3.30 mL of vinegar needs 40.0 mL of 0.150M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.10 qt sample of this vinegar?
mols NaOH = M x L = 0.150 x 0.0400 = ?
mols acetic acid = same as mols NaOH.
That gives you mols acetic acid in 3.30 mL. That times molar mass acetic acid gives you grams acetic acid in 3.30 mL.
g acetic acid in 3.30 mL x (mL in 1 qt/3.30) = ?
1.58
To determine the number of grams of acetic acid in the vinegar sample, we need to calculate the number of moles of acetic acid first.
Given:
Volume of vinegar = 3.30 mL
Volume of NaOH solution = 40.0 mL
Concentration of NaOH solution = 0.150M
Step 1: Convert the volumes from mL to liters:
Vinegar volume = 3.30 mL = 3.30 mL * (1 L / 1000 mL) = 0.00330 L
NaOH volume = 40.0 mL = 40.0 mL * (1 L / 1000 mL) = 0.0400 L
Step 2: Calculate the number of moles of NaOH:
moles of NaOH = concentration of NaOH * volume of NaOH
moles of NaOH = 0.150 M * 0.0400 L = 0.00600 moles
Step 3: Determine the stoichiometry ratio between NaOH and acetic acid.
From the balanced equation, we know that for every one mole of NaOH, there is one mole of acetic acid. Therefore, the moles of acetic acid will also be 0.00600 moles.
Step 4: Convert from moles to grams.
The molar mass of acetic acid is 60.05 g/mol.
grams of acetic acid = moles of acetic acid * molar mass
grams of acetic acid = 0.00600 moles * 60.05 g/mol = 0.360 g
Therefore, there are 0.360 grams of acetic acid in a 1.10 qt sample of this vinegar.