Net Ionic Equations and Spectator Ions

This video tutorial explains how to write a balanced net ionic equation for the reaction between lead(II) nitrate and magnesium sulfate. It begins with balancing the molecular equation, then discusses the solubility of the reactants and products, identifying which are aqueous and which form a precipitate. The tutorial proceeds to split the strong electrolytes into ions to form the complete ionic equation, and finally derives the net ionic equation by removing spectator ions. The video concludes by verifying the conservation of charge and mass in the net ionic equation.

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

• Identify spectator ions

• Balance the molecular equation

• Determine the solubility of compounds

• Split strong electrolytes into ions

Which of the following compounds is generally soluble?

• Silver chloride

• Calcium carbonate

• Magnesium sulfate

• Lead(II) sulfate

What is the solubility state of lead(II) sulfate in the reaction?

• Aqueous

• Gaseous

• Solid

• Liquid

What is the charge of the nitrate ion?

• 1+

• 1-

• 2+

• 2-

Which ion is not split in the net ionic equation?

• Magnesium ion

• Nitrate ion

• Lead(II) ion

• Lead(II) sulfate

What is the purpose of identifying spectator ions?

• To simplify the net ionic equation

• To calculate molar mass

• To balance the equation

• To determine solubility

Which ions are considered spectator ions in this reaction?

• Nitrate ions

• Lead(II) ions

• Sulfate ions

• Magnesium ions

What is the net charge on both sides of the balanced net ionic equation?

• 1+

• 0

• 2-

• 2+

Why is it important to ensure that the atoms are balanced in the net ionic equation?

• To conserve mass

• To predict the products

• To identify spectator ions

• To determine solubility

What is the final product formed as a precipitate in this reaction?

• Magnesium sulfate

• Magnesium nitrate

• Lead(II) nitrate

• Lead(II) sulfate