Question

3.) if a 3.45 ml sample of vinegar needs 42.5 ml of 0.115 m naoh to be neutralized, how many grams of acetic acid are in a 1.00 l bottle of this vinegar? include the correct number of significant figures and the correct unit.

Explanation:

Step1: Find moles of NaOH

Use the formula \(n = M\times V\).
\(M = 0.115\space M\), \(V=42.5\space mL=42.5\times10^{- 3}\space L\)
\(n_{NaOH}=0.115\space M\times42.5\times10^{-3}\space L = 0.0048875\space mol\)

Step2: Relate moles of NaOH to moles of acetic acid (\(CH_3COOH\))

The reaction is \(CH_3COOH+NaOH
ightarrow CH_3COONa + H_2O\). Mole ratio \(CH_3COOH:NaOH = 1:1\). So \(n_{CH_3COOH}=n_{NaOH}=0.0048875\space mol\) in \(3.45\space mL\) of vinegar.

Step3: Find moles of acetic acid in \(1.00\space L\)

\(1.00\space L = 1000\space mL\).
Let \(n\) be moles in \(1000\space mL\). Using proportion \(\frac{n}{1000\space mL}=\frac{0.0048875\space mol}{3.45\space mL}\)
\(n=\frac{0.0048875\space mol\times1000\space mL}{3.45\space mL}\approx1.4167\space mol\)

Step4: Convert moles of acetic acid to grams

Molar mass of \(CH_3COOH=(12\times2)+(1\times4)+(16\times2)=60\space g/mol\)
Mass \(m=n\times M\)
\(m = 1.4167\space mol\times60\space g/mol\approx85.0\space g\)

Answer:

\(85.0\space g\)