Question

What is the percentage of acetic acid (60.05 g/mol, monoprotic acic) in a sample of vinegar purchased in a supermarket if 40.00 mL of the vinegar required 31.42 m solution for neutralization? Assume that the vinegar solution has a density of 0.960 g/ml. 1. L of 0.8010 M KOH

I think the answer is 3.94% but I don’t know if that is right, and I don’t have any work to support it

Answer 1

Reaction between KOH and acetic acid is:

KOH + CH₃COOH ---> CH₃COOK +H₂O

Thus, 1 mol of acetic acid reacts with 1 mol of KOH

Volume of KOH = 31.42 mL = 0.03142 L

Molarity of KOH= 0.8010 M

Moles of KOH = Molarity * volume = 0.8010 M * 0.03142 L = 0.0252 mol = moles of acetic acid

Molar mass of acetic acid = 60.052 g/mol

Mass of acetic acid = Moles * molar mass = 0.0252 mol * 60.052 g/mol = 1.51 g

Volume of vinegar solution = 40 mL

density of solution =0.960 g/mL

Mass of vinegar solution = volume * density = 40 mL * 0.960 g/mL = 38.4 g

So, Mass percent of acetic acid = ( mass of acetic acid/ Mass of vinegar solution )*100% = (1.51 g/ 38.4 g)*100% = 3.93 %