Results Sheet, Experiment 13 Brand of Vinegar used:________________ Acetic Acid % from bottle label_____________ | Initial NaOH Reading(estimate to 0.1 mL) | Final NaOH Reading(estimate to 0.1 mL) | Volume NaOH used(Final – Initial) | Trial 1 | 9.5 | 1.3 | 8.2 | Trial 2 | 9.6 | 1.4 | 8.2 | Trial 3 | 9.7 | 1.4 | 8.3 | Average Volume of NaOH used | 8.2 | Calculations: Normality of the acetic acid: .82 Mass of acetic acid: 49.2 % acetic acid (divide the grams, above, by the volume you used to get g/L): 4.92 Answer questions A-G in the lab manual: A. What is the average % acetic acid in your vinegar sample? Standard vinegar is 5% acetic acid. How does your result compare with the standard? The average was 4.92 which is pretty close comparably to the standard.
The melting point of the unknown was determined to be between 90-110 degrees. Based on the melting point we identified the identity of the unknown as Acetanilide. Acetanilide is an odorless, white flake solid or crystalline powder (pure form) which is soluble in hot water alcohol, ether, chloroform, acetone, glycerol, and benzene. Acetanilide is used as an inhibitor of peroxide sand as a stabilizer for cellulose ester varnishes. Acetanilide is also used as an intermediate for the synthesis of rubber accelerators, dyes, and is used as a precursor in penicillin synthesis.
A binding material can be separated by dissolving the tablet in ether since three active ingredients are soluble. Then, aqueous K2HPO4 and KOH are used to extract the ether solution with the conjugate base of the Aspirin layer and the conjugate base of the Acetaminophen, respectively. The caffeine can be isolated from the left over layer by evaporating the ether. Two separated solutions form solid by acidifying with HCl and the solid residue can be collected by vacuum filtration. According to the Revell's lab experiment, aspirin, acetaminophen, caffeine recoveries are 60% or higher, <10% and <5%, respectively.
One mixture will contain vinegar, and the other will contain distilled water. What I expect to find it that the mixtures containing the vinegar’s pH will remain relatively constant, while the distilled water mixture will change drastically. Results and Observations: 1) The colors of the mixtures varied from drop to drop. One mixture turned blue, while the other turned a gold-ish color. Data Table 1: Add 0.1 M HCl| Drops|pH|Color| 0|6.0|Yellow| 1|6.0 |Yellow| 2|6.0|Yellow| 3|6.0|Yellow| 4|6.0|Yellow| 5|6.0|Yellow| 6|6.0 |Yellow| 7|6.0 |Yellow| 8|6.0 |Yellow| 9|6.0 |Yellow| 10|6.0 |Yellow| 11|6.0 |Yellow| 12|6.0 |Yellow| 13|6.0 |Yellow| 14|6.0 |Yellow| 15|6.0 |Yellow| Data Table 2: Add 0.1 M NaOH| Drops|pH|Color| 0|6.0|Yellow| 1|6.0|Yellow| 2|6.0|Yellow| 3|6.0|Yellow| 4|6.0|Yellow| 5|6.0|Yellow| 6|6.0|Yellow| 7|6.0|Yellow| 8|6.0|Yellow| 9|6.0|Yellow| 10|6.0|Yellow| 11|6.0|Yellow| 12|6.0|Yellow| 13|6.0|Yellow| 14|6.0|Yellow| 15|6.0|Yellow| Data Table 3: Add 6M HCl| Drops|pH|Color| 0|6.0|Yellow| 1|5.8|Yellow Orange| 2|5.6|Yellow Orange| 3|5.4|Light Orange| 4|5.0|Orange| 5|4.5|Orange| Data Table 4: Add 6M NaOH| Drops|pH|Color| 0|6.0 |Yellow| 1|6.2|Yellow Green| 2|6.4|Yellow Green|
This is used when a weak acid is being used in the reaction as it acid does not fully dissociated and there is an A- value. This is the equation used for reaction (ii) as it used ClCHCOOH as the acid. Results, Observations and Discussion: (ii) 0.25M HCl | (iii) 0.5M Cl2CHCOOH | (iv) 0.125M HCl | Time(mins) | Vol (Na2S2O3) | Time(mins) | Vol(Na2S2O3) | Time(min) | Vol (Na2S2O3) | 0 | 0.46 | 0 | 0.4 | 0 | 0.33 | 6 | 0.44 | 8 | 0.4 | 6 | 0.28 | 12 | 0.4 | 16 | 0.36 | 12 | 0.24 | 18 | 0.36 | 24 | 0.29 | 18 | 0.22 | 24 | 0.32 | 32 | 0.26 | 24 | 0.2 | 30 | 0.28 | 40 | 0.2 | 30 | 0.19 | Calculations: Finding Kobs - (ii) -dI2dt=0.005
Recrystallise the crude ester from methanol using a quantity of hot solvent approximately twice the minimum required for complete solution (This is to ensure that the phenyl benzoate does not separate until the temperature of the solution has fallen below the melting point of the ester). Phenyl benzoate separates as colourless crystals. 5. Determine the melting point of the product. Results : Yield (mass in gram) = __________________0.128 g_____________________ Melting point of pure phenyl benzoate (OC) = ___________70O______________ Melting point of your phenyl benzoate (OC) = ___________69O______________ Questions : 1.
