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Chapter 15: Problem 30

Balance each of the following equations; then change them into balanced net ionic equations: (a) \(\mathrm{Mg}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{Cu}(s)+\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{K}_{2} \mathrm{SO}_{3}(a q)+\mathrm{HClO}_{4}(a q) \longrightarrow\) \(\mathrm{KClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{2}(g)\)

Short Answer

Expert verified
(a) Balanced: \[ \mathrm{Mg}(s) + \mathrm{Cu}(\mathrm{NO}_3)_2(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}(\mathrm{NO}_3)_2(aq) \] Net Ionic: \[ \mathrm{Mg}(s) + \mathrm{Cu}^{2+}(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}^{2+}(aq) \] (b) Balanced: \[ \mathrm{H}_2 \mathrm{SO}_4(aq) + 2 \mathrm{NaOH}(aq) \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4(aq) + 2 \mathrm{H_2O}(l) \] Net Ionic: \[ \mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \longrightarrow \mathrm{H_2O}(l) \] (c) Balanced: \[ \mathrm{K}_2 \mathrm{SO}_3(aq) + 2\mathrm{HClO}_4(aq) \longrightarrow 2 \mathrm{KClO}_4(aq) + \mathrm{H_2O}(l) + \mathrm{SO}_2(g) \] Net Ionic: \[ \mathrm{SO}_3^{2-}(aq) + 2\mathrm{H}^{+}(aq) \longrightarrow \mathrm{H_2O}(l) + \mathrm{SO}_2(g) \]

Step by step solution

01

- Write down the unbalanced equation for part (a)

The equation is: \[ \mathrm{Mg}(s) + \mathrm{Cu}(\mathrm{NO}_3)_2(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}(\mathrm{NO}_3)_2(aq) \]
02

- Balance the equation for part (a)

Check the number of atoms of each element on both sides and balance. Here, it is already balanced: \[ \mathrm{Mg}(s) + \mathrm{Cu}(\mathrm{NO}_3)_2(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}(\mathrm{NO}_3)_2(aq) \]
03

- Write the complete ionic equation for part (a)

Dissociate all strong electrolytes: \[ \mathrm{Mg}(s) + \mathrm{Cu}^{2+}(aq) + 2 \mathrm{NO}_3^{-}(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}^{2+}(aq) + 2 \mathrm{NO}_3^{-}(aq) \]
04

- Write the net ionic equation for part (a)

Remove the spectator ions (\(\mathrm{NO}_3^{-}\)): \[ \mathrm{Mg}(s) + \mathrm{Cu}^{2+}(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}^{2+}(aq) \]
05

- Write down the unbalanced equation for part (b)

The equation is: \[ \mathrm{H}_2 \mathrm{SO}_4(aq) + \mathrm{NaOH}(aq) \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4(aq) + \mathrm{H_2O}(l) \]
06

- Balance the equation for part (b)

Balance the elements. The balanced equation is: \[ \mathrm{H}_2 \mathrm{SO}_4(aq) + 2 \mathrm{NaOH}(aq) \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4(aq) + 2 \mathrm{H_2O}(l) \]
07

- Write the complete ionic equation for part (b)

Dissociate all strong electrolytes: \[ 2\mathrm{H}^{+}(aq) + \mathrm{SO}_4^{2-}(aq) +2 \mathrm{Na}^{+}(aq) + 2 \mathrm{OH}^{-}(aq) \longrightarrow 2 \mathrm{Na}^+(aq) + \mathrm{SO}_4^{2-}(aq) + 2 \mathrm{H_2O}(l) \]
08

- Write the net ionic equation for part (b)

Remove the spectator ions (\(\mathrm{Na}^+\) and \(\mathrm{SO}_4^{2-}\)): \[ 2 \mathrm{H}^{+}(aq) + 2 \mathrm{OH}^{-}(aq) \longrightarrow 2 \mathrm{H_2O}(l) \] This can be simplified to: \[ \mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \longrightarrow \mathrm{H_2O}(l) \]
09

- Write down the unbalanced equation for part (c)

