Volumetric Vinegar Analysis

Experiment 9 and 10: Volumetric/Vinegar Analysis Abstract: The goal of the experiment that was conducted was to figure out both the molar concentration of NaOH and the standard mole ratio of the NaOH solution. In order to find the concentration of the NaOH solution, volumetric analysis was used. In volumetric analysis, a titration mechanism was utilized in order to find the reaction that the base will end up having with KHC8H4O4. , also known as KHP. Phenolphthalein, which is the indicator that was used in this experiment, assisted in figuring out at exactly what point was there neutralization.The indicator turns the solution into a bright pink color once neutralization has occurred. In experiment 10, the average molarity of NaOH that was found in experiment nine was used in order to find out if the vinegar that was being used in the experiment contained around the same percent mass of acetic acid that is found in regular vinegar. The experimental value of NaOH that was used was 1. 0 425 grams and the molarity of NaOH was found to be 0. 089 m/L of NaOH. Towards the conclusion of the experiment, the average percent mass of acetic acid was calculated and found to be 1. 695%.Regular house hold vinegar’s average percent mass of acetic acid usually ranges to 4-5%. Based on the percent mass of acetic acid obtained in the experiment, the vinegar that was used in experiment 10 was clearly not house hold vinegar. The hypothesis for this experiment was, if the average percent mass of acetic acid ranged between 4-5%, then it is house hold vinegar. However, due to the results from the experiments conducted, this hypothesis was rejected. In order to obtain the results that the groups were searching for, titration was used in both experiments to find the answer.The method of titration involves the measurement of KHP and NaOH. Afterwards, the volumetric analysis was used, with the indicator included. The experiment starts by finding the measurements of KHP. The indicato r was added later on, and then the titration began with the NaOH solution. It was apparent once the solution was neutralized because the indicator caused the solution to turn bright pink. The experiment also required the utilization of volumetric mass in order to find the percent mass of acetic acid in vinegar.The mass of vinegar is then titrated along with the indicator endpoint with the sodium hydroxide solution. In order to find the average acetic percent mass of vinegar, the concentration found in NaOH in experiment 9 was utilized together with the known volume of NaOH. Materials: Please refer to Experiment 9 and 10 on pages 127-136 and 137-142, of Laboratory Manual for Principles of General chemistry 9th Edition by J. A. Beran. The only deviation that was performed during this experiment was the two to three extra drops of the indicator phenolphthalein in order to distinguish a titration point.Results: Experiment 9: Data: |Table 1: Measurement |Trial 1 |Trial 2 | |Mass of KHC8H 4O4. (g) |. 509 g |. 501 g | |Buret Reading of NaOH (mL) |28. 3 mL |26. 7 mL | Table 1 shows the measurements recorded for experiment 9, volumetric analysis Table 2: Calculations |Trial 1 |Trial 2 | |Moles of KHC8H4O4 (mol) |. 000303 |. 0002485 | |Volume of NaOH Dispensed (L) |. 0034 |. 0032 | |Molar Concentration of NaOH (mol/L) |. 089 |. 089 | Table 2 shows the calculations derived from experiment 9, volumetric analysis Calculations:Moles of KHC8H4O4 x 1 mol KHC8H4O4/ Molar Mass KHC8H4O4: 0. 089 m/L NaOH x 0. 0034 L= . 000303 moles NaOH 0. 089 m/L NaOH x 0. 0032 L= 0. 0002485 NaOH Volume of NaOH Dispensed (mL): Buret Reading of NaOH= 28. 3 mL, 26. 7 mL Molar Concentration Concentration of NaOH: 2. 45 x 10 -3 mol OH-/. 0275 L NaOH = 0. 089 M/L NaOH Results: Experiment 10 |Table 3: Measurement |Trial 1 |Trial 2 | |Mass of Vinegar (g) |1. 048 g |1. 37 g | |Buret Reading of NaOH (mL) |3. 4 mL |3. 2 mL | Table 3 shows the measurements recorded for experiment 10, vinegar analysis |Table 4: Calculations |Trial 1 |Trial 2 | |Volume of NaOH Used (mL)(L) |3. 4(. 0034) |3. 2(. 0032) | |Molar Concentration of NaOH (mol/L) (given) |0. 089 |0. 89 | |Molar Mass of Acetic Acid (g/mol) |. 0182 |. 0171 | |Mass of Acetic Acid in Vinegar (g) |1. 048 g |1. 037 g | |Avg. Percent Mass of Acetic Acid in Vinegar (%) |1. 695% | | Table 4 shows the calculations derived from experiment 10, vinegar analysis. Calculations: 1. Molar Concentration of NaOH (mol/L) Given (. M Solution) 2. Mass of Acetic Acid in Vinegar (g): Moles of Acetic Acid (mol) x Molar Mass of Acetic Acid (g/mol): 3. 026 x 10 -4moles of acetic acid x 60. 05 g/mol= . 0182 g 2. 848 x 10 -4moles of acetic acid x 60. 05 g/mol= . 0171 g 3. Avg. Percent Mass of Acid in Vinegar (%): 1. 65%+1. 74%/2= 1. 