrium concentrations of the F_{Red} and F_{Ox} be [F_{Red}] and [F_{Ox}], respectively. Since the F_{Ox} is derived from F_{Red}, 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 [F_{Red}]_{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 F_{Red} and F_{Ox} species.

So, [F_{Red}] + [F_{Ox}] = 5.70 x 10^{-5} M – Equation E11.1

Using Beer-Lambert’s Law equation, A = e C L, the absorbances of the two forms of the flavoprotein can be given as-

Abs of F_{Red} (A_{Red}) = 3.82 x 10^{3} M^{-1} cm^{-1} x [F_{R-E}] x 1.0 cm = 3.82 x 10^{3} [F_{R-E}] M^{-1}

Abs of F_{Ox} (A_{Ox}) = 1.12 x 10^{4} M^{-1} cm^{-1} x [F_{O-E}] x 1.0 cm = 1.12 x 10^{4} [F_{O-E}] M^{-1}

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

A_{Red} + A_{Ox} = 3.82 x 10^{3} [F_{Red}] M^{-1} + 1.12 x 10^{4} [F_{Ox}] M^{-1} = 0.500

So, 3.82 x 10^{3} [F_{Red}] + 1.12 x 10^{4} [F_{Ox}] = 0.500 M – Equation E11.2

**Step 2:** (Equation E11.1 x 3.82 x 10^{3}) – Equation E11.2

b. Following the stoichiometry of the balanced reaction, 1 mol of the reduced form (F_{Red}) reacts with 1 mol substrate (S) to form 1 mol reduced substrate (S^{–}). So, at equilibrium, the concentration of [S^{–}] is equal to the [F_{Ox}] formed. And, the equilibrium [S] is equal to the remaining [S] in the solution.

Hence, Equilibrium [S^{–}] = [F_{Ox}] = 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 25.0^{0}C or 298.15 K (standard temperature).

When a half-reaction is reversed, the numeric sign of the resultant reaction is also reversed. Hence, of E^{0’} 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 e_{Co} = 36400 M^{-1} cm^{-1} and e_{Ni} = 5520 M^{-1} cm^{-1} at 510 nm, and e_{Co }= 1240 M^{-1} cm^{-1} and e_{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.

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**

At 656 nm: Total abs = Abs of Co + Abs of Ni

Or, 0.228 = (1240 M^{-1} cm^{-1} x X M x 1.0 cm) + (17500 M^{-1} cm^{-1} x Y M x 1.0 cm)

Hence, **1240X + 17500Y = 0.228 – Equation E12.2**

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

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’s 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’s 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 C_{1}V_{1} (solution B) = C_{2}V_{2} (solution A)

[Co] in solution A, C_{2} = (C_{1}V_{1}) / V_{2} = (7.639 x 10^{-6} M x 100 mL) / 50 mL = 1.5278 x 10^{-5} M

[Ni] in solution A, C_{2} = (C_{1}V_{1}) / V_{2} = (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

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

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

= [(Mol x atomic mass) x 10^{6} mg g^{-1}] / Mass of sample in grams

= [(1.5278 x 10^{-6} mol x 58.933 g mol^{-1}) x 10^{6} mg g^{-1}] / 0.635 g

= [(9.0038 x 10^{-5} g) x 10^{6} mg g^{-1}] / 0.635 g

= 90.038 mg / 0.635 g = 141.79 mg g^{-1} = **141.79 ppm**

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

= [(Mol x atomic mass) x 10^{6} mg g^{-1}] / Mass of sample in grams

= [(2.498 x 10^{-6} mol x 58.69 g mol^{-1}) x 10^{6} mg g^{-1}] / 0.635 g

= [(9.0038 x 10^{-5} g) x 10^{6} mg g^{-1}] / 0.635 g

= 146.61 mg / 0.635 g = 141.79 mg g^{-1} = **230.88 ppm**

**Example 14:** Molar absorptivity data for the cobalt and nickel complexes with 2,3-quinoxalinedithiol are e_{Co} = 36400 M^{-1} cm^{-1} and e_{Ni} = 5520 M^{-1} cm^{-1} at 510 nm, and e_{Co }= 1240 M^{-1} cm^{-1} and e_{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.

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, 0.446 = (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 = 0.446 – Equation E13.1**

At 656 nm: Total abs = Abs of Co + Abs of Ni

Or, 0.326 = (1240 M^{-1} cm^{-1} x X M x 1.0 cm) + (17500 M^{-1} cm^{-1} x Y M x 1.0 cm)

Hence, **1240X + 17500Y = 0.326 – Equation E13.2**

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

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’s 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’s 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 C_{1}V_{1} (solution B) = C_{2}V_{2} (solution A)

[Co] in solution A, C_{2} = (C_{1}V_{1}) / V_{2} = (9.530 x 10^{-6} M x 50 mL) / 25 mL = 1.906 x 10^{-5} M

[Ni] in solution A, C_{2} = (C_{1}V_{1}) / V_{2} = (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 mg / Mass of sample in grams

