Net Ionic Equations and Spectator Ions

The video tutorial explains how to write and balance net ionic equations, using the reaction between strontium nitrate and sodium phosphate as an example. It covers balancing the molecular equation, determining the solubility of compounds, and writing both complete and net ionic equations. The tutorial emphasizes the importance of identifying spectator ions and ensuring charge and atom balance in the final net ionic equation.

What is the first step in writing a net ionic equation?

• Identify spectator ions

• Balance the molecular equation

• Write the complete ionic equation

• Determine solubility

How many strontium atoms are needed to balance the molecular equation?

• Two

• Three

• Four

• One

Which of the following compounds is generally insoluble?

• Strontium nitrate

• Sodium phosphate

• Sodium nitrate

• Strontium phosphate

What happens to strontium phosphate in the reaction?

• It dissolves completely

• It forms a precipitate

• It forms a gas

• It remains as ions

What is the role of spectator ions in a reaction?

• They participate in forming the precipitate

• They change the charge balance

• They form new compounds

• They do not participate in the reaction

Which ions are considered spectator ions in this reaction?

• Nitrate ions

• Sodium ions

• Phosphate ions

• Strontium ions

What is the net charge of the final net ionic equation?

• Depends on the reaction

• Zero

• Negative

• Positive

Why is it important to ensure charge balance in a net ionic equation?

• To predict the color of the solution

• To ensure the reaction is balanced

• To determine the solubility

• To identify spectator ions

What is the final product formed in the reaction between strontium nitrate and sodium phosphate?

• Sodium phosphate

• Strontium nitrate

• Strontium phosphate

• Sodium nitrate

What is the significance of a net ionic equation?

• It highlights the formation of gases

• It shows all compounds in the reaction

• It focuses on the ions that participate in the reaction

• It predicts the color change in the reaction