Titration Experiment Report

Rolla Tyas Amalia- Grade 10

Aim:

- To find how many NaOH needed for titrating 10 ml of vinegar

- To find concentration of the vinegar

Materials:

1. Vinegar 6. 2 beakers

2. Conical flask 7. Phenophtalyn

3. Measuring pipette 8. Dropper

4. 2 grm of NaOH (0,1 M) 9. 50 ml burette

5. Retord stand 10. 2 volumetric flasks (500 ml)

Procedure:

1. Fill a beaker with 2 grm of NaOH granule

2. Add water to the beaker

3. Stir the water until the NaOH granule dissolved

4. Put the dissolved NaOH to a 500 ml volumetric flask (1st volumetric flask)

5. Set the retord stand, conical flask, and burette like the picture below

6. Fill another 500 ml volumetric flask (2nd volumetric flask) by water and

10 ml of vinegar, then mix them (to make concentration of the vinegar

lesser)

7. Pour the conical flask by 10 ml of vinegar from the 2nd volumetric flask, by

measuring pipette

8. Add 3 small drop of phenophtalyn to the conical flask

9. Fill the 50 ml burette with 50 ml of NaOH from the 1st volumetric flask

10. Titrate the NaOH in the burette to 10 ml of vinegar in the conical flask

11. Observe until the solution in the conical flask change color into light

purple (not dark purple)

12. Measure how many ml of NaOH take to make the solution change into

purple color

13. Repeat step 7 until 12 for second and third times

Data/Experiment result:

Titrant: NaOH

Concentration: 0,1 M

Volume:

1st Experiment result: 49, 1 ml

2nd Experiment result: 49,3 ml

3rd Experiment result: 49, 5 ml

Average volume: 49,1 ml + 49,3 ml +49,5 ml

_______________________________ = 49,3 ml

3

Analyte: Vinegar

Volume: 10 ml

Concentration: V1 x M1 = V2 x M2

49,3 ml x 0,1 M = 10 ml x M2

4,93 : 10 = M2

0,493 =M2

0,5 M = M2

Analysis:

It takes 49,3 ml of NaOH in average, for titrating 10 ml of vinegar, and the

concentration of the vinegar is 0,5 M.