The equation is: \[ \mathrm{K}_2 \mathrm{SO}_3(aq) + \mathrm{HClO}_4(aq) \longrightarrow \mathrm{KClO}_4(aq) + \mathrm{H_2O}(l) + \mathrm{SO}_2(g) \]
10

- Balance the equation for part (c)

Balance the elements. The balanced equation is: \[ \mathrm{K}_2 \mathrm{SO}_3(aq) + 2 \mathrm{HClO}_4(aq) \longrightarrow 2 \mathrm{KClO}_4(aq) + \mathrm{H_2O}(l) + \mathrm{SO}_2(g) \]
11

- Write the complete ionic equation for part (c)

Dissociate all strong electrolytes: \[ 2 \mathrm{K}^{+}(aq) + \mathrm{SO}_3^{2-}(aq) + 2 \mathrm{H}^{+}(aq) + 2 \mathrm{ClO}_4^{-}(aq) \longrightarrow 2 \mathrm{K}^{+}(aq) + 2 \mathrm{ClO}_4^{-}(aq) + \mathrm{H_2O}(l) + \mathrm{SO}_2(g) \]
12

- Write the net ionic equation for part (c)

Remove the spectator ions (\(\mathrm{K}^+\) and \(\mathrm{ClO}_4^{-}\)): \[ \mathrm{SO}_3^{2-}(aq) + 2 \mathrm{H}^{+}(aq) \longrightarrow \mathrm{H_2O}(l) + \mathrm{SO}_2(g) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Net Ionic Equations
Net ionic equations simplify chemical reactions by showing only the species that actually participate in the reaction. This means, you need to write only the ions and molecules directly involved in the chemical change.
For example, in reaction (a):
\[ \mathrm{Mg}(s) + \mathrm{Cu}^{2+}(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}^{2+}(aq) \]
We leave out the nitrate ions (\(\mathrm{NO}_3^{-}\) ) because they do not change during the reaction.
To write a net ionic equation, first write the balanced molecular equation, then write the complete ionic equation where all strong electrolytes are dissociated into ions. Finally, remove the spectator ions.
Spectator Ions
Spectator ions are ions in a solution that do not participate in a chemical reaction. They appear unchanged on both sides of a chemical equation.
For example, in reaction (a):
\[ \mathrm{Mg}(s) + \mathrm{Cu}(\text{NO}_3)_2(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}(\text{NO}_3)_2(aq) \]
The nitrate ions (\(\mathrm{NO}_3^{-}\) ) are spectator ions. In the ionic form, nitrate ions appear as:
\[ \mathrm{NO}_3^{-}(aq)\]
Since nitrate ions do not change during the reaction, they can be removed to form the net ionic equation.
Acid-Base Reactions
Acid-base reactions involve transfer of protons (H+) between reactants.
For example, in reaction (b):
\[ \mathrm{H}_2 \mathrm{SO}_4(aq) + 2 \mathrm{NaOH}(aq) \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4(aq) + 2 \mathrm{H_2O}(l) \]
Sulfuric acid (H2SO4) donates protons (H+) to the hydroxide ions (OH-) from sodium hydroxide (NaOH), forming water.
The net ionic equation for this can be written as:
\[ \mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \longrightarrow \mathrm{H_2O}(l) \]
Notice that sodium (\text{Na}^{+}) and sulfate (\text{SO}_4^{2-}) are spectators in this case.
Oxidation-Reduction Reactions
Oxidation-reduction or redox reactions involve the transfer of electrons between species. These reactions contribute to changes in oxidation states.
For example, in reaction (a):
\[ \mathrm{Mg}(s) + \mathrm{Cu}(\text{NO}_3)_2(aq) \longrightarrow \mathrm{Cu}(s) + \mathrm{Mg}(\text{NO}_3)_2(aq) \]
Magnesium goes through oxidation by losing electrons and changing from \mathrm{Mg}(s) to \mathrm{Mg}^{2+}(aq). Copper undergoes reduction by gaining electrons and converting from \mathrm{Cu}^{2+}(aq) to \mathrm{Cu}(s).
The electrons transferred in this process are not seen in the net ionic equation but are integral to the reaction occurring.

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