Net Ionic Equations and Spectator Ions

The video tutorial explains how to write a balanced net ionic equation for the reaction between barium bromide (BaBr2) and potassium sulfate (K2SO4). It begins by balancing the molecular equation, then discusses the solubility rules to determine the states of the compounds involved. The tutorial proceeds to write the complete ionic equation by splitting strong electrolytes into ions and identifies the spectator ions. Finally, it presents the net ionic equation, ensuring that both atoms and charges are balanced.

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

• Determine the states of the compounds

• Split the compounds into ions

• Balance the molecular equation

• Identify spectator ions

Why is it important to balance the molecular equation first?

• To ensure the number of atoms is equal on both sides

• To identify spectator ions

• To ensure the reaction is possible

• To determine the states of the compounds

Which of the following compounds is insoluble in water?

• Potassium sulfate

• Barium bromide

• Barium sulfate

• Potassium bromide

What does the 's' symbol indicate in the context of this reaction?

• Sublimation

• Solid

• Soluble

• Solution

What is the charge of the barium ion in the complete ionic equation?

• 1+

• 2-

• 2+

• 1-

How are strong electrolytes treated in the complete ionic equation?

• They are split into their constituent ions

• They are left as whole compounds

• They are combined into a single ion

• They are ignored

Which ions are considered spectator ions in this reaction?

• Barium and bromide ions

• Barium and sulfate ions

• Potassium and sulfate ions

• Potassium and bromide ions

What is the purpose of crossing out spectator ions?

• To balance the charges

• To simplify the equation

• To identify the precipitate

• To determine the solubility

What is the net charge of the reactants in the net ionic equation?

• 2+

• 2-

• 0

• 1+

What ensures that the net ionic equation is correctly balanced?

• Equal number of compounds

• Equal number of spectator ions

• Presence of a precipitate

• Equal number of atoms and charges on both sides