Lab 6: Calibration Methods PDF

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Lab 6: Calibration Methods
Studying calibration methods, creating calibration curves with standardized solutions....

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Lab 6 Calibration Methods and Introduction to UV-Visible Spectroscopy Prelab Questions 1. a. (5) You will create 1.000L of 100.0 ppm stock solution of Acetylsalicylic acid (ASA, from the pure powder) in a 2 (v/v) % HCl in this lab. 2 (v/v) % HCl means the volume of concentrated HCl and the volume of the diluted HCl is 2:100. Determine the mass of the solute and volume of conc. HCl you are going to weigh out or measure with a graduated cylinder. Show your calculation. HCl: 1.00L × 0.02 = 0.0200L HCl 0.0200L HCl ×

1000mL

Solute: 1 ppm =

1L 1mg L

= 20.00mL HCl 100ppm =

100.0mg L

× 1.00L = 100.0mg Acetylsalicylic Acid

100.0mg ASA = 0.0001g ASA b. (5) Describe exactly how you will make the 1.000 L of 100.0 ppm ASA. When this is weighed out, you must get the mass you have found exactly! Rinse a 1.00L volumetric flask with deionized water and add 100.0mg of acetylsalicylic acid to the flask. Add the deionized water until halfway full then add the 20.00mL of the HCl-. Mix until solutions are dissolved completely. Once solutions have dissolved, add deionized water to the calibration mark. Once all solution is added into the volumetric flask, invert the flask three times to mix promptly. 2. (5) Create dilution tables to create your standards for the external calibration method given a 100.0 ppm solution and a 10.00 mL volumetric flask and a micropipette with a range from 100.0 to 1000. µL. Use C1V1=C2V2. C1V1=C2V2 0 ppm: (100 ppm) (V1) = (0 ppm) (10.00mL) V1 = 0 4 ppm: (100 ppm) (V1) = (4 ppm) (10.00mL) V1 = 0.400mL 6 ppm: (100 ppm) (V1) = (6 ppm) (10.00mL) V1 = 0.600mL 8 ppm: (100 ppm) (V1) = (8 ppm) (10.00mL) 1

V1 = 0.800mL 10 ppm: (100 ppm) (V1) = (10 ppm) (10.00mL) V1 = 1.000mL 15 ppm: (100 ppm) (V1) = (15 ppm) (10.00mL) V1 = 1.500mL 20 ppm: (100 ppm) (V1) = (20 ppm) (10.00mL) V1 = 2.000mL 100 ppm: (100 ppm) (V1) = (100 ppm) (10.00mL) V1 = 10.000mL 3. (5) Create a procedure and dilution for a 325 mg/tablet ASA to be placed in a 1.000 L volumetric flask and then diluted to 10.00 ppm in a 10.00 mL volumetric using a micropipette. 325mg 1L

= 325 ppm

(325 ppm) (V1) = (10 ppm) (10.00mL) V1 = 0.308mL 0.308mL x

1000mL 1µL

= 308µL

Break down the aspirin tablet and prepare the ASA and HCl solution in a 1.000L flask. We will substitute the aspirin tablet for the stock ASA solution. Add the 308µL of solution to a 10.00mL volumetric flask using a micropipette. Add deionized water to the flask to the calibration line and mix well by inverting the flask three times.

Table 1. Calibration Data.

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Concentration (ppm) Blank 4 ppm 6 ppm 8 ppm 10 ppm 15 ppm 20 ppm 100 ppm Unknown Unknown + 10 ppm Spike

Concentration (ppm) Blank 4 ppm 6 ppm 8 ppm 10 ppm 15 ppm 20 ppm 100 ppm Unknown Unknown + 10 ppm Spike

Group 1 Absorbance 0 0.1027 0.1742 0.246 0.3133 0.5006 0.6992 2.6090 0.3313 0.671

Group 5 Absorbance 0.0002 0.1288 0.2014 0.2638 0.3306 0.5039 0.6581 2.6136 0.5705 0.9258

Concentration (ppm) Blank 4 ppm 6 ppm 8 ppm 10 ppm 15 ppm 20 ppm 100 ppm

Group 2 Absorbance 0.0001 0.1046 0.1649 0.2231 0.2902 0.4495 0.6007 2.5844 0.3096 0.6941

