Problem 101 - Chemistry Matter and Change

Question

Write complete ionic and net ionic equations for each of the following reactions. $$\begin{array}{l}{\text { a. } \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+3 \mathrm{RbOH}(\mathrm{aq}) \rightarrow 3 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{Rb}_{3} \mathrm{PO}_{4}(\mathrm{aq})} \\ {\text { b. } \mathrm{HCl}(\mathrm{aq})+\mathrm{NH}_{4} \mathrm{OH}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})} \\ {\text { c. } 2 \mathrm{HI}+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+2 \mathrm{NH}_{4} \mathrm{I}(\mathrm{aq})} \\ {\text { d. } \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{KCN}(\mathrm{aq})+\mathrm{HCN}(\mathrm{g})+\mathrm{KNO}_{3}(\mathrm{aq})}\end{array}$$

Step-by-Step Solution

Answer

Net ionic equations: (a) \(3\mathrm{H}^{+}(aq) + \mathrm{PO}_4^{3-}(aq) + 3\mathrm{OH}^{-}(aq) \rightarrow 3\mathrm{H}_2\mathrm{O}(l)\), (b) \(\mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)\), (c) \(2\mathrm{H}^{+}(aq) + \mathrm{S}^{2-}(aq) \rightarrow \mathrm{H}_{2}\mathrm{S}(g)\), (d) \(\mathrm{H}^{+}(aq) + \mathrm{CN}^{-}(aq) \rightarrow \mathrm{HCN}(g)\).

Step 1: Write the Balanced Equation (Part a)

For reaction a, the chemical equation is \( \mathrm{H}_{3} \mathrm{PO}_{4} (aq) + 3 \mathrm{RbOH} (aq) \rightarrow 3 \mathrm{H}_{2} \mathrm{O} (l) + \mathrm{Rb}_{3} \mathrm{PO}_{4} (aq) \). Ensure that it is properly balanced by checking the number of each type of atom on both sides.

Step 2: Complete Ionic Equation (Part a)

Break down all the aqueous compounds into ions: \( \mathrm{H}_{3} \mathrm{PO}_{4} (aq) \rightarrow 3\mathrm{H}^{+} (aq) + \mathrm{PO}_{4}^{3-} (aq) \) and \( 3\mathrm{RbOH} (aq) \rightarrow 3\mathrm{Rb}^{+} (aq) + 3\mathrm{OH}^{-} (aq) \). The complete ionic equation is \( 3\mathrm{H}^{+} (aq) + \mathrm{PO}_{4}^{3-} (aq) + 3\mathrm{Rb}^{+} (aq) + 3\mathrm{OH}^{-} (aq) \rightarrow 3\mathrm{H}_{2} \mathrm{O} (l) + \mathrm{Rb}_{3} \mathrm{PO}_{4} (aq) \).

Step 3: Net Ionic Equation (Part a)

Identify and cancel spectator ions. Spectator ions are \( \mathrm{Rb}^{+} \). The net ionic equation is \( 3\mathrm{H}^{+} (aq) + \mathrm{PO}_{4}^{3-} (aq) + 3\mathrm{OH}^{-} (aq) \rightarrow 3\mathrm{H}_{2} \mathrm{O} (l) + \mathrm{PO}_{4}^{3-} (aq) \).

Step 4: Write the Balanced Equation (Part b)

For reaction b, the equation is \( \mathrm{HCl} (aq) + \mathrm{NH}_{4} \mathrm{OH} (aq) \rightarrow \mathrm{H}_{2} \mathrm{O} (l) + \mathrm{NH}_{4} \mathrm{Cl} (aq) \). Confirm it is balanced by ensuring equal numbers of each atom on both sides.

Step 5: Complete Ionic Equation (Part b)

Dissociate all aqueous compounds: \( \mathrm{HCl}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{Cl}^{-}(aq) \) and \( \mathrm{NH}_{4} \mathrm{OH}(aq) \rightarrow \mathrm{NH}_{4}^{+} (aq) + \mathrm{OH}^{-} (aq) \). The complete ionic equation is \( \mathrm{H}^{+}(aq) + \mathrm{Cl}^{-}(aq) + \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{NH}_{4}^{+}(aq) + \mathrm{Cl}^{-}(aq) \).

Step 6: Net Ionic Equation (Part b)

Cancel spectator ions (\( \mathrm{Cl}^{-} \) and \( \mathrm{NH}_{4}^{+} \)) to find the net ionic equation as \( \mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \).

Step 7: Write the Balanced Equation (Part c)

Write the chemical equation as \( 2 \mathrm{HI}(aq) + (\mathrm{NH}_{4})_{2} \mathrm{S}(aq) \rightarrow \mathrm{H}_{2} \mathrm{S}(g) + 2 \mathrm{NH}_{4} \mathrm{I}(aq) \). Make sure that all atoms are balanced.

