Answer:
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!
