Sean DeSilva Chem 1A Section3961 Locker Number A518 Dr. Cornett Dec Experiment # 12 Titrimetric Analysis of an Antacid Preparation

Transcription

1 Sean DeSilva Chem 1A Section3961 Locker Number A518 Dr. Cornett Dec Experiment # 12 Titrimetric Analysis of an Antacid Preparation

2 Objective To find the number of moles of acid neutralized by a tums tablet. We found the number of moles of acid neutralized in the titration by subtracting it from the moles of acid in the initial solution. This was achieved by through a standardization of a HCL and NaOH solution. Procedure The procedures of all the experiments that where preformed can be found in the Santa Rosa Junior College Chem 1A laboratory manual from the fall of The only deviation from these procedures was the doing three trails on section and section instead of the recommended one. Data Please see the attacked sheet marked Data. Calculations and Results : Calculated molarity of HCL solution through Titration (Grams of Tris ) divided (Grams per mol of Tris) x (Ratio of HCL to Tris) = mols HCL Take mols HCL and divide it by amount of HCL added for titration to occur. That will give you the Molarity of your HCL Solution g (Tris) x (1 mol/ g (Tris per mol)) x (1 mol HCL/1 mol Tris) = mols HCL mols HCL /.039 L = mols HCL per Liter Trial2:.2205 mols HCL per Liter Trial3:.2195 mols HCL per Liter Calculated molarity of HCL solution through given values of concentration. (Amount of HCL added) x (Mols of concentrate) = (mols of HCL) Take (mols of HCL) divide (total volume of HCL+ H2O) = Molarity of Solution (.014L of HCL) x (6 mols/1 L) =.084 mol HCL (.084 mol HCL)/ (.414L of solution)=.2029 mols HCL per Liter Error Calculations: Part 1

3 Standard Deviation SD = SquareRoot( (( )^2 + ( )^2 + ( )^2 ) / 2) SquareRoot( /2) = /.2029 = x 100 = 9% Standard Deviation = 9% Mean = (MolarityTrial1+ MolarityTrial2+ MolarityTrial3) / (number of trials) ( ) / 3 = Mean = RSD= (SD/Mean) x 100 ( / ) = x 100 = 8.4% RSD = 8.4% : Calculated molarity of NaOH solution (Grams of KHP) divide (Grams per mol of KHP) x (Ratio of HCL to KHP) = mols NaOH Take mols NaOH and divide it by amount of NaOH added for titration to occur. That will give you the Molarity of your NaOH Solution..8001g (KHP) x (1 mol/ g (KHP per mol)) x (1 mol NaOH /1 mol KHP) = mols NaOH mols NaOH /.0374 L = mols NaOH per Liter Trial2:.1051 mols NaOH per Liter Trial3:.1056 mols NaOH per Liter Calculated molarity of NaOH solution through given values of concentration. (Amount of NaOH added) x (Mols of concentrate) = (mols of NaOH) Take (mols of NaOH) divide (total volume of NaOH + H2O) = Molarity of Solution (.020L of NaOH) x (2 mols/1 L) =.04 mol NaOH (.04 mol NaOH)/ (.395L of solution)= mols NaOH per Liter Error Calculations: Part 2 Standard Deviation SD = SquareRoot( )^2 + ( )^2 + ( )^2 ) / 2) SquareRoot( /2) = / = x 100 =.0023%

4 Standard Deviation =.0023% Mean = (MolarityTrial1+ MolarityTrial2+ MolarityTrial3) / (number of trials) ( ) / 3 = Mean = RSD= (SD/Mean) x 100 ( / ) = x 100 =.022% RSD =.022% Total moles of HCL added Liters of HCL added x (Molarity found in for HCL (mol/l)) = Total moles of HCL added.05l x ( mols/1l) = moles of HCL added Trial 2: moles of HCL added Trail 3: moles of HCL added Moles of HCL reacted with NaOH NaOH L x (Molarity found in for NaOH (mol/l)) x (1 mol HCL/1 mol NaOH).0374L x ( M) x (1 mol HCL/1 mol NaOH) = moles of NaOH react Trial2: Trial3: Moles of HCL reacted with CaCO3 nacid neutralized by CaCO3 = nacid initially in flask nacid neutralized by NaOH CaCo3 react with HCL = moles of HCL added moles of NaOH react CaCo3 react with HCL = moles Trial2:.0071 moles Trial3: moles Mass of CaCO3 neutralized in g Moles of CaCO3 x (1mol CaCo3 / 2 mols of HCL) x grams per mole of CaCO3

