Question

Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate): (a) \(\mathrm{HBr}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) (c) \(\mathrm{HClO}_{4}(a a)+\mathrm{Me}(\mathrm{OH})_{2}(s)\)

Short answer

(a) Net ionic: \( \mathrm{H}^{+} + \mathrm{NH}_{3} \rightarrow \mathrm{NH}_{4}^{+} \). (b) Net ionic: \( 3\mathrm{Ba}^{2+} + 2\mathrm{PO}_4^{3-} \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2 \). (c) Net ionic: \( 2\mathrm{H}^{+} + \mathrm{Mg}(\mathrm{OH})_2 \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{H}_2\mathrm{O} \).

Step-by-step solution

Balance Equation (a)

We need to balance the chemical reaction first. The reaction is \[ \mathrm{HBr}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{NH}_{4}Br(aq) \]This equation is already balanced with one bromine, one hydrogen, and one nitrogen on both sides.

Write the Ionic Equation for (a)

For the balanced reaction \( \mathrm{HBr}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{NH}_{4}Br(aq) \), we write the ionic form:\[ \mathrm{H}^{+}(aq) + \mathrm{Br}^{-}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{NH}_{4}^{+}(aq) + \mathrm{Br}^{-}(aq) \]

Write the Net Ionic Equation for (a)

To get the net ionic equation, we remove spectator ions (species that appear unchanged on both sides of the equation). In this case, \( \mathrm{Br}^{-} \) is a spectator ion:\[ \mathrm{H}^{+}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{NH}_{4}^{+}(aq) \]

Balance Equation (b)

The unbalanced reaction is:\[ \mathrm{Ba}(\mathrm{OH})_2(aq) + \mathrm{H}_3\mathrm{PO}_4(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2(s) + \mathrm{H}_2\mathrm{O}(l) \]To balance:- Balance barium: 3 Ba on right, so 3 on left: \( 3\mathrm{Ba}(\mathrm{OH})_2 \)- Balance phosphorus: 2 P on right, use 2 phosphoric acids: \( 2\mathrm{H}_3\mathrm{PO}_4 \)- Balance water: 12 H on left; so 6 \( \mathrm{H}_2\mathrm{O} \)Balanced equation:\[ 3\mathrm{Ba}(\mathrm{OH})_2(aq) + 2 \mathrm{H}_3\mathrm{PO}_4(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2(s) + 6\mathrm{H}_2\mathrm{O}(l) \]

Write the Ionic Equation for (b)

For the balanced equation:\[ 3\mathrm{Ba}^{2+}(aq) + 6\mathrm{OH}^{-}(aq) + 6\mathrm{H}^{+}(aq) + 2\mathrm{PO}_4^{3-}(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2(s) + 6\mathrm{H}_2\mathrm{O}(l) \]

Write the Net Ionic Equation for (b)

Remove spectator ions. Hydroxide and hydrogen ions form water, which is not present as ions in the net ionic equation:\[ 3\mathrm{Ba}^{2+}(aq) + 2\mathrm{PO}_4^{3-}(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2(s) \]

Balance Equation (c)

The initial reaction is:\[ \mathrm{HClO}_4(aq) + \mathrm{Mg}(\mathrm{OH})_2(s) \rightarrow \mathrm{Mg}(\mathrm{ClO}_4)_2(aq) + \mathrm{H}_2\mathrm{O}(l) \]To balance:- 2 HClO4: \( 2\mathrm{HClO}_4 \)- Yields 2 water: \( 2 \mathrm{H}_2\mathrm{O} \)The balanced equation is:\[ 2\mathrm{HClO}_4(aq) + \mathrm{Mg}(\mathrm{OH})_2(s) \rightarrow \mathrm{Mg}(\mathrm{ClO}_4)_2(aq) + 2\mathrm{H}_2\mathrm{O}(l) \]

Write the Ionic Equation for (c)

For:\[ 2\mathrm{H}^{+}(aq) + 2\mathrm{ClO}_4^{-}(aq) + \mathrm{Mg}(\mathrm{OH})_2(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2\mathrm{ClO}_4^{-}(aq) + 2\mathrm{H}_2\mathrm{O}(l) \]

Write the Net Ionic Equation for (c)

After cancelling out the spectator ions \( \mathrm{ClO}_4^{-} \), we find the net ionic equation:\[ 2\mathrm{H}^{+}(aq) + \mathrm{Mg}(\mathrm{OH})_2(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2\mathrm{H}_2\mathrm{O}(l) \]

Key concepts

Ionic Equations

Ionic equations are a way of showing chemical reactions with greater detail by emphasizing the ions that participate in the reaction process. This type of equation is useful for reactions happening in aqueous solutions, where substances dissolve in water and dissociate into individual ions. For example, when hydrobromic acid \( \mathrm{HBr} \) dissociates in water, it separates into \( \mathrm{H}^{+} \) and \( \mathrm{Br}^{-} \) ions. Ionic equations express these ions separately, giving a clearer picture of how the actual substances interact during a reaction.

When writing ionic equations, follow these simple steps:

• Start by writing the balanced molecular equation of the reaction.
• Identify the compounds that are soluble in water and separate them into their constituent ions.
• Insoluble compounds or substances that do not dissociate remain unchanged.

Using this approach helps chemists and students focus on the chemistry that occurs, without the distracting presence of "spectator ions," which are ions not involved in the actual chemical change.

Net Ionic Equations

Net ionic equations take the process of writing ionic equations a step further by simplifying them to only show ions and molecules directly involved in the reaction. This provides a much clearer depiction of the chemical process. To write a net ionic equation:

• First, write the complete balanced ionic equation.
• Next, identify and remove the spectator ions. These are ions present on both sides of the equation unchanged in the reaction and don't affect the outcome.
• What remains is the net ionic equation, showing only the entities actually undergoing a chemical change.

For example, in the reaction between ammonium bromide \( \mathrm{NH}_{4} \mathrm{Br} \) and \( \mathrm{NH}_3 \), the spectator ion \( \mathrm{Br}^{-} \) is omitted, leaving \( \mathrm{H}^{+}(aq) + \mathrm{NH}_3(aq) \rightarrow \mathrm{NH}_{4}^{+}(aq) \).This adjustment reveals that the transfer of a hydrogen ion is the main event, eliminating complexities that could distract from understanding the core reaction happening.

Acid-Base Reactions

Acid-base reactions are a fundamental type of chemical interaction where an acid donates a proton (\( \mathrm{H}^{+} \)) to a base. The quintessential model follows the Brønsted-Lowry concept, which defines acids as proton donors and bases as proton acceptors.

In an acid-base reaction, such as \( \mathrm{HBr} + \mathrm{NH}_{3} \rightarrow \mathrm{NH}_{4} \mathrm{Br} \), the hydrobromic acid (\( \mathrm{HBr} \)) acts as an acid by donating a proton to ammonia (\( \mathrm{NH}_{3}\)), which acts as a base. As a result, ammonium bromide (\( \mathrm{NH}_{4} \mathrm{Br} \)) is formed.

To balance a given acid-base reaction, consider the following steps:

• Identify each substance as either an acid or a base.
• Write the balanced molecular equation of the reaction.
• Convert it into ionic and net ionic equations if necessary to clearly see the proton transfer.

Understanding acid-base reactions is crucial in various fields, including biology, environmental science, and industrial applications, due to their widespread occurrence in nature and human activity.