Experiment No.14- Acid/Base Titrations

By: Magaly MartinezGeneral Chemistry 1412 –P01

Objective: An unknown acid (either a sample of vinegar or an acid salt ) will be analyzed by the process of titration, using a standard sodium hydroxide solution.

Procedure: A. Preparation of Burets and PipetFor precise quantitative work, volumetric glassware must be scrupulously clean. Rinse the burets and the pipet with distilled water to see if they are clean

B. Preparation of the Sodium Hydroxide Solution.Clean and rinse the 1-L bottle and stopper. Label the bottle “Approx. 0.1 M NaOH.” Put about 500 mL of distilled water into the bottle. Transfer approximately 4 grams of sodium hydroxide pellets to the 1-L . Stopper and shake the bottle to dissolve the pellets. Add additional distilled water to the bottle until the water level is approximately 1 inch from the top. This sodium hydroxide solution is the titrant for the analyses to follow. Set up one of the burets in the buret clamp. See picture 14-1. Rinse and fill the buret with the sodium hydroxide solution.

C. Standardization of The Sodium Hydroxide SolutionClean and dry a small beaker. Pour a few grams of KHP into the beaker. Clean 3 250-mL flasks, and rinse them with distilled water. Weigh three samples of KHP between 0.6 and 0.8, one for each flask. Record the exact weight. Add 100 mL of water to KHP sample 1, and 2-3 drops of phenolphthalein indicator solution. Record the initial reading of the NaOH solution in the buret to the nearest 0.2 mL. Begin adding NaOH solution from the buret to the sample in the Erlenmeyer flask, swirling the flask constantly during the addition. The titration should require at least 20 mL of NaOH solution causes a permanent pale pink color that does not fade on swirling .One single drop of NaOH causes a permanent pale pink color. Repeat the titration of the remaining KHP samples.

D. Analysis of the Unknown Acid Sample.Vinegar is a dilute solution of acetic acid and can be effectively titrated with NaOH using the phenolphthalein endpoint. Obtain a 20 – 30 mL of the unknown vinegar solution, and record the code number of the sample 1, 2, and 3. Using rubber safety bulb to provide suction, rinse the 5 mL pipet with small portions of the vinegar solution and discard the rinsing. Add 100 mL of distilled water to each flask, 5 mL of vinegar and 2 or 3 drops of phenolphthalein in solution. Refill the buret with NaOH solution and record the initial reading of the buret. Titrate sample 1 of vinegar in the same manner as in the standardization until one drop of NaOh causes the appearance of a pale pink color. Record and repeat the titration.

Data Tables:

Sample 1 Sample 2 Sample 3Weight of KHP taken

.844 .700 .740

Initial NOH buret reading

.02 0 .02

Final NaOH buret treading

24.1 16.3 16.8

Volume NaOH used 23.9 16.1 16.6Moles KHP present 0.0041 .0034 0.036Molarity of NaOH solution

.000, .38 M .008, 27 M .008, 65 M

Mean Molarity 8.43 x 10 -3Average deviation .0014

Analysis of Vinegar Solutions:

Sample 1 Sample 2 Sample 3

Initial NaOH buret reading

0 0 0

Volume NaOH used 25.6 31.4 25.1

Quantity of Vinegar Taken

5 mL 5 mL 5 mL

Final NaOH buret reading

25.6 31.4 25.1

Molarity of vinegar 3.28 M 2.68 M 3.35 MMean Molarity of Vinegar

3.10 M

Average Deviation .28 M deviation

% by Mass Acetic Acid Present

10%

Calculations:

Calculate moles: Divide the mass of KHP by its molar Mass

Mols= Mass of KHP/ Molar Mass=8.44/204.2=.0041 moles of KHP

Calculate the Concentration of NaOH: Divide the moles by the volume to find the Molarity.

M= mols / VolumeM= 0.0041/ (23.9mL+25mL)/1000M= 0.00838

Calculate mean and average deviation.Mean Value: Add up all the numbers, then divide by how many numbers there are.

=(.00838+.00827+.00865)/ 3=.00843 M

Average Deviation: Find the difference between each value and the mean, then divide by 3.

= (0.5+.16+.22)/3=.0014 M deviation

Calculate Molarity:Divide the mols of Vinegar by the amount in Liters of the solution.

M= mols/ Liters of solutionM= .0841 mol /.0258 LM= 3.28 M

Calculate the average deviation :Find the difference between each value, and the mean, then divide by 3

= (.18 + .6 + .07)/3= .28 M deviation

Mass %:Divide the mass of Acetic acid bt the mass of vinegar then multiply by 100

= (Mass of Acetic Acid/ Mass of Vinegar) x 100= (.506 g / 5.01 g) x 100= 10.0 %

Conclusion: The acid base titration provides key insights into the behavior of both acid and the base in their equilibrium with each other and with water. It was observed that when a strong acid and a strong base react with each other in equivalent concentrations the pH of the solution is neutral. (1:1) We got a mean molarity of vinegar of 3.10 and 10% of mass of the acetic acid in the solution.