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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

1. Unknown mass of compound was obtained from the rack

2. 0.8g of the unknown compound was measured into the 250 mL Erlenmeyer flask

3. 50mL of 0.6 M HCL solution from the auto pipette bottle was added

4. The mixture was swirled until the solid completely dissolved

5. 20 drops of Bromocresol green indicator were added

6. The burette was rinsed with the 0.5M sodium hydroxide solution

7. The solution was titrated by slowly adding NAOH with continued swirling until the solution turned green

8. The procedure was repeated thrice to get three data sets.

Results

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

Çalık, M., Ültay, N., Kolomuç, A., & Aytar, A. (2015). A cross-age study of science student

teachers' chemistry attitudes. Chemistry Education Research and Practice, 16(2),

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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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