September 18, 2012
To study the affect of pH on a food preservative.
C6H5COONa + HCL ------ C6H5COOH + NaCl
This experiment was started with a clear solution of sodium benzoate and HCl was added to it, ultimately producing benzoic acid. First, .3395 g of sodium benzoate was weighed, then it was dissolved in water, causing it to disassociate into ions. Next, 3M of HCl were added drop wise to the solution until it reached a pH of 2, thus introducing the hydronium ion. This addition caused a white, solid benzoic acid to precipitate out of the solution. A vacuum filtration system was used to separate the solid from the liquid. What was once water soluable became insoluable. Calculations:
Theoretical yield and Percent yield
0.3995 g of sodium benzoate x 144.1 g/mol sodium benzoate= 0.0028 mole of
benzoate = 0.0028 mole benzoic acid 0.0028 mole sodium benzoate x 122.1g/mole benzoic acid = 0.34 g benzoic acid
% yield = (0.45g benzoic acid /0.34g benzoic acid) x 100 = 132.352% Results:
A clear solution became a precipitate when HCl was added. What was once water soluable became insoluable. Sodium benzoate was converted into benzoic acid. The Theoretical yield of benzoic acid was found to be .34 g and the percent yield was 132.352%. Conclusion:
To study the affect of pH on a food preservative. Sodium benzoate changed into benzoic acid once the 3 M of HCl was added making it a pH of 2, making the solution acidic. The hydronium ion concentration was high enough that it gave a yield of benzoic acid, which, is very insoluble in water and it formed a precipitate in the solution. The end percent yield was 132.352%, this was due to a few flaws in the experiment. During the filtration process of the experiment the funnel was not working properly and not all the...