MORGAN Titration Of The Acetic Acid In Vinegar Chemistry Lab 5 Report

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MORGAN Titration Of The Acetic Acid In Vinegar Chemistry Lab 5 Report
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MORGAN Titration Of The Acetic Acid In Vinegar Chemistry Lab 5 Report
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EXPERIMENT #5: Titration of the Acetic Acid in Vinegar General Chemistry Laboratory, Spencer Hall 207 Morgan State University, Baltimore, MD March 21, 2019. 8:00 a.m. - 10:50 a.m Title: Titration of the Acetic Acid in Vinegar Abstract: Introduction: Experimental details Objective -To titrate vinegar using a standard NaOH solution. -Determine the concentration in both the methods of preparation that are used. Materials -Electronic balance -1 Beaker (450ml) -Cleaning solution -Cleaning brush -Lab coat/ apron -Safety goggles -Standard NaOH solution -50 mL Buret - 10 mL Volumetric pipet - 100 mL Volumetric flask -Buret Clamp and stand - Vinegar - Phenolphthalein - NaOH (solid) Experiment : Procedures: (Liquid) Method #2 Dissolution of the solute WK#1 -Using a cleaning solution, ali appartus was throughly cleaned. (Volumetric flasks, beaker, pipet) -All equipment was dried using papaertowels. -150mL of the liquid solution was transferred to a beaker -Next, the volume of the solution is calculated. M1V1=M2V2 M1 = 0.98M M2=0.40M V1= ? V2=100mL (DILUTED) M2=(0.40M) V2 (100ML) =40m/mL m1(0.98) V1= 40 m/mL -This is to calculate the mass and volume of the solution(stock) m1(0.98) v1(41mL)= 40m/mL -Next, a 50mL erlenmeyer flask was filled with 41mL of the stock solution. -41mL of the stock solution was transferred into a volumetric flask using a pipet. -Next, the liquid solution is dilutled using distilled water, from 41mL, to the v2 value of 100mL (59ml distilled water) -A cap is now placed on the volumetric flask, and it is shaken to mix throughly. -Next, a pinch of the solution is added to the buret to coat the inside, then it is shoved around to evenly cover it. It was then poured out of the buret. -Afterwards, 50mL the stock solution is added to the buret. -10ml of vinegar was then poured into a 50mL. -After that 2 drops of phenolphthalein was added to each erlenmeyer flask. -The 10mL of vinegar was transferred to an erlenmeyer flask afterwards. (3) Each flask contains: -10ml distilled water, 10mL vinegar and 2 drops of phenolphthalein -Next, the stock solution is added to each flask containing the mixture. A purple color was produced, then faded away upon mixing. -After multiple insertions of the stock solution, the solution turned a pale-purple tint 18 mL was added in the flask before permanent color change. Procedures: (Solid) Method #1 Dissolution of the solute WK#2 -50mL beaker was filled with 25mL cleaning solution -All apparatus were cleaned using solution, distilled water , and a cleaning brush, then distilled dried with paper towels. -First, the amount of the solute (NaOH) needed was calculated 0.4 moles x 0.1 =0.04 x 39.77g =1.59g -This concluded that 1.59g of the solute was needed; Three samples of 1.59g of NaOH was weighted of an electronic balance. -Next, the solid NaOH was added to a volumetric flask, then 2 drops of distilled water was added to the flask to dissolve the solute. -The volumetric flask containing NaOH was filled to 100mL with distilled water; this was repeated twice for 3 trials with 3 different volumetric flasks, NaOH samples, and distilled H20. -The flask was then mixed using a lid, then transferred into a 50mL Biunet. -In 3 different erlenmeyer flasks, 10mL of vinegar , 10 mL of distilled water, and 2 drops of phenolphthalein was prepared. Results and Calculations: Table #1 containing the Results CH3COOH + NaOH → CH3COONa + H2OCH3COOH+NaOH→CH3COONa+H2O WEEK 1 METHOD 2 Flasks Burnet Trants Volume Titrant Volume (Bunet) Final Used Trial #1 50 mL 32 mL 18 mL Trial #2 32 mL 15 mL 17 mL Trial #3 43 mL 28 mL 15 mL Flask Intial 1. 