Question

The titration of 25.00mL of a vinegar sample with 0.500N sodium hydroxide solution required 44.60mL. Assume the density of the vinegar sample is 1.00g/mL.

a) Calculate the number of gram-equivalents of acetic acid in the vinegar sample.

b) Calculate the number of grams of acetic acid in the sample.

c) what is the percent acetic acid in the vinegar sample.

d) if 16 oz of the vinegar cost 53 cents, what is the cost per oz of vinegar solution.

e) what is the cost per oz of acetic acid.

Answer 1

a) CH3COOH + NaOH --------> CH3COONa + H2O

1:1 molar reaction

Equivalent weight = molecular waight

no of gram - equivalent of NaOH = (0.500mol/1000ml)*44.60ml =0.0223

No of gram equivalents of acetic acid = 0.0223

b) molar mass of acetic acid = 60.05

mass of acetic acid = 60.05g/mol * 0.0223mol = 1.3391g

c) Volume of vinegar sample = 25ml

density of vinegar sample =1.00g/ml

mass of vinegar sample = 25g

Percent acetic acid = (1.3391g/25g)*100 = 5.356%

d)

cost per Oz = (53cents/16 oz) * 1oz = 3.3125 cents

e)

oz of acetic acid in 1oz of vinegar = (5.356/100)*1=0.05356 oz

0.0536oz of acetic acid cost = 3.3125cents

cost of acetic acid per oz = (3.3125cents/0.0536oz)*1oz = 61.8 cents