Problem 67 - Chemistry An Atoms First Approach

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. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

Step-by-Step Solution

Answer

a. Balanced formula: \(2\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg(ClO_4)_2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\) Complete ionic: \(\mathrm{2H^{+}}(a q)+\mathrm{2ClO_4^{-}}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg^{2+}}(a q)+\mathrm{2ClO_4^{-}}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\) Net ionic: \(\mathrm{2H^{+}}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg^{2+}}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\) b. Balanced formula: \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCN}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\) Complete ionic: \(\mathrm{H^{+}}(a q)+\mathrm{CN^{-}}(a q)+\mathrm{Na^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{Na^{+}}(a q)+\mathrm{CN^{-}}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\) Net ionic: \(\mathrm{H^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{H}_{2}\mathrm{O}(l)\) c. Balanced formula: \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\) Complete ionic: \(\mathrm{H^{+}}(a q)+\mathrm{Cl^{-}}(a q)+\mathrm{Na^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{Na^{+}}(a q)+\mathrm{Cl^{-}}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\) Net ionic: \(\mathrm{H^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{H}_{2}\mathrm{O}(l)\)

Step 1: Predict the products and balance the equation

When an acid reacts with a base, the products formed are salt and water. In this case, \(\mathrm{HClO}_{4}(a q)\) reacts with \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\) to form a salt (\(\mathrm{Mg(ClO_4)_2}\)) and water. The balanced equation is: \(\mathrm{2HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg(ClO_4)_2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\)

Step 2: Write the complete ionic equation

We separate the aqueous species into their respective ions, while solid, liquid, and gas species remain unchanged. \(\mathrm{2H^{+}}(a q)+\mathrm{2ClO_4^{-}}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg^{2+}}(a q)+\mathrm{2ClO_4^{-}}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\)

Step 3: Write the net ionic equation

We remove the spectator ions that are present on both sides of the complete ionic equation to obtain the net ionic equation. \(\mathrm{2H^{+}}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg^{2+}}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\) b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

Step 1: Predict the products and balance the equation

In this case, \(\mathrm{HCN}(a q)\) reacts with \(\mathrm{NaOH}(a q)\) to form a salt (\(\mathrm{NaCN}\)) and water. The balanced equation is: \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCN}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\)

Step 2: Write the complete ionic equation

\(\mathrm{H^{+}}(a q)+\mathrm{CN^{-}}(a q)+\mathrm{Na^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{Na^{+}}(a q)+\mathrm{CN^{-}}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\)

Step 3: Write the net ionic equation

\(\mathrm{H^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{H}_{2}\mathrm{O}(l)\) c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

Step 1: Predict the products and balance the equation

In this case, \(\mathrm{HCl}(a q)\) reacts with \(\mathrm{NaOH}(a q)\) to form a salt (\(\mathrm{NaCl}\)) and water. The balanced equation is: \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\)

Step 2: Write the complete ionic equation

\(\mathrm{H^{+}}(a q)+\mathrm{Cl^{-}}(a q)+\mathrm{Na^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{Na^{+}}(a q)+\mathrm{Cl^{-}}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\)

Step 3: Write the net ionic equation

\(\mathrm{H^{+}}(a q)+\mathrm{OH^{-}}(a q) \rightarrow\mathrm{H}_{2}\mathrm{O}(l)\)