A 31.43 mL volume of 0.108 M NaOH is required to reach the phnolphthalein endpoint in the titration of a 4.441 g sample of vinegar. Calculate % acetic acid in the vinegar.
A 30.84 mL volume of 0.128 M NaOH is required to reach the phnolphthalein endpoint in the titration of a 5.441 g sample of vinegar. Calculate % acetic acid in the vinegar.
To calculate the percent acetic acid in the vinegar, we need to determine the number of moles of NaOH used in the titration.
Step 1: Calculate the number of moles of NaOH used.
We can use the equation:
Moles = (concentration) × (volume in liters).
Given:
Volume of NaOH used = 31.43 mL = 31.43/1000 = 0.03143 L
Concentration of NaOH = 0.108 M
Moles of NaOH = (0.108 M) × (0.03143 L)
Moles of NaOH = 0.00342844 mol
Step 2: Convert the grams of acetic acid to moles.
Given:
Mass of vinegar sample = 4.441 g
We need to use the molar mass of acetic acid to convert the mass to moles.
The molar mass of acetic acid (CH3COOH) is:
(12.01 g/mol) + (1.01 g/mol × 3) + (16.00 g/mol) + (1.01 g/mol + 2) = 60.05 g/mol.
Moles of acetic acid = (mass) / (molar mass)
Moles of acetic acid = (4.441 g) / (60.05 g/mol)
Moles of acetic acid = 0.073979 mol
Step 3: Calculate the percent acetic acid.
To find the percent, we use the equation:
Percent = (moles of acetic acid / moles of NaOH) × 100
Percent acetic acid = (0.073979 mol / 0.00342844 mol) × 100
Percent acetic acid = 21.57%
Therefore, the percent acetic acid in the vinegar is approximately 21.57%.
To find the percent of acetic acid in the vinegar, we need to determine the number of moles of acetic acid in the 4.441 g sample of vinegar. Then we can calculate the percent by mass using the equation:
% Acetic acid = (moles of acetic acid / mass of vinegar sample) x 100
First, we need to calculate the number of moles of NaOH that reacted in the titration. We know the volume and concentration of NaOH, so we can use the equation:
moles of NaOH = volume of NaOH (L) x concentration of NaOH (mol/L)
Given:
Volume of NaOH = 31.43 mL = 0.03143 L
Concentration of NaOH = 0.108 M
moles of NaOH = 0.03143 L x 0.108 mol/L = 0.0033924 mol
Since the stoichiometric ratio between NaOH and acetic acid is 1:1 (from the balanced chemical equation of the reaction), the number of moles of acetic acid is also 0.0033924 mol.
To convert the mass of the vinegar sample to grams, we multiply by the molar mass of acetic acid, which is 60.05 g/mol.
mass of vinegar sample = 4.441 g
Next, we can calculate the percent acetic acid in the vinegar:
% Acetic acid = (0.0033924 mol / 4.441 g) x 100
% Acetic acid = 0.0764 x 100 = 7.64 %
Therefore, the percent acetic acid in the vinegar is 7.64%.
The reaction is:
CH3COOH + NaOH = CH3COONa + H2O
Thus one mole of NaOH reacts with one mole of acetic acid.
31.43 ml of 0.108M NaOH contains 0.03143*0.108=0.003394M.
1M of acetic acid weighs (12+3+12+2*16+1)=60 g
0.003394M contains .003394*60=0.2037g
This amount is contained in about 4,441 g of water. Therefore the percentage is 0.2037/4.441=4.59%