Question

The titration process is used to analyze the content of some food product. Let's say you are analyzing the concentration of acetic acid in the vinegar. Answer the following question:

• What would you categorize the vinegar, an acid or a base?

You have 5 mL of vinegar in a flask and 2 drops of phenolphthalein indicator. You titrate exactly 36.9 mL of 0.1 mol/L NaOH from a burette directly into the flask which gives you a pale pink color. This means the solution has reached neutralization.

• Calculate n of the NaOH.

• Calculate c of the vinegar in mol/L.

• Depending on your answer for the previous question complete the following sentence:

one liter of vinegar contains …………………………. mol of acetic acid.

Answer 1

Ans. 1. Vinegar is categorized as acid because it has acetic acid.

Ans. 2. CH3COOH + NaOH ----> CH3COONa + H2O - balanced equation

1 mol acetic acid reacts with 1 moles NaOH. So, reach the equivalence point, the number of moles of acetic acid must be equal to that of NaOH solution.

Number of moles of NaOH = molarity x volume (in L) of NaOH solution.

                                    = 0.1 M x 0.0369 L

= 0.1 (moles/ L) x 0.0369 L                      [ 1 M = 1 mole/ L]

                                    = 0.00369 moles

From balanced reaction, 1 mol CH3COOH reacts with 1 mol NaOH.

Thus, number of moles of acetic acid = number of moles of NaOH

Hence, number of moles of acetic acid = 0.00369 moles

Now, volume of acetic acid = 5 mL. Therefore, 0.00369 moles are present in 5 ml vinegar sample.

Moles of acetic acid in 1 L (1000 mL) = (1000 mL/ 5 mL) x 0.00369 moles = 0.738 moles

Thus, 1 L of vinegar has 0.738 moles acetic acid.

Molarity, M = number of moles / volume in L

                        = 0.738 moles / 1 L

                        = 0.738 M

Hence, concentration of vinegar = 0.738 M

# One liter of vinegar contains 0.738 moles of acetic acid.