Assuming the density of 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
moles acid needed= .025*.1 moles.
grams acetic acid= molesacid*molmassAcid
Now at this point, I am stupified on the density. The given density is the same as water. So how can it be acetic acid solution? Vinegar has a density of about 1.01g/ml, that is 5 percent acetic acid.
I am not certain you can work this without better data.
To determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH, we first need to use the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
The balanced chemical equation is as follows:
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
Given that the molarity (M) of NaOH is 0.10 M, we can calculate the number of moles of NaOH that are in 25.0 mL of the solution.
Using the equation:
moles = volume x molarity
moles of NaOH = 25.0 mL x 0.10 M
moles of NaOH = 2.5 mmol (millimoles)
Since 1 mole of NaOH reacts with 1 mole of acetic acid, the moles of acetic acid needed to neutralize the specified amount of NaOH is also 2.5 mmol.
Now, we need to convert 2.5 mmol of acetic acid to grams, using the molar mass of acetic acid.
The molar mass of CH3COOH is:
(1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (4 x 16.00 g/mol) + 1.01 g/mol
= 60.05 g/mol
Using the equation:
mass = moles x molar mass
mass of acetic acid = 2.5 mmol x 60.05 g/mol
mass of acetic acid = 150.13 mg
Since the density of the acetic acid solution is given as 1.0 g/mL, we can convert the mass of acetic acid to volume.
Using the equation:
volume = mass / density
volume of acetic acid solution = 150.13 mg / 1.0 g/mL
volume of acetic acid solution = 150.13 mL
Therefore, the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH is 150.13 mL.