Problem 1 - Chemistry: Principles and Reactions

Question

Write a net ionic equation for the reaction between aqueous solutions of (a) ammonia and hydrofluoric acid. (b) perchloric acid and rubidium hydroxide. (c) sodium sulfite and hydriodic acid. (d) nitric acid and calcium hydroxide.

Step-by-Step Solution

Answer

Question: Write the net ionic equation for each reaction. a) A reaction between ammonia and hydrofluoric acid. b) A reaction between perchloric acid and rubidium hydroxide. c) A reaction between sodium sulfite and hydriodic acid. d) A reaction between nitric acid and calcium hydroxide. Answer: a) NH3(aq) + HF(aq) → NH4+(aq) + F-(aq) b) H+(aq) + OH-(aq) → H2O(l) c) SO3^2-(aq) + 2H+(aq) → H2O(l) + SO2(g) d) 2H+(aq) + 2OH-(aq) → 2H2O(l)

Step 1: Write the balanced molecular equation.

The balanced molecular equation for the reaction between ammonia (NH3) and hydrofluoric acid (HF) is: NH3(aq) + HF(aq) → NH4F(aq)

Step 2: Write the total ionic equation.

To write the total ionic equation, break down the aqueous compounds into their respective ions: NH3(aq) + HF(aq) → NH4+(aq) + F-(aq)

Step 3: Identify and eliminate the spectator ions.

In this reaction, there are no spectator ions. All ions are involved in the reaction.

Step 4: Write the net ionic equation.

Since all ions are involved in the reaction, the net ionic equation is the same as the total ionic equation: NH3(aq) + HF(aq) → NH4+(aq) + F-(aq) (b) Perchloric acid and rubidium hydroxide

Step 1: Write the balanced molecular equation.

The balanced molecular equation for the reaction between perchloric acid (HClO4) and rubidium hydroxide (RbOH) is: HClO4(aq) + RbOH(aq) → RbClO4(aq) + H2O(l)

Step 2: Write the total ionic equation.

Break down the aqueous compounds into their respective ions: H+(aq) + ClO4-(aq) + Rb+(aq) + OH-(aq) → Rb+(aq) + ClO4-(aq) + H2O(l)

Step 3: Identify and eliminate the spectator ions.

The spectator ions in this reaction are Rb+ and ClO4-. Eliminate them from the total ionic equation:

Step 4: Write the net ionic equation.

The net ionic equation for this reaction is: H+(aq) + OH-(aq) → H2O(l) (c) Sodium sulfite and hydriodic acid

Step 1: Write the balanced molecular equation.

The balanced molecular equation for the reaction between sodium sulfite (Na2SO3) and hydriodic acid (HI) is: Na2SO3(aq) + 2HI(aq) → 2NaI(aq) + H2O(l) + SO2(g)

Step 2: Write the total ionic equation.

Break down the aqueous compounds into their respective ions: 2Na+(aq) + SO3^2-(aq) + 2H+(aq) + 2I-(aq) → 2Na+(aq) + 2I-(aq) + H2O(l) + SO2(g)

Step 3: Identify and eliminate the spectator ions.

The spectator ions in this reaction are Na+ and I-. Eliminate them from the total ionic equation:

Step 4: Write the net ionic equation.

The net ionic equation for this reaction is: SO3^2-(aq) + 2H+(aq) → H2O(l) + SO2(g) (d) Nitric acid and calcium hydroxide

Step 1: Write the balanced molecular equation.

The balanced molecular equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2) is: 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O(l)

Step 2: Write the total ionic equation.

Break down the aqueous compounds into their respective ions: 2H+(aq) + 2NO3-(aq) + Ca^2+(aq) + 2OH-(aq) → Ca^2+(aq) + 2NO3-(aq) + 2H2O(l)

Step 3: Identify and eliminate the spectator ions.

The spectator ions in this reaction are Ca^2+ and NO3-. Eliminate them from the total ionic equation:

Step 4: Write the net ionic equation.

The net ionic equation for this reaction is: 2H+(aq) + 2OH-(aq) → 2H2O(l)

Key Concepts

Molecular Equations

Molecular equations represent chemical reactions in their entirety. They show all the reactants and products in their complete, undissociated forms. This means that the compounds in the solution are written as neutral molecules. Such equations are essential for giving us a clear initial picture of what substances are reacting.
For instance, the reaction between ammonia (NH₃) and hydrofluoric acid (HF) can be represented as:

• NH₃(aq) + HF(aq) → NH₄F(aq)

This shows ammonia and hydrofluoric acid as reactants and ammonium fluoride as a product without indicating their ionic nature or the dissociation of the products or reactants.
Molecular equations are particularly helpful when initially identifying the substances participating in a chemical reaction.

Total Ionic Equations

Total ionic equations dig deeper into a chemical reaction than molecular equations by expressing the ions that each compound forms in an aqueous solution. This focuses on the reality that, in solution, ions separate and move independently. It provides a clearer picture of the actual species present in the solution.
Consider the reaction between sodium sulfite (Na₂SO₃) and hydriodic acid (HI). The total ionic equation is:

• 2Na⁺(aq) + SO₃^2-(aq) + 2H⁺(aq) + 2I^-(aq) → 2Na⁺(aq) + 2I^-(aq) + H₂O(l) + SO₂(g)

This equation shows all ions that are present in the solution, highlighting the disassociated nature of the compounds being reacted. The detailed breakdown in a total ionic form helps in identifying which ions are actually participating in the chemical change versus those that remain unchanged.

Spectator Ions

Spectator ions are the ions present in a solution that do not participate in the actual chemical reaction. They are present on both sides of a total ionic equation unchanged. Identifying them is crucial for obtaining a net ionic equation since these ions do not contribute to the reaction outcome. Removing them simplifies the equation to focus only on the species that undergo change.
For example, in the reaction between perchloric acid (HClO₄) and rubidium hydroxide (RbOH), the total ionic equation includes spectator ions: Rb⁺(aq) and ClO₄^-(aq).
By eliminating these, you obtain the net ionic equation:

• H⁺(aq) + OH^-(aq) → H₂O(l)

This demonstrates the fundamental reaction occurring, which is the formation of water from hydrogen and hydroxide ions, leaving the spectators out. Understanding the concept of spectator ions helps students clarify and streamline net ionic equations in chemical reactions.