Of the two remaining, one turns slightly yellow, while the other remains blue. What can you assume about the acid/base content of these particular flavors of Vitamin Water? * You could come to the conclusion that the first three will have a balance while the other two do not. The one that turned yellow, would be an acid. The one that stayed blue is a base.
Aim Prepare standard NaOH and HCl to determine the concentration of acetic acid in vinegar as well as ammonia in window cleaner via titration using suitable indicators. Results and Calculations 1.1. Titration Set 1: Standardization of NaOH 1.2.1 Results of Titration of KHC8H4O4 against NaOH Titration | 1 | 2 | 3 | Final Volume of Titrant/mL | 8.60 | 13.55 | 18.50 | Initial Volume of Titrant/mL | 3.20 | 8.60 | 13.55 | Volume of Titrant used/mL | 5.40 | 4.95 | 4.95 | | | | | Indicator used | Phenolphthalein | Colour Change at Endpoint | Colourless to Pink | 1.2.2 Calculations Mass of KHC8H4O4 used = 1.0291 g Molar Mass of KHC8H4O4 = 204.22 g/mol Molarity of KHC8H4O4 = (1.0291 ÷ 204.22) ÷ 0.1000 = 0.05039 M No. of moles of KHC8H4O4 in 10.0 mL = (10.0 ÷ 1000) × 0.05039 = 5.039 ×10-4 mol Average volume of NaOH titrant used = 4.95 mL Chemical Equation of Reaction: KHC8H4O4 (aq) + NaOH (aq) KHC8H4O3-Na+ (aq) + H2O (l) KHC8H4O4 reacts with NaOH in a 1:1 ratio, thus No. of moles of NaOH in 4.95 mL = 5.039 ×10-4 mol Concentration of NaOH = (5.039 ×10-4) ÷ (4.95÷ 1000) = 0.1018 M 1.2.
Calculate the molarity of the original vinegar solution and its concentration in gdm-3, given that it reacts with NaOH in a 1:1 ratio. 7. 2.5 g of a sample of ethanedioic acid, H2C2O4.nH2O, was dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 15.8 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Given that ethanedioic acid reacts with NaOH
Because the reaction is set in motion by the backside attack of the nucleophile, the nucleophile and the electrophile are in anti-configuration (chemwiki.org). Nerolin is part of a group of alkyl aryl ethers, compounds vastly used in the production of perfumes- most specifically as perfume additives and scent fixatives. These scents include rose, lavender and lemon (prelab-notes). Calculation of percent yield Iodoethane (MW=155.97., d=1.940 g/ml) Amount used: 0.30 ml d=mv m= 1.940g/ml x 0.30 ml 0.582g | 0.5 g 2-naphthol .1 mol of 2-naphthol144.17g= 0.00347 mol of naphthol 0.582 g iodoethane .1 mol of iodoethane155.97g= 0.00373 mol of iodoethane KOH (MW=56.11.) 0.200 g KOH .1 mol of KOH56.11 g= 0.00356 mol of KOH Given the 1:1 stoichiometry of the reaction, 2-naphthol is the limiting reagent Theoretical yield = 0.5 g Percent yield=actual yieldtheoretical yield .100% 0.3189 g0.5g .100% 63.8 % | Conclusion A 63.8% yield of nerolin was recovered in the reaction of 2-naphthol and ethyl iodide.