Volumetric Analysis: Lab Report

Floyd Askew 3/19/13 CHEM 1211L Lab Report Introduction The purpose of this lab is to use volumetric analysis to determine the concentration of unknown substances. A sodium hydroxide solution is standardized to assist in finding the concentration of an acetic acid. An indicator must be used to pin point the equivalence point, the point in which 1 mole of a substance is equal to 1 mole of another. When that is found, we can determine the concentration. HC2H3O2 (aq) + NaOH (aq) H2O (l) + NaC2H3O2 (aq) The above equation is used to neutralize the acetic acid.The acid reacts with a base to produce water and a salt. Because there’s a 1:1 ratio, the moles of the acid must equal the moles of the base in order to reach the equivalence point. As far as the indicators go, an acid-base indicator will be used to show when we are close to the end point. For example, when HIn is dissociated In – is produced and it is pink. (See equation below) HIn + H2O H3O + +In – Procedure St andardization of NaOH Solution 1. A known amount of KHP is transferred to an Erlenmeyer flask and an accurately measured amount of water is added to make up a solution. . NaOH solution is carefully added to the KHP solution from a buret until we reach the equivalence point. At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic. Titration of an unknown 1. A measured amount of an acid of unknown concentration is added to a flask using a buret. An appropriate indicator such as phenolphthalein is added to the solution. The indicator will indicate, by a color change, when the acid and base has been neutralized). 2. Base (standard solution) is slowly added to the acid. 3. The process is continued until the indicator shows that neutralization has occurred. This is called the END P OINT. The end point is usually signaled by a sharp change in the color of the indicator in the acid solution. In acid-base titrations, indicators are substances that have distinct different colors in acid and base (Phenolphthalein pink in base, colorless in acid). 4. At the equivalence point, both acid and base have been completely neutralized and the solution is still colorless.However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic. This slight excess of NaOH is not much beyond the end point. The volume of the base is recorded and used to determine the molarity of the acetic acid solution. Experimental Data Standardization of NaOH solution | Trial 1| Trial 2| Trial 3| Mass of KHP| 0. 297 g| 0. 325 g| 0. 309 g| Initial buret reading, NaOH| 0. 00 mL| 0. 50 mL| 7. 70 mL| Final buret reading, NaOH| 32. 0 mL| 34. 0 mL| 38. 7 mL| Volume used, NaOH| 32. 0 mL| 33. mL| 31. 0 mL| Molarity of NaOH soluti on| 0. 0454 M| 0. 0475 M| 0. 0488 M| Average molarity of NaOH| 0. 0472 M| Titration of unknown | Trial 1| Trial 2| Trial 3| Initial buret reading, NaOH| 2. 70 mL| 19. 9 mL| 0. 00 mL| Final buret reading, NaOH| 19. 9 mL| 36. 2 mL| 19. 8 mL| Volume used, NaOH| 17. 2 mL| 16. 3 mL| 19. 8 mL| Molarity of acetic acid solution| 0. 0780 M| 0. 0769 M| 0. 0935 M| Average molarity of acetic acid solution| 0. 0828 M| Sample Calculations The following calculations were used for each Trial, but only inputs for Trial 1 will be shown below.Volume = Final buret reading – Initial buret reading i. Volume of NaOH = Final buret reading of NaOH – Initial buret reading of NaOH ii. Volume of NaOH = 32. 0 mL NaOH – 0. 00 mL NaOH iii. Volume of NaOH = 32. 0 mL Molarity = Moles/Liters i. Molarity of NaOH solution = (mass of KHP/molar mass of KHP) / Volume of NaOH ii. Molarity of NaOH solution = (0. 2966 g/204. 22 g)/0. 032 L iii. Molarity of NaOH solution = 0. 0454 M Molarity of acetic ac id = (Molarity NaOH * Volume NaOH) / Volume Acetic Acid i. Molarity of acetic acid = (0. 0472 M * 0. 0172 L)/ 0. 1 L ii. Molarity of acetic acid = 0. 0780 M Percent Error = Experimenal value-Accepted valueAccepted value*100 i. Percent Error of Molarity of NaOH = 0. 0472 M-0. 05 M0. 05 M*100 ii. Percent Error of Molarity of NaOH = 5. 6% i. Percent Error of Molarity of acetic acid = 0. 078 M-0. 080 M0. 080 M*100 ii. Percent Error of Molaarity of acetic acid = 2. 5% Discussion The results obtained from the experiment proved to the principle that using the indictor we can find the end point, which is very close to the equivalence point of an acidic solution.Then using that point we were able to calculate the unknown molarity which was one of the goals of the experiment. The calculations also verify Boyle’s Theory. When we calculated the molarity of the acetic solution, an average value of 0. 078 M was obtained. The true value of the molarity of the acetic acid solution was 0. 08 M. Although it isn’t right on, it is very close to the true value which leads me into discussing the percent error. We found the percent error of the molarity of NaOH to be 5. 6%, and the percent error of the molarity of acetic acid to be 2. 5%, which are both pretty small.The error may have occurred when adding NaOH solution. Occasionally slightly more pressure was put on tilts of the piece on the buret to allow the solution to flow through. This means that more of the solution may have been used than needed. Overall, experiment agrees with the formulated hypothesis. Pre-Lab and Post Lab Questions Pre-Lab 1. Molarity of NaOH solution = (mass of KHP/molar mass of KHP) / Volume of NaOH a. Molarity = (0. 2816 g/204. 22 g)/29. 68 mL Molarity = 4. 64*10-5 M 2. Molarity of acetic acid = (Molarity NaOH * Volume NaOH) / Volume Acetic Acid b.Molarity = ((4. 64*10-5 M)*20. 22 mL)/10. 06 mL Molarity = 9. 34*10-5 M Post Lab 1. A. TD B. TD 2. A graduated cylinder with calibration type TD could be used to deliver a certain amount of a liquid into another container. A graduated cylinder marked TC could be used to contain an accurate volume of a liquid that is to be mixed with another solution, where the experiment is to be done inside of that graduated cylinder. 3. 50g * 1mol /49. 997g = 1 mol 100g * 1mL / 1. 53g = 1L / 15. 3 1mol / (1L / 1. 53) = 1mol* 1. 53 / 1L = 15. 3 mol/L= 15. 3 M