Answer to Question #146496 in General Chemistry for Gusty

Question

1. Assuming the density of a 5% by the mass acetic acid solution is 1.0 g/mL, determine the volume of acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH.

2. A 22.0 mL volume of 0.095 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 3.05 g sample of vinegar.

a. Calculate the number of moles of acetic acid in the vinegar sample.

b. Calculate the mass of acetic acid in the vinegar sample.

c. Calculate the percent by mass o f acetic acid in the vinegar sample. Assume the density of the vinegar is 1.0 g/mL.

3. In part B if the end point of the titration is overshot! Does this technique error result in an increase, a decrease, or have no effect on the reported percent acetic acid in the vinegar?

Explain.

Answer 1

1) No. of moles of NaOH = 0.10*0.025

Therefore, Required no. of moles of acetic acid is 2.5*10^-3

Moles of acetic acid per mL of solution = (1*0.05)/60, where 60 is molar mass of Acetic Acid

required volume of acetic acid soln. = 2.5*60*10^-3/0.05

=3 mL

2)

a) No. of moles of NaOH = 0.095*0.022

Therefore, Required no. of moles of acetic acid is 2.5*10^-3 for complete neutralization at titration endpoint.

Moles of acetic acid in vinegar sample = 2.09*10^-3

b)

mass of acetic acid in vinegar sample = 2.09*10^-3*60 g

= 0.125 g

c) Mass percent =(0.125/3.05)*100

= 4.11%

3) Phenolphthalein is an indicator of alkaline medium. If the end point is overshot, that implies more alkaline solution is used than required. When the volume of NaOH used is reported, it will be more than the actual amount that should ideally be used. This will lead to an increase in percent of acetic acid in given sample of Vinegar, as the only acidic component assumed in the given sample is acetic acid.