The equivalence point of titration occurred at a volume of 31 mL 0.081M NaOH (aq). The halfway volume of titration was thus 15.5 mL and the pH of the titration system at this point, as determined graphically, was 5.08. After identification of half point, Henderson–Hasselbalch equation pH=pKa, was used to determine the value of Ka. The resulting Ka value and the concentration of KHP made it possible to calculate the initial pH of the acid, which were 3.19. How it was expected the pH at equivalence point was 9.17 this is because of the domination of hydroxide ion in solution.
Or Reaction w/ hot H20: None Litmus test: R-turned blue, B-None Dilute HCI: Bubbling Dilute NaOH: None Substance: CuC03 Color: Green Odor: None Effect of Heat: Turned black Solub. Or Reaction w/cold H20: None Solub. Or Reaction w/ hot H20: None Litmus test: R-None, B-None Dilute HCI: Turned lime green and bubbled Dilute NaOH: Turned teal Substance: Cu(NO3)2 Color: Blue Odor: None Effect of Heat: None Solub. Or Reaction w/cold H20: Dissolves Solub. Or Reaction w/ hot H20: None Litmus test: R-None, B-Turned orange Dilute HCI: Turned Green Dilute NaOH: None Substance: NaCl Color: White Odor: None Effect of Heat: None Solub.
The NMR spectrum does contain impurities including methanol (4.80 ppm), methyl oleate (5.40 ppm), CHCL3 (7.20 ppm), acetone (2.00 ppm), and water (1.60 ppm). The methyl oleate may be present due to incomplete reaction with hydrogen and the CHCL3 may be present due to the contamination of CDCL3 NMR solvent. Acetone was used as a cleaning agent on the apparatus before starting the experiment. The experimental NMR spectrum does match that of an authentic spectrum except the experimental spectrum contains a greater number of peaks due to
Let’s call this number X. Then the structural formula of BaCl2 hydrate can be written as BaCl2•XH2O. The reaction of dehydration is |BaCl2•XH2O ( BaCl2 + XH2O |(5) | According to the stoichiometry of the reaction (5) |[pic] |(6) | Where N1 is the number of moles of BaCl2 formed in the reaction (5) and N2 is the number of moles of water lost in the reaction (5). To find out N1 you need to divide the mass of BaCl2 after the reaction by its molar mass. To find out N2 you have to determine the mass of water produced in the reaction.
In this mixture, there are no enzymes present and therefore no products are formed. The reaction rate was rated 0, just as we anticipated. 2. The 2nd experiment was a mixture of 2 mL H2O2 and 0.1g Manganese dioxide. We can use the same principal we used for the first experiment, Manganese dioxide is not a protein, not an enzyme, it is a catalyst.
Overall reaction of the acid-catalyzed dehydration of 2-methylcyclohexanol Figure 2. Reactions for the unsaturation tests Table of Reactants and Products: Table 1. Molecular weight, density, melting point, and boiling point for all reagents used in Experiment #7 – Dehydration of 2-Methylclyclohexanol, Tests for Unsaturation, and Gas Chromatography. Name: | MW (g/mol): | Density (g/mL): | MP (C): | BP (C): | MSDS: | 2-methylcyclohexanol | 114.19 | 0.93 | -38.00 | 165-168 | Irritant, flammable | 95% phosphoric acid | 97.994 | 1.69 | 42.35 | 158 | Corrosive, irritant | 1-methylcyclohexene | 96.17 | 0.81 | -120.4 | 110 | Irritant, flammable | 3-methylcyclohexene | 96.17 | 0.81 | -124.00 | 104 | Irritant, flammable | Procedure: Part One – Dehydration of 2-methylcyclohexanol A microdistillation apparatus was assembled by securing a microdistillation flask to a ring stand and submerging it in an empty heating mantle. A thermometer and Teflon adaptor were tightly sealed into the top of the flask with the thermometer bulb being below the side arm.
Experiment 5: Preparation of Cyclohexene- Acid Catalyzed Elimination of Cyclohexanol Ball State University Organic Chemistry Lab 231 (Sec. 5) Dr. Bock/Daniel Miller Tiffany Raber 5 November 2012 Purpose: To prepare an alkene by an elimination reaction of an alcohol in the presence of sulfuric acid as a catalyst to produce a successful yield and desired results for the purity of the alkene. Introduction: The dehydration of an alcohol such as cyclohexanol is a useful technique for generating alkenes. The success of this reaction relies on a dehydration, such as the presence of a strong acid (sulfuric acid). Because this reaction is reversible, a fractional distillation is necessary to produce cyclohexene, water, and some impurities and a simple distillation is followed to further purify the product.
Record several points of pH and NaOH added (especially near equivalence point) to be use later to prepare a titration curve. Observations and Results Part I: Solution | pH | 0.1 M HCl | .70 | 0.1 M NaOH | 13.30 | Part II: Volume of 0.1 M NaOH at equivalence point: 35mL pH at equivalence point: 11.45 Molarity of the Unknown Acid A (HCl): 2.0 x 10-4 Discussion In this lab, we found out that water self ionizes itself into hydrogen ion and hydroxide ion naturally to a very small extent. An indicator, in an acid base reaction, is a substance whose color changes over a particular pH range. Phenolphthalein is an example of an indicator which changes from colorless to pink as pH goes from 8 to 10. We plotted the pH against the amount of base added producing a
The reaction is a synthesis. The Kc for this reaction is Kc = 49.7 at 458oC [Answer: [H2] = [I2] = 1.2 × 10-2 M, [HI] = 8.6 × 10-2 M] 2. Iodine and bromine react to give iodine monobromide, IBr. What is the equilibrium composition of a mixture at 150oC that initially contained 0.0015 mol each of iodine and bromine in a 5.0 L vessel? The equilibrium constant Kc for this reaction at 150oC is 1.2 102.
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.