Question

Ten grams of vinegar, which contains acetic acid, HC₂H₃O₂, is titrated with 65.40 mL of a 0.150 M NaOH solution. a) How many moles of acetic acid are present in ten grams of vinegar?b) How many grams of acetic acid are present in ten grams of vinegar?

Answer 1

Explanation:

The acetic acid will react with sodium hydroxide to give sodium acetate and water according to this reaction:

HC₂H₃O₂ + NaOH ---> NaC₂H₃O₂ + H₂O

First we can find the number of moles of NaOH that were used to titrate the acetic acid in the vinegar.

Molarity = moles of solute/volume of solution in L

Volume = 65.40 mL = 65.40 mL * 1 L/(1000 mL)

Volume = 0.06540 L

Moles of NaOH = Molarity * Volume of solution in L

Moles of NaOH = 0.150 mol/L * 0.06540 L

Moles of NaOH = 0.00981 mol

HC₂H₃O₂ + NaOH ---> NaC₂H₃O₂ + H₂O

According to the coefficients of the reaction 1 mol of NaOH will neutralize 1 mol of HC₂H₃O₂. Since the molar ratio between them is 1 to 1. We will have the same number of moles.

1 mol of NaOH = 1 mol of HC₂H₃O₂

moles of HC₂H₃O₂ = 0.00981 mol of NaOH * 1 mol of HC₂H₃O₂/(1 mol of NaOH)

moles of HC₂H₃O₂ = 0.00981 moles

Answer: