Separation Scheme Lab Report

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The separation scheme was created by centrifuging a controlled sample of one of the ions or anions combined with one of the reagents such as NaCl. It was found that it could be separated out after centrifuging and would result in a sample without the ions or anions if it precipitated. With the data from week one a method of adding the reagents in a certain order would result in the ability to separate all ions or anions out of a solution. Overall, knowing how the cations or anions reacted with certain reagents benefited greatly to determine a way to create a separation scheme. In the unknown sample T it was determined that Pb, Ba, and Cu were present. Adding NaCl resulted in a white precipitate and based on our data from week one it meant that Pb was present. We centrifuged and separated the mixture from the Pb precipitate. Then by adding H2SO4, another white precipitate was formed. It was found to be Ba due to Pb already been taken out of the solution from before. After once again centrifuging and separating, adding NH3 resulted in a half blue and clear solution. Based on data from week one, Cu had to be present, we then tested the clear solution with Na2C3O4 which resulted in no reaction. Through process of elimination and which reagents resulted in a precipitate being formed it was determined that Pb, Ba, and …show more content…
By adding AgNO3, a white precipitate was formed. According to data from week one NaCl had to be present because the other anions would have created a precipitate in a different color. After centrifuging and separating the sample we added BaCl2 which resulted in another white precipitate. The reaction showed that Na2SO4 must have been present and not Na2CO3 because there was no brown precipitate formed in the previous reaction and Na2SO4 would have been the only anion to create a white precipitate with BaCl2. After adding both reagents it was clear that unknown sample 1 contained NaCl and

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