Assuming the density of a 5%acetic acid solution is

1.0g/mL, determine the volume of the aceticacid solution necessary

to neutralize 25.0mL of 0.10M NaOH.

3.0 mL

Explanation:

Let's consider the neutralization between acetic acid and NaOH.

CH₃COOH + NaOH → CH₃COONa + H₂O

The moles of NaOH are:

25.0 * 10⁻³ L * 0.10 mol/L = 2.5 * 10⁻³ mol

The molar ratio of CH₃COOH to NaOH is 1:1. The moles of CH₃COOH are 2.5 * 10⁻³ mol.

The molar mass of CH₃COOH is 60.05 g/mol. The mass of CH₃COOH is:

2.5 * 10⁻³ mol * (60.05 g/mol) = 0.15 g

The mass percent of acetic acid is 5%. The mass of the solution of acetic acid is:

0.15 g CH₃COOH * (100 g solution / 5 g CH₃COOH) = 3.0 g solution

The density of the acetic acid solution is 1.0 g/mL. The volume of the solution is:

3.0 g * (1 mL/1.0 g) = 3.0 mL