Question

Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic acid in the vinegar using the average of three good trials used to titrate 5.00 mL of the vinegar. You may assume that the density of your vinegar sample is 1.01 g mL^-1. Drain off unused NaOH solution into a clean container and place it into a designated container. Wash the buret two times with regular tap water, then rinse with deionized water, and clamp it upside down. Solutions from the Erlenmeyer flasks may be poured down the sink. Wipe up your work area. Wipe off and return the safety goggles to the cabinet (or place it in your drawer if it belongs to you). Note that once you put away your safety goggles, you must leave the lab. Remember to wash your hands before leaving. If 2.50 mL of vinegar requires 34.90 mL of 0.0960 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in 500.0 mL of this vinegar? (density of vinegar = 1.01 g mL^-1) Calculate the mass % of acetic acid in the vinegar in the previous exercise (question#1 above). Assume that the density of the vinegar is 1.01 g mL^-1. If phenolphthalein is used as indicator, what color will the solution be at the endpoint? In your reading, it talks about errors that arc possible when doing titrations, what are three of those possible sources of error? What would happen if you forgot to add phenolphthalein indicator solution to the vinegar in step 3? Explain using full sentences. Find % error in your experiment if theoretical molarity of HC_2H_3O_2 is 0.975 M.

Answer 1

Calculate the moles of NaOH:

Molarity =moles/volume of NaOH

0.096*0.0349 L= Moles of NaOH

Moles of NaOH = 3.3504*10^-3 moles

The acetic acid present in the vinegar reacts with NaOH with a stiochiometric ratio of 1.

so moles of NaOH = moles of acetic acid

Therefore, the moles of acetic acid in the vinegar solution are 3.3504*10^-3 .

Grams of acetic acid = 60 g/mol * 3.3504*10^-3 =0.2010 g

b)

grams of vinegar solution= density of vinegar solution x volume of vinegar solution

=1.01 g/mL *500 mL =505 g

  % of acetic acid = grams of acetic acid/grams of vinegar solution*100 =(0.2010g/505 g)*100=0.0398 %