Question

5, Vinegar should contain more than 4% acetic acid (CHCOOH, molar mass 60.0 g mor) 25.0 mL of vinegar are diluted to 100.0 mL and 10.0 mL samples of this solution are titrated against 0.100 mol L-NaOH solution. An average of 18.1 mL NaOH are required to neutralize the vinegar. You will need to calculate the concentration of the diluted solution and hence the percentage concentration of acetic acid in the original vinegar via the following questions: CH COOH(aq) + NaOH(aq) CH COONa(aq) + H:00) Is the known substance NaOH or acetic acid? Find the number of moles of the known substance. a. According to the balanced reaction, what is the mol-to-mol ratio of acetic acid to sodium hydroxide? b. Use your answer in part B to determine how many moles of the unknown substance were utilized in this titration. c.

a-c

Answer 1

Hi,

a) We know the concentration of NaOH, hence it is the known.

The concentration of NaOH = 0.100 mol/L = 0.100 M

The volume of NaOH used up = 18.1 mL

0.100 mols in 1 L that is 1000mL NaOH solution.

In 1 mL = 0.100 / 1000 = 0.0001 mols

In 18.1 mL = 0.0001 x 18.1 = 1.81 x 10^-3 mols

No. of moles of NaOH = 1.81 x 10^-3 mols

b) From the balanced chemical reaction given,

1 mole of acetic acid reacts with 1 mole of sodium hydroxide, hence the ratio is 1:1.

c) The volume of acetic acid = 10 mL

Hence Molarity of acetic acid = M₂V₂ / V₁ = 0.100 M x 18.1 mL / 10 mL = 0.181 M = 0.181 mol/L

Since we used 10 mL of 0.181 mol/L concentrated acetic acid solution,

No. of moles of acetic acid used in titration = ( 0.181 / 1000 ) x 10 = 1.81 x 10^-3

As we know 1 mol of acetic acid reacts with 1 mol of NaOH, we have the same no. of moles of acetic acid reacted in titration.

( Note: Molarity of original Vinegar = 0.181 mol/L x (100mL / 25mL) = 0.724 mol/L )

Best Wishes.