The equilibrium constant was found by titrating a series of reactions containing H2O‚ HCl‚ and the unknown ester with only the last solution containing the unknown alcohol. Equilibrium constants can be easily found once the concentrations of all reactants are known using Keq= [ acid][alcohol]/[ester][water]. Procedure: Part A: Prepare six bottles with the following volumes in each: Bottle # 3M HCl in mL H2O in mL Ester in mL Alcohol in mL 1 5 5 0 0 1/A 5 5 0 0 2 5 0 5 0 3 5 1 4 0 4
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University of Detroit Mercy Quantitative Analysis Lab CHM 3880 Fall 2011 Partner: Edwin Gay Abstract The PH at each point during the titration of sodium carbonate unknown sample was determined. An Unknown sample of Na2CO3 was titrated with a standard HCL solution. In addition to titration‚ the pH at each point of titration was measured using PH meter. The % of the unknown Na2CO3 was 25.83% Introduction1 The purpose of this experiment was to determine the actual pH at each point during the titration
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sample was converted to its equivalent neutralizing capacity in terms of sodium carbonate. A mass of the impure sample was dissolved and diluted in distilled water. 3 drops of indicator was mixed and the solution was titrated with the standard acid (HCl) to the endpoint. Volumetric analysis is a quantitative analytical technique which employs a titration in comparing an unknown with a standard. A titration reaction may use two indicators‚ as in the analysis of carbonate-bicarbonate mixture. The two
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Aim: To compare the foaming capacities of five different commercial soaps. Apparatus: 5 test tubes‚ 5 conical flasks (100 ml)‚ test tube stand‚ Bunsen burner and stop watch. Materials Required: 5 different samples of soap and distilled water Theory: The foaming capacity of a soap sample depends upon the nature of soap and its concentration. This can be compared for various samples of soaps by taking the same concentration of solution and shaking them.The foam is formed and the time taken
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starch (cooled) + 3 ml saliva (cooled) in ice bath; Tube 4 = 3 ml starch + 3 ml saliva 5 drops conc. HCL in water bath. An incubation period of 1 hour was followed and each was tested for starch and maltose. A similar procedure was repeated with pepsin‚ with the test tubes prepared as follows: Tube 1 = 5 ml pepsin 5% soln + 5 ml HCL (0.5)%; Tube 2 = 5 ml pepsin (5% soln)+ water 5 ml; Tube 3 = 5 ml HCL (0.5%) + 5 ml of water; Tube 4 = 5 ml pepsin 5% soln) + 5 ml NAOH (0.5%). Lastly‚ the above procedure
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the acid and the base. In this experiment acid-base titration will be used to determine the concentration of HCl at equilibrium when it reacts with the NaOH H+ (aq)+ Cl- (aq) + Na+ (aq) + OH- →H2O(l) + Na+ (aq)+ Cl- (aq) Procedure When performing this experiment one must first obtain and wear goggles. Next add 40mL of distilled water to a 100mL beaker‚ then add 5.00mL of HCl to the beaker.Then obtain 40mL of 0.1M NaOH. Place the NaOH in a 60mL reagent reservoir and drain a small amount
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and also my merging with a company called AXON. Different Ways Messaging Was Used Instead of holding a live Blueprint meeting‚ HCL decided to have all three hundred of their managers record their business plans‚ which would then be posted on the MyBlueprint portal. Once all of the business plans were posted‚ they’d be open for review by another eight thousand HCL managers. The idea was to transform the planning process into a peer-to-peer review rather than a top-down judgment. (Nayar‚ 2010)
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Prof. T.D. Macasil BSE Physical Science Reactions of amines Reaction of amines with acids – acids and amines form ammonium salts. R – NH2 + HCl → R – NH3+ + OH – Amine acid amine salt Example 1 CH3 – NH2 + HCl → CH3 – NH3 + Cl- Methylamine Methylammonium Example 2 CH3CH2 – NH3 + Cl → CH3CH2NH2 . HCl Ethylammonium Ethylamine Reaction of amine with water – when amines react with water‚ they produce hydroxide ( OH ) R – NH2 + H2O
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5.00 mL HCl soln. x 1.217g HCl soln = 6.085g HCl soln. 1 mL HCl soln 3.0 mol HCl x 36.5g HCl x 1mL soln. = 0.0899g HCl x 5mL = 0.4495 g HCl 1000mL soln mol HCL 1.217g HCl soln mL 6.085g HCl soln. 0.4495 g HCl = 5.63 g H2O 5.63 g H2O x 1 mole H2O = 0.312 moles H2O from HCl solution 18.02 g H2O Then the product amounts at equilibrium were calculated
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yield versus the theoretical yield of NaCl when HCl is titrated into Na2CO3 and NaHCO3. When 0.15g of both NaHCO3 and Na2CO3 are titrated with HCl‚ then 0.165g of NaCl should form from the NaHCO3‚ and 0.104g of NaCl should form from the 0.15g of Na2CO3. Procedure: Weigh 2 samples of 0.15g of dried unknown each‚ and dissolve each into 50mL of distilled water. Add 0.5 to 1mL of bromocresol green indicator until the solution turns blue. Titrate the HCl until it turns green. Gently heat and boil out
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