Sandeep Voleti-Period 4/5 Honors Chemistry Titration Lab Write Up
Sandeep Voleti- Period 4/5 Honors Chemistry Titration Lab Writeup
Introduction
Titration is a method, which is meant to find the concentration of either an acid or a base by adding a measured amount of it to a known volume and concentration of an acid or base1. Titration starts with a beaker or Erlenmeyer flask containing a very precise volume of the known concentration solution and a small amount of indicator, which is put underneath a burette containing the solution with unknown concentration1. Small drops of the titrant are then added to the known solution and indicator until the indicator changes which means the endpoint has been reached. Single drops of the titrant can sometimes make a permanent or temporary change in the indicator2. …show more content…
Results for Titration of 3 Trials of HCl and 1 Trial of Acetic Acid (Vinegar) Trial # | Va | Vb | 1 | 10.00 mL | Start:0.000 mLFinish:3.700 mLVb:3.700 mL | 2 | 10.00 mL | Start:0.000 mLFinish:3.750 mLVb:3.750 mL | 3 | 10.00 mL | Start:0.000 mLFinish:3.750 mLVb:3.750 mL | Vinegar | 5.00 mL | Start:0.000 mLFinish:12.100 mLVb:12.100 mL |
Figure 1
Calculations Involved With Experiment Mb Value using 0.1 M HCL and measured Va and VbTrial 1:(0.1)(10.00)=(3.700)(x) x=0.270 | Mb Value using 0.1 M HCL and measured Va and VbTrial 2: (0.1)(10.00)=(3.750)(x) x=0.267 | Mb Value using 0.1 M HCL and measured Va and VbTrial 3: (0.1)(10.00)=(3.750)(x) x=0.267 …show more content…
10 ml of acid was used for all the three trials to keep a constant. During the first trial it took 3.700 ml of NaOH to titrate the HCl, in the second trial it took 3.750 ml, and in the third trial it took 3.750 ml. These differences in amounts of NaOH required to titrate the acid led to different molarities being calculated for the NaOH. The first being 0.270M and the other to being 0.267M. In order to calculate the total volume of base used, the readings from the burette from before and after titration were subtracted from each other. The average of the three molarities was found to be 0.268M for NaOH (Figure 2).
Figure 1 also displays the results of the titration of Acetic Acid in vinegar with the NaOH. As seen in the table, it took much more NaOH to titrate the vinegar than it did with the HCL, a difference of almost 9 mL. Furthermore, the Acetic Acid was also determined to have a higher concentration compared to the HCl, with a molarity of 0.649 M. (Figure 2)
Figure 2 shows the calculations that were used in order to determine the molarities of the NaOH, the total amount of base used, and the percent error for the molarity of the acetic acid. The titration formula used was: MaVa=MbVb.) The percent error calculated was 18.9%. (Figure
Introduction
Titration is a method, which is meant to find the concentration of either an acid or a base by adding a measured amount of it to a known volume and concentration of an acid or base1. Titration starts with a beaker or Erlenmeyer flask containing a very precise volume of the known concentration solution and a small amount of indicator, which is put underneath a burette containing the solution with unknown concentration1. Small drops of the titrant are then added to the known solution and indicator until the indicator changes which means the endpoint has been reached. Single drops of the titrant can sometimes make a permanent or temporary change in the indicator2. …show more content…
Results for Titration of 3 Trials of HCl and 1 Trial of Acetic Acid (Vinegar) Trial # | Va | Vb | 1 | 10.00 mL | Start:0.000 mLFinish:3.700 mLVb:3.700 mL | 2 | 10.00 mL | Start:0.000 mLFinish:3.750 mLVb:3.750 mL | 3 | 10.00 mL | Start:0.000 mLFinish:3.750 mLVb:3.750 mL | Vinegar | 5.00 mL | Start:0.000 mLFinish:12.100 mLVb:12.100 mL |
Figure 1
Calculations Involved With Experiment Mb Value using 0.1 M HCL and measured Va and VbTrial 1:(0.1)(10.00)=(3.700)(x) x=0.270 | Mb Value using 0.1 M HCL and measured Va and VbTrial 2: (0.1)(10.00)=(3.750)(x) x=0.267 | Mb Value using 0.1 M HCL and measured Va and VbTrial 3: (0.1)(10.00)=(3.750)(x) x=0.267 …show more content…
10 ml of acid was used for all the three trials to keep a constant. During the first trial it took 3.700 ml of NaOH to titrate the HCl, in the second trial it took 3.750 ml, and in the third trial it took 3.750 ml. These differences in amounts of NaOH required to titrate the acid led to different molarities being calculated for the NaOH. The first being 0.270M and the other to being 0.267M. In order to calculate the total volume of base used, the readings from the burette from before and after titration were subtracted from each other. The average of the three molarities was found to be 0.268M for NaOH (Figure 2).
Figure 1 also displays the results of the titration of Acetic Acid in vinegar with the NaOH. As seen in the table, it took much more NaOH to titrate the vinegar than it did with the HCL, a difference of almost 9 mL. Furthermore, the Acetic Acid was also determined to have a higher concentration compared to the HCl, with a molarity of 0.649 M. (Figure 2)
Figure 2 shows the calculations that were used in order to determine the molarities of the NaOH, the total amount of base used, and the percent error for the molarity of the acetic acid. The titration formula used was: MaVa=MbVb.) The percent error calculated was 18.9%. (Figure