in the ot NCI by the w 12.54 Naotlo 173 11. 0.100 froiNaOFI × : Nao answer (M HC
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Question
in the ot NCI by the w 12.54 Naotlo 173 11. 0.100 froiNaOFI × : Nao answer (M HC) are c tion in t of the NaOH s mot HC o.otooo 22 87ml ofa 0.158 M KOH of NaOH and e magnitude of titration of a C. HCOOH aw all ion(s/moleculeis) that would be present in each beaker if the substance Type of electrolyte: type of electrolyte. Also, list or dra as diselved in water in the 23. Determine e approximate ratios that they would be present A Na Type of electrolyte Type of electrolyte You are given an unknown solution and told that it has one or more of the following cations, Ag'. Ba+ and Zn2. You add hydrochloric acid and a white precipitate forms. You remove the solid precipitate (and all of the ion that formed the precipitate) and d sulfuric acid to the remaining solution. Nothing happens (no solid forms). Then you add sodium hydroxide to the solution unti the solution is strongly basic and a white precipitate forms (that is a differeat compound than the first white precipitate) Of the three possible ions in the solution, which ion or ions are present? Justify your answer with net ionic equations showing the formation of any precipitates involving the ions that are present. A. If you place 1.55 g of aluminum in a beaker with 125 mL of 0.125 M KOH, what mass of KAI(OH), will be produced? B. How many g of Al will remain after the reaction? How many moles of KOH will remain after the reaction? 25. You can dissolve an aluminum soft drink can in an aqueous base such as potassium hydroxide. 2 Al(s) + 2 KOHaq) + 6 H20 (I)--2 KAI(OH)4(aq) + 3 H2(g)Explanation / Answer
Qualitative analysis of metal cations
24.
Adding HCl gave white precipitate : AgCl formed, thus Ag+ present
adding NaOH gave white precipitate : Zn(OH)2 formed, thus Zn2+ present
So, of the four possible cations, Ag+, Zn2+ and Ba2+, upon qualitative analysis we found that the ions present in solution are, Ag+ and Zn2+. Ba2+ is absent.
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