Question

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) $\mathrm{HBr}(aq)+\mathrm{Ca}(\mathrm{OH}{)}_2(aq)⟶$ (b) $\mathrm{Cu}(\mathrm{OH}{)}_2(s)+{\mathrm{HClO}}_4(aq)⟶$ (c) $\mathrm{Al}(\mathrm{OH}{)}_3(s)+{\mathrm{HNO}}_3(aq)⟶$

Short answer

The net ionic equations for the given reactions are: (a) $2H^{+}(aq)+2O{H}^{-}(aq)⟶2H_{2}O(l)$ (b) $Cu(OH{)}_2(s)+2H^{+}(aq)⟶C{u}^{2+}(aq)+2H_{2}O(l)$ (c) $Al(OH{)}_3(s)+3H^{+}(aq)⟶A{l}^{3+}(aq)+3H_{2}O(l)$

Step-by-step solution

Balance the molecular equation

The balanced molecular equation is: $2\mathrm{HBr}(aq)+\mathrm{Ca}(\mathrm{OH}{)}_2(aq)⟶{\mathrm{CaBr}}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(l)$

Separate the equation into individual ions

$2{\mathrm{H}}^{+}(aq)+2{\mathrm{Br}}^{-}(aq)+{\mathrm{Ca}}^{2+}(aq)+2{\mathrm{OH}}^{-}(aq)⟶{\mathrm{Ca}}^{2+}(aq)+2{\mathrm{Br}}^{-}(aq)+2{\mathrm{H}}_2\mathrm{O}(l)$

Cancel out common ions

The common ions to be canceled out are $2{\mathrm{Br}}^{-}(aq)$ and ${\mathrm{Ca}}^{2+}(aq)$. After canceling the common ions, we obtain: $2{\mathrm{H}}^{+}(aq)+2{\mathrm{OH}}^{-}(aq)⟶2{\mathrm{H}}_2\mathrm{O}(l)$

Write the net ionic equation

The net ionic equation is: $2{\mathrm{H}}^{+}(aq)+2{\mathrm{OH}}^{-}(aq)⟶2{\mathrm{H}}_2\mathrm{O}(l)$ (b) $\mathrm{Cu}(\mathrm{OH}{)}_2(s)+{\mathrm{HClO}}_4(aq)⟶$

Balance the molecular equation

The balanced molecular equation is: $\mathrm{Cu}(\mathrm{OH}{)}_2(s)+2{\mathrm{HClO}}_4(aq)⟶\mathrm{Cu}({\mathrm{ClO}}_4{)}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(l)$

Separate the equation into individual ions

We don't separate solid and liquid substances, so we only separate ${\mathrm{HClO}}_4(aq)$: $\mathrm{Cu}(\mathrm{OH}{)}_2(s)+2{\mathrm{H}}^{+}(aq)+2{\mathrm{ClO}}_4^{-}(aq)⟶{\mathrm{Cu}}^{2+}(aq)+2{\mathrm{ClO}}_4^{-}(aq)+2{\mathrm{H}}_2\mathrm{O}(l)$

Cancel out common ions

The common ions to be canceled out are $2{\mathrm{ClO}}_4^{-}(aq)$. After canceling the common ions, we obtain: $\mathrm{Cu}(\mathrm{OH}{)}_2(s)+2{\mathrm{H}}^{+}(aq)⟶{\mathrm{Cu}}^{2+}(aq)+2{\mathrm{H}}_2\mathrm{O}(l)$

Write the net ionic equation

The net ionic equation is: $\mathrm{Cu}(\mathrm{OH}{)}_2(s)+2{\mathrm{H}}^{+}(aq)⟶{\mathrm{Cu}}^{2+}(aq)+2{\mathrm{H}}_2\mathrm{O}(l)$ (c) $\mathrm{Al}(\mathrm{OH}{)}_3(s)+{\mathrm{HNO}}_3(aq)⟶$