Group 3 Absorbance -0.0003 0.0844 0.1544 0.2247 0.2737 0.4353 0.6165 2.6146 0.3361 0.6051

Group 4 Absorbance -0.0001 0.1417 0.2473 0.2846 0.3453 0.5043 0.6811 2.6891 0.3363 0.6864

Group 5 Absorbance 0.0002 0.1288 0.2014 0.2638 0.3306 0.5039 0.6581 2.6136 0.5705 0.9258

Group 6 Absorbance

Group 7 Absorbance

Group 8 Absorbance

Group 9 Absorbanc

0.0001 0.0527 0.1064 0.2229 0.2371 0.4943 0.5298 2.4095 0.4509 0.6661

0 0.1882 0.2159 0.297 0.3585 0.547 0.7216 2.6407 0.4763 0.8015

0 0.1198 0.1835 0.2626 0.316 0.4771 0.6034 2.5416 0.343 0.6639

0.0001 0.1997 0.2618 0.3266 0.3936 0.6876 0.7878 2.697 0.5273 0.8622

Group 10 Absorbance 0.0001 0.1419 0.2778 0.3466 0.3506 0.4894 0.6904 2.6923

Average 0.00002 0.12645 0.19876 0.26979 0.32089 0.5089 0.65886 2.6092 3

Standard Deviation 0.000139841 0.044657194 0.053081681 0.043655404 0.045393501 0.06996056 0.073475895 0.086348233

Unknown Unknown + 10 ppm Spike

0.546 0.9515

0.4227 0.75276

0.102262875 0.122715572

Graph of External Standard Calibration of ASA solution to 20 ppm:

External Standard Calibration (to 20ppm) 1 0.9 0.8 y = 0.1023x - 0.1114 R² = 0.9631

0.7

Absorbance

0.6 0.5 0.4 0.3 0.2 0.1 0 Blank -0.1

4 ppm

6 ppm

8 ppm

10 ppm

Concentration of ASA solution (ppm)

Graph of External Standard Calibration of ASA solution to 100 ppm:

4

15 ppm

20 ppm

External Standard Calibration (to 100ppm) 2.8 2.4

y = 0.2608x - 0.587 R² = 0.5737

2

Absorbance

1.6 1.2 0.8 0.4 0 Blank

4 ppm

6 ppm

8 ppm

10 ppm

15 ppm

20 ppm

100 ppm

-0.4 -0.8

Concentration of ASA solution (ppm)

Concentration 3 4 6 7

Average Absorbance 1 2 3 4

Standard Deviation 0.4 0.2 0.3 0.4

1. Plot the data again and provide the graph, this time showing all of the raw data instead of averaging the replicates into one point. To do this you will need all of the data in single (x,y) data matrix. Please ask your instructor if you need help. Using this method, you will not have error bars. Does this change the R 2 value? How about the linear equation? How does averaging change the results on the calibration?

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Graph of Raw Data Concentration of ASA Solution to 20 ppm.

Raw Data Concentration of ASA Solution to 20 ppm 0.9 0.8

y = 0.0331x + 0.0006 R² = 0.9495

0.7

Absorbance

0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1

0

2

4

6

8

10

12

14

16

18

20

22

Concentration of ASA solution (ppm)

Graph of Raw Data Concentration of ASA Solution to 100 ppm.

Raw Data Concentration of ASA Solution to 100 ppm 3

y = 0.0256x + 0.0684 R² = 0.9922

2.6

Absorbance

2.2 1.8 1.4 1 0.6 0.2 -0.2 0

20

40

60

80

100

120

Concentration of ASA solution (ppm)

Does this change the R2 value? Yes, it does change the R2 value, but not significantly. The raw data graph has an R2 value of 0.9495 and the average data graph has an R 2 value of 0.9631. How about the linear equation?

