How will this affect the test for the presence of ammonium ion in the solution?

Question

How will this affect the test for the presence of ammonium ion in the solution? Explain. Part D.1. What is    the fate of Zn^2+ in the experiment? Explain. Part D.2. Instead    of 6 M HNO_3 being added to the solution, 6 M HC1 is added (both are    strong acids).    How    will this affect the test for the presence of copper(II) ion in the solution? Explain. Part D.3. Instead    of cone NH_3 being added to the solution, 6 M NAOH is added (both arc bases).    How    will this affect the test for the identification of copper(II) ion in the solution ? Explain. Part E.1. Instead of 6 M NH_3 being added to the solution. 6 M  NAOH is added (both arc bases) before the addition of the K_2C_2O_4. What would be the appearance of the solution? Explain. When NH_3is used,the cu^2+ ions will from complex with the NH_3.The complex is soluble and makes the solution dark blue if LM NAOH is used a Blue Precipitate is formed

Explanation / Answer

(7)Cu2+ ions react with HCl to form CuCl2 but they donot react with HNO3, no double displacement reaction is occured.

(9) Cu2+ when treated with NH3 forms complex ion where as when treated with NaOH forms a blue precipitate.

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