Date: Concentration of the NaOH solution 0.495 M Volume of NaOH used during trial titration Jo. mL TITRATIONS Final buret reading 16.8 mL Initial buret reading O mL HI IT Volume of NaOH solution mL M Concentration of HC2H302 (show calculations) Average concentration Standard deviation -3 L mol Naot Imol NOCH step 1, (0.495 mol Nach ) (16.8x1032)=8-32x10 step 2:8.22x10 mol Neoh x mol Heahp?8.32 xbmc slep3: 8.32xl03 mol HC a H302 = 0.832 M H C 2 H 3 02 10.08/03/ step 4: 8.32 mol Hea H₃ oq x 60.09 H C 2 H 309 Imol HCq H302 Experiment 6 113 02 X - = 0.2199 29 HC2H = 0.499 G H C 2 H 3 or e.4999 C 2 H 302 2.09 C2 H202 C) 2 x 100 = 4.99% Experiment #6 Pre-Lab Questions 1. Define the terms: (a) titration (b) standard solution (c) end-point 2. Why is the meniscus kept at eye level when reading the volume of a buret? To get accurate reading in it has to be reeding in ao degree. 3. What is the purpose of adding a few drops of phenolphthalein solution to the vinegar in this experiment? Experiment 6 111 1. Experiment #6 Problems Write (a) the molecular equation and (b) the net ionic equation for the reaction of H2SO4 with NaOH. 2. Exactly 25.00 mL of 0.1522 M H2SO, were needed to titrate 45.25 mL of NaOH according to the balanced equation in problem 1 above. (a) Calculate the moles of NaOH needed for the reaction. (b) Calculate the molarity of the NaOH solution. 001 gradova 3. One Group of students made an error by using HCl instead of H2SO4 to titrate an unknown solution of NaOH. Would this error change the calculated molarity of the NaOH solution? Explain your answer. Experiment 6 115 Calculate the grams of Na CO2 needed to react with 25.0 mL of 0.220 M HCl solution if one mole of Na2CO3 reacts with 2 moles of HCI. 5. What volume of 0.10 M NaOH is required to titrate 12 mL of vinegar that contains 5% by weight of acetic acid. The density of the vinegar is 1.00 g/mL. LO 116 Titration of Vinegar
Purchase answer to see full attachment