Chemistry Lab: How to Prepare Silver Acetate and Determine Precipitate Formation

[ChemDunce]

A quick question that would request a quick response.. please help.

Prepare, by mixing proper quantities of silver nitrate and sodium acetate solutions, approximately 0.3 millimole of silver acetate. Wash this precipitate with a small quantity of distilled water and transfer it to a test tube containing 10 mL of distilled water. Warm the test tube and its contents for 15 minutes, using a water bath held at 40 degrees C (+/-) 5 degrees C, stirring intermittently. Cool the contents to 0 degrees C. Then filter. Divide the filtrate into two approximately equal portions. To one portion add 0.5 mL of 3 M sodium acetate solution; to the other add 0.5 mL of 3 M acetic acid solution. Wait. Why did a precipitate form in one case and not in the other?

Thanks a bunch

 

[chemisttree]

common ion effect.

 

[Blueshift5]

for reaching out! It seems like you are conducting a very interesting experiment in your chemistry lab. Silver acetate is a compound commonly used in organic synthesis and it is important to understand how to prepare it and determine precipitation formation. From your experiment, it seems like you have successfully prepared silver acetate by mixing silver nitrate and sodium acetate solutions.

To answer your question, in one case a precipitate formed and in the other, it did not. This is because when you added sodium acetate solution to one portion of the filtrate, it caused the formation of a new compound, sodium acetate, which is insoluble in water. This resulted in the formation of a precipitate. On the other hand, when you added acetic acid solution to the other portion of the filtrate, it reacted with the sodium acetate present in the solution, forming a soluble compound, sodium acetate. As a result, no precipitate was formed.

I hope this helps to clarify the results of your experiment. Keep up the good work in your chemistry lab!