Write the formula of the anion present (i.e. anions are charged; do not write compounds) in these unknowns as indicated by these test results.
1) An unknown was mixed with ammonium hydroxide and then barium chloride. Nothing happened. A fresh portion was mixed with nitric acid and silver nitrate and again nothing happens. A third portion was treated with sulfuric acid and no odor was detected. Upon addition of ferrous sulfate and concentrated sulfuric acid, a purple-brown discolorization was seen.
2) An unknown was mixed with ammonium hydroxide and barium chloride. Nothing happened. A fresh portion was mixed with nitric acid and silver nitrate, whereupon a precipitate formed. A third portion was treated with chlorine water, followed by mixing with hexane. The hexane layer turned yellow, then gradually light brown upon further mixing.
Please help!!!
1 nitrate ion, NO3^-
2. AgNO3 test shows either Cl^-, Br^-, or I^- present. Oxidation with Cl2 water and mixing with hexane shows either Br^- or I^-. The hexane solvent is purple if I^- so the anion must be Br^-.
To determine the formula of the anion present in the unknowns based on the given test results, we need to analyze the observations and reactions. Let's discuss each case separately:
1) In the first case, when the unknown was mixed with ammonium hydroxide and barium chloride, and nothing happened, it indicates the absence of sulfate ions (SO4^2-) since barium forms an insoluble white precipitate with sulfate ions.
When the unknown was mixed with nitric acid and silver nitrate, and nothing happened, it suggests the absence of chloride ions (Cl-) because silver forms a white precipitate with chloride ions.
Furthermore, when treated with sulfuric acid, and no odor was detected, it indicates the absence of sulfide ions (S^2-) since the reaction of sulfide ions with acid produces a distinct odor.
However, when ferrous sulfate and concentrated sulfuric acid were added, a purple-brown discoloration was seen. This suggests the presence of chromate ions (CrO4^2-). The purple-brown color is indicative of the formation of chromium(III) sulfate (Cr2(SO4)3) when chromate ions react with ferrous sulfate and concentrated sulfuric acid.
Therefore, the formula of the anion present in this unknown is CrO4^2- (chromate ion).
2) In the second case, when the unknown was mixed with ammonium hydroxide and barium chloride and nothing happened, it indicates the absence of sulfate ions (SO4^2-).
When the unknown was mixed with nitric acid and silver nitrate, a precipitate formed, indicating the presence of chloride ions (Cl-). The formation of a white precipitate with silver nitrate confirms the presence of chloride ions.
When treated with chlorine water, followed by mixing with hexane, the hexane layer turned yellow, then gradually light brown upon further mixing. This indicates the presence of iodide ions (I-). The yellow coloration is produced due to the reaction between iodide ions and chlorine water, forming iodine, which dissolves in hexane, giving the yellow color.
Therefore, the formula of the anion present in this unknown is I- (iodide ion).
In summary:
1) Anion present in the first unknown: CrO4^2- (chromate ion).
2) Anion present in the second unknown: I- (iodide ion).