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ANALYTICAL CHEMISTRY LAB REPORT CHEM604 ANNIE SUDEEPTHI URE 18045586

Name: Annie Sudeepthi Ure ID: 18045586 Lab: Tuesday 12- 2pm Week 1: 25. Feb. 2020 DETERMINATION OF PH AND TITRATABLE ACIDITY OF WINE USING A PH ELECTRODE AIM: To determine the titratable acidity of grape juice and concentration (g/L) of acid content using a pH electrode. REFERENCE: Analytical chemistry laboratory manual page(s) 7&8. EXPERIMENTAL VALUES: Volume in (ml)

Burette readings Initial burette reading Final burette reading Titre value CALCULATIONS: Titratable acidity = 0.75 x Titre value(ml)

0.31 ml 10.9 ml 10.59 ml

= 0.75 x 10.59ml =7.94 g/L DISCUSSION: This acid – base titration is a method of analysis used to find the acidity which can be measured as pH or titratable acidity. In this, the concentration of an unknown acid or base is determined by titrating it with a given volume of acid or base of known concentration which is called a titrant. Titration is done till the end point is reached. Once the end point is reached, the volume of titrant added, is used to determine the concentration of unknown acid or base. In grape juice and wine which normally has pH range between 2.8 to 4.2, Tartaric acid and Malic acid are the two main acids present and rest of them are produced by yeast. Tartaric acid and Malic acid in grape juice are usually present at a range of 4 to 9 g/L concentration. CONCLUSION: Till the end point where pH reaches 8.2, NaOH, a strong base is added, and the obtained titrant value is 10.59. The concentration of Tartaric acid present in the grape juice is calculated using the titrant value, which is approximately 7.94g/L. QUESTION(S): 1.

Where does the numerical factor of 75 comes from in the titratable acidity calculation? The numerical factor 75 in the Titratable acidity calculation is constant. 75 is a value based on various conversion factors (i.e. molecular weight, reaction ratio, etc.). The units of the constant 75 is worked out and it is found to be g mol -1 , which corresponds to the unit of the Molar mass. However, the molar mass of the tartaric acid is 150.09 g mol -1 . Therefore, it is concluded that the 75 is the equivalent weight of the tartaric acid which will supply 1 mole of the H+ions.

2.

The pH electrode used is a “combination pH electrode”. What does this mean? Combination pH electrode means it is a combination of two features that are of a reference electrode, and glass membrane electrode in a single cylinder (Bialkowski, 2004). The combination pH electrode combines both the pH and reference electrodes along with a temperature sensing probe in a single unit. Hence the it measures the value of pH by considering the activity of hydronium ions in the solution, in the presence of the reference electrode which already has a stable and known electrode potential.

Name: Annie Sudeepthi Ure ID: 18045586 Lab: Tuesday 12- 2pm Week 2: 03. Mar. 2020 DETERMINATION OF SODIUM BY FLAME EMISSION AIM: To determine the concentration of sodium in tap water and mineral water by comparing the flame emission intensities of the water samples with a series of standard sodium solutions. REFERENCE: Analytical chemistry laboratory manual page(s) 10&11. EXPERIMENTAL VALUES:

Tube

Results Na 589.582 mg L -1

Samples

1 2 3 4 5 6 7

Blank Standard 1 (5 ppm) Standard 2 (10 ppm) Standard 3 (15 ppm) Standard 4 (20 ppm) Tap Water Mineral Water (Sample)

0 5 ppm 10 ppm 15 ppm 20 ppm unknown unknown

CALCULATIONS: NaCl / Na = 58.43 x 1 22.98 = 2.54 = 0.254 g/100 ml is the mass of NaCl which gives 1 gram of Sodium. Calculating the volume of 100 ppm Na needed to add in each 100 mL flask using C 1 V 1 =C 2 V 2 V 1=C2V 2/ C1 5 ppm standard = (5 ppm x 100 mL) / 100 ppm = 5 mL 10 ppm standard = (10 ppm x 100 mL) / 100 ppm = 10 ml 15 ppm standard = (15 ppm x 100 mL) / 100 ppm = 15 ml 20 ppm standard = (20 ppm x 100 mL) / 100 ppm = 20 ml

Concentration vs Intensity 14000 y = 638.36x + 201.28

20, 13000

12000

Intensity

10000

15, 9670

8000 10, 6503

6000

7.52, 5000 4000

5, 3753

2000 0

0, 0 0

5

10

15

Concentration (ppm)

