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

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(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

01

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
02

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
03

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
04

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
05

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
06

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
07

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
08

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
09

Net Ionic Equationc.

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

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