CHEMISTRY CHM256 EXPERIMENT 3 PDF

Title CHEMISTRY CHM256 EXPERIMENT 3
Course Applied Sciences
Institution Universiti Teknologi MARA
Pages 6
File Size 231 KB
File Type PDF
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Summary

LAB REPORTCHM256 BASIC ANALYTICAL CHEMISTRYExperiment No. Experiment 3TitleACID – BASED TITRATION II (BACKTITRATION)Determination the percent (w/w) of the activeingredient in aspirin tabletName Kaiyisah binti KamaruzzamanMuhamad Amar bin ZakariaStudent ID 20192005362019289758Course Applied Science (...


Description

LAB REPORT CHM256 BASIC ANALYTICAL CHEMISTRY Experiment No.

Experiment 3 ACID – BASED TITRATION II (BACK TITRATION) Determination the percent (w/w) of the active ingredient in aspirin tablet Kaiyisah binti Kamaruzzaman Muhamad Amar bin Zakaria

Title Name Student ID

2019200536 2019289758

Course

Applied Science (AS120)

Group

A4AS1204_5

Date of Experiment

19 April 2021

Date of Submission

April 2021

Lecturer’s Name

1.0

OBJECTIVE

Madam Umi Kalsum binti Abdul Karim

To determine the percent (w/w) of the active ingredient, acetylsalicylic acid (C 9H8O4) in Aspirin tablet.

2.0

PROCEDURE

A. Standardization of HCL solution 1. The burette was filled with the prepared diluted HCl and the initial burette reading was recorded and written down in a table accordingly. 2. A 25.0 mL of the standard Na2 CO 3

that has been prepared earlier was pipetted into

a 250 mL of conical flask. 2-3 drops of methyl orange that act as an indicator was added. 3.

Na2 CO 3

Was titrated in the conical flask with the HCl from the burette slowly

while the flask was shaken until the indicator color changes from yellow to red. The reading of the end point of the titration was recorded. 4. Step 1 until 3 was repeated 2–3 times and all of the readings was recorded in a table. 5. The exact molarity of HCl solution was calculated. 6. The remaining acid solution was then transferred into a clean reagent bottle.

B. Preparation of 1.0 M NaOH (Standardization of NaOH solution with HCl) 1. Fill the burette with NaOH solution. HCl solution is pipette into 250mL of conical flask and 2-3 drop of phenolphthalein is added to act as indicator 2. HCl in conical flask is titrated slowly with NaOH in the burette while the flask is shaken until the indicator color changes. 3. From the titration and calculation in lab, 1.0 M of NaOH from HCl is prepared as secondary standard solution

C. Hydrolysis of aspirin 1. A commercial aspirin tablet is collected and the brand name and calculate content of the active ingredient is record in table A 2. The aspirin tablet is weight accurately in weighing paper and then then tablet is transfer into a 250 mL conical flask

3. By using a pipette 25 mL of 1.0M NaOH solution and 25 ml of water is added to the tablet. The mixture then heated until boiled for 10 minutes to hydrolyze the aspirin. 4. The solution is cooled at room temperature and then transferred quantitatively into 250mL volumetric flask. The solution is dilute to the marker carefully using a dropper when closed to the calibration mark.

D. Back titration 1. 25 mL of Aspirin solution is pipetted into 250 mL conical flask 2. 1-2 drop of phenolphthalein indicator is added. 3. The aspirin solution in the conical flask is titrated with the HCl solution from the burette slowly while the flask is shaken until the indicator color changed. The reading at the end point of titration is recorded. 4. Step 2-3 is repeated 2-3 times and the reading is recorded in table 3. 5. The % w/w of acetylsalicylic in aspirin tablets is calculated.

3.0

RESULT / DATA ANALYSIS

Table 1 Millisprin

Commercial/brand name of aspirin tablet Name of manufacturer

KCK Pharmaceutical

Mass per tablet of active ingredient (acetylsalicylic acid)

900 mg

stated on the label Mass of aspirin tablet

1.1567 g

Calculated % w/w

77.81 %

Table 2 Standardization of HCl solution. Titration Final burette reading Initial burette reading Volume of HCl used

1 23.70 0.00 23.70

2 23.70 0.00 23.70

3 23.80 0.00 23.80

a)

Average volume of HCl: 23.73 mL

b) Volume of Na2CO3 used: 25.00 mL c)

Molarity of Na2CO3: 0.05 M

Standardization of NaOH solution with HCl Molarity of NaOH: 1.0 M

Table 3 Titration of aspirin solution with HCl Titration Final burette reading Initial burette reading Volume of HCl used

1 13.90 0.00 13.90

2 13.10 0.00 13.10

3 13.10 0.00 13.10

a) Average volume of HCl: 13.37 mL b) Initial volume of NaOH used to hydrolyse aspirin: 25.00 mL

4.0

CALCULATION

1. a) Based on the data obtained in Table 2 calculate the exact molarity of HCl. Use (MaVa /MbVb) = a/b MaVa/MbVb = a/b (Ma × 0.02373) / (0.05 × 25.00) = 2 / 1 Ma × 0.02373 = 2 (0.05×0.025) Ma × 0.02373 = 0.0025 Ma = 0.0025 / 0.02373 = 0.105 M of HCl

b) Using back titration method, calculate the % (w/w) ASA in the aspirin tablet. i.

Calculate the initial mole of NaOH used for hydrolysis.

NaOH + HCl → NaCl + H2O Moles of NaOH = (0.025L) (1.0M) = 0.025 mol (excess)

ii.

Based on Equation (2), calculate the mole of excess (unreacted) NaOH. (Remember in C: Hydrolysis of aspirin, solution mix with NaOH is diluted 10 x [25 mL into 250 mL]. Need to find mol NaOH before dilution)

No of moles of NaOH = molarity of NaOH x volume of HCL used

= 0.11 M x 0.01337 L

= 0.0015 moles

iii.

Calculate the mole of NaOH that has actually reacted with acetylsalicylic acid.

Mole of NaOH that reacted with acetylsalicylic acid (initial mol - excess mol) = Initial mole of NaOH used in hydrolysis - Mole of unreacted NaOH = 0. 025mol–0. 0015mol =0. 024mol e s

iv.

Based on Equation (1), calculate the mole of reacted acetylsalicylic acid (ASA) in the solution.

Mole of acetylsalicylic acid (from Eqn. 1) =

1 2

× 0.024 mol

= 0.012 mol

v.

Calculate the mass then the % w/w of acetylsalicylic acid in the tablet. Ma s sofa c e t y l s a l i c y l i ca c i d=

1 2

c×MW

=0. 012×180.158 g/mol = 2.16 g

% w/ w=ma s sofa c e t y l s a l i c y l i ca c i dx100 ma s sofa s pi r i n % w/ w=

2.16 g 1.1567 g

×100

=186. 7% vi.

Compare the % w/w calculated in Table 1 and the result obtained in (v). Calculate the % error. (% Error =

measured /calc value – true value/ standard true value /standard

%e r r or=

186.7 −77.81 77.81

=1. 4%

5.0

CONCLUSION

6.0

REFERENCE

×100 %

× 100%...


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