A vinegar sample was analyzed and found to be 5.88% acetic acid by volume. The density of pure acetic acid is 1.049g/mL. What volume of 0.200M NaOH would be required to titrate a 2ml aliquot of this vinegar?
Some help would be greatly appreciated!
thanks
My assumption, when you say 5.88% by volume that you mean 5.88% v/v or 5.88 mL acetic acid/100 mL solution. If that is so, then each mL solution contains 0.0588 mL acetic acid.
mass = volume x density so
mass = 0.0588 mL x 1.049 g/mL = ??
Change that to moles. moles = g/molar mass acetic acid.
Then L x M = moles. Plug in moles acetic acid and M of NaOH and calculate L NaOH, then convert to mL.
Check my thinking. I get something like 5 mL but check this carefully.
To determine the volume of 0.200M NaOH required to titrate a 2ml aliquot of the vinegar sample, we need to use the concept of stoichiometry and the equation of 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 one mole of acetic acid reacts with one mole of sodium hydroxide.
First, we need to determine the number of moles of acetic acid in the 2ml aliquot of vinegar. We can use the given concentration and density of acetic acid to do this:
Density of acetic acid = 1.049g/mL
Volume of acetic acid in the aliquot = 2ml
Mass of acetic acid in the aliquot = Density x Volume
Mass = 1.049 g/mL x 2 mL = 2.098 g
Next, we calculate the moles of acetic acid using its molar mass:
Molar mass of acetic acid = 60.05 g/mol
Moles of acetic acid = Mass / Molar Mass
Moles = 2.098 g / 60.05 g/mol = 0.03494 mol
Since the reaction is 1:1, we also know that the number of moles of sodium hydroxide required is equal to the number of moles of acetic acid in the vinegar.
Now, we can calculate the volume of 0.200M NaOH required using the given concentration:
Molarity of NaOH solution = 0.200M
Moles of NaOH = Moles of acetic acid = 0.03494 mol
Volume of NaOH solution = Moles of NaOH / Molarity of NaOH
Volume = 0.03494 mol / 0.200 mol/L = 0.1747 L = 174.7 mL
Therefore, the volume of 0.200M NaOH required to titrate a 2ml aliquot of the vinegar sample is 174.7 mL.
To determine the volume of 0.200M NaOH required to titrate a 2ml aliquot of the vinegar, we first need to calculate the moles of acetic acid present in the 2ml aliquot.
Step 1: Calculate the mass of acetic acid in the 2ml aliquot.
Density = mass/volume
Given that density of pure acetic acid is 1.049g/mL, we can use this to calculate the mass of acetic acid in 2ml.
mass = density * volume
mass = 1.049g/mL * 2ml
Step 2: Calculate the moles of acetic acid using the mass and molar mass.
Molar mass of acetic acid (CH3COOH) = (12.01g/mol * 2) + (1.01g/mol * 4) + (16.00g/mol) = 60.05g/mol
moles = mass / molar mass
moles = (1.049g/mL * 2ml) / 60.05g/mol
Step 3: Calculate the volume of 0.200M NaOH required to react with the calculated moles of acetic acid.
The balanced chemical equation between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, it can be seen that the mole ratio between acetic acid and NaOH is 1:1.
Therefore, moles of NaOH required = moles of acetic acid
Now, we can use the volume and molarity of NaOH to calculate the required volume:
moles = volume (in liters) * molarity
moles = volume (in liters) * 0.200M
Since the mole ratio is 1:1, the volume of NaOH required will be the same as the volume of acetic acid.
Thus, the volume of 0.200M NaOH required to titrate a 2ml aliquot of this vinegar is also 2ml.