Assuming the density of a 5% acetic acid solution is 1.0 g/mL,determine the volume of the acetic acid solution necessary toneutralize 25.0 mL of 0.10M NaOH.
L x M = moles NaOH used.
moles acetic acid = moles NaOH
g acetic acid = moles x molar mass.
1 g/cc x #cc x 0.05 = grams acetic acid.
Solve for #cc.
Check my thinking. Check my work.
To determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10M NaOH, we need to use the concept of stoichiometry and understand the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that the molar ratio between acetic acid and NaOH is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of NaOH.
First, we need to find the number of moles of NaOH in the 25.0 mL solution.
Number of moles = concentration (M) × volume (L)
Number of moles of NaOH = 0.10 mol/L × 0.0250 L
Number of moles of NaOH = 0.0025 moles
Since the reaction is 1:1, this means that we need an equal amount of moles of acetic acid to neutralize the NaOH.
Now, let's calculate the mass of acetic acid needed using the density of the solution:
Mass = density × volume
Given that the density of the acetic acid solution is 1.0 g/mL and we have 0.0025 moles of acetic acid, we can calculate the mass:
Mass = 1.0 g/mL × volume
0.0025 moles × (60.052 g/mole) (molar mass of acetic acid)
Since we have the mass of acetic acid, we can use the density to calculate the volume of the solution:
Volume = mass / density
Now we can plug in the values to calculate the volume of the acetic acid solution:
Volume = (0.0025 moles × 60.052 g/mole) / 1.0 g/mL
Volume = 0.1501 mL
Therefore, the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10M NaOH is approximately 0.1501 mL.