20 mL 2. 20 mL 3. 20 mL Flask Final (Used ) (+20 mL) 38 mL 37 mL 35 mL WEEK 2 METHOD 1 Flask Titrant vol Final Used Initial Final Trial 1 52mL 33mL 18mL 20mL 38mL Trial 2 33mL 10mL 23mL 20mL 43mL Trial 3 31mL 10mL 21mL 20mL 41mL Flask #1 38 mL - 20mL =18 mL NaOH =0.08L NaOH CH 3 CooH ⇒ CH 3 CooNa +H20 (1=0) Balanced raio n=MV 0.4 mol NaOH x 0.018 =0.0072 mol NaOH 1 mol NaOH = 1 mol CH3CooH 0.0072 NaOH 0.0072 mol NaOH x 1 mol CH3CooH =0.0072 mol CH3CooH 1 mol NaOH 10ml H20 + 10 mL Vinegar = 20 mL ⇒ Pink = 6.8 mL NaOH (Biuret) Flask #1 (10 mL) Flask pt.2 (20mL) ConC = 0.0072 mL CH3 CooH of 0.01L ConC = 0.0072 mL CH3 CooH of 0.01L Flask Pt. 3 (Turns Pink) (30)mL ConC = 0.0072 mL of 0.038 L CH3 CooH =0.72 M CH 3CooH =0.36M CH3CooH =0.189 M CH3CooH Mass of CH3 CooH M=moles x MW Mass of Vinegar Density of vinegar g/mL Mass of CH3CooH used x 100 mass of vinegar 0.432 g x 100= 4.32% 10g Molecular weight of Acetic Acid calculations C= 12.04 x 2 = 24.022 H= 1.00794 x 4 = 4.03176 O= 15.9994 x 2 = 31.9988 Mass of CH3C00H Used 60.05g/mol x 0.0072 mol =0.432g Ch3CooH Density of vinegar 1g/mL D=m/v M=D x v M= 1g/ml x 10 ml =10g vinegar Flask #2 37mL - 20mL =17mL NaOH =0.017L NaOH CH3CooH ⇒ CH3CooNa +H20 (1:1) Balanced Ratio n=MV 0.4mol NaOH x 0.017L =0.0068mol NaOH 1mol NaOH = 1mol CH3CooH 0.0068 mol NaOH x 1mol CH3CooH = 0.0068 mol CH3CooH 1 mol NaOH 10mL H20 + 10mL Vinegar = 20mL ⇒Pink =6.8mL nAoh (Buret) Flask #2 pt.1 (10mL) Flask #2 pt. 2 (20mL) ConC = 0.0068 mL CH3 CooH of 0.01L ConC = 0.0068 mL CH3 CooH of 0.02L =0.68 M CH 3CooH =0.34 M CH 3CooH Flask #2 pt.3 (37mL) 0.0068 mol CH3CooH 0.038L =0.184 M CH3CooH Mass of CH3CooH m=moles x MV 0.0072 x g Mass of vinegar Density of vinegar g/mL Mass of CHC00H Used 60.05g/mol x 0.0068 mol = 0.408g CH3CooH M=1g/mL X 10mL =10g Vinegar 0.408 x100 = 4.081 10g Flask #1 Relative error True Value of CH3CooH =5% [4.32-5] = (-0.38) x 100 =136 % 5 5 4.32 + 4.08 + 3.6 = 12 / 3 = 4% Method #2 (Liquid) Dilution From stock solution 4 - 5 = (1) x 100 = 20% 5 5 Method #1 (Solid) Dilution From stock 4.96- 5 = (-0.04) x 100 = 0.8% 5 5 Flask #1 (Method 1, Wk 2) 38 mL- 20mL =18mL NaOH =0.018L NaOH N=MV 0.4mol NaOH x 0.018L =0.0072mol NaOH 1mol NaOH = 1 Mol CH3CooH 0.0072 mol NaOH x 1 mol CH3CooH = 0.0072 mol CH3CooH 1 mol NaOH 10mL H20 + 10mL vinegar = 20mL⇒ PINK = 7.2 mL NaOH (Burnet) Flask #1 (Method 1, wk 2) (10ml) Flask #2 (20ml) ConC = 0.0072 mL CH3 CooH of 0.01L ConC = 0.0072 moL CH3 CooH of 0.02L =0.72 M CH 3CooH =0.36 M CH 3CooH Flask #1 (pt.3) (38mL) Mass of CH3CooH M= moles x MW ConC = 0.0072 moL CH3 CooH of 0.038L 60.05 g/mol x 0.0072 mol =0.432 CHCooH =0.189 M CH 3CooH M=1g/ml x 10 ml =10g Vinegar 0.432 x 100 = 4.32 % 10g Flask #3 (Method 1, Wk #2) 41mL -20mL =21mL NaOH =0.021L NaOH N=MV 0.4mol NaOH x 0.021L =0.0084mol NaOH 1mol NaOH = 1 Mol CH3CooH 0.0084 mol NaOH x 1 mol CH3CooH = 0.0084 mol CH3CooH 1 mol NaOH 10mL H20 + 10mL vinegar = 20mL⇒ PINK = 8.4 mL NaOH (Burnet) Flask #3 (PT. 1) (10ml) Flask #3 PT.2 (20ml) ConC = 0.0084 mL CH3 CooH of 0.01L ConC = 0.0084 moL CH3 CooH of 0.02L =0.84 M CH 3CooH =0.42 M CH 3CooH Flask #1 (pt.3) (41mL) Mass of CH3CooH M= moles x MW ConC = 0.0084 moL CH3 CooH of 0.041L 60.05 g/mol x 0.0084 mol =0.504 CHCooH =0.184 M CH 3CooH M=1g/ml x 10 ml =10g Vinegar 0.504 x 100 = 5.04 .” % 10g Discussion: . Conclusion: References: General Chemistry Laboratory Manual, An integrated Critical Thinking Approach, By Angela J Winstead, Morgan State University, McGraw Hill 2013. ...
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Dr_PaulSharp
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EXPERIMENT #5: Titration of the Acetic Acid in Vinegar
General Chemistry Laboratory, Spencer Hall 207
Morgan State University, Baltimore, MD
March 21, 2019. 8:00 a.m. - 10:50 a.m

