Question

Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all salts formed are soluble. Acid salts are possible. (a) $$ \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow $$ (b) $$ 2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow $$ (c) $$ \mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow $$ (d) $$ \operatorname{Sr}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \longrightarrow $$

Short answer

Each reaction forms water; net ionic equations focus on \( 2 \text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O}\).

Step-by-step solution

Complete the Molecular Equation for Reaction (a)

To complete the equation, we need to balance reactants with products. The reaction is between Calcium hydroxide \( \text{Ca(OH)}_2 \) and sulfuric acid \( \text{H}_2\text{SO}_4 \).Calcium hydroxide neutralizes sulfuric acid to form calcium sulfate and water:\[\text{Ca(OH)}_2(aq) + 2 \text{H}_2\text{SO}_4(aq) \rightarrow \text{CaSO}_4(aq) + 2 \text{H}_2\text{O}(l) + \text{H}_2\text{SO}_4(aq).\]

Write the Net Ionic Equation for Reaction (a)

Identify the ions in the solution: \( \text{Ca}^{2+} \), \( \text{OH}^- \), \( \text{H}^+ \), and \( \text{SO}_4^{2-} \). The net ionic reaction focuses on the formation of water from hydroxide and hydrogen ions:\[2 \text{OH}^- (aq) + 2 \text{H}^+(aq) \rightarrow 2 \text{H}_2\text{O}(l).\]

Complete the Molecular Equation for Reaction (b)

The reaction involves phosphoric acid \( \text{H}_3\text{PO}_4 \) and calcium hydroxide \( \text{Ca(OH)}_2 \).The balanced molecular equation is:\[2 \text{H}_3\text{PO}_4(aq) + \text{Ca(OH)}_2(aq) \rightarrow \text{Ca}\text{HPO}_4(aq) + 4 \text{H}_2\text{O}(l).\]

Write the Net Ionic Equation for Reaction (b)

Focus on the ions participating in water formation. However, note we'll also consider possible formation of salts:\[2 \text{H}^+(aq) + 2 \text{OH}^-(aq) \rightarrow 2 \text{H}_2\text{O}(l).\]

Complete the Molecular Equation for Reaction (c)

This equation involves sodium hydroxide \( \text{NaOH} \) and sulfuric acid \( \text{H}_2\text{SO}_4 \). It neutralizes to form sodium sulfate and water:\[2 \text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2 \text{H}_2\text{O}(l).\]

Write the Net Ionic Equation for Reaction (c)

Focus on the net ionic equation involving water formation:\[2 \text{OH}^- (aq) + 2 \text{H}^+ (aq) \rightarrow 2 \text{H}_2\text{O}(l).\]

Complete the Molecular Equation for Reaction (d)

The reaction is between strontium hydroxide \( \text{Sr(OH)}_2 \) and carbonic acid \( \text{H}_2\text{CO}_3 \). Strontium carbonate and water are formed:\[\text{Sr(OH)}_2(aq) + 2 \text{H}_2\text{CO}_3(aq) \rightarrow \text{SrCO}_3(aq) + 3 \text{H}_2\text{O}(l) + \text{CO}_2(g).\]

Write the Net Ionic Equation for Reaction (d)

The net ionic equation focuses on forming water with elimination of one hydrogen for each carbonate after decomposition:\[2 \text{OH}^-(aq) + 2 \text{H}^+(aq) \rightarrow 2 \text{H}_2\text{O}(l).\]

Key concepts

Molecular Equations

Molecular equations are balanced chemical equations that show the complete chemical formulas of reactants and products. In these equations, you write all compounds involved in the reaction without breaking them into ions. This provides a detailed view of what compounds are interacting and what new compounds are produced.
In an example from the given solutions, calcium hydroxide reacts with sulfuric acid forming calcium sulfate and water. The molecular equation is written as: \[\text{Ca(OH)}_2(aq) + 2 \text{H}_2\text{SO}_4(aq) \rightarrow \text{CaSO}_4(aq) + 2 \text{H}_2\text{O}(l) + \text{H}_2\text{SO}_4(aq).\]
This clearly represents all compounds as they are in a solution. Understanding molecular equations helps to visualize the overall reaction process.

Acid-Base Reactions

Acid-base reactions involve an acid and a base reacting to form water and a salt. In these reactions, the acid donates protons (\text{H}^+) to the base, which typically contains hydroxide ions (\text{OH}^-).
A simple example of such a reaction from the exercise is the combination of sodium hydroxide and sulfuric acid:\[2 \text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2 \text{H}_2\text{O}(l).\]
During the reaction, the acid donates \text{H}^+ ions that combine with the \text{OH}^- ions from the base to form water, a neutral substance. These neutralization reactions are foundational in chemistry because they demonstrate how acids and bases can neutralize each other.

Balancing Chemical Equations

Balancing chemical equations involves making sure that all atoms present in the reactants are also present in the products in the same quantity. This is crucial because atoms are neither created nor destroyed in a chemical reaction.
To balance a chemical equation, you adjust the coefficients of the reactants and products. For example, balancing the reaction between strontium hydroxide and carbonic acid: \[\text{Sr(OH)}_2(aq) + 2 \text{H}_2\text{CO}_3(aq) \rightarrow \text{SrCO}_3(aq) + 3 \text{H}_2\text{O}(l) + \text{CO}_2(g),\]
we ensure the same number of each type of atom exists on both sides of the equation. Balancing equations is a fundamental skill in chemistry that ensures the law of conservation of mass is upheld.

Solubility Rules

Solubility rules help determine whether a compound is soluble or insoluble in water. These rules are essential for writing net ionic equations, which only include the ions that participate in the reaction.
For instance, while completing the net ionic equation for calcium hydroxide and sulfuric acid, solubility rules tell us that calcium sulfate remains in solution: \[2 \text{OH}^- (aq) + 2 \text{H}^+(aq) \rightarrow 2 \text{H}_2\text{O}(l).\]
Knowing this helps you understand which compounds dissociate into ions and which form precipitates. Applying solubility rules allows for precision in predicting products' states and determining the correct ions to include in net ionic equations.