= [(Mol x atomic mass) x 10^{6} mg g^{-1}] / Mass of sample in grams

= [(9.530 x 10^{-6} mol x 58.933 g mol^{-1}) x 10^{6} mg g^{-1}] / 0.435 g

= [(5.616 x 10^{-5} g) x 10^{6} mg g^{-1}] / 0.435 g

= 56.16 mg / 0.435 g = 129.11 mg g^{-1} = **129.11 ppm**

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

= [(Moles x atomic mass) x 10^{6}] mg / Mass of sample in grams

= [(1.795 x 10^{-6} mol x 58.69 g mol^{-1}) x 10^{6} mg g^{-1}] / 0.435 g

= [(1.054 x 10^{-4} g) x 10^{6} mg g^{-1}] / 0.435 g

= 105.37 mg / 0.435 g = 305.41 mg g^{-1} = **305.44 ppm**

**2. (External) Standard Addition Methods**

**2.A. Single Aliquot **__Standard Addition__

__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 Cu^{2+} gave an absorbance of 0.262 in an atomic absorption analysis. Then 1.0 mL of the solution containing 100.0 ppm Cu^{2+} 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 [Cu^{2+}] in the unknown.

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

**Step 1: For the original unknown solution.**

Using Beer-Lambert’s Law A = e C L

A_{1} = e C_{1} L

Where, A_{1} = Absorbance of the original unknown solution = 0.262

C_{X} = [Cu^{2+}] in the original unknown solution

In the above expression, A_{1} is a known quantity, C_{1} is the unknown to be determined, e and L are unknown but remain constant for a specified analyte under the same set of experimental conditions. Since e 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 e and L.

That is, **e**** L = A _{1} / C_{x}**

**– Equation 1E.1**

**Step 2:** **For the spiked and finally diluted solution**

To derive the expression in terms of eL, we first need to determine the total [Cu^{2+}] in the spiked aliquot. There are two sources of [Cu^{2+}] in the spiked aliquot- one [Cu^{2+}] from the original unknown solution, and the other [Cu^{2+}] from the standard solution. We need to calculate [Cu^{2+}] in the final diluted solution from both the sources separately, and then add them to get the total [Cu^{2+}] in the spiked solution.

In the finally diluted spiked aliquot,

using C_{1}V_{1} (original solution) = C_{2}V_{2} (spiked solution)

[Cu^{2+}] from standard solution = C_{1}V_{1} / V_{2} = (100.0 ppm x 1.0 mL) / 100.0 mL = 1.0 ppm

[Cu^{2+}] from unknown solution = (C_{x} x 95.0 mL) / 100.0 mL = 0.95 C_{x}

And,

Total [Cu^{2+}] in the spiked solution, [Cu^{2+}]_{T} = [Cu^{2+}] from (standard +unknown) solution

Or, [Cu^{2+}]_{T} = 1.0 ppm + 0.95 C_{x}

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

A_{2} = e C_{T} L

Where, A_{2} = Absorbance of the spiked and diluted solution = 0.500

C_{T} = [Cu^{2+}] in the spiked and diluted solution = 1.0 ppm + 0.95 C_{x}

Or, **e**** L = A _{2} / C_{T}**

**– Equation 1E.2**

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

Now, comparing equations 1 and 2-

**e****L =** **A _{1} / C_{1} = A_{2} / C_{T}^{ }**

**Or, A _{1} / C_{1} = A_{2} / C_{T}^{ } – 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 / C_{x} = 0.500 / (1.0 ppm + 0.95 C_{x})

Or, 0.262 x (1.0 ppm + 0.95 C_{x}) = 0.500 C_{x}

Or, 0.262 ppm + 0.2489 C_{x} = 0.500 C_{x}

Or, 0.500 C_{x} – 0.2489 C_{x} = 0.262 ppm

Or, 0.2511 C_{x} = 0.262 ppm

So, C_{x} = 0.262 ppm / 0.2511 = 1.043 ppm

Hence, [Cu^{2+}] in the original unknown solution, C_{x} = 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 [Cu^{2+}] in the original unknown solution be C_{x}.

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 C_{1}V_{1} (original solution) = C_{2}V_{2} (spiked solution)

[Cu^{2+}] from standard solution = C_{1}V_{1} / V_{2} = (100.0 ppm x 1.0 mL) / 100.0 mL = 1.0 ppm

[Cu^{2+}] from unknown solution = (C_{x} x 95.0 mL) / 100.0 mL = 0.95 C_{x}

And,

Total [Cu^{2+}] in the spiked solution, [Cu^{2+}]_{T} = [Cu^{2+}] from (standard +unknown) solution

Or, [Cu^{2+}]_{T} = 1.0 ppm + 0.95 C_{x}

Now, increase in [Cu^{2+}] due to spiking = [Cu^{2+}]_{T} – [Cu^{2+}] of unknown soln.

