Question

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid–base reactions. a. \(\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow\) b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\) c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

Short answer

(a) Balanced formula: \(2\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Mg}(\mathrm{ClO}_{4})_{2}(a q)\) Complete ionic: \(\mathrm{2H^{+}(a q) + 2ClO_{4}^{-}(a q) + Mg(OH)_{2}(s) \rightarrow 4H_{2}O(l) + Mg^{2+}(a q) + 2ClO_{4}^{-}(a q)}\) Net ionic: \(\mathrm{2H^{+}(a q) + Mg(OH)_{2}(s) \rightarrow 2H_{2}O(l) + Mg^{2+}(a q)}\) (b) Balanced formula: \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{NaCN}(a q)\) Complete ionic: \(\mathrm{H^{+}(a q) + CN^{-}(a q) + Na^{+}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l) + Na^{+}(a q) + CN^{-}(a q)}\) Net ionic: \(\mathrm{H^{+}(a q) + CN^{-}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l)}\) (c) Balanced formula: \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{NaCl}(a q)\) Complete ionic: \(\mathrm{H^{+}(a q) + Cl^{-}(a q) + Na^{+}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l) + Na^{+}(a q) + Cl^{-}(a q)}\) Net ionic: \(\mathrm{H^{+}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l)}\)

Step-by-step solution

Balanced Formula Equationa.

When an acid reacts with a base, they neutralize each other, resulting in water and a salt. The products of this reaction will be water (H2O) and magnesium perchlorate (Mg(ClO4)2). The balanced formula equation is: \(\mathrm{2HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Mg}(\mathrm{ClO}_{4})_{2}(a q)\) Step 2: Complete Ionic Equation

Complete Ionic Equationa.

Magnesium hydroxide is not soluble in water, so we do not break it into ions. The other soluble ionic compounds are broken into their constituent ions: \(\mathrm{2H^{+}(a q) + 2ClO_{4}^{-}(a q) + Mg(OH)_{2}(s) \rightarrow 4H_{2}O(l) + Mg^{2+}(a q) + 2ClO_{4}^{-}(a q)}\) Step 3: Net Ionic Equation

Net Ionic Equationa.

Cancel out the spectator ions (those that appear on both sides of the equation): \(\mathrm{2H^{+}(a q) + Mg(OH)_{2}(s) \rightarrow 2H_{2}O(l) + Mg^{2+}(a q)}\) #b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\)# Step 1: Balanced Formula Equation

Balanced Formula Equationb.

The acid-base reaction between hydrogen cyanide (HCN) and sodium hydroxide (NaOH) results in water (H2O) and sodium cyanide (NaCN). The balanced formula equation is: \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{NaCN}(a q)\) Step 2: Complete Ionic Equation

Complete Ionic Equationb.

Break the soluble ionic compounds into their constituent ions: \(\mathrm{H^{+}(a q) + CN^{-}(a q) + Na^{+}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l) + Na^{+}(a q) + CN^{-}(a q)}\) Step 3: Net Ionic Equation

Net Ionic Equationb.

Cancel out the spectator ions: \(\mathrm{H^{+}(a q) + CN^{-}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l)}\) #c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)# Step 1: Balanced Formula Equation

Balanced Formula Equationc.

The acid-base reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) results in water (H2O) and sodium chloride (NaCl). The balanced formula equation is: \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{NaCl}(a q)\) Step 2: Complete Ionic Equation

Complete Ionic Equationc.

Break the soluble ionic compounds into their constituent ions: \(\mathrm{H^{+}(a q) + Cl^{-}(a q) + Na^{+}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l) + Na^{+}(a q) + Cl^{-}(a q)}\) Step 3: Net Ionic Equation

Net Ionic Equationc.

Cancel out the spectator ions: \(\mathrm{H^{+}(a q) + OH^{-}(a q) \rightarrow H_{2}O(l)}\)

Key concepts

Ionic Equations

Ionic equations help us understand chemical reactions at a deeper level by showing how compounds dissociate into ions in solution. These equations are especially useful for reactions taking place in aqueous environments, where many compounds will split into their respective ions. For acids and bases dissolved in water, they usually dissociate into hydrogen ions (\(\mathrm{H^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)) respectively.

The complete ionic equation differs from the balanced formula equation as it breaks down aqueous compounds into ions. For instance, in the reaction between hydrochloric acid, \(\mathrm{HCl(aq)}\), and sodium hydroxide, \(\mathrm{NaOH(aq)}\), the complete ionic equation presents the compounds in their ionic forms:

• \(\mathrm{H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq)}\)
• reacting to form water and the ions \(\mathrm{Na^+(aq) + Cl^-(aq)}\)

This representation allows us to see which ions remain unchanged throughout the reaction and are known as spectator ions. These are later removed when writing the net ionic equation, leaving only those that participate directly in the transformation.

Net Ionic Equations

Net ionic equations further simplify the complete ionic equation by removing the spectator ions. These equations highlight the actual chemical change occurring in a reaction. For example, in the reaction between \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), the net ionic equation leaves out the sodium and chloride ions because they do not alter during the reaction.

Net ionic equations are particularly helpful when studying acid-base reactions. They show the neutralization process directly:

• \(\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}\)

This clearly shows the formation of water from the hydrogen and hydroxide ions. By focusing only on the ions that undergo change, net ionic equations offer a concise view of what physically takes place during the reaction. This is invaluable in understanding and predicting the outcomes of chemical reactions.

Chemical Balancing

Chemical balancing is a crucial step in writing chemical equations. It ensures that the same number of each type of atom is present on both sides of the equation, respecting the conservation of mass. Balancing an equation involves adjusting the coefficients in front of each compound or element until the number of atoms for each element is the same across reactants and products.

For example, consider the equation without coefficients for the reaction between \(\mathrm{HClO_4}\) and \(\mathrm{Mg(OH)_2}\):

• We start with:\(\mathrm{HClO_4(aq) + Mg(OH)_2(s)}\)
• to produce \(\mathrm{H_2O(l) + Mg(ClO_4)_2(aq)}\)

Balancing involves ensuring that we have equal numbers of each atom on both sides. Here, two molecules of \(\mathrm{HClO_4}\) and two \(\mathrm{H_2O}\) molecules are necessary to keep the balance:

• \(\mathrm{2HClO_4(aq) + Mg(OH)_2(s) \rightarrow 2H_2O(l) + Mg(ClO_4)_2(aq)}\)

This balancing process is vital in accurately depicting the stoichiometry of the reaction, essential for quantitative chemical analysis.