695% Discussion: The experiment began by adding NaOH to the mixture of deionized water and KHP in the beaker. The H+ ion that is found in KHP, reacted to the OH- ions that are found in the NaOH solution, even as more of the Na OH continued to be added into the mixture.When there turned out to be an abundance of NaOH, there were no longer any H+ to be added to KHP. As a solution, the extra OH-ions were found in the NaOH solution was used to make the indicator activate and make the solution turn pink. It was imperative that the solution be mixed the correct way. If it was not mixed the correct way, the results from the experiment will be inaccurate. If the reading had proven to be inaccurate because of that mistake, the volume of the NaOH solution mixed with the KHP will eventually get neutralized to a point where the numbers in the results would be very off.Two trials were done in this experiment in order to ensure that that mistake never happened and the volume of NaOH was found. Once the solution had finally been able to neutralize, the moles of the KHP were found and ended up being equal to the moles of NaOH. This information allowed for the molarity to be found. The average molarity that was in NaOH ha d been found in experiment 9, it was . 089 M. Both experiments 9 and 10 seemed to have similar traits because both of them involved titration. The titration was used in order to find the number of moles that was found in the acetic acid of the vinegar solution that was used.The normal amount of acetic acid found in household vinegar is between 4-5%. The experiments helped determine that household vinegar was definitely not the vinegar that was being used since the acetic amount that was found was 1. 695%. Conclusion The hypothesis was proven in the first experiment because the base of NaOH did end up neutralizing KHP’s acids. The indicator turned the solution pink; therefore the hypothesis in the first experiment was not rejected. The experiment involving the molarity of NaOH was very close in numbers. The molarity that was given was . 1 M, and the molarity that was found in the experiment was . 89 M. The hypothesis for the second experiment was â€Å"If the average percent mass of acetic acid ranged between 4-5%, then the vinegar that was being used for the experiment was household vinegar. † However, since the average percent mass of acetic acid resulted as 1. 695%, which was lower than household vinegar; this caused the hypothesis to be rejected. Works Cited Beran, Jo A. Laboratory Manual for Principles of General Chemistry. Hoboken, NJ: Wiley, 2011. Print. Tro, Nivaldo J. Principles of Chemistry: A Molecular Approach. Upper Saddle River, NJ: Prentice Hall, 2010. Print. Volumetric Vinegar Analysis Experiment 9 and 10: Volumetric/Vinegar Analysis Abstract: The goal of the experiment that was conducted was to figure out both the molar concentration of NaOH and the standard mole ratio of the NaOH solution. In order to find the concentration of the NaOH solution, volumetric analysis was used. In volumetric analysis, a titration mechanism was utilized in order to find the reaction that the base will end up having with KHC8H4O4. , also known as KHP. Phenolphthalein, which is the indicator that was used in this experiment, assisted in figuring out at exactly what point was there neutralization.The indicator turns the solution into a bright pink color once neutralization has occurred. In experiment 10, the average molarity of NaOH that was found in experiment nine was used in order to find out if the vinegar that was being used in the experiment contained around the same percent mass of acetic acid that is found in regular vinegar. The experimental value of NaOH that was used was 1. 0 425 grams and the molarity of NaOH was found to be 0. 089 m/L of NaOH. Towards the conclusion of the experiment, the average percent mass of acetic acid was calculated and found to be 1. 695%.Regular house hold vinegar’s average percent mass of acetic acid usually ranges to 4-5%. Based on the percent mass of acetic acid obtained in the experiment, the vinegar that was used in experiment 10 was clearly not house hold vinegar. The hypothesis for this experiment was, if the average percent mass of acetic acid ranged between 4-5%, then it is house hold vinegar. However, due to the results from the experiments conducted, this hypothesis was rejected. In order to obtain the results that the groups were searching for, titration was used in both experiments to find the answer.The method of titration involves the measurement of KHP and NaOH. Afterwards, the volumetric analysis was used, with the indicator included. The experiment starts by finding the measurements of KHP. The indicato r was added later on, and then the titration began with the NaOH solution. It was apparent once the solution was neutralized because the indicator caused the solution to turn bright pink. The experiment also required the utilization of volumetric mass in order to find the percent mass of acetic acid in vinegar.The mass of vinegar is then titrated along with the indicator endpoint with the sodium hydroxide solution. In order to find the average acetic percent mass of vinegar, the concentration found in NaOH in experiment 9 was utilized together with the known volume of NaOH. Materials: Please refer to Experiment 9 and 10 on pages 127-136 and 137-142, of Laboratory Manual for Principles of General chemistry 9th Edition by J. A. Beran. The only deviation that was performed during this experiment was the two to three extra drops of the indicator phenolphthalein in order to distinguish a titration point.Results: Experiment 9: Data: |Table 1: Measurement |Trial 1 |Trial 2 | |Mass of KHC8H 4O4. (g) |. 