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Double Replacement Reaction Lab Report

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In this experiment, double replacement reactions amongst six aqueous compounds were witnessed. (Find better place for this sentence->)The six known solutions were NaOH, K2SO4, Cu (NO3)2, Pb(NO3)2, Co(NO3)2, and BaCl2. Following this, deductive reasoning was used to determine the identity of two unknown solutions, based off of how they reacted with the six standard compounds. In order to set up the experiment, firstly each of the six compounds were poured onto six makeshift "wells " on a mirrored sheet, with each well containing two drops of a single compound. The original color of each compound was recorded. After this, two more drops of the compounds were added to the individual wells so that each well would contain a unique combination of two compounds. Once this was completed, it was determined which solutions had reacted by whether or not a precipitate formed in the solution. If a precipitate had formed, the new color of the solution was recorded. The result of this portion of the experiment where as follows: The clear solution of …show more content…

Because the first unknown’s cation formed a precipitate with the OH- from NaOH, the cation cannot be Barium, as BaOH is a soluble compound. The unknown substance also did not react with BaCl2, and so the cation cannot be lead, as lead always forms a precipitate with chlorides. However, it formed no precipitate with either Cu(NO3)2 or , Co(NO3)2, which means the cation is either copper or lead. Because the unknown formed a precipitate with BaCl2, and Barium can only form an insoluble compound with sulfate, the anion of the compound must be SO42. Finally, because the resulting precipitate was blue, it was finally deduced the cation is Cu2+, as blue solutions are a characteristic of copper. Therefore, the identity of the first unknown is CuSO4, or Copper (II)

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