Question

assuming the density of a 5% acetic acid solution is 1.0g/ml, determine the volume of the acetic acid solution necessary to neutralize 25.0 ml of 0.10 m NaOH. also record this calculation on your report sheet.

Answer 1

5% means 5.00g acetic acid in 100 mls of solution. That is equivalent to 50.0 g acid/L.
The molar mass of acetic acid is about 60.05g/mole
(50.0 g acid/L.)(1 mol / 60.05g) = 0.8326 mol/L
The reaction is:
HC2H3O2 + NaOH --> NaC2H3O2 + H2O
based on the reaction,
moles of HC2H3O2 = moles of NaOH
(0.025 L)(0.10 mol/L) = 0.0025 moles NaOH
moles of acid = 0.0025 moles HC2H3O2
V*M = 0.0025 moles
Substitute to molarity of the acid and solve for V to get the volume in liters. You may want to convert the volume in liters to milliliters at the end.

Answer 2

3.001lL