From the balanced molecular equations, write the complete ionic and net ionic equations for the following: (a) \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow 2 \mathrm{KOH}(a q)+\mathrm{BaC}_{2} \mathrm{O}_{4}(s)\) (b) \(\operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\)

In chemistry, understanding reactions in aqueous solutions is often facilitated by writing out complete ionic equations. These are representations of molecular reactions where soluble ionic compounds are expressed as individual ions.

For instance, if we consider the balanced molecular equation for the reaction between potassium oxalate (\( K_2C_2O_4 \text{(aq)} \) and barium hydroxide (\( Ba(OH)_2 \text{(aq)} \)), the complete ionic equation, following Step 1, will involve splitting the soluble compounds into their constituent ions:

• \( 2 K^+ \text{(aq)} + C_2O_4^{2-} \text{(aq)} \)
• \( Ba^{2+} \text{(aq)} + 2 OH^- \text{(aq)} \)

On mixing the two solutions, the ions remain in their dissociated form. This is a powerful way to visualize the underlying changes during the reaction process, making the complex processes at the molecular level more graspable.

When balancing chemical reactions, not all ions will undergo change. These are known as 'spectator ions.' Spectator ions do not participate in the actual reaction; instead, they remain in the solution unchanged.

In Step 3 of the solution process, spectator ions are identified and removed to simplify the equation. For example, in the reaction between potassium oxalate and barium hydroxide, potassium (\( K^+ \text{(aq)} \)) and hydroxide (\( OH^- \text{(aq)} \)) ions do not form a precipitate or gas; they remain in the solution and are thus spectator ions. By eliminating them, we focus on the ions that contribute to the formation of new products, which draws attention to the heart of the chemical change.

Precipitation reactions are a cornerstone of solubility discussions in chemistry. They occur when two soluble salts react in solution to form one or more insoluble products, known as a precipitate. This solid precipitate can be seen falling out of the solution.

Referring to our example reactions, when barium (\( Ba^{2+} \text{(aq)} \)) and oxalate (\( C_2O_4^{2-} \text{(aq)} \)) ions come together, they form barium oxalate (\( BaC_2O_4 \text{(s)} \) ), a solid precipitate that is visible as a fine suspension or sediment in the reaction mixture. Recognizing and predicting such reactions is critical for understanding chemical reactivity and is particularly relevant in fields such as environmental science, medicine, and industrial chemistry.

The law of conservation of mass dictates that matter is neither created nor destroyed in a chemical reaction. To reflect this principle, chemical equations must be balanced, with an equal number of atoms of each element on both sides of the equation.

In the textbook solution, Step 1 involves writing the balanced molecular equations where the coefficients are used to indicate the stoichiometric amounts of reactants and products. Balancing chemical equations is an essential skill in chemistry that ensures the stoichiometry of the reaction is correct, which is vital for quantitative analytical methods and industrial applications. By understanding this principle, students can accurately predict the amounts of reactants needed and products formed in a chemical reaction.