Question

In: Chemistry

Consider how each of the situations described below would affect the experimental value determined for the...

Consider how each of the situations described below would affect the experimental value determined for the molecular weight of an unknown solid acid. explain why the experimental molecular weight would be too large, too small, or unaffected because of the situation described.

A) During the titration the student use deionized water to rinse a drop of NaOH into their flask.

b) after standardizing the NaOH the student allows the container of NaOH to stand open in the air. this allows carbon dioxide from the air to react with the water in the solution to make a small amount of carbonic acid.

C) before beginning the titration of the unknown solid acid, the student incorrectly records the initial buret reading as 50.00 ml instead of 0.00ml . He correctly records the final buret reading as 32.81 ml. He then calculates the volume of NaOH used as 50.00 ml -- 32.81ml.

Solutions

Expert Solution

a) During the titration, if deionized water is added to rinse a drop of NaOH into the flask, the experimental molecular weight is unaffected. Because, this rinses down the drop of NaOH at tip of buret only but not effect the titre value. This also rinses down any contains that has splashed onto the walls of the flask.

b) The formed carbonic acid neutralises some of the starndardized NaOH solution and causes to decrease in the concentration of NaOH than that it was standrardized. So, the required amount of NaOH solution is increases to neutralise the acid in the flask and titre value increases. Then the concentration of acid will be found greater than the original.

Since, Concentration = (Wt./M.Wt) /V, concentration is inversly proportional to molecular weight.

Thus the molecular weight found experimentally is smaller than original.

c) If the student record the initial reading correctly as 0.00 ml, the volume of NaOH consumed is 32.81mL. Since he recorded it as 50.00ml, the volume of NaOH consumed will be (50.00 - 32.81) = 17.19 mL. So, it shows lesser amount of NaOH is required to neutralize the acid in flask. Then the concentration of acid will be found lesser than the original.

Since, Concentration = (Wt./M.Wt) /V, concentration is inversly proportional to molecular weight.

Thus the molecular weight found experimentally is greater than original.


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