Balance the molecular equation

The balanced molecular equation is: $\mathrm{Al}(\mathrm{OH}{)}_3(s)+3{\mathrm{HNO}}_3(aq)⟶\mathrm{Al}({\mathrm{NO}}_3{)}_3(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$

Separate the equation into individual ions

We don't separate solid and liquid substances, so we only separate ${\mathrm{HNO}}_3(aq)$: $\mathrm{Al}(\mathrm{OH}{)}_3(s)+3{\mathrm{H}}^{+}(aq)+3{\mathrm{NO}}_3^{-}(aq)⟶{\mathrm{Al}}^{3+}(aq)+3{\mathrm{NO}}_3^{-}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$

Cancel out common ions

The common ions to be canceled out are $3{\mathrm{NO}}_3^{-}(aq)$. After canceling the common ions, we obtain: $\mathrm{Al}(\mathrm{OH}{)}_3(s)+3{\mathrm{H}}^{+}(aq)⟶{\mathrm{Al}}^{3+}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$

Write the net ionic equation

The net ionic equation is: $\mathrm{Al}(\mathrm{OH}{)}_3(s)+3{\mathrm{H}}^{+}(aq)⟶{\mathrm{Al}}^{3+}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$

Key concepts

Balancing Chemical Equations

Balancing chemical equations is akin to ensuring both sides of a scale have equal weight. It's necessary because the Law of Conservation of Mass states that matter cannot be created or destroyed in an ordinary chemical reaction. Thus, each type of atom on the reactant side must be present in the same number on the product side.

For instance, take the reaction between hydrobromic acid (HBr) and calcium hydroxide (Ca(OH)₂). First, we must ensure that the number of each type of atom on the reactants' side equals that on the products' side. If we count the atoms before and after the reaction, we see that balancing requires two molecules of HBr to react with one molecule of Ca(OH)₂ to form one molecule of calcium bromide (CaBr₂) and two molecules of water (H₂O). This balancing act ensures that each hydrogen, bromine, calcium, and oxygen atom is accounted for in the products.

Molecular Equations

A molecular equation provides a macroscopic view of a chemical reaction where compounds are expressed as molecules, not showing individual ions. For our example reactions, the balanced molecular equations for the interaction of acids and bases provide a bird’s-eye view.

Take the reaction of copper(II) hydroxide and perchloric acid. We write down the reactants and the expected products without indicating the ionic nature of the compounds. This step allows us to anticipate the products (in this case, copper(II) perchlorate and water) and balance the equation as outlined in our previous section. However, molecular equations don't tell the whole story as they do not give insight into which compounds dissolve to form ions in solution.

Solubility Rules

To understand a reaction's ionic behavior, knowing the solubility rules is essential. These rules predict whether a compound will dissolve in water to form ions—this is crucial for writing net ionic equations.

Based on solubility rules, compounds like CaBr₂ dissolve and dissociate into ions in water, whereas others like Cu(OH)₂ do not. The rules tell us that all nitrates are soluble, so aluminum nitrate will dissolve, whereas hydroxides are generally not, except for those of alkali metals and some alkaline earth metals, which is why Cu(OH)₂ remains a solid in the equation and does not separate into ions. Understanding solubility helps us identify spectator ions—the ones present in solution but do not participate in the chemical change.

Strong Electrolytes

Substances that completely dissociate into ions in solution are known as strong electrolytes. They allow electricity to flow through the solution effectively, hence the name 'electrolyte'. These typically include strong acids, strong bases, and most salts.

For example, HBr is a strong acid and therefore a strong electrolyte, meaning it fully dissociates into H⁺ and Br⁻ ions when dissolved in water. Calcium hydroxide, being a strong base, also dissociates completely. It's important to recognize which compounds are strong electrolytes because, in a net ionic equation, we only focus on the ions from strong electrolytes that undergo a chemical change and exclude the spectator ions, thereby streamlining our equation to show the essence of the chemical reaction.