Title: Titration of the Acetic Acid in Vinegar

Abstract:
The following work will demonstrate a step-by-step process of titrating acetic acid in
vinegar. The materials that will be required for the process are listed as well as the main
objectives which are titrating vinegar through a standard NaOH solution and establishing the
concentration in the two methods used in preparation. A guide on how the process will be
done is provided as well as the results of the experiment. Typically, three trials will be
conducted during the titration process to get best answers with reduced errors.
Introduction:
Vinegar is produced through the process of oxidizing ethanol. Vinegar usually has
acetic acid levels of between 4 percent and 8 percent. Therefore, titration can be used as a
method of determining the concentration of acetic acid by use of a strong base like aqueous
sodium hydroxide solution. An acid-base indicator will be used at the end of the experiment
as a way of detecting the end point of the titration process. Acetic acid is a carbon compound
which has a single ionized proton, thus making it an organic acid known as Carboxylic Acid.
The following work presents the titration procedure and the outcome of the process.

Experimental details
Objective
-To titrate vinegar using a standard NaOH solution.
-Determine the concentration in both the methods of preparation that are used.
Materials
-Electronic balance
-1 Beaker (450ml)
-Cleaning solution
-Cleaning brush
-Lab coat/ apron
-Safety goggles
-Standard NaOH solution

-50 mL Buret
- 10 mL Volumetric pipet
- 100 mL Volumetric flask
-Buret Clamp and stand
- Vinegar
- Phenolphthalein
- NaOH (solid)

Experiment :
Procedures: (Liquid) Method #2 Dissolution of the solute WK#1
-Using a cleaning solution, ali appartus wa...

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