Formula: for NaoH N NaoH = grm KHP / ml NaoH x KHp /1000ml Formula: for HCL (NV) HCL = (NV) NaoH or N HCL V HCL = N NaOH V NaoH N HCL = VNAOH (NNaoH) / VHCL (for more info page 62-63 of the photocopy) KHP = Potassium hydrogen phthalate = KHC8H4O4 = Mw - 204.23 Trial 1: NNaoH = 1grm / 20.6 x 204.23 /1000 = 1grm / 4.21 = 0.24 normal NHCL = 20.6 (0.24) / 39.5 = 4.94 / 39.5 = 0.125 or 0.13 Trial 2: NNaoH = 1grm / 20.8 x 204.23 /1000
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NaOH Standardization and Titration of an Unknown Organic Acid Overview: Methods for counting the number of molecules in a sample is a major emphasis of laboratory work. In this experiment we will use the method of titration to count the number of acid molecules in a solution. Measuring mass is a relatively easy procedure to do in the lab (although a balance is expensive). Counting the number of particles requires more effort. Molecular counting can be done by setting an unknown amount of a substance
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EXPERIMENT 10 Volumetric Analysis I Standardization of NaOH Solution Outcomes After completing this experiment‚ the student should be able to: 1. 2. 3. 4. Demonstrate the concept of quantitative analysis. Make solution and standardize it. Explain the difference between primary and secondary standard solutions. Quantitatively determine the concentration of a base. Introduction Titration is a common method of quantitative analysis used to determine the concentration of an unknown substance in a solution
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Experiment 5- Standardization of NaOH and determination of Molarity of an unknown Acid Objectives 1. Preparation and standardization of a 0.1M NaOH solution 2. To learn the technique of titration 3. Determination of the concentration of an unknown diprotic acid. Introduction Titration can be traced to the origins of volumetric analysis‚ which began in the late eighteenth century. Study of analytical chemistry began in France and the first burette was made by Francois Antoine Henri
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| Preparation and standardization of 0.1 M NaOH using KHP | Analysis of unknown solid containing KHP | | Steanie Rodriguez | 6/7/2011 | CHEM-C311 Analytical Chemistry Laboratory July 7‚ 2011 The main purpose of this experiment is to identify an unknown organic acid by conducting various experiments to determine the acid’s unique properties. By determining selected constant properties of the unknown and then comparing these properties to the constant properties of known substances
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Standardization of a NaOH Solution with Potassium Hydrogen Phthlate (KHP) Name:________________________________________________________________ Period:_____ Prelab 1. A 0.8234-g sample of "KHP" required 38.76 mL of NaOH for titration to the phenolphthalein endpoint. What is the exact molarity of the NaOH solution? 2. A 25.00-mL aliquot of an unstandardized HCl solution is titrated with the previously standardized NaOH solution from #1 above. If 32.55 mL of NaOH titrant
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Chemistry SL! Research Question! How is the solubility of sodium hydroxide (NaOH) in water affected by temperature? ! ! Introduction ! Sodium hydroxide is categorised as metal halide salt‚ composed of sodium and chlorine.! The ions present in the solid crystals of potassium chloride dissolve and gain mobility in water. When potassium chloride is dissolved in water‚ the following endothermic reaction occurs:! NaOH(s) + H2O(l) → Na+(aq) + OH–(aq) + H2O(l)! This reaction takes place because
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Number of Drops of HCl and NaOH on the Net Change in the pH of Plant‚ Animal‚ and Nonbiological Solutions Research Question: How do plants and animals respond to changes in pH? Analysis Questions: Summarize the effects of HCl and NaOH on the tap water. HCl is an acid and when in tap water‚ dissociates into H+ and Cl-. Since it releases the H+ ion in the tap water‚ it raises the concentration of H+‚ which lowers the pH of the tap water. Unlike HCl‚ NaOH is a base. NaOH is a base because when in the
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Preparation of the NaOH Solution Mass of NaOH before standing = NaOH before standing (g) – beaker (g) = 111.490 g – 110.970 g = 0.520 g Mass of NaOH after standing = NaOH after standing (g) – beaker (g) = 111.500 g – 110.970 g = 0.530 g Standardization of the NaOH solution TRIAL I Volume of NaOH solution = Final reading of buret - Initial reading of buret = 13.80 mL – 0.00 mL = 13.80 mL Converting the Volume (mL) to Volume (L) Volume (liters) = Volume (mL) x
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Standardization of international marketing strategy by firms from a developing country Standardization of marketing strategy 107 Shaoming Zou University of Missouri‚ Columbia‚ Missouri‚ USA David M. Andrus and D. Wayne Norvell Kansas State University‚ Manhattan‚ Kansas‚ USA A major debate in the international marketing literature deals with the globalization of markets and the extent to which a company’s international marketing strategy can be standardized (Buzzell‚ 1968; Cavusgil
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