Data for calibration curve 1. Hydrolysis of acetylsalicylic acid a. Work in pair
ID: 1053165 • Letter: D
Question
Data for calibration curve 1. Hydrolysis of acetylsalicylic acid a. Work in pairs for this part of experiment. Weigh 0.50 ± g of acetylsalicylic acid, ASA, (pure) and transfer it to a 125mLmenmeyer flask. ..49-. 5. -4979-ASA Add 10 mL of 1.0 ± M sodium hydroxide and 10 ml of distilled water. Heat the flask gently, with hot plate, until the contents boil, to hydrolyze the d. ASA. i1 ml Cool to room temperature and add 15-20 mL of 0.10 ± M nitric acid. 2. Dilution of salicylic acid solution a. Quantitatively transfer the contents of the flask to a 250 mL volumetric flask and bring up to volume. Important: Use 0.10 M nitric acid for rinsing the Erlenmeyer flask and making the dilution. Carefully follow the directions given by the lab instructor for correct use of a volumetric flask. b. Pipet 10 mL of the solution and transfer it to a 100 mL volumetric flask. Pre-rinse the pipet.) Make up to volume with 0.10 M nitric acid (Reminder: and mix the solution thoroughly ompleted and the solution thoroughly mixed, transfer C. half of the solu ind ean dry beaker and, continue working as an 3. Concentration Series Using a buret filled with your diluted salicylic acid solution, measure 2.4,6,8 and 10 mL of solution into your large test tubes. a. Use buret and add the volume of 0.10 M nitric acid needed to give a total sample volume of 10 ml. (8,6,4,2, and 0 mL). MIX b. Pipet 1.0 mL of the Fes color reagent (2% Fe(NO3); . 9H20 in 0.10 M nitric acid) into each 10 mL sample. Mix well. Using the most dilute solution and the spectrophotometer determine the operating wavelength for the iron complex solution. d. 139Explanation / Answer
0.00276 moles ASA in 250 mL solution. Molarity = 0.00276 *1000mL/250mL = 0.011 M
10mL of this solution is diluted to 100mL (part 2a)
Molarity of the diluted solution = 0.011*10/100 = 0.0011 M
(a) 2:11 dilution means that 2 Ml of the 0.011M solution is diluted to 11mL.
Molarity = 0.011 M* 2/11mL=2*10^-3 M
Similarty molarity ofther diluted solutions are as follows :
(b) Molarity = 0.011 M* 4/11mL= 4*10^-3 M
(c) Molarity = 0.011M* 6mL/11mL = 6*10^-3 M
(d) Molarity = 0.011M*8mL/11mL = 8*10^-3M
(e) Molarity = 0.011M*10mL/11mL = 1*10^-2 M
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