Question

A titration of vinegar with a solution of NaOH was performed. If 3.45 mL of vinegar needs 44.0 mL of 0.140 M NaOH to reach the equivalence point in a titration, Calculate the mass of acetic acid present in the vinegar sample: mastering chemistry answers

Answer 1

Answer: The mass of acetic acid present in the vinegar sample is 0.370 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of NaOH solution = 0.140 M

Volume of solution = 44.0 mL

Putting values in above equation, we get:

$0.140M=\frac{\text{Moles of NaOH}\times 1000}{44.0}\\\\\text{Moles of NaOH}=\frac{0.140\times 44}{1000}=0.00616mol$

The chemical equation for the reaction of acetic acid and NaOH follows:

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of acetic acid

So, 0.00616 moles of NaOH will react with = $\frac{1}{1}\times 0.00616=0.00616mol$ of acetic acid

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of acetic acid = 60 g/mol

Moles of acetic acid = 0.00616 moles

Putting values in above equation, we get:

$0.00616mol=\frac{\text{Mass of acetic acid}}{60g/mol}\\\\\text{Mass of acetic acid}=(0.00616mol\times 60g/mol)=0.370g$

Hence, the mass of acetic acid present in the vinegar sample is 0.370 grams