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Chapter 4: Problem 84

The distinctive odor of vinegar is due to aceticacid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide according to: $$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow_{\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{COO}(a q)}$$ If 3.45 \(\mathrm{mL}\) of vinegar needs 42.5 \(\mathrm{mL}\) of 0.115 \(\mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 -qt sample of this vinegar?

Short Answer

Expert verified
In the given titration, 3.45 mL of vinegar containing acetic acid reacts with 42.5 mL of 0.115 M NaOH to reach equivalence. The balanced chemical equation for the reaction is \(CH_3COOH(aq) + NaOH(aq) \longrightarrow H_2O(l) + NaCH_3COO(aq)\). After calculating the moles of NaOH and acetic acid based on the titration data, we find the concentration of acetic acid in the vinegar sample to be 1.417 M. In a 1.00-qt (0.946353 L) sample of this vinegar, there are approximately 1.341 moles of acetic acid. Therefore, the mass of acetic acid in a 1.00-qt sample is roughly 80.51 g.

Step by step solution

01

Moles of NaOH

Calculate the moles of sodium hydroxide used in the titration using the given volume and concentration: Moles = Volume (L) × Concentration (M) Note that volume should be converted from mL to L: 42.5 mL = 42.5 / 1000 L = 0.0425 L Moles of NaOH = 0.0425 L × 0.115 M Moles of NaOH = 0.0048875 moles
02

Moles of Acetic Acid

Since the stoichiometry of the reaction is 1:1, the moles of acetic acid are equal to the moles of NaOH. Moles of Acetic Acid = Moles of NaOH Moles of Acetic Acid = 0.0048875 moles
03

Concentration of Acetic Acid

Find the concentration of acetic acid in the 3.45 mL vinegar sample by dividing the moles of acetic acid by the volume (in L): Concentration = Moles / Volume (L) Convert 3.45 mL to L: 3.45 mL = 3.45 / 1000 L = 0.00345 L Concentration of Acetic Acid = 0.0048875 moles / 0.00345 L Concentration of Acetic Acid = 1.417 M
04

Volume of 1.00-qt Sample

Convert the volume of the 1.00-qt vinegar sample to liters: 1 qt = 0.946353 L
05

Moles of Acetic Acid in 1.00-qt Sample

Calculate the moles of acetic acid in the 1.00-qt vinegar sample using the concentration: Moles = Volume (L) × Concentration (M) Moles = 0.946353 L × 1.417 M Moles = 1.341 moles
06

Mass of Acetic Acid in 1.00-qt Sample

Calculate the mass of acetic acid in the 1.00-qt vinegar sample by multiplying the moles of acetic acid by the molar mass of acetic acid: Mass = Moles × Molar Mass Molar Mass of Acetic Acid = 60.05 g/mol Mass of Acetic Acid = 1.341 moles × 60.05 g/mol Mass of Acetic Acid = 80.51 g Therefore, there are approximately 80.51 grams of acetic acid in a 1.00-qt sample of this vinegar.

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