Question

A 30.84ml volume of 0.128M NaOH is required to reach

thephenolphthalein endpoint in the titration of a 5.441g sample
ofvinegar. Calculate the Percent acetic acid in the vinegar.

Answer 1

Percent acetic acid in the vinegar is 4.35%.

Step-by-step explanation:

`Moles (n)=Molarity(M)* Volume (L)`

Moles of NaOH = n

Volume of NaOH solution = 30.84 mL = 0.03084 L

Molarity of the NaOH = 0.128 M

`n=0.128 M* 0.03084 L=0.003948 mol`

`NaOH+CH_3COOH\rightarrow CH_3COONa^++H_2O`

According to reaction ,1 mole of NaOH reacts with 1 mol of acetic acid.

Then ,0.03948 mol of NaIOH will recat with:

`(1)/(1)* 0.003948 mol=0.003948 mol` 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 :

`=(0.2368 g)/(5.441 g)* 100=4.35\%`