How would you prepare the following substances by a precipitation reaction? (a) PbSO4

Verified step by step guidance

1. Step 1: Identify two soluble compounds that contain the ions you need. In this case, we need Pb2+ and SO4^2-. A possible choice could be Pb(NO3)2 and Na2SO4, both of which are soluble in water.
2. Step 2: Write down the balanced chemical equation for the reaction. Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq).
3. Step 3: Mix the two solutions together. The Pb2+ ions from the Pb(NO3)2 will react with the SO4^2- ions from the Na2SO4 to form PbSO4, which is insoluble and will precipitate out of the solution.
4. Step 4: Allow the reaction to proceed until no more precipitate forms. This indicates that all of the Pb2+ and SO4^2- ions have reacted.
5. Step 5: Separate the precipitate from the solution. This can be done by filtration. The solid PbSO4 will remain on the filter, while the soluble NaNO3 will pass through.

Key Concepts

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in solution to form an insoluble product, known as a precipitate. This process typically involves the exchange of ions between the reactants, leading to the formation of a solid that separates from the liquid phase. Understanding the solubility rules is crucial for predicting whether a precipitate will form.

Solubility Rules

Solubility rules are guidelines that help predict the solubility of various ionic compounds in water. For example, most sulfates are soluble, but lead(II) sulfate (PbSO4) is an exception and is insoluble in water. Familiarity with these rules allows chemists to determine which combinations of reactants will yield a precipitate during a reaction.

Ionic Equations

Ionic equations represent the species involved in a chemical reaction, highlighting the ions that participate in the formation of a precipitate. In the case of preparing PbSO4, the net ionic equation would show the lead ions (Pb²⁺) and sulfate ions (SO4²⁻) combining to form the solid PbSO4. Writing and balancing these equations is essential for understanding the stoichiometry of the reaction.