= (1.0 ppm + 0.95 C_{x}) – C_{x} = 1.0 ppm – 0.05 C_{x}

**Step 2: Using Equation 1E.5-**

C_{x} / 0.262 = (1.0 ppm – 0.05 C_{x}) / 0.238

Or, 0.238 C_{x} = 0.262 x (1.0 ppm – 0.05 C_{x}) = 0.262 ppm – 0.0131 C_{x}

Or, 0.238 C_{x} + 0.0131 C_{x} = 0.262 ppm

Or, C_{x} = 0.262 ppm / 0.2511 = 1.043 ppm

Hence, [Cu^{2+}] in the original unknown solution, C_{x} = 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 C_{1}V_{1} (original soln.) = C_{2}V_{2} (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.

Using C_{1}V_{1} (original soln.) = C_{2}V_{2} (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,

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

Now, Using Equation 1E.5-

([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 mg 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 mg 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.

Using C_{1}V_{1} (original soln.) = C_{2}V_{2} (spiked soln.) for the spiked solution-

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

[Cu] from Std. soln. = (1000.0 mg mL^{-1} x 250.0 mL) / 400.0 mL = 625.0 mg mL^{-1}

So, Total [Cu] in the spiked soln., [Cu]_{T} = 0.375 X + 625.0 mg 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 mg mL^{-1}) – X

= -0.625X + 625.0 mg 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 mg mL^{-1}) / 0.576

Or, 0.576 X = 0.451 x (-0.625X + 625.0 mg mL^{-1}) = -0.2819 X + 281.875 mg mL^{-1}

Or, 0.576 X + 0.2891 X = 281.875 mg mL^{-1}

So, X = 281.875 mg mL^{-1} / 0.8579 = 328.57 mg mL^{-1}

Hence, Initial or Uncorrected [Cu] in the original sample water = 328.57 mg mL^{-1}

**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 mg mL^{-1}) / 0.576

Or, 0.576 X = 0.372 x (-0.625X + 625.0 mg mL^{-1}) = -0.2325 X + 232.500 mg mL^{-1}

Or, 0.576 X + 0.2325 X = 232.500 mg mL^{-1}

So, X = 232.500 mg mL^{-1} / 0.8085 = 287.57 mg mL^{-1}

Hence, the Corrected [Cu] in the original sample water = 287.57 mg mL^{-1}

**Now, Account Error in [Cu]:**

Error in [Cu] = Corrected [Cu] – Initial [Cu] = (287.57 – 328.57) mg mL^{-1} = -41.00 mg mL^{-1}

Now,

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

= (-41.00 mg mL^{-1} / 287.57 mg 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 a solution 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 spiked standard measured as 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 C_{1}V_{1} (diluted sample) = C_{2}V_{2} (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

Now, Using Equation 1E.5-

([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 absorbds 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.

Using C_{1}V_{1} (original soln.) = C_{2}V_{2} (spiked soln.) for the spiked solution-

[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

Now, Using Equation 1E.4-

(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 Cd^{2+} content using graphite furnace atomic absorption spectrometry (GFAAS). The chemist dried the clams at 95^{0}C 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 (mg L^{-1}) Cd^{2+} standard were added to each flask, which were then brought to volume with 0.1 M HCl. The Cd^{2+} content of the solution was then analyzed using GFAAS, using the absorbance data given in the table.

### Determine the amount of Cd^{2+} per gram of dry clam. Express your final result as milligrams of Cd^{2+} per gram dry clam.

Ans. Method 1: By determining __[Cd ^{2+}]__ in the un-spiked aliquot: Create a data table as shown below to calculate [Cd

^{2+}] from various standard additions. Plot absorbance vs [Cd

^{2+}] 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, Cd^{2+} content in dry clam (dry mass) = 0.137 mg Cd^{2+} per gram dry clam.

**Method 2:** **By the mass of Cd^{2+} in the un-spiked aliquot:** Create a data table as shown below to calculate the mass (mg) of Cd

^{2+}from various standard additions. Plot absorbance vs mass (mg) of Cd

^{2+}added from standard solutions of various aliquots, label the axes appropriately. Generate the trendline equation or linear regression equation for the standard graph.

Mass (mg) of Cd^{2+} added from standard solution = C1 x (V1 / 1000)

Where, C1 = [Cd^{2+}] of the standard solution

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

V1 / 1000 = mL / 1000 = liters

ppb x L = μg L^{-1} x L = μg

Once the trendline equation is obtained, the required calculations can be done in following two steps-

Hence, Cd^{2+} content in dry clam (dry mass) = 0.137 mg Cd^{2+} 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*)à 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-

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.

**Example 22:** Students performed experment 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-

### Calculate sodium ion concentration and its uncertainty in the original serum sample. Calculate 95% confidence interval.

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.

**Example 23:** Tooth enamel consists mainly of the mineral calcium hydroxyapatite, Ca_{10}(PO_{4})_{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.

### 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 (mg 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.

**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.

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