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Chapter 23: Problem 43

Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fluoride is added, a green precipitate is formed. When aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what is happening in these two cases.

Short Answer

Expert verified
Copper(II) sulfate reacts with potassium fluoride and potassium chloride to form copper(II) fluoride and copper(II) chloride respectively, both of which are green but have different shades. The change in color symbolizes these reactions.

Step by step solution

01

Understanding the Components

Firstly, examine each compound. Copper(II) sulfate (CuSO4) is a blue compound. Potassium fluoride (KF) and potassium chloride (KCl) are both colorless salts.
02

Reaction with Potassium Fluoride

When CuSO4 reacts with KF, a double displacement reaction occurs which results in the formation of copper(II) fluoride (CuF2), a green compound, and potassium sulfate (K2SO4). Here is the balanced chemical equation: \(CuSO_4 (aq) + 2KF (aq) \rightarrow CuF_2 (s) + K_2SO_4 (aq)\) . The green color indicates the presence of copper(II) fluoride precipitate.
03

Reaction with Potassium Chloride

On the other hand, when CuSO4 reacts with KCl, another double displacement reaction occurs which results in the formation of copper(II) chloride (CuCl2), a bright green compound, and potassium sulfate (K2SO4). Here is the balanced chemical equation: \(CuSO_4 (aq) + 2KCl (aq) \rightarrow CuCl_2 (aq) + K_2SO_4 (aq)\). The bright green color indicates the formation of a solution of copper(II) chloride.

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