Baking Soda and Vinegar Lab
Purpose: To determine the amount of baking soda necessary in terms of a ratio to vinegar to complete a chemical reaction with vinegar with the products of water, carbon dioxide, and sodium acetate.
Hypothesis: A ratio exists such that, with baking soda and vinegar as the reactants, it is possible to define a chemical reaction that uses all of the available reactants and yields carbon dioxide, water, and sodium acetate as the products.
(I'm such a mathematician.)
Materials: A beaker to complete the reaction in, a stirring rod to stir, and baking soda and vinegar to use as the reactants.
1) Measure 1 gram of baking soda and put it in the bottom of the beaker.
2) Measure and record an estimated amount of vinegar and add it to the beaker with the baking soda.
3) Allows the reactants to react and occasionally stir the beaker.
4) When reaction finishes, check to see if any baking soda or vinegar remains. If so, add more of the substance that was used up until you completely use up both reactants.
5) Record the amount of vinegar required to completely react with 1 gram of baking soda.
When we completed the lab, we found approximately 14.29 grams of vinegar to be sufficient in using up all of the vinegar.
It is possible to solve for the amount of vinegar necessary in the reaction. Start by balancing the equation:
NaHCO3 + HC2H3O2 -----> CO2 + H2O + NaC2H3O2
This shows us that there should be one mole of acetic acid to one mole of baking soda. The molar mass of acetic acid (HC2H3O2) is about 60 grams per mole, and the molar mass of baking soda (NaHCO3) is about 84 grams per mole. Since we have a 1:1 ratio of acetic acid to baking soda, we can solve for how many grams of acetic acid are necessary for this equation to work with one gram of baking soda--it's about .7 g. Since vinegar is a 5% aqueous solution of acetic acid, we should multiply this by twenty to learn that it should be necessary to use slightly over 14 grams (the error comes from my rounding for ease of calculation). The experiment needed slightly more than this, because the percent yield of the equation was less than 100%.