Problem 39 - Chemistry The Central Science

Question

Complete and balance the following molecular equations, and then write the net ionic equation for each: \begin{equation} \begin{array}{l}{\text { (a) } \operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow} \\ {\text { (b) } \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow} \\\ {\text { (c) } \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow}\end{array} \end{equation}

Step-by-Step Solution

Answer

The net ionic equations for the given molecular equations are: (a) \(2\mathrm{H}^{+}(a q)+ 2\mathrm{OH}^{-}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+2\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{H}^{+}(a q) \longrightarrow 3\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Al}^{3+}(a q)\)

Step 1: (a) Identify reactants and products

For the reaction \(\operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\), the reactants are HBr (hydrobromic acid) and Ca(OH)₂ (calcium hydroxide), which is an acid-base reaction. The products will be water H₂O (l) and a salt, calcium bromide CaBr₂ (aq).

Step 2: (a) Determine balancing coefficients

To balance the equation, we need 2 moles of HBr for every mole of Ca(OH)₂, 2 moles of water H₂O, and 1 mole of CaBr₂.

Step 3: (a) Write balanced molecular equation

The balanced molecular equation is: \[ 2\operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CaBr}_{2}(a q) \]

Step 4: (a) Write complete ionic equation

The complete ionic equation is: \[ 2\mathrm{H}^{+}(a q)+2\mathrm{Br}^{-}(a q)+\mathrm{Ca}^{2+}(a q)+ 2\mathrm{OH}^{-}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Ca}^{2+}(a q)+ 2\mathrm{Br}^{-}(a q) \]

Step 5: (a) Net ionic equation

Cancel out spectator ions (Br⁻ and Ca²⁺) from both sides: \[ 2\mathrm{H}^{+}(a q)+ 2\mathrm{OH}^{-}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l) \]

Step 6: (b) Identify reactants and products

For the reaction \[\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\], the reactants are Cu(OH)₂ (copper(II) hydroxide) and HClO₄ (perchloric acid), which is another acid-base reaction. The products will be water H₂O (l) and a salt, copper(II) perchlorate Cu(ClO₄)₂ (aq).

Step 7: (b) Determine balancing coefficients

To balance the equation, we need 1 mole of Cu(OH)₂, 2 moles of HClO₄, 2 moles of water H₂O, and 1 mole of Cu(ClO₄)₂.

Step 8: (b) Write balanced molecular equation

The balanced molecular equation is: \[ \mathrm{Cu}(\mathrm{OH})_{2}(s)+2\mathrm{HClO}_{4}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cu}(\mathrm{ClO}_{4})_{2}(a q) \]

Step 9: (b) Write complete ionic equation

The complete ionic equation is: \[ \mathrm{Cu}(\mathrm{OH})_{2}(s)+2\mathrm{H}^{+}(a q)+2\mathrm{ClO}_{4}^{-}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cu}^{2+}(a q)+2\mathrm{ClO}_{4}^{-}(a q) \]

Step 10: (b) Net ionic equation

Cancel out spectator ions (ClO₄⁻) from both sides: \[ \mathrm{Cu}(\mathrm{OH})_{2}(s)+2\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cu}^{2+}(a q) \]

Step 11: (c) Identify reactants and products

For the reaction \[\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\], the reactants are Al(OH)₃ (aluminum hydroxide) and HNO₃ (nitric acid), which is another acid-base reaction. The products will be water H₂O (l) and a salt, aluminum nitrate Al(NO₃)₃ (aq).

Step 12: (c) Determine balancing coefficients

To balance the equation, we need 1 mole of Al(OH)₃, 3 moles of HNO₃, 3 moles of water H₂O, and 1 mole of Al(NO₃)₃.