5 Trial1: moles x (1mol CaCo3 / 2 mols of HCL) x grams per mol of CaCO3 =.3367 grams CaCO3 Trial2:.3578 grams CaCO3 Trial3:.3551 grams CaCO3 Mass Percent CaCO3 in antacid: Mass Percent = (Grams of CaCO3 in above / Grams of antacid added) x 100 Trial1: (.3367 grams CaCO3 /.865 (g) ) x 100 = 38.92% Trial2: 41.36% Trial3: 41.05% Average Mass Percent = (38.92% %+41.05%) / 3 = 40.44% Average Mass of one tablet = (The mass of four tablets/4) One Tablet = g/ 4 tablets = g per tablet Average Mass of CaCO3 per tablet g per tablet x 40.44% =.526 g CaCO3 per tablet Standard Deviation On the bottle it says 500mg of active ingredient Average weight of one tablet equals g or mg 500mg / mg = 38.4% SD = SquareRoot( )^2 + ( )^2 + ( )^2 ) / 2) SquareRoot( /2) = /.384 = x 100 =.209% Standard Deviation =.209% Mean = (MolarityTrial1+ MolarityTrial2+ MolarityTrial3) / (number of trials) ( ) / 3 =.4044 Mean = RSD= (SD/Mean) x 100 (..0008/. 4044) = x 100 =.1978% RSD =.1978% Day 1 Discussion

6 First we had to make a solution of HCL and NaOH this in its self was not exciting, but knowing that this solution would be used in further steps added to its importance. After the monkey work of mixing the solutions we need to find the Molarity of that solution this was a very important step. It was nice to be able to get a theoretical value of the solution and then match it up against the values that I received through the process of titration. With my largest percent error being less then 10% deviation that this was an efficient method for finding molarity. Can I just mention the crazy process that we had to go through to find the exact amount of Tris and KHP to use for Titration; we weighted the same substance three different ways with in the three times we had to weight each. This is done of course to get the most accurate results because some of the Tris or KHP could have stayed on our weighting paper. The second half of the day we made a solution of NaOH3 that we used to titrate KHP. The titration process was pretty much identical in both of process for finding molarity of each solution. The NaOH3 how every was much more accurate according to my percent errors, getting a RSD of less then one percent. The molarity figures for NaOH3 and HCL were in our Santa Rosa Junior College Lab manual and each value was found on the container. With out these figure we would never be able to find an actual molarity. After the completion we had a long two weeks to wait until we could complete our experiment this lead to forgetting some of what and why we did what we did on Day 1. On the second day our goal was to calculate the grams of CaCO3 found in a tablet of tums; we would then take our calculations and compare it to the amount of active ingredient, which was 500mg according to the package that contained the tablet. It was an interesting process back titrating to find the mols of the acid HCL, which reacted with the base NAOH3. This process was very accurate seeing that my Standard Deviation and my Relative Standard Deviation were both under a half of a percent. The hardest part of this section was doing all of the calculations because there were a lot of values that we had to keep track of. It doesn t help that the lab manual doesn t give any help on the equations, luckily you showed us the procedure but the number where a little different. We also had the milligrams of active ingredient per tablet, unfortunately it took 20 minutes of calculations before we could check our answer and then came the challenge of finding the mistake. I was able to figure it out and had some good results. Conclusion Despite the fact that my calculation for the molarity of HCL was around eight percent different from actual molarity, it did not affect my final results. The molarity of HCL might have been thrown off because of the HCL dispenser it seemed to stick a little bit. I remember having to add in a little extra after I thought it had fully dispensed. I would have said that weighting out the TRIS and KHP might have caused a mistake but we where very careful about those measurements. I think processes like the weighting procedure, allowed for few chances to mess up the chance to get good numbers.

7 The largest chance for error came in the titration processes; to stop adding solution when the point of equilibrium had been reached was hard to do. I know that during the titration of CaCO3 that I had three different hues of pink in each of my trials. I could see the solution start to titrate so I would stop adding NaOH and start gentle agitating the flask, the solution would immediately change from a faint pink back to the white of our original solution. Just a drop or two extra of NaOH would severely chance the color. I was able to the see the reaction that was occurring between the Chemical solution added and the titrate solution. It was a successful lab my numbers were very consistent, it was doing the math behind these numbers that I think is going to be most beneficial going forward especially on the exam. Being able to find out the amount to active ingredient in a substance through calculating the molarity of a substance by titration is a process that is often used by chemist and calculating mole will be important moving forward in chemistry. Post Lab Questions: 1. a. Heart burn is caused by excess acid that is in the stomach, that excess acid travels out of the stomach and into the esophagus. As we saw in our lab CaCO3 is very good at neutralizing acid. b. When acid reacts with marble; the material breaks down into H2O, carbon dioxide and calcium carbonate. This is why acid rain has lead to the erosions of many famous statues throughout the world. 2. There is no doubt a pipet is the most accurate way to measure a liquid, but I can imagine being fairly accurate with a graduated cylinder. The last formal lab we had to add 50ml of water and then had to weight the calorimeter, it was within.5 grams of 50g each time. Yes you will most likely get a 5 or 6 % add error on average using a graduated cylinder. 3. You can see for the equations below that the hydrogen carbonate is much more effective at neutralizing acid. 1.7 grams NaHCO3 (1 mole NaHCO3/ grams NaHCO3)(1 mole HCl/ 1 mole NaHCO3) =.0202 moles HCl neutralized..5 grams CaCO3 (1 mole CaCO3/ g CaCO3)(2 moles HCl/ 1 mole CaCO3) = moles HCL neutralized 4. NaOH is very soluble in water and dissociates almost completely because it is such a strong base. I am sure that such a strong chemical would not be good in anyone s stomach