Question

Calculate the volume of 1:5 diluted vinegar which would be titrated by 35.00mL of 0.1000 M NaOH. The molarity of acetic acid in vinegar is 0.83M

Answer 1

Ans:

Vinegar = Diluted acetic acid (CH₃COOH)

Reaction: 1 NaOH + 1 CH₃COOH --> CH₃COONa + H₂O

Millimoles of NaOH used for titration = Molarity x Volume(mL) = 0.1 M x 35.0 mL = 3.5 mmol.

To react with 3.5 mmol of NaOH, we should take that volume of 1:5 diluted vinegar which consists of 3.5 mmol of CH₃COOH.

Molarity of acetic acid in vinegar = 0.83 M

Molarity = mmol/Volume(mL)

Volume(mL) of 1:5 diluted 0.83 M vinegar needed = mmol/Molarity = 3.5mmol/0.83M = 4.2169 mL

Volume(mL) of 1:5 diluted 0.83 M vinegar needed to react with 35.0 mL of 0.1 M NaOH is 4.2169 mL