Assuming the density of a 5% acetic acid solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH. Record this calculation on your report sheet.

5% acetic acid means 5 g/100 g soln.

How many moles NaOH will you need to neutralize. M x L = ??

You will need that many moles of aceitic acid. Convert 5% acetic acid to molarity.
5g/100 g. Convert 100 g soln to volume using the density. Convert 5 g acetic acid to moles. moles = grams/molar mass.

Post your work if you get stuck.

4 ml

To determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH, you need to use the concept of stoichiometry and the balanced equation for the neutralization reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).

The balanced equation for the reaction is:

CH3COOH + NaOH → CH3COONa + H2O

From the equation, you can see that 1 mole of CH3COOH reacts with 1 mole of NaOH. This means that the ratio between the moles of CH3COOH and NaOH is 1:1.

First, let's determine the number of moles of NaOH in the 25.0 mL of 0.10 M NaOH solution.

moles of NaOH = volume (L) × concentration (mol/L)
= 25.0 mL × (1 L / 1000 mL) × 0.10 mol/L
= 0.0025 mol

Since the stoichiometry of the reaction is 1:1, this means that you also need 0.0025 mol of CH3COOH to neutralize the NaOH.

Next, we will use the density of the acetic acid solution to find the volume required.

mass of acetic acid = density × volume
mass of acetic acid = 1.0 g/mL × volume (mL)

To convert the mass to moles, we need the molar mass of acetic acid, which is approximately 60.05 g/mol.

moles of acetic acid = mass of acetic acid (g) / molar mass of acetic acid (g/mol)
= (1.0 g/mL × volume (mL)) / 60.05 g/mol

Equating the moles of acetic acid to the moles of NaOH:

0.0025 mol = (1.0 g/mL × volume (mL)) / 60.05 g/mol

Rearranging the equation and solving for the volume of the acetic acid solution:

volume (mL) = (0.0025 mol × 60.05 g/mol) / 1.0 g/mL
volume (mL) = 0.1501 mL

Therefore, the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH is approximately 0.1501 mL.