Question

The acidic substance in vinegar is acetic acid, HC2H3O2. When 4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH, 25.56 mL of base had to be added to reach the equivalence point. What percent by mass of this sample of vinegar is acetic acid?

Answer 1

Answer:

7.7%

Explanation:

the balanced equation for the neutralization reaction is:

CH3COOH+NaOH→CH3COONa+H2O

from the balanced equation above,the mole ratio of CH3COOH to NaOH is 1:1

no of mole of CH3COOH= no of mole of NaOH

no of mole of NaOH= concentration(M) × Volume(L)

=0.2 M× 0.02556=0.005112 mol

since, no of mole of CH3COOH= no of mole of NaOH

no of mole of CH3COOH=0.005112 mol

However ,no of mole =mass/molar mass

molar mass of CH3COOH=60.052g/mol

mass of CH3COOH=0.005112×60.052

=0.307g

percentage by mass of acetic acid in the vinegar=0.307/4

=0.07675×100%

=7.675%