a 50ml aliquot sample of vinegar is titrated with a 0.078 M NaOH sol'n. The phenolpthalein end point was reached after the addition of 35.6 ml.
Equation:
HC2H3O2 + NaOH _____> NaC2H3O2 + H20
Calculate the % by mass/vol. of acetic acid in the sample.
mols NaOH used = M x L = estimated 0.0028 but you need a more accurate answer.
Convert mols NaOH to mols acetic acid and that is 1:1; therefore, mols acetic acid (vinegar) = estimated 0.0028.
g acetic acid = mols x molar mass = approx 0.0028 x about 60 = about 0.17g.
So you had about 0.17g vinegar in 50 mL or about 0.34g in 100 mL and that is % w/v or about 0.34%.
Note: commercial vinegar is about 3-5% w/v.
To calculate the percentage by mass/volume of acetic acid in the sample, we need to determine the amount of acetic acid that reacted with NaOH during the titration.
Step 1: Calculate the number of moles of NaOH used.
First, calculate the moles of NaOH used by multiplying the volume of NaOH solution (35.6 ml) by the molarity of the NaOH solution (0.078 M):
Moles of NaOH = Volume of NaOH solution (L) × Molarity of NaOH (mol/L)
Moles of NaOH = 35.6 ml × (1 L / 1000 ml) × 0.078 mol/L
Step 2: Determine the moles of acetic acid.
According to the balanced equation, the mole ratio between NaOH and acetic acid is 1:1. Therefore, the moles of acetic acid would be equal to the moles of NaOH used.
Step 3: Calculate the mass of acetic acid.
To determine the mass of acetic acid, multiply the moles of acetic acid by its molar mass. The molar mass of acetic acid is 60.052 g/mol.
Mass of acetic acid (g) = Moles of acetic acid × Molar mass of acetic acid
Step 4: Calculate the percentage by mass/volume.
The percentage by mass/volume can be calculated by dividing the mass of acetic acid by the volume of the vinegar sample and multiplying by 100.
Percentage by mass/volume = (Mass of acetic acid / Volume of vinegar sample) × 100
Now, let's calculate each step:
Step 1:
Moles of NaOH = 35.6 ml × (1 L / 1000 ml) × 0.078 mol/L
Moles of NaOH = 0.002775 moles
Step 2:
Since the mole ratio between NaOH and acetic acid is 1:1, the moles of acetic acid would also be 0.002775 moles.
Step 3:
Mass of acetic acid (g) = Moles of acetic acid × Molar mass of acetic acid
Mass of acetic acid = 0.002775 moles × 60.052 g/mol
Mass of acetic acid = 0.1667 g
Step 4:
Percentage by mass/volume = (Mass of acetic acid / Volume of vinegar sample) × 100
Percentage by mass/volume = (0.1667 g / 50 ml) × 100
Percentage by mass/volume = 0.3334%
To calculate the percentage by mass/volume of acetic acid in the vinegar sample, we need to determine the number of moles of acetic acid present in the aliquot titrated with NaOH.
Step 1: Calculate the moles of NaOH used in the titration.
Moles of NaOH = Molarity x Volume (in liters)
Moles of NaOH = 0.078 M x 35.6 ml = 0.078 M x 0.0356 L = 0.00278 moles
Step 2: Use the stoichiometry of the balanced equation to determine the moles of acetic acid.
From the balanced equation:
1 mole of HC2H3O2 reacts with 1 mole of NaOH
Therefore, the number of moles of HC2H3O2 (acetic acid) in the sample is equal to the moles of NaOH used.
Moles of HC2H3O2 = 0.00278 moles
Step 3: Calculate the mass of acetic acid.
To calculate the mass of acetic acid, we need to know the molar mass of HC2H3O2, which is 60.05 g/mol.
Mass of HC2H3O2 = Moles of HC2H3O2 x Molar mass of HC2H3O2
Mass of HC2H3O2 = 0.00278 moles x 60.05 g/mol = 0.1667 g
Step 4: Calculate the percentage by mass/volume.
Percentage by mass/volume = (Mass of HC2H3O2 / Volume of sample) x 100
Volume of sample = 50 ml = 0.05 L
Percentage by mass/volume = (0.1667 g / 0.05 L) x 100 = 333.4%
Therefore, the percentage by mass/volume of acetic acid in the vinegar sample is 333.4%. Note that this value exceeds 100% due to an error in the calculation caused by a possible typo or error in the provided data.