20

25

➢ From the slope equation y = 638.34x + 201.8 concentration of sodium in mineral water sample with 5000 intensity is; 5000 = 638.34 x + 201.8 x = 5000 – 201.8 / 638.34 = 7.52ppm ➢ From the slope equation y = 638.34x + 201.8 concentration of sodium in tap water sample with 6300 intensity is; 6300 = 638.34 x + 201.8 x = 6300 – 201.8 / 638.34 = 9.55 ppm

DISCUSSION: The ICP-AES (Inductively coupled plasma atomic emission spectroscopy) was used to determine the emission intensities of the standard Na solutions, tap water and mineral water. The wavelength is set to 330 nm. An emission intensity of about 6300 was obtained from the graph for tap water whilst mineral water had an emission intensity value of about 5000. Mineral goes through a treatment process to get rids of unwanted contaminants and thus this could be the cause as to why the sodium concentration is lower than that of tap water. CONCLUSION: The concentration of sodium in tap water was 9.55ppm whilst mineral water’s sodium concentration was only 7.52ppm. This can be due to the fact that tap water contains a lot of minerals such as fluoride, chlorine and sodium where the concentration of sodium found is about 10 ppm. QUESTION(S):

Why is distilled water used to dilute the solutions? Distilled water is appropriate for dilution because its pH is slightly less than 7, suitable since dilution does not affect the pH of the solutions. Additionally, distilled water is used instead of pure water owing the absence of ions in the former. Pure water contains ions like Na + and Cl -. This will introduce a greater number of ions in the solutions, hence the actual concentration of ions in the solution is altered, which will affect the value of the emission intensity.

Wh at w weigh eigh eightt of so sodiu diu dium m ca carrb on onaa te wo wouu ld be nnee ee eeded ded to pr prepa epa eparr e a 1000 pp ppm m solu solution tion of sodiu sodium m in a 1 -liter fa falsk? lsk? Na₂CO₃ = 4.6102 g Na

Use C 1 V 1 =C 2 V 2 equation to calculate the volume of a 50 ppm Mg solution required to prepare the following concentrations of Mg in 50 ml volumetric flasks: a. 5 ppm = (5 ppm x 50 mL) / 50 ppm = 5 mL b. 10 ppm = (10 ppm x 50 mL) / 50 ppm = 10 mL c. 15 ppm = (15 ppm x 50 mL) / 50 ppm = 15 mL d. 20 ppm = (20 ppm x 50 mL) / 50 ppm = 20 mL

Name: Annie Sudeepthi Ure ID: 18045586 Lab: Tuesday 12- 2pm Week 3: 10. Mar. 2020

determination of phospho phosphor r ic a ac c id in cola bev bever er erag ag ages es colorime colorimetr tr trical ical ically. ly. AIM: To measure the amount of Phosphorus present in the cola beverage and obtain the molarity of the phosphoric acid in the beverage. REFERENCE: Analytical chemistry lab manual page(s) 12. EXPERIMENTAL VALUES: SAMPLE Blank Sample (Solution B) 2.5 5.0 7.5 10

ABSORBANCE 0.000 0.216 0.141 0.267 0.406 0.554

12

Absorbance vs Concentration y = 18.197x + 0.0211

Concentration (ppm)

10

0.554, 10

8 0.406, 7.5 6 0.267, 5

4

0.216, 3.95 0.141, 2.5

2

0

0, 0 0

0.1

0.2

0.3

0.4

Absorbance

CALCULATIONS: Concentration of solution B from slope equation y = 18.196x + 0.0215 is; Concentration = 18.196 x 0.216 + 0.0215 = 3.95 ppm = 0.395 mg L -1 Phosphorus in solution B = 0.395 mg P Phosphorus in solution A = 0.395mg x 250 25 = 3.95 mg P Phosphorus in 1 L of cola = 3.95 x 1000 20 = 197.5 mg L-1

0.5

0.6

Moles of P = Moles of H3PO4 (197.5 x 10 -3 g L -1) / 31 g mol -1 = 6.37 x 10 -3 mol L -1 H3PO4

DISCUSSION: The concentration of the solution B was derived by using the known x and y values along with the gathered absorbance value of the solution B. Through this, Solution B obtained 0.395 from the calibration curve. Therefore, solution A contains 0.395mg x 250 / 25 = 3.95 mg P. CONCLUSION: From the calculation the amount of phosphorous present in the cola beverage is 197.5 mg L-1 which has phosphoric acid with a calculated molarity of 6.37 x 10 -3 mol L -1 H3PO4.

