CHM2045L. Practice Exam 2 -Titration Analysis, Spectrophotometry, Dumas h OK Det
ID: 570594 • Letter: C
Question
CHM2045L. Practice Exam 2 -Titration Analysis, Spectrophotometry, Dumas h OK Determination Analogy to the Vinegar Analysis We analyze a vitamin C tablet for ascorbic acid, a weak acid with a molar mass of 176.13 g/mole. Once crushed and dissolved in water, it titrates to a nice phenolphthalein/peach-colored endpoint in a 1:1 reaction with sodium hydroxide. A single tablet of 0.607 g is titrated to endpoint against 0.1120 M sodium hydroxide solution. The initial volume of sodium hydroxide titrant is 0.34 mL, and at endpoint the volume is 26.38 ml. Use this information to answer the following questions. HAsc + NaOH NaAsc,aq) + H2O Asc-ascorbate ion (-1 charge) HAsc = ascorbic acid 1. Determine the number of moles of sodium hydroxide dispensed. 2. Determine the number of moles of ascorbic acid (vitamin C) present in the tablet 3. Determine the number of grams of ascorbic acid in the tablet. Is the manufacturer honest when it reports there are 500 mg per tablet? 4. Determine the percent by mass of ascorbic acid in the tablet. 5, Why do you rinse the clean buret several times with a small amount of titrant solution before starting the experiment?Explanation / Answer
Extimation of ascorbic acid in Vit.C tablet
1. Volume of NaOH dispensed = 26.38 - 0.34 = 26.04 ml = 0.02604 L
moles of NaOH dispensed = molarity x volume = 0.1120 M x 0.02604 L = 0.00292 mol
2. moles Ascorbic acid in the tablet = 0.00292 mol
3. grams ascorbic acid in the tablet = 0.00292 mol x 176.13 g/mol = 0.514 g
4. percent mass ascorbic acid in the tablet = (0.514/0.607 g) x 100 = 84.68%
5. The buret was rinsed several times with the titrant to ensure no water is left in the system which could dilute the titrant and results of moles of NaOH dispensed would change giving an error in measurement.