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Ashley Campoverde Ramsey Salcedo March 11, 2020 Lab 12: Heartburn and “AUNTY ACID” Introduction Gastric acid, formed in our stomach, is composed of hydrochloric acid, potassium chloride, and sodium chloride. This acid plays an essential part in the digestion of proteins activating digestive enzymes. If too much hydrochloric acid if formed (a part of gastric acid), then one may develop heartburn or indigestion. Heartburn can lead to ulcer formation in the stomach lining. In this lab, we are going to use titration to determine which antacid is more effective. Titration is a technique used in labs to determine the concentration/quantity of an unknown substance (called the analyte) by adding a reagent whose concentration is known (titrant). In the experiment, we will be adding an indicator to the titrant and this will change the color of the titrant once it reached its endpoint. If the reaction between an analyte and titrant is slow, a technique called back titration is used. This process involves two reagents: the first being the one that reacts with the first reagent and the second being the one that reacts with the first reagent. Because we know the amount of the first reagent that was used from titration, we are able to calculate how much reagent was used for the reaction. Antacids contain a base, and, in that way, help neutralize the excess acid. In this lab, we will use titration to see how well different brands of antacids neutralize HCl and how efficient each one is. Materials 1. Tums 2. Weighing machine 3. Weighing paper 4. Mortar and pestle 5. Erlenmeyer Flasks 6. 0.1M Hydrochloric Acid 7. Graduated Cylinder 8. Hot Plate 9. Heat gloves 10. Burette 11. 0.1M Sodium Hydroxide 12. Phenolphthalein Observations and Experimental Table 1: Our group data Antacid

T1 (g)

Tums

0.51 g

Calculations: Trial 1:

T1 (NaOH ml) 3.05 ml

T2 (g) 0.51 g

T2 (NaOH ml) 3.11 ml

T3 (g) 0.53

T3 (NaOH ml) 4.55 ml

Ashley Campoverde Ramsey Salcedo March 11, 2020 Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.00305 mol NaOH=0.04695 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.04695 mol x 2.20 g/ 0.51 g = 0.20 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.20 mol/ 2.20 g= 0.090 mol/g Cost effectiveness Price per tablet, ȼ: ($8.37/bottle)(100ȼ/$1)(1 bottle/150 tablets)=5.58 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.20 mol/5.58 ȼ= 0.036 mol/ȼ Trial 2: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.00311 mol NaOH=0.04689 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet=0.04689 mol x 2.20 g/ 0.51 g = 0.20 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.20 mol / 2.20 g= 0.090 mol/g Cost effectiveness Price per tablet, ȼ: ($8.37/bottle)(100ȼ/$1)(1 bottle/150 tablets)=5.58 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.20 mol/5.58 ȼ= 0.036 mol/ȼ Trial 3: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.00455 mol NaOH=0.04545 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.04545 mol x 2.20 g/ 0.53 g = 0.19 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.19 mol / 2.20 g= 0.086 mol/g Cost effectiveness Price per tablet, ȼ: ($8.37/bottle)(100ȼ/$1)(1 bottle/150 tablets)=5.58 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.00069 mol/5.58 ȼ= 0.034 mol/ȼ Average moles HCl neutralized per tablet: 0.197 mol Average mass effectiveness: 0.089 mol/g Average cost effectiveness: 0.035 mol/ȼ

Ashley Campoverde Ramsey Salcedo March 11, 2020

Table 2: Our class data Antacid

T1 (g)

Alka-S Alka-S Moore/Al Moore/Al Tums Tums

0.51 0.51 0.36 0.37 0.51 0.52

T1 (NaOH ml) 42* 39* 11.3* 14* 3.05 2.4

T2 (g) 0.50 0.52 0.39 0.36 0.53 0.52

T2 (NaOH ml) 41.2 39.5 9.7* 17.5* 3.11 2.0

T3 (g) 0.52 0.52 0.40 0.40 0.51 0.52

T3 (NaOH ml) 42.3 39.23 9.3 12.0 4.55* 2.0

*- overshot Calculations for Average of Alka-S: Trial 1: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.0405 mol NaOH=0.0095 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.0095 mol x 3.20 g/ 0.51 g = 0.060 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.060 mol/ 3.20 g= 0.019 mol/g Cost effectiveness Price per tablet, ȼ: ($8.75/bottle)(100ȼ/$1)(1 bottle/36 tablets)=24.3 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.060 mol/24.3ȼ= 0.0025 mol/ȼ Trial 2: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.0426 mol NaOH=0.0074 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.0074 mol x 3.20 g/ 0.51 g = 0.046 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.046 mol/ 3.20 g= 0.014 mol/g Cost effectiveness Price per tablet, ȼ: ($8.75/bottle)(100ȼ/$1)(1 bottle/36 tablets)=24.3 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.060 mol/24.3ȼ= 0.0019 mol/ȼ

