Question

1. the titration of 25.00 ml of a vinegar sample with 0.500 m naoh required 44.60 ml. assume the density of the vinegar sample is 1.00 g/ml. (hint: what is the equation?)a. calculate the number of grams of acetic acid in the vinegar sample.b. what is the percent acetic acid in the vinegar sample?

Explanation:

Step1: Write the balanced chemical equation

The reaction between acetic acid (\(CH_3COOH\)) and \(NaOH\) is \(CH_3COOH + NaOH
ightarrow CH_3COONa + H_2O\). From the equation, the mole ratio of \(CH_3COOH\) to \(NaOH\) is \(1:1\).

Step2: Calculate the moles of \(NaOH\) used

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume). Given \(C_{NaOH}=0.500\space M\) and \(V_{NaOH}=44.60\space mL = 44.60\times10^{- 3}\space L\). Then \(n_{NaOH}=0.500\space mol/L\times44.60\times 10^{-3}\space L = 0.0223\space mol\).

Step3: Determine the moles of acetic acid

Since the mole ratio of \(CH_3COOH\) to \(NaOH\) is \(1:1\), \(n_{CH_3COOH}=n_{NaOH} = 0.0223\space mol\).

Step4: Calculate the mass of acetic acid

The molar mass of \(CH_3COOH\) is \(M=(2\times12)+(4\times1)+(2\times16)=60\space g/mol\). Use the formula \(m = n\times M\). So \(m_{CH_3COOH}=0.0223\space mol\times60\space g/mol = 1.34\space g\).

Step5: Calculate the mass of the vinegar sample

Use the formula \(m=
ho\times V\). Given \(
ho = 1.00\space g/mL\) and \(V = 25.00\space mL\). Then \(m_{vinegar}=1.00\space g/mL\times25.00\space mL=25.00\space g\).

Step6: Calculate the percent of acetic acid

Use the formula \(\%\text{acetic acid}=\frac{m_{acetic acid}}{m_{vinegar}}\times100\%\). Substitute \(m_{acetic acid}=1.34\space g\) and \(m_{vinegar} = 25.00\space g\). Then \(\%\text{acetic acid}=\frac{1.34\space g}{25.00\space g}\times100\%=5.36\%\)

Answer:

a. The mass of acetic acid is \(1.34\space g\).
b. The percent of acetic acid is \(5.36\%\)