Complete Ionic Equations: Videos & Practice Problems

Understanding ionic equations is crucial for grasping the nuances of chemical reactions in aqueous solutions. In a molecular equation, all reactants and products are listed, including aqueous compounds, solids, liquids, and gases. However, only aqueous compounds dissociate into ions. Using solubility rules helps identify which compounds are aqueous. A complete ionic equation, derived from the molecular equation, breaks down aqueous compounds into their respective ions, applying coefficients to determine the correct number of ions present. The net ionic equation further refines this by eliminating spectator ions—ions that appear unchanged on both the reactant and product sides of the reaction—highlighting the ions that actively participate in the chemical transformation. This progression from molecular to complete to net ionic equations reveals the underlying ionic activity driving the observable chemical changes.

Complete Ionic Equations

In chemistry, understanding the transition from a molecular equation to a complete ionic equation is crucial for analyzing reactions in aqueous solutions. A molecular equation presents all reactants and products in their molecular form, including solids, liquids, and gases, while a complete ionic equation breaks down only the aqueous compounds into their constituent ions.

When dealing with complete ionic equations, it is essential to recognize that only aqueous compounds dissociate into ions. Solids, liquids, and gases remain intact and do not break apart. To determine whether a compound is aqueous, solubility rules are applied, which help identify which substances will dissolve in water and form ions.

To derive a complete ionic equation from a molecular equation, one must distribute the coefficients of each compound correctly. This ensures that the number of ions represented reflects the stoichiometry of the reaction. For example, if a compound has a coefficient of 2, it will produce twice the number of ions in the complete ionic equation.

In summary, the complete ionic equation provides a clearer picture of the actual species present in solution during a chemical reaction, allowing for a better understanding of the interactions between ions. This approach is fundamental in predicting the outcomes of reactions and understanding the behavior of different compounds in aqueous environments.

Complete Ionic Equation Example

To convert a molecular equation into a complete ionic equation, it is essential to recognize which compounds are aqueous and which are not. In this case, we have the reaction of 3 moles of calcium bromide (CaBr₂) aqueous with 2 moles of lithium phosphate (Li₃PO₄) aqueous, resulting in the formation of 6 moles of lithium bromide (LiBr) aqueous and 1 mole of calcium phosphate (Ca₃(PO₄)₂) solid.

Only the aqueous compounds will dissociate into their respective ions in the complete ionic equation. Therefore, we will break down the calcium bromide, lithium phosphate, and lithium bromide into their ions, while the calcium phosphate remains intact as a solid.

Starting with calcium bromide, the dissociation yields:

3 CaBr₂ (aq) → 3 Ca²⁺ (aq) + 6 Br^- (aq)

Next, for lithium phosphate, the dissociation gives:

2 Li₃PO₄ (aq) → 6 Li⁺ (aq) + 2 PO₄^3- (aq)

Finally, lithium bromide dissociates as follows:

6 LiBr (aq) → 6 Li⁺ (aq) + 6 Br^- (aq)

Since calcium phosphate is a solid, it does not dissociate:

Ca₃(PO₄)₂ (s)

Combining all these components, the complete ionic equation is:

3 Ca²⁺ (aq) + 6 Br^- (aq) + 6 Li⁺ (aq) + 2 PO₄^3- (aq) → 6 Li⁺ (aq) + 6 Br^- (aq) + Ca₃(PO₄)₂ (s)

In summary, when converting a molecular equation to a complete ionic equation, remember to only break apart the aqueous compounds and distribute the coefficients to the respective ions formed.

Net Ionic Equations

A net ionic equation is a simplified representation of a chemical reaction that highlights the ions directly involved in the reaction while omitting the spectator ions. Spectator ions are those that appear unchanged on both sides of the equation, meaning they do not participate in the actual chemical change. To derive a net ionic equation, one must first start with the molecular equation, which represents the reactants and products in their molecular form.

From the molecular equation, the next step is to write the complete ionic equation. This equation breaks down all soluble ionic compounds into their respective ions, showing all species present in the reaction. Finally, by removing the spectator ions from the complete ionic equation, you arrive at the net ionic equation, which succinctly illustrates the essential chemical changes occurring during the reaction.

This process of transitioning from a molecular equation to a net ionic equation is crucial for understanding the specific interactions between ions in a solution, allowing for a clearer insight into the underlying chemistry at play.

Molecular and Complete Ionic Equations Example

In a reaction between ammonium sulfate and calcium chloride, we can derive both the molecular equation and the complete ionic equation. The molecular equation begins with the reactants:

Ammonium sulfate: (NH₄)₂SO₄ and Calcium chloride: CaCl₂.

