The organic compounds will be separated from the aqueous mixture by extraction with an organic solvent that is immiscible with water. Therefore, they will form two layers when they are mixed together. The denser liquid will form the bottom layer. (Conversely, miscible liquids are soluble in each other.) The organic compounds would dissolve in extraction solvent and the inorganic impurities would remain dissolved in the water.
Purpose: The purpose of this lab is to titrate an unknown solid acid (KH2PO4) with a standardized sodium hydroxide solution. After recording and plotting the data, the acid’s equivalence point will be recorded once the color changes. Using the equivalence point, the halfway point will be calculated, which is used to determine the acid’s equilibrium constant. The acid’s calculated equilibrium constant will be compared with the acid’s established pKa value.
In addition, phenolphthalein was added as an indicator. The aliquots were titrated against sodium hydroxide (NaOH) solution until end point was reached, after which volume of NaOH consumed was recorded. The value of the rate constant, k, obtained was 0.0002 s-1. The experiment was then repeated with 40/60 V/V isopropanol/water mixture and a larger value of k = 0.0007 s-1 was obtained. We concluded that the rate of hydrolysis of (CH3)3CCl is directly proportional to water content in the solvent mixture.
A titration is the precise addition of a solution from a buret into an accurately measured volume of a sample solution. A titrant is the solution in the buret that is used for the titration, and the volume of the solution is known. The titrants used in this lab were 0.1M hydrochloric acid and 0.1M sodium hydroxide (the reactions can be seen in figure 4). A Bronsted-Lowry acid is a compound that donates a proton. A Bronsted-Lowry base is a compound that accepts a proton.
One of the reactions you observed resulted in this product: NaCl + H2O + CO2 (g)? What well did this reaction occur in? Describe how the observations for this reaction support your answer. B BoldI ItalicsU Underline Bulleted list Numbered list Superscript Subscript70 Words A reaction I observed in number 1.)
1) Percentage yield experiment: First we measured 20cm3 of sulphuric acid into a beaker using a measuring cylinder, this will help us determine the percentage yield at the end of the experiment. We then heated the beaker containing the sulphuric acid using a Bunsen burner in order to heat it up for the copper oxide to mix with. We then weighed out 1.02g of copper oxide and added it to the acid and stirred it whilst doing so, we did that until the liquid turned blue, this proves that the chemicals have mixed together. We then weighed this liquid which will help us determine the percentage yield. We then filtered the liquid off which gave us the amount we obtained.
ABSTRACT The experiment aims to find the concentration of acetic acid in the vinegar sample by titrating it to a standardized NaOH solution, a base solution. A standardized 0.100 M NaOH solution was prepared from an available concentration of NaOH specifically a 1.00 M NaOH. The volume of the standardized solution (titrant) used which is 0.100 L was calculated using the M 1 V 1 = M 2 V 2 relationship.
Hanusaiodine solution, chloroform, aqueous KI solution, Na2S2O3 and starch solution is used. Iodine values are calculated from the difference between the blankaand the test sample. For peroxide value; solvent mixture (composed of glacial acetic acid and chloroform), saturated KI solution, starch solution and Na2S2O3 soluiton is used and peroxideavalues are calculated. A) Iodine Value: Hanus Method In this experiment, iodine value of sun floweraoil was determined with Hanusamethod.
SPEEK Synthesis SPEEK was prepared through the via of sulfonation reaction by using concentrated sulfuric acid at desired temperature. The dried PEEK pellets were ground well with the help of a martter for reducing dissolution time of the PEEK polymer. 5
In a small 125ml Erlenmeyer flask, dry the ether solution over anhydrous calcium chloride. Add sufficient calcium chloride so that it no longer clumps to pellets added earlier on the bottom of the flask. Remove the solvent using a rotary evaporator and weigh product. Results 1 mole of benzoic acid (C6H5COOH = 122.12grams) reacts with 1 mole of methanol (CH3OH = 32grams/mole) to produce 1 mole of methyl benzoate (C6H5COOCH3 = 136.15grams) and 1 mole of water.