I need a final draft for this Lab Report. Please read this and make it perfect. My Professor wrote something on the paper. Please fix it and add it. and please add more information on it. Please
Junho Lee
Chem 152
Dr. Gilbert
Evaluation of compounds used in anti-acids.
Purpose:
Measure the amount of acid that can be neutralised by an unknown sample
Introduction and theory
When alkalis and acids reacts quantitatively, a product that is neither acidic nor alkaline is formed. This is known as neutralisation reaction. It is possible to measure the amount of known concentration of acid that can be neutralised by an unknown sample and hence deduce the concentration of the unknown sample. This is done through a lab experiment called titration.
Chemicals
Antacids eg Axid or Pepcid
0.6 M HCL solution
Bromocresol green indicator
0.5m Sodium Hydroxide solution
Apparatus
Weighing boat
250-mL Erlenmeyer flask
50mL burette
Clamp
Procedure
Unknown mass of compound was obtained from the rack
0.8g of the unknown compound was measured into the 250 mL Erlenmeyer flask
50mL of 0.6 M HCL solution from the auto pipette bottle was added
The mixture was swirled until the solid completely dissolved
20 drops of Bromocresol green indicator were added
The burette was rinsed with the 0.5M sodium hydroxide solution
The solution was titrated by slowly adding NAOH with continued swirling until the solution turned green
The procedure was repeated thrice to get three data sets.
Results
First | Second | Third | |
Mass of sample(g) | 0.8 | 0.81 | 0.79 |
Final volume of NAOH, mL | 23.10 | 22.86 | 23.40 |
Initial volume of NAOH, mL | 0.00 | 0.00 | 0.00 |
volume of NAOH used, mL | 23.10 | 22.86 | 23.40 |
Moles HCL initial | 0.60 | 0.60 | 0.60 |
Moles of NAOH Back titrated | 0.01155 | 0.01143 | 0.0117 |
Calculations and discussion
The reaction equation NAOH+ HCl= NaCl+ H2O
From the reaction equation above, the reaction ratio is 1:1
Initial number of moles of HCl in the 50mL HCl solution is given as (0.6M*50)/1000=0.03moles
Moles of NAOH Back titrated is calculated as
Titration 1
(0.5*23.1)1000=0.01155
(0.5*22.86)1000=0.01143
(0.5*23.4)1000=0.0117
Since the reaction ratio of NAOH:HCl is 1:1, the number of moles of HCl Back titrated is equal to the values above. Therefore, the remaining number of moles is given by initial moles of HCl – Moles back titrated which becomes
0.03-0.01155= 0.01845
0.03-0.01155= 0.01845
0.03-0.01143= 0.01857
0.03-0.0117= 0.0183
The moles of HCl neutralised per gram or the acid-neutralising capacity is given by Number of moles used in neutralisation divided by the number of grams of solid neutralised. Which is
0.01845/0.8=0.023
0.01845/0.81=0.023
0.0183/0.79=0.023
Therefore, from the experiment, the acid-neutralising capacity of the HCl used is 0.023
Conclusion
To get the amount of HCl acid that can be neutralised by the unknown sample you take (0.023*1000)/0.6= 38.3mL
This is the amount of HCl acid that can neutralise one gram of the acid 13 38.3 mL
References
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Lin, C., Maddocks, G., Lin, J., Lancaster, G., & Chu, C. (2004). Acid neutralising capacity of
two different bauxite residues (red mud) and their potential applications for treating
acid sulfate water and soils. Soil Research, 42(6), 649-657.
Miller, G. L. (1959). Use of dinitrosalicylic acid reagent for determination of reducing
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