Ashley Campoverde Ramsey Salcedo March 11, 2020 Trial 3: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.0408 mol NaOH=0.0092 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.0092 mol x 3.20 g/ 0.52 g = 0.057 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.057 mol/ 3.20 g= 0.018 mol/g Cost effectiveness Price per tablet, ȼ: ($8.75/bottle)(100ȼ/$1)(1 bottle/36 tablets)=24.3 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.060 mol/24.3ȼ= 0.0023 mol/ȼ Average moles HCl neutralized per tablet: 0.0087 mol Average mass effectiveness: 0.017 mol/g Average cost effectiveness: 0.0022 mol/ȼ Calculations for Average of Moore/Al: Trial 1: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.01265 mol NaOH=0.03735 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.03735 mol x 0.40 g/ 0.365 g = 0.041 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.041 mol/ 0.40 g= 0.10 mol/g Cost effectiveness Price per tablet, ȼ: ($23.06/bottle)(100ȼ/$1)(1 bottle/500 tablets)=4.61 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.041 mol/4.61ȼ= 0.0089 mol/ȼ Trial 2: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.0272 mol NaOH=0.0228 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.0228 mol x 0.40 g/ 0.375 g = 0.024 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.024 mol/ 0.40 g= 0.06 mol/g Cost effectiveness Price per tablet, ȼ: ($23.06/bottle)(100ȼ/$1)(1 bottle/500 tablets)=4.61 ȼ

Ashley Campoverde Ramsey Salcedo March 11, 2020 C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.024 mol/4.61ȼ= 0.0052 mol/ȼ Trial 3: Moles of HCl (aq) neutralized= moles of HCl (aq) added to antacid- moles of NaOH required to back titrate Moles of HCl (aq) neutralized= 0.050 mol HCl- 0.01065 mol NaOH=0.03935 mol Moles of HCl (aq) neutralized per tablet= moles of HCl (aq) neutralized x m(tablet)/ m(sample) Moles of HCl (aq) neutralized per tablet= 0.03935 mol x 0.40 g/ 0.40 g = 0.040 mol Mass effectiveness E= Moles of HCl (aq) neutralized per tablet/ mass of tablet, g E= 0.040 mol/ 0.40 g= 0.10 mol/g Cost effectiveness Price per tablet, ȼ: ($23.06/bottle)(100ȼ/$1)(1 bottle/500 tablets)=4.61 ȼ C= Moles of HCl (aq) neutralized per tablet/ cost of tablet, ȼ C=0.040 mol/4.61ȼ= 0.0087 mol/ȼ Average moles HCl neutralized per tablet: 0.035 mol Average mass effectiveness: 0.087 mol/g Average cost effectiveness: 0.0076 mol/ȼ Discussion and Conclusion When adding the titrant to analyte, there is a point when the endpoint is reached. The endpoint is when all of the acid is neutralized by the base. Because the indicator we used in this lab, phenolphthalein, turns hot pink in basic solution we know that once our analyte reaches a faint pink color, endpoint has been reached. A very important part of performing the titration procedure is to go slow. For our first trial, we added just the right amount of the titrant but for our third trial, we weren’t as careful, and we went a little above endpoint, it was hot pink meaning we overshot. For this experiment, there were many sources of error that may have influenced our results. For example, because the antacid come packaged, there may have been expired tablets or the tablets may have been uneven in concentration. For the titration part of the lab, a source of error is not going slow with it. Focus Questions: 1. Which antacid is most effective in terms of mass? The averages for effectiveness in terms of mass are 0.017 mol/g for Alka-Seltzer, 0.089 mol/g for Tums, and 0.087 mol/g for Moore-Alcalak. Out of these three averages, the most effective would be Alka-S because it neutralized the most HCl. 2. Which antacid is most effective in terms of cost? The averages for effectiveness in terms of cost are 0.035 mol/cent for Tums, 0.0022 mol/ȼfor Alka-Seltzer, and 0.0076 mol/ȼ for Moore-Alcalak. Out of these three, the most effective cost-wise would be Alka-S because it neutralized more HCl per cent as opposed to the other two.

Ashley Campoverde Ramsey Salcedo March 11, 2020 References Smeureanu, G., Geggier, S. General Chemistry Laboratory, New York City, Hunter College Publishing, 2018, p.89-94

Post-Lab Questions: 1. I would take Alka-S if needed because it took the fastest to neutralize and was most effective mass and cost wise. 2. CaCO3 + 2HCL→CaCl2 + H2O + CO2 0.300 grams of CaCO3 / 100g = 0.003 moles of CaCO3. This is a 1:2 ratio so 0.003 x 2 = 0.006 moles of HCl. 3. Based on the balanced equation, 1 mole CaCO3 neutralizes 2 moles of HCl. When the pH value is 1, the concentration of HCl is 0.1M. When the pH value is 2, the concentration of HCl is 0.01M. 0.1-0.01=0.09. The ratio is 1:2, so 0.9/2=0.45 moles of CaCO3 required to change the pH value. 0.45 moles is equal to 4.5 grams CaCO3. 4. Mg(OH)2 relieves the pain caused by excess stomach acid because it is a base and HCl is an acid. In a reaction, Mg(OH)2 neutralizes the stomach acid. Mg(OH)2 + 2HCL→ MgCl2 + H2O 5. A. The moles of HCl neutralized by the tablet: (0.026 L)(0.650 M) = 0.0169 moles of HCl (0.01124 L)(1.05 M) = 0.011802 moles of NaOH 0.0169-0.011802 = 5.0x10-3 moles of HCl neutralized by the tablet B. Mass effectiveness of antacid: E=5.0x10-3/0.0834g = 0.060 mol/g C. Cost effectiveness of antacid: Price per tablet = ($3.99 bottle)(100 cents/ $1)(1 bottle/150 tablets) = 2.66 cents C = 5.0x10-3/2.66 = 1.9x10-3 mol/cent...