Pb(CH₃COO)₂(aq) + CuSO₄(aq) → PbSO₄(s) + Cu(CH₃COO)₂(aq)
Here's a breakdown of the reaction:
* Reactants:
* Lead(II) acetate (Pb(CH₃COO)₂): This is a white crystalline solid that is soluble in water.
* Copper sulfate (CuSO₄): This is a blue crystalline solid that is also soluble in water.
* Products:
* Lead(II) sulfate (PbSO₄): This is a white solid that is insoluble in water, so it will precipitate out of the solution.
* Copper(II) acetate (Cu(CH₃COO)₂) : This is a blue-green solid that is soluble in water.
Explanation:
This reaction is a double displacement reaction, where the cations and anions of the reactants switch places. The lead(II) ions (Pb²⁺) from lead(II) acetate combine with the sulfate ions (SO₄²⁻) from copper sulfate to form the insoluble lead(II) sulfate. The copper(II) ions (Cu²⁺) from copper sulfate combine with the acetate ions (CH₃COO⁻) from lead(II) acetate to form soluble copper(II) acetate.
The balanced equation ensures that the number of atoms of each element on both sides of the equation is equal. In this case, there are:
* 1 lead atom (Pb)
* 2 copper atoms (Cu)
* 4 carbon atoms (C)
* 6 hydrogen atoms (H)
* 4 oxygen atoms (O)
* 2 sulfur atoms (S)
on both the reactant and product sides of the equation.
