Potassium sulfate would be colorless; copper iodide is probably blue or green (I don't know which off the top of my head, but copper salts are generally either blue or green).
In the laboratory, copper(I) Iodide is prepared by simply mixing an aqueous solutions of potassium iodide and a soluble copper(II) salt such copper sulphate. : :: Cu2+ + 2I− → CuI2 The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. : :: 2 CuI2 → 2 CuI + I2
When copper sulfate solution is mixed with potassium iodide, a solid precipitate of copper iodide is formed, while potassium sulfate remains in solution. This reaction is a double displacement reaction. The balanced chemical equation is CuSO4 + 2KI → CuI2 + K2SO4.
2K(I) + CuCl2 => 2KCl + Cu(I)2 The (I) is iodide, because the capital I and lowercase l look the same. This is a double replacement reaction, where copper (II) iodide comes out of solution, because it's not soluble, assuming that both of those are in solution.
No one copper compound is purple, the Cuprous oxide Cu2O is red and ammonical copper sulphate is blue the combination of these compounds may produce purple colour.
Copper iodide (CuI) is a white to pale-yellow solid.
In the laboratory, copper(I) Iodide is prepared by simply mixing an aqueous solutions of potassium iodide and a soluble copper(II) salt such copper sulphate. : :: Cu2+ + 2I− → CuI2 The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. : :: 2 CuI2 → 2 CuI + I2
The amount of excess potassium iodide depends on the stoichiometry of the reaction between potassium iodide and copper sulfate. One equivalent of potassium iodide is needed to react with one equivalent of copper sulfate. Excess potassium iodide would be any amount added beyond this stoichiometric ratio.
The balanced chemical equation for the reaction between potassium iodide and copper sulfate is: 2KI(aq) + CuSO4(aq) → CuI2(s) + K2SO4(aq). In this reaction, a double displacement occurs leading to the formation of insoluble copper iodide and soluble potassium sulfate.
The color of aqueous iron(II) sulfate is typically pale green. When potassium iodide is added, it reacts with iron(II) ions to form a dark brown precipitate of iron(II) iodide.
The endpoint color of sodium hypochlocrite and potassium iodide is not listed.
When copper sulfate solution is mixed with potassium iodide, a solid precipitate of copper iodide is formed, while potassium sulfate remains in solution. This reaction is a double displacement reaction. The balanced chemical equation is CuSO4 + 2KI → CuI2 + K2SO4.
2K(I) + CuCl2 => 2KCl + Cu(I)2 The (I) is iodide, because the capital I and lowercase l look the same. This is a double replacement reaction, where copper (II) iodide comes out of solution, because it's not soluble, assuming that both of those are in solution.
No one copper compound is purple, the Cuprous oxide Cu2O is red and ammonical copper sulphate is blue the combination of these compounds may produce purple colour.
Copper iodide (CuI) is a white to pale-yellow solid.
React a solution of Copper(II) sulfate with Sodium Iodide. It will produce Copper(II) Iodide and Sodium sulfate. This Copper(II) iodide, due to instability, spontaneously decomposes, producing Iodine and Copper(I) Iodide.The Copper(I) iodide comes out as a yellowish brown precipitate. So the overall reaction seems to be:2 CuSO4 + 4 NaI ----> 2 CuI + 2 Na2SO4 + I2
Copper iodide is sparingly soluble in water, meaning it dissolves only to a small extent. It is more soluble in concentrated aqueous ammonia and potassium iodide solutions.
Cuprous iodide can be formed from potassium iodide by mixing a solution of copper(II) sulfate with potassium iodide. The reaction between the two compounds results in the precipitation of cuprous iodide, which is a solid compound. This solid can then be filtered and dried to obtain cuprous iodide.