When these compounds react, they form ammonium chloride (NH₄Cl) and calcium sulfate (CaSO₄). The balanced molecular equation is:

(NH₄)₂SO₄ (aq) + CaCl₂ (aq) → 2 NH₄Cl (aq) + CaSO₄ (s)

Next, we break down the aqueous compounds into their ionic forms for the complete ionic equation. Ammonium sulfate dissociates into 2 ammonium ions (2 NH₄⁺) and 1 sulfate ion (SO₄^2-), while calcium chloride dissociates into 1 calcium ion (Ca²⁺) and 2 chloride ions (2 Cl^-). The complete ionic equation is:

2 NH₄⁺ (aq) + SO₄^2- (aq) + Ca²⁺ (aq) + 2 Cl^- (aq) → 2 NH₄Cl (aq) + CaSO₄ (s)

In this equation, we identify the spectator ions, which are ions that do not participate in the actual precipitation reaction. Here, the ammonium ions (NH₄⁺) and chloride ions (Cl^-) are spectator ions. By removing these from the complete ionic equation, we arrive at the net ionic equation:

SO₄^2- (aq) + Ca²⁺ (aq) → CaSO₄ (s)

This net ionic equation highlights the essential components of the reaction, showing that the sulfate ions react with calcium ions to form solid calcium sulfate, indicating a successful reaction has occurred.

Provide the net ionic equation that occurs when the following aqueous compounds are mixed together: 
Copper (II) Bromide and Lithium Hydroxide

• A. Cu⁺ (aq) + OH^– (aq) → CuOH (s)
• B. Cu²⁺ (aq) + 2 OH^– (aq) → Cu(OH)₂ (s)
• C. Cu²⁺ (aq) + 2 H₂O (aq) → Cu(OH)₂ (s) + 2 H⁺ (aq)
• D. Cu²⁺ (aq) + OH^– (aq) → Cu(OH)₂ (s)

Answer: (B)

Which of the following reagents could be used to separate the two anions from a solution containing magnesium nitrate and cesium hydroxide?

• A. NH₄CN
• B. NaCl
• C. KNO₃
• D. ZnBr₂
• E. CsBrO₃

Answer: (D)

Which of the following reagents could be used to separate the two cations from a solution containing Lead (IV) acetate and cesium permanganate?

• A. Sr(NO₃)₂
• B. TiC₂H₃O₂
• C. K₂S
• D. NaClO₄
• E. KNO₃

Answer: (C)

Here’s what students ask on this topic:

What is a complete ionic equation?

A complete ionic equation is a chemical equation that shows all of the species as they are actually present in solution. This means that compounds that dissociate and separate into their constituent ions in a solution are written as those individual ions. For example, if sodium chloride (NaCl) is dissolved in water, it separates into Na⁺ and Cl^- ions, and the complete ionic equation would reflect this dissociation.

It's important to note that the complete ionic equation does not include the spectator ions, which are ions that do not participate in the actual chemical reaction. To write a complete ionic equation, you start with a balanced molecular equation, then rewrite it showing all soluble strong electrolytes as ions. This provides a more detailed picture of the chemical reaction by illustrating which ions are involved in the reaction and which ones are merely present in the solution.

How do you write a complete ionic equation?

Writing a complete ionic equation involves representing all strong electrolytes (substances that dissolve in water to form a solution that conducts electricity) as ions. Here's how to do it step by step:

1. Start with a balanced molecular equation. Write down the chemical reaction with all the reactants and products in their molecular form, ensuring that the number of atoms of each element is the same on both sides of the equation.
2. Identify the strong electrolytes. Strong acids, strong bases, and most salts are considered strong electrolytes. They will dissociate completely into ions in solution.
3. Separate the strong electrolytes into ions. Write out all the ions that each strong electrolyte would form in solution. For example, NaCl would be written as Na⁺ and Cl^-.
4. Indicate the states of matter. Beside each compound or ion, write (s) for solid, (l) for liquid, (g) for gas, or (aq) for aqueous (dissolved in water).
5. Cancel out the spectator ions. Spectator ions are ions that appear unchanged on both the reactant and the product side of the equation. They do not participate in the actual chemical reaction and can be omitted to write the net ionic equation.

Here's an example

What information does the complete ionic equation give?

A complete ionic equation provides detailed information about the species present in a chemical reaction that takes place in an aqueous solution. Unlike a standard chemical equation that shows the reactants and products as whole compounds, a complete ionic equation breaks down soluble ionic compounds into their individual ions. It indicates which substances dissociate into ions in water and shows the actual forms of the reactants and products as they exist in solution.

For example, in the reaction between sodium chloride (NaCl) and silver nitrate (AgNO₃), the complete ionic equation would show the sodium (Na⁺), chloride (Cl^-), silver (Ag⁺), and nitrate (NO₃^-) ions separately. This level of detail helps you understand which ions are spectator ions (ions that do not participate in the reaction) and which ions combine to form the precipitate or the products of the reaction. It's particularly useful for predicting the outcomes of precipitation reactions and for understanding the movement of individual ions in electrochemical cells.