What happens when Copper Sulphate reacts with KI and KCl each?

I need explanation with the chemical equation.

1 Answer
Meave60
Mar 17, 2018

#"CuSO"_4("aq") + "KI(aq)"##rarr##"no reaction"#

#"CuSO"_4("aq") + "KCl(aq)"##rarr##"no reaction"#

Explanation:

There will be no reaction. This is a supposed double replacement reaction (metathesis), in which one product must be an insoluble (solid) precipitate, an insoluble gas, or water.

The two possible products from copper sulfate and potassium iodide are copper(II) iodide and potassium sulfate. Both products are soluble in water (aqueous), and neither is a gas. What you would have is a mixture of ions:

#"Cu"^(2+)("aq") + "SO"_4^(2+)("aq") + "K"^(-)("aq") + "I"^(-)("aq")#

It would be written as:

#"CuSO"_4("aq") + "KI(aq)"##rarr##"no reaction"#

The two possible products from copper sulfate and potassium chloride are copper(II) chloride and potassium sulfate. Both products are soluble in water (aqueous), and neither is a gas. What you would have is a mixture of ions:

#"Cu"^(2+)("aq") + "SO"_4^(2+)("aq") + "K"^(-)("aq") + "Cl"^(-)("aq")#

It would be written as:

#"CuSO"_4("aq") + "KCl(aq)"##rarr##"no reaction"#

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