Answer:
Percent acetic acid in the vinegar is 4.35%.
Explanation:
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
Moles of NaOH = n
Volume of NaOH solution = 30.84 mL = 0.03084 L
Molarity of the NaOH = 0.128 M
[tex]n=0.128 M\times 0.03084 L=0.003948 mol[/tex]
[tex]NaOH+CH_3COOH\rightarrow CH_3COONa^++H_2O[/tex]
According to reaction ,1 mole of NaOH reacts with 1 mol of acetic acid.
Then ,0.03948 mol of NaIOH will recat with:
[tex]\frac{1}{1}\times 0.003948 mol=0.003948 mol[/tex] of acetic acid.
Mass of 0.03948 moles of acetic acid = 0.003948 mol × 60 g/mol = 0.2368 g
Mass of vinegar solution = 5.441 g
Percent acetic acid in the vinegar :
[tex]=\frac{0.2368 g}{5.441 g}\times 100=4.35\%[/tex]
