25 ml of commercial vinegar was diluted to 250 ml water, 22 ml of 0.1 M sodium hydroxide was needed for the titration of 25 ml dilute vinegar solution. Determine the %(w/v) of acetic acid in commercial vinegar.

5,28 % (W/V)

Explanation:

Comercial vinegar is a natural solution of acetic acid (CH₃COOH; MW: 60,052). The reaction of CH₃COOH with sodium hydroxide (NaOH) is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

Moles of 22mL 0,1M NaOH are:

0,022L × (0,1mol / 1L) = 2,2x10⁻³ moles of NaOH. Based on the reaction, 1mole of NaOH reacts with 1 mole of CH₃COOH, That means moles of CH₃COOH are 2,2x10⁻³ moles

As the volume titrated of the solution was 25mL, molar concentration of the dilution is:

2,2x10⁻³mol / 0,025L = 0,088M

The dilution was from 25mL to 250mL, thus, concentration of commercial vinegar is:

0,088M × (250mL / 25mL) = 0,88M.

Using molar weight to convert moles of acetic acid in grams. And converting Liters in Mililiters:

0,88 moles / L × (60,052g / mol) × (1L / 1000mL) × 100 = 5,28g/mL% ≡ 5,28 % (W/V)

I hope it helps!

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8