Prelab question #4

Step 1 calculate the molar concentration of a 5% acetic acid in vinegar, remember that a 1% solution is 1 g in 100mL

5.0 % = 50g/L x 1 mol/60.0g = .83 mol/L

Step 2. Convert 1.00 g of solution to a volume, given the density.

1.00g soln/1.006 g/mL = .994 mL = 9.94x10^-4 L

Step 3 calculate the volume of 0.10 M NaOH base required to neutralize 1.00g of the 5% solution

9.94x10^-4 L x 0.83mol/1L x 1 molNaOH/1 mol HC2H3O2 x 1 L NaOH/0.10 mol NaOH = 8.25x10-3 L

or 8.254 mL of base for 1 mL of vinegar.

One pipet full of the NaOH solution will not be nearly enough to titrate a single sample of full strength vinegar. This is why the vinegar needs to be diluted 1:10.