Step 8: Complete Ionic Equation (Part c)

Break down the aqueous substances: \( 2 \mathrm{HI}(aq) \rightarrow 2\mathrm{H}^{+}(aq) + 2\mathrm{I}^{-}(aq) \) and \( (\mathrm{NH}_{4})_{2} \mathrm{S}(aq) \rightarrow 2\mathrm{NH}_{4}^{+}(aq) + \mathrm{S}^{2-}(aq) \). The complete ionic equation is \( 2\mathrm{H}^{+}(aq) + 2\mathrm{I}^{-}(aq) + 2\mathrm{NH}_{4}^{+}(aq) + \mathrm{S}^{2-}(aq) \rightarrow \mathrm{H}_{2} \mathrm{S}(g) + 2\mathrm{NH}_{4}^{+}(aq) + 2\mathrm{I}^{-}(aq) \).

Step 9: Net Ionic Equation (Part c)

Cancel spectator ions, which are \( \mathrm{I}^{-} \) and \( \mathrm{NH}_{4}^{+} \). The net ionic equation is \( 2\mathrm{H}^{+}(aq) + \mathrm{S}^{2-}(aq) \rightarrow \mathrm{H}_{2} \mathrm{S}(g) \).

Step 10: Write the Balanced Equation (Part d)

The reaction equation is \( \mathrm{HNO}_{3}(aq) + \mathrm{KCN}(aq) \rightarrow \mathrm{HCN}(g) + \mathrm{KNO}_{3}(aq) \). Look over and ensure it is balanced.

Step 11: Complete Ionic Equation (Part d)

Dissociate all aqueous compounds: \( \mathrm{HNO}_{3}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{NO}_{3}^{-}(aq) \) and \( \mathrm{KCN}(aq) \rightarrow \mathrm{K}^{+}(aq) + \mathrm{CN}^{-}(aq) \). The complete ionic equation is \( \mathrm{H}^{+}(aq) + \mathrm{NO}_{3}^{-}(aq) + \mathrm{K}^{+}(aq) + \mathrm{CN}^{-}(aq) \rightarrow \mathrm{HCN}(g) + \mathrm{K}^{+}(aq) + \mathrm{NO}_{3}^{-}(aq) \).

Step 12: Net Ionic Equation (Part d)

Cancel the spectator ions (\( \mathrm{NO}_{3}^{-} \) and \( \mathrm{K}^{+} \)) to obtain the net ionic equation: \( \mathrm{H}^{+}(aq) + \mathrm{CN}^{-}(aq) \rightarrow \mathrm{HCN}(g) \).

Key Concepts

Complete Ionic Equations

In chemistry, understanding the complete ionic equation is essential. This type of equation reveals the full details of the ions involved in a chemical reaction when substances are dissolved in water. When writing a complete ionic equation, we dissociate aqueous compounds into their respective ions. For example, if we have hydrochloric acid ( HCl) in an aqueous solution, it dissociates into hydrogen ions ( H^+(aq)) and chloride ions ( Cl^-(aq)). Similarly, ammonium hydroxide ( NH_4OH) dissociates into ammonium ions ( NH_4^+(aq)) and hydroxide ions ( OH^-(aq)).

Thus, the complete ionic equation for these components would display all the ions explicitly, like so:
\[ \mathrm{H}^+(aq) + \mathrm{Cl}^-(aq) + \mathrm{NH}_4^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l) + \mathrm{NH}_4^+(aq) + \mathrm{Cl}^-(aq) \].

In this manner, each entity in the reaction that exists as ions in the aqueous state is shown, capturing the real dynamic of reactions in solution.

Net Ionic Equation

The net ionic equation takes the complete ionic equation a step further by simplifying it to highlight only the components that actively participate in the chemical reaction. This process involves identifying and removing the spectator ions. Spectator ions are those present on both sides of the equation that do not change during the course of the reaction. They do not participate in forming the product.

For example, in the previous complete ionic equation, the spectator ions are  Cl^- and  NH_4^+. Removing them leaves the net ionic equation: \[ \mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l) \].

This equation concisely shows that hydrogen and hydroxide ions combine to form water, simplifying our understanding of the essential chemical changes happening in the reaction.

Chemical Reactions

Chemical reactions are the heart of chemistry. They are processes where reactants transform into products, involving rearrangement of atoms and breaking and forming of bonds. Reactions can include synthesis, decomposition, single replacement, or double replacement types. Each type involves unique rearrangements and outcomes.

For instance, the reaction: \( \mathrm{HCl}(aq) + \mathrm{NH_4OH}(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l) + \mathrm{NH_4Cl}(aq) \) is a neutralization reaction, where an acid ( HCl) reacts with a base ( NH_4OH) to produce water and a salt ( NH_4Cl). This example represents a typical double replacement reaction.

Understanding the types and mechanisms of chemical reactions helps predict the products formed and the conditions needed for these reactions to occur, enriching our comprehension of how substances interact.

Balanced Chemical Equations

A balanced chemical equation is crucial for accurately representing a chemical reaction. Balancing ensures that the law of conservation of mass is followed. Each side of the equation must have the same number of atoms of each element involved in the reaction. This principle is important because atoms are neither created nor destroyed in a chemical reaction.

To balance an equation, one adjusts the coefficients—the numbers in front of molecules/atoms—without changing the actual formula of the compounds. For instance, the simple reaction \( \mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow \mathrm{H_2O}(l) \) is unbalanced until we recognize the need for coefficients and adjust it to \( 2 \mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow 2 \mathrm{H_2O}(l) \).

Balancing chemical equations ensures we understand the correct stoichiometry of reactants and products, providing a clear depiction of how much of each substance is involved in the reaction.