1a) You mix 125.5mL of 4.5x10^-6 M cobalt (II) iodide with 525.5mL of 3.8x10^-5

Question

1a) You mix 125.5mL of 4.5x10^-6 M cobalt (II) iodide with 525.5mL of 3.8x10^-5 M potassium carbonate. When you mix the two solutions you do not observe any reaction. Explain, with relevant calculations shown. b) Even with your analysis above you become concerned that there is something wrong with your original solutions. Suggest a method to test for the presence of one of the ions present in either the cobalt (II) iodide or the potassium carbonate solution.

Explanation / Answer

a) CoI2 + K2CO3 > 2 I- + 2K+ + CoCO3 n = mv = 125.5mL x 4.5x10^-6 M = 0.000565mol n = mv = 525.5mL x3.8x10^-5 M = 0.01997 mol The ratio of moles of potassium carbonate to cobalt iodide is about 35:1 so you don't observe a reaction. b) To test for the presence of an ion, add a solution that you know will form a precipitate. For example, to test for iodine, add lead nitrate, and that should react with the cobalt iodide to precipitate out lead iodide. If a precipitate forms, you know iodine is present.

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