Question

The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion:
CH3COOH(aq)+NaOH(aq)→H2O(l)+NaC2H3O2(aq)

1.) If 3.55 mL of vinegar needs 43.0 mL of 0.150 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.10 qt sample of this vinegar?

Answer 1

Given, the balanced chemical reaction between acetic acid and NaOH,

CH₃COOH(aq) + NaOH(aq) H₂O(l) + NaC₂H₃O₂(aq)

Also given,

Concentration of NaOH = 0.150 M

Volume of NaOH = 43.0 mL x ( 1L /1000 mL) = 0.043 L

Now, calculating the number of moles of NaOH from the given concentration and volume,

We know, the formula

Molarity = Number of moles / L of solution

Rearranging the formula,

Number of moles = Molarity x L of solution

The number of moles = 0.150 M x 0.043 L

The number of moles = 0.00645 mol NaOH

Now, at equivalence point moles of acetic acid = Moles of NaOH

moles of acetic acid in the titrated sample = 0.00645 mol

Thus, the concentration of acetic acid in the 3.55 mL vinegar sample is, 0.00645 mol / 3.55 mL

Now, converting 1.10 qt to mL,

= 1.10 qt x (946.353 mL/ 1 qt)

= 1040.988 mL

The number of moles of acetic acid present in 1040.988 mL vinegar sample is,

= 1040.988 mL x (0.00645 mol / 3.55 mL)

= 1.89 mol acetic acid

Converting moles to grams,

= 1.89 mol acetic acid x (60.052 g / 1 mol)

= 113.6 g acetic acid

Thus, 114 g [ 3S.F] acetic acid are in a 1.10 qt sample of this vinegar