In this experiment, two cations in an unknown solution were found using qualitative analysis. For Part A, the solution was to contain either silver or lead, while in Part B, the solution that was set on the side form Part A was identified to contain either calcium or barium. Part C explored a different method through the use of cation flame tests to observe colored flames from six metal chloride solutions.
In Part A, the initial step was to add HCl to the unknown sample. This was necessary because hydrochloric acid helped separate the cations, which is what was needed in order to distinguish the lead or silver from the calcium or barium. It did this because the chlorine ion of HCl was able to bond with silver or lead, leaving the other cation in the liquid solution that was decanted and set aside for Part B.1 In Step 5, the sample with either silver or lead was heated in a water bath. This step was essential as heating the solution broke any ionic bonds within the …show more content…
K2CrO4 was added in Step 12 and this was what indicated whether or not the sample from Part B contained barium or not. The chloride ion in potassium chromate bound to the barium cation and formed a precipitate, if there were any calcium cations present, they would have stayed in the liquid solution. If the substance in Step 12 remained cloudy, it would have been a sign for a false positive. This refers to a result that may not be correct, and if it were the case, a precipitate of barium may have been formed, but that does not mean it has to. After that step, the precipitate was dissolved in 6M HCl and 6M H2SO4 was added resulting in a white precipitate forming. That confirmed the presence of barium but one last step was taken for the purpose of identifying a false positive that shows the presence of calcium. Once that last step was performed and nothing happened after adding K2C2O4 to the basic yellow solution, Part B was
In Part A, the initial step was to add HCl to the unknown sample. This was necessary because hydrochloric acid helped separate the cations, which is what was needed in order to distinguish the lead or silver from the calcium or barium. It did this because the chlorine ion of HCl was able to bond with silver or lead, leaving the other cation in the liquid solution that was decanted and set aside for Part B.1 In Step 5, the sample with either silver or lead was heated in a water bath. This step was essential as heating the solution broke any ionic bonds within the …show more content…
K2CrO4 was added in Step 12 and this was what indicated whether or not the sample from Part B contained barium or not. The chloride ion in potassium chromate bound to the barium cation and formed a precipitate, if there were any calcium cations present, they would have stayed in the liquid solution. If the substance in Step 12 remained cloudy, it would have been a sign for a false positive. This refers to a result that may not be correct, and if it were the case, a precipitate of barium may have been formed, but that does not mean it has to. After that step, the precipitate was dissolved in 6M HCl and 6M H2SO4 was added resulting in a white precipitate forming. That confirmed the presence of barium but one last step was taken for the purpose of identifying a false positive that shows the presence of calcium. Once that last step was performed and nothing happened after adding K2C2O4 to the basic yellow solution, Part B was