Question

A 2.40 g sample of vinegar was added to an Erlenmeyer flask along with 100 mL of deionized water and 3 drops of phenolphthalein indicator. It took 22.15 mL of 0.0981 M NaOH (aq) to reach the faint pink endpoint. The following balanced chemical equation represents the chemistry of the titration: NaOH (aq) + CH3COOH (aq) CH3COO–Na+ (aq) + H2O (l) • Calculate the mass % of acetic acid (CH3COOH) in the vinegar sample.

Answer 1

The mass % of acetic acid is 5.42 %

Step-by-step explanation:

Step 1: Data given

Mass of sample of vinegar = 2.40 Grams

Volume of water = 100 mL = 0.100 L

Volume of NaOH = 22.15 mL = 0.02215 L

Molarity of NaOH = 0.0981 M

Step 2: The balanced equation

NaOH (aq) + CH3COOH (aq) ⇆ CH3COO–Na+ (aq) + H2O (l)

Step 3: Calculate moles NaOH

Moles NaOH = molarity * volume

Moles NaOH = 0.0981 M * 0.02215 L

Moles NaOH = 0.00217 moles

Step 4: Calculate molesCH3COOH

For 1 mol NaOH we need 1 mol CH3COOH to produce 1 mol CH3COONa and 1 mol H2O

For 0.00217 moles NaOH we need 0.00217 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00217 moles * 60.05 g/mol

Mass CH3COOH = 0.130 grams

Step 6: Calculate mass % of acetic acid

Mass % = (0.130 grams / 2.40 grams) * 100 %

Mass % = 5.42 %

The mass % of acetic acid is 54.2 %

Answer 2

5.42%

Step-by-step explanation:

Let's consider the neutralization reaction between NaOH and acetic acid.

NaOH(aq) + CH₃COOH(aq) → CH₃COONa(aq) + H₂O(l)

It took 22.15 mL of 0.0981 M NaOH(aq) to reach the faint pink endpoint. The reacting moles of NaOH are:

0.02215 L × 0.0981 mol/L = 2.17 × 10⁻³ mol

The molar ratio of NaOH to CH₃COOH is 1:1. The reacting moles of CH₃COOH are 2.17 × 10⁻³ moles.

The molar mass of acetic acid is 60.05 g/mol. The mass of acetic acid is:

2.17 × 10⁻³ mol × 60.05 g/mol = 0.130 g

0.130 g of acetic acid are in a 2.40 g sample of vinegar. The percent by mass of acetic acid in vinegar is:

0.130 g/2.40 g × 100% = 5.42%