509 g |. 501 g | |Buret Reading of NaOH (mL) |28. 3 mL |26. 7 mL | Table 1 shows the measurements recorded for experiment 9, volumetric analysis Table 2: Calculations |Trial 1 |Trial 2 | |Moles of KHC8H4O4 (mol) |. 000303 |. 0002485 | |Volume of NaOH Dispensed (L) |. 0034 |. 0032 | |Molar Concentration of NaOH (mol/L) |. 089 |. 089 | Table 2 shows the calculations derived from experiment 9, volumetric analysis Calculations:Moles of KHC8H4O4 x 1 mol KHC8H4O4/ Molar Mass KHC8H4O4: 0. 089 m/L NaOH x 0. 0034 L= . 000303 moles NaOH 0. 089 m/L NaOH x 0. 0032 L= 0. 0002485 NaOH Volume of NaOH Dispensed (mL): Buret Reading of NaOH= 28. 3 mL, 26. 7 mL Molar Concentration Concentration of NaOH: 2. 45 x 10 -3 mol OH-/. 0275 L NaOH = 0. 089 M/L NaOH Results: Experiment 10 |Table 3: Measurement |Trial 1 |Trial 2 | |Mass of Vinegar (g) |1. 048 g |1. 37 g | |Buret Reading of NaOH (mL) |3. 4 mL |3. 2 mL | Table 3 shows the measurements recorded for experiment 10, vinegar analysis |Table 4: Calculations |Trial 1 |Trial 2 | |Volume of NaOH Used (mL)(L) |3. 4(. 0034) |3. 2(. 0032) | |Molar Concentration of NaOH (mol/L) (given) |0. 089 |0. 89 | |Molar Mass of Acetic Acid (g/mol) |. 0182 |. 0171 | |Mass of Acetic Acid in Vinegar (g) |1. 048 g |1. 037 g | |Avg. Percent Mass of Acetic Acid in Vinegar (%) |1. 695% | | Table 4 shows the calculations derived from experiment 10, vinegar analysis. Calculations: 1. Molar Concentration of NaOH (mol/L) Given (. M Solution) 2. Mass of Acetic Acid in Vinegar (g): Moles of Acetic Acid (mol) x Molar Mass of Acetic Acid (g/mol): 3. 026 x 10 -4moles of acetic acid x 60. 05 g/mol= . 0182 g 2. 848 x 10 -4moles of acetic acid x 60. 05 g/mol= . 0171 g 3. Avg. Percent Mass of Acid in Vinegar (%): 1. 65%+1. 74%/2= 1. 695% Discussion: The experiment began by adding NaOH to the mixture of deionized water and KHP in the beaker. The H+ ion that is found in KHP, reacted to the OH- ions that are found in the NaOH solution, even as more of the Na OH continued to be added into the mixture.When there turned out to be an abundance of NaOH, there were no longer any H+ to be added to KHP. As a solution, the extra OH-ions were found in the NaOH solution was used to make the indicator activate and make the solution turn pink. It was imperative that the solution be mixed the correct way. If it was not mixed the correct way, the results from the experiment will be inaccurate. If the reading had proven to be inaccurate because of that mistake, the volume of the NaOH solution mixed with the KHP will eventually get neutralized to a point where the numbers in the results would be very off.Two trials were done in this experiment in order to ensure that that mistake never happened and the volume of NaOH was found. Once the solution had finally been able to neutralize, the moles of the KHP were found and ended up being equal to the moles of NaOH. This information allowed for the molarity to be found. The average molarity that was in NaOH ha d been found in experiment 9, it was . 089 M. Both experiments 9 and 10 seemed to have similar traits because both of them involved titration. The titration was used in order to find the number of moles that was found in the acetic acid of the vinegar solution that was used.The normal amount of acetic acid found in household vinegar is between 4-5%. The experiments helped determine that household vinegar was definitely not the vinegar that was being used since the acetic amount that was found was 1. 695%. Conclusion The hypothesis was proven in the first experiment because the base of NaOH did end up neutralizing KHP’s acids. The indicator turned the solution pink; therefore the hypothesis in the first experiment was not rejected. The experiment involving the molarity of NaOH was very close in numbers. The molarity that was given was . 1 M, and the molarity that was found in the experiment was . 89 M. The hypothesis for the second experiment was â€Å"If the average percent mass of acetic acid ranged between 4-5%, then the vinegar that was being used for the experiment was household vinegar. † However, since the average percent mass of acetic acid resulted as 1. 695%, which was lower than household vinegar; this caused the hypothesis to be rejected. Works Cited Beran, Jo A. Laboratory Manual for Principles of General Chemistry. Hoboken, NJ: Wiley, 2011. Print. Tro, Nivaldo J. Principles of Chemistry: A Molecular Approach. Upper Saddle River, NJ: Prentice Hall, 2010. Print.