Titration lab report - determine the percentage of acetic acid present in vinegar

• Use hot water to clean a 5 mL pipet:
• Place a rubber propipette bulb on the end of the pipet and squeeze “A” to push the air out.
• Press “S” to draw hot water up slightly over the calibration mark.
• Press “E” to drain the water.
• Rinse the pipet with 2 successive 5 mL portions of colorless vinegar.
• Draw the vinegar solution up until the bottom of the meniscus coincides with the calibration mark.
• Deliver the 5 mL aliquot into a 125 mL Erlenmeyer flask. Touch off the last drop on the inside wall.
• Fill the buret with the standardized NaOH solution
• Record the initial buret reading.
• Add 2 drops of phenolphthalein indicator to the vinegar solution.
• Titrate to a faint pink end point.
• Record final buret reading.
• Repeat the titration at least once more until the volume of NaOH solution used in 2 different titrations agrees within 0.2 mL.

Chemical equations:

H2C2O4.2H2O+2NaOH→Na2C2O4.2H2O+2H2O

CH3COOH+NaOH→CH3COONa+H2O

 

DATA and CALCULATION:

Standardization of NaOH solution:

Sample calculation for Titration 1:

H2C2O4.2H2O+2NaOH→Na2C2O4.2H2O+2H2O          (1)

mole of H2C2O4.2H2O= 0.1594 g 126.068gmol=0.001264 mol

From equation (1):

mole of NaOH=2×mole of H2C2O4.2H2O=2×0.001264 mol=0.002528 mol

        
MNaOH=mole of NaOHVolume of NaOH used=0.002528 mol24.62 ×10-3 L=0.1027 M

   

Titration of vinegar with standardized NaOH solution:

CH3COOH+NaOH→CH3COONa+H2O                             (2)

From the equation (2):

mol of CH3COOH=mol of NaOH=VNaOH (used)×MNaOH (average)

Mvinegar=mol of CH3COOHVvinegar

% of CH3COOH in vinegar=mass of CH3COOHmass of vinegar×100%=mol of CH3COOH×molar mass of CH3COOHdensity of vineger×Vvineger×100%=mol of CH3COOH×60.052 g/mol1 g/mL×5mL×100%=mol of CH3COOH×12.0104 mol-1×100%

Titration 1:

mol of CH3COOH=mol of NaOH=42.00×10-3L×0.1022molL=4.292×10-3 mol

Mvinegar=4.292×10-3 mol5×10-3L=0.8584 M

% of CH3COOH in vinegar=4.292×10-3 mol×12.0104 mol-1×100%=5.155%

Titration 2 and 3:

mol of CH3COOH=mol of NaOH=42.10×10-3L×0.1022molL=4.303×10-3 mol

Mvinegar=4.303×10-3 mol5×10-3L=0.8605 M

% of CH3COOH in vinegar=4.303×10-3 mol×12.0104 mol-1×100%=5.168%

CONCLUSION:

We did the experiment to standardize the NaOH solution in order to get the exact concentration of NaOH solution to be 0.1022 M that agrees with the given concentration of NaOH solution, 0.1 M.

We did the titrations successfully in order to get the percentages of acetic acid in vinegar to be approximately to be 5.2%. The value is acceptable for the table vinegar that was given in the lab.