Name: Annie Sudeepthi Ure ID: 18045586 Lab: Tuesday 12- 2pm Week 4: 17. Mar. 2020

Polarimetry – De Deter ter terminati minati mination on of Sucrose in a Soft Dr Drii nk. AIM: To determine the concentration of the Sucrose in sprite (soft drink) using Biots Law from Polarimeter. REFERENCE: Analytical chemistry lab manual page(s) 15&16. EXPERIMENTAL VALUES: Sample Observed Optical Rotation (α) Blank (Deionized Water) 0.00 5 g in 100 mL 3.36 10 g in 100 mL 6.61 25 g in 100 mL 16.8 50 g in 100 mL 33.6 Sprite Sample 3.09 Standard Solution measured in a 2 dm polarimeter cell 25 g in 100 mL

33.6

Rotation vs concentration Concentration of Sucrose( g/100 ml )

60 50

yyy===1.4869x 1.4869x 1.4869x+++0.0469 0.0469 0.0469

33.6, 50

25

35

40 30 16.8, 25 20 10

6.61, 10 3.09, 4.64

0

0, 0 0

5

10

15

20

30

Observed rotation(°)

CALCULATIONS:

Specific rotation of sucrose: Slope of the equation = 1.4869 Specific rotation of Sucrose = Slope x 100 = 1.4869 x 100 = 148.69

Concentration of sucrose in the lemonade from graph: From the slope equation y = 1.4869x + 0.0474

40

Concentration = 1.4869 x 3.09 + 0.0474 = 4.64 g / 100 ml = 0.0464 g / ml

Concentration of sucrose in the lemonade from Biots law: Specific rotation = Observed rotation x 100 Path length x concentration concentration = Observed rotation x 100 path length x specific rotation = 2.08 g/100 ml The Biots Law was followed, since the plot of observed rotation against concentration is a straight line passing through origin (no y intercept), with the slope giving the value of the specific rotation. It shows that the rotation is directly proportional to concentration.

DISCUSSION: Biots law is very similar to Beers Law, the only difference being that the former involves specific rotation and observed rotation, while Beers law involves absorbance and molar absorptivity. Another difference could be that the path length used in Biots law is much greater that is measured in basic unit of path-length is the decimeter (1 dm = 10 cm = 100mm) whereas in Beers law which measures path-length in centimeter (cm).

CONCLUSION: From the slope equation the concentration of Sucrose in sprite is 21.6 g / ml. QUESTION(S): The wavelength of 589 nm is used most often in polarimetry for historical reasons, but other wavelengths

may also be used depending on the capabilities of the instrument. Why was 589 nm light used historically and what advantage might be gained using a different wavelength e.g. 436 nm? When the polarimeter is used the extent of optical rotation is measured by matching the intensity of the split’s fields, hence the D-line of the sodium lamp at the visible wavelength of 589 nm is most commonly used. However, a much lower wavelength gives a much better sensitivity. The observed rotation at436 nm is about twice the observed rotation at 589nm.

Name: Annie Sudeepthi Ure ID: 18045586 Lab: Tuesday 12- 2pm Week 5: 24. Mar. 2020

Quantitative I nf nfr r ared A An n alysis of Ethano Ethanol l AIM: Using the Nicolet infrared spectrophotometer to absorbance of the ethanol standards and unknown sample at the peak maximum and plot a graph of the absorbance against their corresponding concentrations to determine the value of the concentration of the unknown sample. REFERENCE: Analytical chemistry lab manual page(s) 19 &20.

EXPERIMENTAL VALUES: Sample Blank (Deionized Water) 2.5% 5.0% 10% 15% Unknown Ethanol

Absorbance 0.000 0.410 0.808 1.623 2.401 0.813

Ethanol % vs Concentration 3 2.5

y = 0.1603x + 0.0068

15, 2.401

Absorbance

2 10, 1.623

1.5 1 5, 0.808 5.03, 0.813

0.5

2.5, 0.41

0

0, 0 0

2

4

6

8

10

12

14

16

Ethanol %

CALCULATIONS: From the slope equation % of Ethanol in the unknown sample is 0.813 = 0.1603x + 0.0067 x = 0.813 – 0.0067 / 0.1603 = 5.03%

DISCUSSION: From the plot obtained, it can be concluded that Beer’s Law was indeed obeyed, since a straight line passing through origin was obtained. The instrument used for this experiment is a FTIR spectrometer, which involves the use of Infra- red radiation. This instrument is very accurate and sensitive. It has a high resolution, with great speed of data acquisition (usually 1 second or lesser). The FTIR spectrometer makes use of both transmittance and absorbance plots to display data, however since this experiment involves quantitative measurement an absorbance plot is used.

