The reaction between potassium iodide (KI) and copper sulfate (CuSO₄) results in a double displacement reaction where the ions exchange partners. Here's the balanced chemical equation:

2KI (aq) + CuSO₄ (aq) → K₂SO₄ (aq) + CuI₂ (s)

Here's a breakdown of the reaction:

* Reactants:

* Potassium iodide (KI) is a soluble ionic compound.

* Copper sulfate (CuSO₄) is also a soluble ionic compound.

* Products:

* Potassium sulfate (K₂SO₄) is a soluble ionic compound that remains dissolved in the solution.

* Copper(II) iodide (CuI₂) is an insoluble ionic compound that precipitates out of the solution as a solid.

Observations:

* When you mix solutions of potassium iodide and copper sulfate, you'll observe the formation of a white precipitate of copper(II) iodide.

* The solution will also turn brown due to the presence of triiodide ions (I₃⁻), which are formed as a side reaction involving the iodide ions and the copper(II) ions.

Important Note:

The formation of the precipitate is a key indicator of this reaction. It demonstrates that a new compound, copper(II) iodide, has been formed and is not soluble in water.