This equation describes the reaction between copper (II) sulfate (CuSO4) and potassium hydroxide (KOH) to form a precipitate of copper (II) hydroxide (Cu(OH)2) and potassium sulfate (K2SO4).
- Reaction conditions
This reaction usually occurs when the two reactants are mixed together in solution at room temperature.
- Reaction process
When the two reactants are mixed together, the copper (II) ions (Cu2+) from CuSO4 will combine with the hydroxide ions (OH-) from KOH to form a precipitate of copper (II) hydroxide (Cu(OH)2). Meanwhile, the potassium ions (K+) from KOH will combine with sulfate ions (SO4^2-) from CuSO4 to form potassium sulfate (K2SO4).
- Phenomena occurring
When the reaction occurs, an important event observable is the formation of a blue precipitate of copper (II) hydroxide (Cu(OH)2). The potassium sulfate (K2SO4) that is formed is a white solid, but it is often retained in the solution.