Titration Lab Essay Example for Free

Titration Lab Essay Method for Control of Variables 1. In order to control the titration of potassium permanganate, we used the pipette to make the titration constant. We also used the same amount of chemical solutions, with the same flasks, to make the titration as constant as possible. 2. To control the amount of redox in the experiment, we have used the pipette to keep on putting the same amount of drop every time. The pipette made the amount of redox constant for every trial, and made the data more accurate. 3. In order to control the amount of potassium permanganate, we have measured the solution with the same plate and same scale. Since we have used the same materials, the amount used was very close. 4. To control the amount of force put in to mix the solution, I used my right hand for every trial and did not vary the length and procedure to mix. I have mixed the solution 5. In order to control the temperature of the liquids, I have used the same room at the same time. I did not used any air conditioning to change the temperature. Method For Data Collection 1. I have weighed out exactly 2.5 grams of AR ammonium iron (II) sulphate crystals, and dissolved them in approximately 40cmà ¯Ã‚ ¿Ã‚ ½ of sulphuric acid solution with a concentration of approximately 1 mol dm-à ¯Ã‚ ¿Ã‚ ½. 2. I did NOT heat the solution to assist the dissolving of the solution. 3. I made up to 100cmà ¯Ã‚ ¿Ã‚ ½ of the solution into a volumetric flask and mixed the solution. 4. I have used the pipette to pipette 10cmà ¯Ã‚ ¿Ã‚ ½ of the solution into a conical flask and added about an equal volume of distilled water. 5. I titrated this with 0.0200mol dm-à ¯Ã‚ ¿Ã‚ ½ potassium manganate (VII) solution to a faint pink color. 6. I repeated the titration for at least 2 times for accuracy in the data and recorded all the results. Table of Raw Results Table Showing Volume of KMnO4 Used to React With the Solution Initial Reading Final Reading Volume of KMnO4 The error or uncertainty of this data is à ¯Ã‚ ¿Ã‚ ½0.05cmà ¯Ã‚ ¿Ã‚ ½. However, the uncertainty for the volume difference is not à ¯Ã‚ ¿Ã‚ ½0.05cmà ¯Ã‚ ¿Ã‚ ½, but is à ¯Ã‚ ¿Ã‚ ½0.10cmà ¯Ã‚ ¿Ã‚ ½. This is because there are two à ¯Ã‚ ¿Ã‚ ½0.05cmà ¯Ã‚ ¿Ã‚ ½ datas, and if you add them up, you will get à ¯Ã‚ ¿Ã‚ ½0.10cmà ¯Ã‚ ¿Ã‚ ½. Evaluation Improvements: In this experiment, there were several inaccurate results and some mistakes I could have improved. 1. Pipette: For each drop from the pipette, there were same amounts of solutions, but since it is made by human hand, there should have been a slight difference in each drop. 2. Temperature: Although I have done this experiment in one class period, there were slight changes in the weather, causing the temperature to change. This could have affected the data results. 3. Force: The force applied to the solution and the flask was approximately the same, but there were some differences since it was done by humans. Humans never can do the same exact thing over, and so there should have been a slight error. Conclusion My hypothesis for this experiment was correct, because it took about 10 drops to titrate. Although there were some slight errors during the experiment, the experiment itself went fairly well, as there werent any big outstanding errors in the data and procedures. The main aim of this experiment, titration of redox with potassium permanganate, was well presented, and overall the experiment was good. Problem Calculation 1. MnO4- + 8H+ + 5Fe- = Mn2+ + 4H2O + Fe3+ 2. 5 moles of Fe2+ is required. 3. 10cm3 = 0.01dm3 0.02 mol dm-3 x 0.01 dm3 = 0.0002 mol 0.0002 moles of KMnO4 is present. 4. MnO4- : Fe2+ 1:5 0.0002 mol : 0.001 mol 0.001 moles of Fe2+ is present in 10cm3 of the solution. 5. 0.001 mol x 10 =0.01 mol 0.01 moles of Fe2+ is present in 100cm3 of the solution. 6. 1 mole of Fe2+ = 56grams 56g x 0.01 mol = 0.56g 0.56 grams of Fe2+ is in the ammonium iron (II) sulfate 7. Number of moles of ammonium iron (II) sulfate = 39 0.56g/(39mol x 0.1) = 14.4% 8. 1 mol of ammonium iron(II) sulfate = (NH4)2Fe(SO4)2 x 6H2O = 2N + 8H+ 1Fe+ 2S + 😠¯ + 12H + 6O = 392 grams 56g/392g = 14.3% 9. My result was 14.4% and it is almost equal to the correct value which is 14.3%. The difference is only 0.1%. 10. The iron (II) salt might evaporate if I heat it, and will affect the result.