If 44.5 mL of 0.0989M sodium hydroxide solution is needed to titrate the amount of acetic acid in a 5.00 mL sample of vinegar, what is the weight/weight percent (w/w %) of acetic acid in the vinegar.

(MM acetic acid = 60.04 g) Do not include units in your answer. Use the correct number of significant figures.

5.28 %

Explanation:

To solve this problem we are going to calculate the moles of NaOH required to titrate the vinegar. Since this reaction is one mole NaOH to 1 mol acetic acid we will have the moles of acetic acid present in the 5 mL sample. By multiplying the number of moles of acetic acid by its molecular weight we obtain the grams of acid.

Now since we are being asked the w/w % we need the weight of solution which is not given in the problem. However we can assume that its value will be close to that of pure water and give it the value of 5 g since the density of water is one g/mL.

moles NaOH = 44.5 mL x 1L/1000 mL x 0.098 mol/L = 0.00440

mol acetic acid = 0.00440

mass acetic acid = 0.00440 mol x 60.04 g/mol = 0.264 g acetic acid

percent (w/w) = (0.264 g / 5 g) x 100 = 5.28 %