Step 13: (c) Write balanced molecular equation

The balanced molecular equation is: \[ \mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{HNO}_{3}(a q) \longrightarrow 3\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Al}(\mathrm{NO}_{3})_{3}(a q) \]

Step 14: (c) Write complete ionic equation

The complete ionic equation is: \[ \mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{H}^{+}(a q)+3\mathrm{NO}_{3}^{-}(a q) \longrightarrow 3\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Al}^{3+}(a q)+3\mathrm{NO}_{3}^{-}(a q) \]

Step 15: (c) Net ionic equation

Cancel out spectator ions (NO₃⁻) from both sides: \[ \mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{H}^{+}(a q) \longrightarrow 3\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Al}^{3+}(a q) \]

Key Concepts

Acid-Base Reactions

Acid-base reactions are a fundamental type of chemical reaction that involves the transfer of protons (H⁺ ions) between the reactants. In these reactions, an acid donates protons to a base. Such reactions are essential in myriad chemical processes and are commonly encountered in both laboratory and everyday chemistry. In an acid-base reaction, acids are substances that produce hydrogen ions, H⁺, when dissolved in water. In contrast, bases produce hydroxide ions, OH⁻.

• When an acid reacts with a base, they typically form water and an ionic compound known as a salt.
• The acid-base reactions presented in the original exercise include HBr with Ca(OH)₂, Cu(OH)₂ with HClO₄, and Al(OH)₃ with HNO₃.
• In each case, the acid (HBr, HClO₄, HNO₃) donates protons to the base’s hydroxide ions (Ca(OH)₂, Cu(OH)₂, Al(OH)₃), resulting in the formation of water (H₂O). The other product is a salt derived from the combination of the remaining ions.

These processes underpin widely used chemical methods for neutralizing acids and bases, providing vital applications in industries from pharmaceuticals to environmental management. Each reaction requires careful balancing to ensure all atoms and charges are conserved, leading us into the next concept.

Net Ionic Equations

Net ionic equations simplify full chemical equations to highlight the active ions and molecules participating directly in a chemical reaction. This provides a clear picture of the actual chemical changes taking place. Whether it's an acid-base reaction or another type, many reactions in aqueous solutions involve ions. However, some ions do not participate in the actual chemical change. These spectator ions remain unchanged throughout the reaction and are left out of net ionic equations.

• To begin, convert the balanced molecular equation into the complete ionic equation, separating all soluble ionic compounds and acids into their respective ions.
• For example, in the reaction between HBr and Ca(OH)₂: - The complete ionic equation is: \[ 2\mathrm{H}^{+}(aq) + 2\mathrm{Br}^{-}(aq) + \mathrm{Ca}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Ca}^{2+}(aq) + 2\mathrm{Br}^{-}(aq) \]
• Simplify by removing the spectator ions (Br⁻ and Ca²⁺) common to both sides of the equation.
• This leaves the net ionic equation, which shows only the substances that actually change:\[2\mathrm{H}^{+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]

Net ionic equations are valuable tools for understanding the driving forces behind a reaction and visualizing the chemical processes activated in a solution.

Balancing Chemical Equations

To balance chemical equations is to achieve equal numbers of each type of atom and charge on both sides of a reaction's equation. This is critical because it illustrates the conservation of mass – meaning, no atoms are lost or gained during a chemical reaction, only rearranged. Each compound in an equation consists of distinct elements and must be balanced by adjusting the coefficients (numbers in front of compounds or elements) instead of altering the compound itself. This step ensures the conservation of both mass and charge.

• Begin by writing out the full chemical equation with chemical formulas of compounds as reactants and products.
• Identify the number of atoms of each element on both sides of the equation.
• Adjust coefficients to make these numbers equal. - For example, in the exercise: the balanced molecular equation for the reaction between HBr and Ca(OH)₂ is: \[ 2\operatorname{HBr}(aq) + \mathrm{Ca(OH)}_2(aq) \rightarrow 2\mathrm{H_{2}O}(l) + \mathrm{CaBr}_2(aq) \] - Here, we see two hydrogen molecules, two bromine atoms, and one calcium atom on both sides.
• Verify the balance by recounting atoms and confirming charge neutrality.

Balancing chemical equations not only reflects the fundamental laws of chemistry but also prepares equations for further manipulation and analysis, such as writing net ionic equations.