fix the laratory Report

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LaboratoryReport.docx

1

One of the biggest problems in chemistry Is to find out the purity of the compound that you have. There are a lot of way to find the purity of your compound depend on the type of the compound. For example, for organic compound using TLC to find if there more than one compound in the solution then ran column then ran NMR (Nuclear Magnetic Resonance) to find if the compound is pure.However, if the compound is acid or base, titration is one of the best ways to determine the purity of the compound. Titration is using known compound to determine the purity of unknown compound in present of phenolphthalein indicator.

CH3COOH(aq)+NaOH(aq)→CH3COONa(aq)+H2O(l)

For example, the balanced equation of Acetic acid and sodium hydroxide show that each one mole of Acetic acid will react completely with one mole of sodium hydroxide. Assuming that the unknown compound is Acetic acid and the known compound is sodium hydroxide. Finding the moles of Acetic acid in the solution by adding sodium hydroxide to the solution in present of phenolphthalein indicator until it reaches equivalence point of the reaction. The pH in the acid will increase by adding base such as sodium hydroxide. If the PH reaches the equivalence point which is 8 pH the solution’s color will change to light pink. By knowing the number of moles of sodium hydroxide that used in the reaction we can know how many moles of the Acetic acid in the solution.

EXPERIMENTAL

Put about 10 mL of 5.09% +_0.02% vinegar in 250 mL flask before that we weight that 250 mL flask. then we weight the flask with the vinegar. The vinegar Mass is 9.79g. We dilute it with 10 mL DI water, and we add 2 drops of phenolphthalein indicator. We run the stopcock with NaOH. Then we add 50 mL of 0.2985+_0.0002 NaOH to stopcock. The NaOH was adding to the vinegar dropwise until the solution turned pink. The processes repeat three times. The solution reaches the egalitarian point that the solution turned pink at 27.79 mL of 0.2985+_0.0002 NaOH.

DATA

the Data Tables 1 and 2, have all the outcome that I get from the procedure above. In table 1 we find the moles of CH3 COOH by measured volumes of NaOH that react with CH3 COOH. Then we convert the volumes of NaOH to moles of NaOH. Than we find the mass of CH3 COOH. According to equation (1) mole of NaOH= mole of CH3 COOH = (mass of CH3 COOH)/ (molar mass of CH3 COOH)

Volume of vinegar (mL)

Moles of Pure

CH3 COOH and of

NaOH (mol)

volume of

NaOH

Delivered (mL)

Molarity of

NaOH Solution

(M)

mass of Pure

CH3 COOH(g)

mass of vinegar (g)

Trial 1

10

0.00832

27.86

0.2985

0.4992

9.82

Trial 2

10

0.00829

27.76

0.2985

0.4974

9.81

Trial 3

10

0.00830

27.80

0.2985

0.4980

9.75

Table 1: Standardization of NaOH Solution

In table 2 we have Statistcal Analysis of CH3 COOH mass present.

Mass present of CH3 COOH (%)

Mean

5.09 %

Standard Deviation

+_ 0.02%

Mean +/- Std. Dev.

5.09+_0.02%

Table 2: Statistcal Analysis of of CH3 COOH mass present

RESULTS AND DISCUSSION

In this experience mass present of CH3 COOH was 5.09+_0.02% that mean the experience right because the true mass present of CH3 COOH is 5.00%. also, the Trial 1 used the biggest amount of NaOH. That mean that Trial 1 may get over-titrated. However, that does not affect the experience a lot because all the trial has closer number.

One of the biggest problems in this experience was over-titrated. It was really difficult to get the perfect titrated. The solution was clear until the last drop. Than the solution become over-titrated. Adding one drop changed the solution from clear to over-titrated.

The second problem was losing some drop of NaOH. Some NaOH stick one the flask and it does not have Chance to participate in the reaction. Also, before the equal point some drop of NaOH when we remove the flask from its places. This losing amount of NaOH would cause inaccurate number of moles that involved in the reaction.

Another problem is inaccurate volume of vinegar. We use graduated cylinder to measure the volume of vinegar. This could cause inaccurate measure. That cause the equally point to be different from Trial to other.

Conclusion

In conclusion, the mass present of CH3 COOH was 5.09+_0.02%. this is not very accurate because the exact mass present of CH3 COOH was 5.00%. this could be more accurate by having more trial. That will decrease the random error. Since the molarity of NaOH which is 0.2985+_0.0002, the mass present of CH3 COOH become more important for this experiment.