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Yes, it greatly changed the linear equation as the average data linear equation was y = 0.2608x - 0.587 and the raw data linear equation was y = 0.0331x + 0.0006. I decided to use the slope of 0.1023 from the external standard calibration to 20 ppm graph because this graph had the best linear regression line. The external standard calibration to 100 ppm graph does not fit the linear regression line, as the 100-ppm concentration absorbance is higher than expected. How does averaging change the results on the calibration? Averaging the results change the results on calibration by increasing the R 2 value. The R2 value was increased in the average results of the data compared to the R2 values from the raw data results. Averaging the results makes it possible for the R 2 value to be equal to or greater than 0.96 2. Fully describe and calculate the figures of merit for the calibration. Calculate the LoD, LoQ, and estimate the linear range. Concentration (ppm) LoD LoQ 0 ppm (blank) 0.0041 0.0137 4 ppm 1.3095 4.3653 6 ppm 1.5566 5.1888 8 ppm 1.2802 4.2732 10 ppm 1.3312 4.4373 15 ppm 2.0516 6.8388 20 ppm 2.1547 7.1824 100 ppm 2.5322 8.4407 Limit of Detection: Minimum detectable concentration =

3s calibration curve slope

Standard deviation of 4 ppm concentrations = 0.044657194 Slope of calibration curve = 0.1023 3 ×0.044657194 0.1023

= 1.3095ppm

The limit of detection is the minimum detectable concentration. This means that smallest change in concentration that can be detected is 1.3095 ppm. 10 s

Lower limit of Quantitation = calibration curve slope

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10 ×0.044657194 0.1023

= 4.3653 ppm

This means the smallest change in concentration that we are able to determine with accuracy is 4.3653 ppm. The calibration curve data fits well with the linear equation provided by the graph. Our data recorded and plotted is appropriate for a calibration curve. There are not many outliers from the linear trendline meaning our values determined from the graph are accurate and precise. Estimated linear range: From the results of the graph I would determine the linear range to go from 0 ppm to 20 ppm because its calibration curve is linear. I wouldn’t say the 100 ppm is in range when looking at the results of the external standard calibration graph since it is much higher than expected so it is not considered inside the linear range. 3. For each group, calculate the concentration of the unknown by the method of external standards using your graph from question 2 and report as average  standard deviation. y = 0.1023m ± 0.102 Group Name James Lester, Maria Vu Patricia G., Mohamed O., Monika P. James Q., Rachel F., Sarah Sydney S, Megan V, Michaela C Azhanae w., Emily, Basirat Jim, Juelian,Matt Hailey, Linda, Garrett Rigo,Sheila, Stori Jose, Ryan Thoren, Peter, Jaeda

External Standards 3.24 3.02

Standard deviation ± ±0.102 ±0.102

3.29

±0.102

3.29

±0.102

5.57 4.41 4.66 3.35 5.15 5.34

±0.102 ±0.102 ±0.102 ±0.102 ±0.102 ±0.102

Sample calculation Group 1: y = 0.1023m ± 0.102

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x= x=

y + 0.102 0.1023 0.3313 0.1023

= 3.24

x = 3.24 ± 0.102 ppm 4. There’re about 7~11 groups in your class. For each group, calculate the concentration of the unknown by the method of standard addition and report as average  standard deviation. Unknown concentration: 10.00ppm ± 0.1227 Group Number Standard Addition 1 4.94 2 4.46 3 5.55 4 4.89 5 6.16 6 6.77 7 5.94 8 5.17 9 6.11 10 5.73 Sample calculation Group 1: [ASA unknown] 10.00mL+[ASA unkown]

=

0.3313 0.671

[ASA unknown] = 4.94

5. Using statistical hypothesis testing (type 3: same sample, different method. Excel calls this t-test: Paired Two Samples for Mean) decide if the methods produce statistically different results or not. Is external standardization different from the standard addition method? You have to put the data in excel following Table 3 below. Add as many rows as needed to include all class data. Remember, the data must be paired, so if one point is discarded as an outlier you must discard both methods for that group. In other words, if a point loses its pair it can’t be included in the t-test.

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Group # 1 2 3 4 5 6 7 8 9 10

Conc. of Unknown by External Standards 3.24 3.02 3.29 3.29 5.57 4.41 4.66 3.35 5.15 5.34

Conc. of Unknown by Standard Addition 4.94 4.46 5.55 4.89 6.16 6.77 5.94 5.17 6.11 5.73

t-Test: Paired Two Sample for Means Variable 1 4.132 0.997906667 10 0.758513808 0 9 -6.973731322 3.25554E-05 1.833112933 6.51108E-05 2.262157163

Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T...