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

In 10 grams of vinegar, there are 0.00981 moles and 0.589 grams of acetic acid. Moles are determined by titration with NaOH, and grams are calculated using the molar mass of acetic acid.

Step-by-step explanation:

To answer how many moles of acetic acid are present in ten grams of vinegar, we need to use titration and molarity calculations. The titration of acetic acid with NaOH is a neutralization reaction, where acetic acid reacts with sodium hydroxide to produce sodium acetate and water. Since NaOH and acetic acid react in a 1:1 molar ratio, the moles of NaOH will be equal to the moles of acetic acid at the endpoint of the titration.

The calculation for the moles of NaOH is:

• Moles of NaOH = Molarity of NaOH × Volume of NaOH in liters = 0.150 mol/L × 0.0654 L = 0.00981 mol

Since the moles of acetic acid are equivalent to the moles of NaOH:

• Moles of acetic acid = Moles of NaOH = 0.00981 mol

To find the grams of acetic acid present, use the molar mass of acetic acid (HC₂H₃O₂), which is approximately 60.05 g/mol:

• Grams of acetic acid = Moles of acetic acid × Molar mass of acetic acid = 0.00981 mol × 60.05 g/mol = 0.589 g

This indicates that in 10 grams of vinegar, there are 0.589 grams of acetic acid. Considering that a commercial vinegar solution typically contains 3.78 g of acetic acid per 100.0 g of solution, provided the density of the solution is 1.00 g/mL, this aligns with expectations.

Answer 2

Step-by-step 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

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