Acid Base Titration. Week 1. Dilute 3 M of NaOH to 600 ml.5m of NaOH. Add about 7 drops of the indicator solution (BTB).

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1 Acid Base Titration Beginning Questions: 1) NaOH has a concentration of what? 2) HCl has a concentration of what? 3) What is the neutralization of the antacid? 4) What is the concentration of OH- in the antacid? 5) What s Back Titration? Why do we use it? 6) How many moles of antacid (weak base) are in one tablet? Safety Considerations: Long pants must be worn at all times including goggles, long pants, close-toed shoes, and hair must be tied back. Also, you must wear gloves when handling the acidic and basic solutions. Be careful when dealing with all glassware. Dispose of all waste in their respected containers after the experiment. Procedures: Week 1 Dilute 3 M of NaOH to 600 ml.5m of NaOH. Fill 250 ml of the diluted solution in a separate bottle for next week. Weigh out 1.2 g of KHP and transfer the material into an Erlenmeyer flask. Add 50 ml of water to the 1.2 g of KHP. Swirl this mixture gently until the KHP is fully dissolved into the water. Add about 7 drops of the indicator solution (BTB). Fill a 50 ml burette with your.5m of NaOH. Begin titrating the NaOH into the dissolved KHP and water. Record the amount of NaOH it takes to turn the KHP and water to BTB Blue endpoint. Dispose of the solution, and repeat the experiment starting from the third bullet.

2 Week 2 Dilute 12 M of HCl to 100 ml 1.2M of HCl. Transfer 10mL of your diluted HCl to an Erlenmeyer flask, and add 2 drops of indicator solution (BTB). Fill a 50 ml burette with the diluted NaOH from last week. Begin titrating the.5m NaOH and the 10 ml of HCl. Record the amount of NaOH it takes to get the HCl to it s blue endpoint. two. Repeat these steps as many times as necessary starting from bullet number Begin your antacid titration. Measure.25 g of the antacid sample assigned to your group. Place the.25 g sample in an Erlenmeyer flask, and add 10 ml of HCl and swirl the solution. The solution should take the color of the indicator. Titrate the mixture with.5m of NaOH. Record the amount of NaOH it takes to get the antacid mixture to it s blue endpoint. Data, Observations, and Calculations: Calculations from Week 1: M = n/v Take 3M to.5m M1V1 = M2V2 600 ml of.5m solution.5(600) = 3V2 V2 = 100 ml add 500 ml of H2O.5M of NaOH Titrate.5M of NaOH with KHP (Weight of substance in (g) / MM (g)) / Vol.

3 Titration 1: 10 ml of solution turned Blue (confirmed) Titration 2: 13 ml of solution turned Blue (confirmed) 16 ml of solution turned Blue (confirmed) Titration 4: 11 ml of solution tunred Blue (confirmed) Standard Deviation: (10 ml = 6.25 (13 ml =.25 (16 ml = (11 ml = = / 4 = 5.25 square root(5.25) = 2.29 KHP 1.2g MKHP = ((1.2 / ) /.05 L) =.1175M Moles of H3O+ = Moles of OH- MKHP x V1 = MNaOH x V x.05 = MNaOH = ( /.012) =.4875 Calculations from Week 2: 1.2M of HCl 12x = 120 x = 10 ml Average of titrated NaOH 12.5 ml.25g of antacid must dissolve with excess acid (10 ml). BTB Titrations with 1.2M of HCl: Titration 1: 25 ml of.5m NaOH

4 Titration 2: 20.5 ml of.5m NaOH 21.5 ml of.5m NaOH Average of titrated HCl 22.3 ml Standard Deviation: (25 ml 22.3) = 7.29 (20.5 ml = 3.24 (21.5 ml = = / 3 = 3.72 square root(3.72) = 1.93 BTB Antacid Titrations:.25g sample of antacid (Green Group) Titration 1: 17.5 ml Titration 2: 19 ml 17 ml Moles of antacid used: n =.25 / 17.5 =.014 moles/g n =.25 / 19 =.013 moles/g n =.25 / 17 =.014 moles/g Average of moles of antacid used Standard Deviation: (17.5 ml 17.83) =.1089 (19 ml = Average of titrated antacid ml

5 (17 ml = = / 3 =.7223 square root(.7223) =.850 Concentration of HCl: MHCl = 1.2 n left over acid = VNaOH x MNaOH Titration 1: (.0175)(.48) =.0084 Moles of excess acid: (.010)(1.2) = =.0036 Titration 2: (.019)(.48) = Moles of excess acid: (.010)(1.2) = =.0029 (.017)(.48) = Moles of excess acid: (.010)(1.2) = =.0038 Average of Standard Deviations.0034 Observations I used the BTB, which had a blue endpoint. Therefore I noticed the change in the color of the solution to blue at the endpoint of the titrations. I thought it was interesting how the antacid changed to the same color at the endpoint as the BTB indicator. Also, I

6 noticed that it took more of the.5m NaOH to titrate the BTB versus the BTB antacid in all three trials. Claims The concentration of the standardized NaOH was.5m. The 3M had to be diluted with water to turn from 3M to.5m. We had to do the same to HCl. The concentration of HCl was 12M and we diluted the sample to 1.2M. The neutralization(s) of the antacid can be seen in the calculations. They came out to be.014,.013, and.014. The average can be seen as You can also see the concentration of the OH- in the solutions in my calculations. The concentrations came out to be.0036,.00192, and The average came out to be Evidence and Analysis Starting with the standardizations of NaOH and HCl, I just used the formula V1M1 = V2M2. For the first values, I substituted.5m and 600 ml for the second values I substituted 3M and found out V2. I found it to be 100 ml and just added 500 ml of water to the solution. For HCl, the first values I substituted in were 1.2M and 100 ml. The second values were 12M and found out V2. I found it to be 10 ml. So in order to take this solution from 12M to 1.2M, I added 90 ml of water to 10 ml of HCl. To find the neutralization of acid I found the moles of antacid used in the solution. I found the moles of antacid by taking the weight of the antacid and dividing by the volume of the solution. Then I just took the average of the all three values. The OHconcentrations can be found through a semi-tedious calculation. Since the Molarity of NaOH is.48, I took the volume in liters and divided by.48. This gives me the moles of left over acid. Then I took 10 ml and multiplied it by the Molarity of HCl, which is 1.2. Then I took 1.2 and subtracted it by the moles of left over acid and this gives us the concentration of OH-. A back titration is similar to a direct titration, but a bit more difficult. When an end point is not easily identified due to no color change, an excess volume of a reactant of KNOWN CONCENTRATION is added to the reactant of unknown concentration. Then the resulting mixture is titrated again (or titrated back) to find the volume of the unreacted reactant, which will tell us the amount that DID react with the solution of unknown concentration. You need to take into account the amount of excess reactant originally added. The relevant calculations can then be taken out. Reflections There was much room for error in this experiment. For example, the readings on the burette when titrating the solutions could have been off even by the slightest mark. (Mine were probably off by a little). Also, there could have been a faulty measurement in the standardizations of the solutions at the beginning stages of each experiment. There

7 were so many calculations in this experiment. The color that the experiment changed to could have been a bit stronger in some than the others. There was room for a miscalculation as well. I double-checked all of my calculations so, hopefully, I do not have a miscalculations. Some new questions I would have would be, What exactly does it mean when the solution turns blue? Another question would have to be, If the NaOH or the HCl wasn t standardized, how the results would be altered, and would they be altered on a large or small scale? We haven t learned anything in class about acid/base titrations yet, so it does not really tie into anything we have learned in class. However, this experiment was the first experiment that we did on our own, and it taught me that I could do any experiment on my own. It showed me have mixing certain substances together can have any sort of affect, including changing color as this experiment did.