Question

1. a vinegar sample contained 4.60% acetic acid. how many ml of 0.450 m sodium hydroxide would be required to titrate 25.00 ml of the vinegar sample? assume the density of the sample to be 1.00 g/ml.

Explanation:

Step1: Calculate mass of vinegar sample

The volume of vinegar is \(25.00\space mL\) and density is \(1.00\space g/mL\). Using \(mass = density\times volume\), we get \(mass = 1.00\space g/mL\times25.00\space mL = 25.00\space g\).

Step2: Find mass of acetic acid

Acetic acid is \(4.60\%\) of the vinegar sample. So mass of acetic acid (\(m\)) is \(4.60\%\) of \(25.00\space g\), i.e., \(m=\frac{4.60}{100}\times25.00\space g = 1.15\space g\).

Step3: Calculate moles of acetic acid

Molar mass of acetic acid (\(CH_3COOH\)) is \(60.05\space g/mol\) ( \(C:12.01, H:1.008, O:16.00\); \(2\times12.01 + 4\times1.008+ 2\times16.00 = 60.05\space g/mol\)). Moles (\(n\)) of acetic acid is \(n=\frac{m}{M}=\frac{1.15\space g}{60.05\space g/mol}\approx0.01915\space mol\).

Step4: Reaction between acetic acid and NaOH

The reaction is \(CH_3COOH + NaOH
ightarrow CH_3COONa + H_2O\). The mole ratio of \(CH_3COOH\) to \(NaOH\) is \(1:1\). So moles of \(NaOH\) needed is also \(0.01915\space mol\).

Step5: Calculate volume of NaOH solution

Molarity of \(NaOH\) is \(0.450\space M\) ( \(M=\frac{n}{V}\), so \(V=\frac{n}{M}\)). Substituting \(n = 0.01915\space mol\) and \(M = 0.450\space mol/L\), we get \(V=\frac{0.01915\space mol}{0.450\space mol/L}\approx0.04256\space L\). Convert to mL: \(0.04256\space L\times1000\space mL/L\approx42.6\space mL\).

Answer:

\(42.6\space mL\)