1) The student conducted a titration experiment to determine the concentration of vinegar by titrating 10 mL of vinegar with a 0.1 M NaOH solution. 2) The student repeated the titration experiment three times and found that on average, it took 49.3 mL of the NaOH solution to titrate the 10 mL vinegar sample. 3) Using the titration data and calculations, the student determined that the concentration of the vinegar was 0.5 M.

3. Conical flask7. Phenophtalyn

4. Measuring pipette8. Dropper

5. 2 grm of NaOH (0,1 M)9. 50 ml burette

7. Add water to the beaker

8. Stir the water until the NaOH granule dissolved

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

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

11. 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)

12. Pour the conical flask by 10 ml of vinegar from the 2nd volumetric flask, by measuring pipette

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

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

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

16. Observe until the solution in the conical flask change color into light purple (not dark purple)

17. Measure how many ml of NaOH take to make the solution change into purple color

18. Repeat step 7 until 12 for second and third timesData/Experiment result:<br />Titrant: NaOH<br />Concentration: 0,1 M<br />Volume:<br />1st Experiment result: 49, 1 ml<br />2nd Experiment result: 49,3 ml<br />3rd Experiment result: 49, 5 ml<br />Average volume: 49,1 ml + 49,3 ml +49,5 ml<br /> _______________________________ = 49,3 ml<br />3<br />Analyte: Vinegar<br />Volume: 10 ml<br />Concentration: V1 x M1 = V2 x M2<br />49,3 ml x 0,1 M = 10 ml x M2<br /> 4,93 : 10 = M2<br />0,493 =M2<br />0,5 M = M2<br />Analysis: <br />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. <br />