CONCLUSION: The concentration of Ethanol present in the unknown sample is 5.03%. QUESTION(S): Suggest a reason why most infrared analyses of organic compounds NaCl cells are used,

yet in this

experiment CaF2 was used as the cell material. Plastic absorbs IR whereas CaF2 cell does not. Organic compounds are not soluble in water hence, they do not dissolve the NaCl cells, as compared to the sucrose solutions prepared in this experiment which would dissolve the NaCl cells. Hence CaF2 cells are used which are not water soluble.

The ethanol absorption band used in the present experiment occurs at ~ 1050 cm -1 . In what region of the infrared is this absorption peak located (i.e. the near, mid of far infrared)? Mid IR, because wavenumber between the range of 400 and 4000 cm-1 and the unit is in cm -.

PART 1 Introduction Questions: 1Ans: Herschel discovered the existence of infrared light by passing sunlight through a glass prism in an experiment similar to the one we describe here. As sunlight passed through the prism, it was dispersed into a rainbow of colors called a spectrum. 2Ans: When a sample of water passes through an Infrared beam, it undergoes different types of transition symmetrical and asymmetrical stretch and scissoring bending movement which occurs due to the change in the dipole moment of the molecule. 3Ans. Near IR wavelength - 0.78–2.5, wave number - 12,800–4,000 Middle IR wavelength -2.5–50, wave number - 4,000–200 Far IR wavelength -50–1,000, wave number - 200–10. 4Ans: IR spectroscopy uses: 1. Dispersive spectrophotometers with a grating monochromator. 2. Fourier Transform Spectrometers employing an interferometer. 5Ans: A polychromic source is used generally in an IR spectrometer. 6Ans: Dispersive IR spectrometer uses monochromator for grating. 7Ans: A detector detects the heat in a spectrometer. 8Ans: An interferometer which acts as a beam splitter is used in the FT spectrometer.

9Ans: Fourier transform is a mathematical process which produces an IR spectrum using an interferogram.

10Ans: Compared with dispersive instruments: FTIR Offers improved signal-to-noise ratio and uses energy from the entire spectrum rather than analyzing a sequence of small wavebands available from a monochromator Precisely reproduces of wavenumber position from one spectrum to the next which allows the averaging of signals from multiple scans Has better frequency, accuracy, and speed.

Name: Annie Sudeepthi Ure ID: 18045586 Lab: Tuesday 12- 2pm Week 6: 31. Mar. 2020

Thin Layer c chr hr hromatog omatog omatography raphy of A An n alge algesi si sics cs AIM: To identify the compound present in an unknown tablet using a thin layer chromatography. REFERENCE: Analytical chemistry lab manual page(s) 20 & 21. EXPERIMENTAL VALUES: Experiment

AM

AS

2.9

4.5 5.2 5.9

CF

1 2.0

2 2.2

3

2.4

3.5 4.0 4.4

1.9

4.6

1.5

CALCULATIONS: Calculating the Rf values for the analgesics. 𝑹𝒇 = 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐬𝐩𝐨𝐭 𝐦𝐨𝐯𝐞𝐝 𝐟𝐫𝐨𝐦 𝐨𝐫𝐢𝐠𝐢𝐧 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐟𝐫𝐨𝐧𝐭 𝐦𝐨𝐯𝐞𝐝 𝐟𝐫𝐨𝐦 𝐨𝐫𝐢𝐠𝐢n For AM(Acetaminophen) = 2.5 / 5.5 = 0.45 For AS(Aspirin) = 4.6 / 5.5 = 0.84 For CF(Caffeine) = 1.8 / 5.5 = 0.33 For the mixture = 3.58 / 5.5 = 0.65 For Unknown sample = 2.5 / 5.5 = 0.46

M 2.0 2.9 4.5 5.1 5.9 1.5 2.1 3.5 3.9 4.4 2.3 4.7 1.6

T

Solvent front

3.1

6.6

2.1 4.9

2.4

4.9

Average Rf values: AM AS CF M T

0.45 0.84 0.33 0.65 0.46

DISCUSSION: The term Analgesics encompasses a class of drugs that are designed to relieve pain without causing the loss of consciousness. Paracetamol, Aspirin, occasionally Caffeine are the analgesics that may be used for pain relief. These act as enzyme inhibitors that catalyze certain biochemical reactions. This technique is used in organic chemistry to determine suitable mobile phase prior to column chromatography. Low polarity compounds often have higher Rf values than higher polarity compounds. The more polar the compounds the stronger it binds to the absorbent (Silica gel).

CONCLUSION: From the compared Rf values and position of the table spots relative to the standard analgesic spots the compound present in the unknown tablet is Acetaminophen. QUESTION(S): No questions....