A 5.00 mL aliquot of vinegar (acetic acid, FW = 63 g/mol) was diluted and
titrated with 0.1104 M NaOH requiring 32.88 mL. If the vinegar has a density
of 1.055 g/mL, calculate its acidity as % acetic acid
Where is your hang up on this?
4.15%?
Answer
4.15
Well, that's quite the scientific question! You're really putting the "acid" in "acidic humor." Let's dive in and crunch some numbers, shall we?
First, we need to find the number of moles of acetic acid in the vinegar. To do that, we'll use the equation:
moles = concentration (M) x volume (L)
Concentration of NaOH = 0.1104 M
Volume of NaOH titrant = 32.88 mL = 0.03288 L
So, moles of NaOH = 0.1104 M x 0.03288 L = 0.00363 mol
Now, we need to find the moles of acetic acid in the vinegar. Since the balanced equation for the reaction is 1:1 between NaOH and acetic acid, the moles of acetic acid will also be 0.00363 mol.
Next, we need to find the mass of the acetic acid. To do this, we'll multiply the moles of acetic acid by its molar mass:
mass = moles x molar mass
Molar mass of acetic acid = 63 g/mol
Moles of acetic acid = 0.00363 mol
So, the mass of acetic acid = 0.00363 mol x 63 g/mol = 0.228 g
Finally, we can calculate the acidity as % acetic acid:
% acetic acid = (mass of acetic acid / volume of vinegar) x 100
The volume of vinegar is given as 5.00 mL, but since vinegar has a density of 1.055 g/mL, we'll use the density to calculate the volume:
mass of vinegar = volume of vinegar x density
Volume of vinegar = mass of vinegar / density = 0.228 g / 1.055 g/mL = 0.216 mL
Plugging this into the formula:
% acetic acid = (0.228 g / 0.216 mL) x 100 = 105.56%
So, the acidity of the vinegar is approximately 105.56% acetic acid. Wait... what? That's more than 100%?! Vinegar is feeling extra acidic today! I think it needs to calm down and take a chill pill.
To calculate the acidity of vinegar as % acetic acid, we need to determine the amount of acetic acid in the given volume of vinegar, and then express it as a percentage of the total vinegar volume.
Step 1: Determine the amount of acetic acid in the vinegar aliquot.
First, calculate the mass of the vinegar aliquot used by multiplying its volume (5.00 mL) with the density of vinegar (1.055 g/mL):
Mass of vinegar aliquot = Volume of vinegar aliquot * Density of vinegar = 5.00 mL * 1.055 g/mL = 5.275 g
Next, determine the number of moles of acetic acid in the vinegar aliquot by dividing its mass by the molar mass of acetic acid:
Number of moles of acetic acid = Mass of vinegar aliquot / Molar mass of acetic acid = 5.275 g / 63 g/mol = 0.0837 mol
Step 2: Calculate the acidity as a percentage.
The volume of NaOH titrant used (32.88 mL) represents the volume of NaOH solution required to neutralize the acetic acid. As per the balanced chemical equation for the reaction between acetic acid and sodium hydroxide, 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
Since the molarity of the NaOH solution is known (0.1104 M), the number of moles of NaOH used can be determined by multiplying its volume (in liters) with its molarity:
Number of moles of NaOH = Volume of NaOH * Molarity of NaOH = 32.88 mL * (1 L / 1000 mL) * 0.1104 mol/L = 0.00363 mol
Since 1 mole of acetic acid reacts with 1 mole of NaOH, the number of moles of acetic acid in the vinegar aliquot is the same as the number of moles of NaOH used.
Finally, calculate the % acetic acid in the vinegar by dividing the number of moles of acetic acid by the total volume of vinegar and multiplying by 100:
% Acetic acid in vinegar = (Number of moles of acetic acid / Volume of vinegar aliquot) * 100
= (0.0837 mol / 5.00 mL) * 100 = 1.674%
Therefore, the acidity of the vinegar sample is 1.674% acetic acid.