Question

) A sample of vinegar is 4.8% acetic acid, HC2H3O2, by weight, andhas a density of 1.00g/mL. Calculate (a) the grams of acetic acidin 1 liter of vinegar (b) the molarity of acetic acid invinegar. 2) calculate the concentration of an unknown acid solution,HA, if 25.24 mL of .1278 M NaOH solution were needed to neutralize 28.20 Ml of the HA solution

Answer 1

4.8 g acetic acid is present in 100g of vinegar

Density of vinegar = 1g/mL which means 1mL vinegar has a weight of 1 g

thus 1000mL vinegar will have a weight of 1000g.

Amount of acetic acid present in 1000g vinegar = 48g

----------------------------------------------------------------------------------------------------------------------------------

moles of acetic acid present = 48g/ 60gmol^-1 =0.8 moles

molarity= moles/ liter

molarity of the acetic acid in vinegae = 0.8 M

--------------------------------------------------------------------------------------------------------------------------

M1V1 = M2V2

where M1 and V1 is the molarity and volume of HA respectively . M2 and V2 is the molarity and volume of NaOH respectively

M1 * 28.20 = 0.1278 * 25.24

M1 = 0.114 M

concentration of HA = 0.114 M