Chemistry and Titrations

Question

In a titration, a student used 25.30 mL of a 0.500 M NaOH solution to reach the equivalence point with 15.80 mL of vinegar. Calculate the molarity of the vinegar.

Answer 1

First, we need to know what the chemical reaction looks like

CH₃COOH_(aq) + NaOH_(aq) —› CH₃COO^-_(aq) + H₂O_(l) + Na⁺_(aq)

The vinegar and the NaOH react in a one to one ratio, meaning that one mole of vinegar reacts with one mole of NaOH. So, if we can find out how many moles of NaOH where used to reach the equivalence point, that amount is also equal to the amount of moles of vinegar present.

Using stoichiometry, we can find the amount of moles of NaOH that where used for this this titration

25.30 mL NaOH 1L 0.500 mol NaOH 1 mol vinegar = 0.0127 mol vinegar (rounded to 3 sig figs)

1000mL 1L 1 mol NaOH

So, 0.0127 moles of NaOH where used. In order to find molarity, we then need to divide the number of moles by the total volume of the solution, which is the volume of NaOH and Vinegar combined (in liters)

0.0127 mol/( 0.02530L + 0.01580L) = 0.309M Vinegar