A Guide on How to Find Spectator Ions in a Chemical Reaction

Spectator ions in a chemical reaction
Spectator ions, as the name suggests, are ions that play a major role when it comes to having equal charges on both sides of a chemical equation. This Buzzle post tells you how to find these ions in a given chemical reaction.
Balancing Act
Spectator ions serve the purpose of balancing the charges of a reaction without affecting the equilibrium of the equation in any way.
In case of total ionic equations, the charges on both sides of the equation need to be balanced. Spectator ions contribute in balancing the charges while not reacting with any elements of the equation. They merely watch over the other ions. And, this is the reason why they are named so. They are atoms or a group of atoms that do not change chemically or physically during the course of a reaction.

In order to understand the concept of spectator ions, we first need to understand the concept of net ionic equations. In relation stoichiometry, Net ionic equations are formed by removing the spectator ions from an entire ionic equation.
Spectator Ions in a Precipitation Reaction
Consider a precipitation reaction between two compounds Pb(NO3)2(aq) and KI(aq) as represented below:

Pb(NO3)2(aq) + 2KI(aq) ➜ PbI2(s) + 2KNO3(aq)
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq) ➜ PbI2(s) + 2 NO3-(aq) + 2 K+(aq)

From the above equation, it can be observed that NO3-(aq) and K+(aq) are present on both; left as well as right side of the equation. They remain unchanged throughout the equation. Therefore, they are termed as 'spectator' ions. PbI2 is the precipitate in this reaction. It is solid, and hence, cannot be separated into ions.

Since NO3-(aq) and K+(aq) remain unaltered during and after the equation, they can be eliminated from the net ionic equation.

The equation below is the final net ionic equation after elimination of the spectator ions, which yields a solid precipitate.

Pb2+(aq) + 2 I-(aq) ➜ PbI2(s)

Note : (aq) and (s) indicate aqueous and solid state respectively.

The term 'net ionic' indicates that the electric charge on both sides of the equation needs to be balanced and equal. For all the charges to be balanced, all the products must not be in aqueous form. Because if all the products of the equation are aqueous, then all the ions will cancel out as spectator ions. Thus, no precipitation reaction occurs.
Spectator Ions in a Neutralization Reaction
Consider a precipitation reaction between two compounds KOH and HNO3 as represented below:

KOH(aq) + HNO3 ➜ KNO3(aq) + H2O(l)
K+(aq) + OH-(aq) + H+(aq) + NO3-(aq) ➜ K+(aq) + NO3-(aq) + H2O(l)

From the above equation, it can be observed that K+(aq) and NO3-(aq)are present on both; left as well as right side of the equation. They remain unchanged throughout the equation. Therefore, they are termed as 'spectator' ions.

Since K+ and NO3- remain unaltered during and after the equation, they can be eliminated from the net ionic equation.

The equation below is the final net ionic equation after elimination of the spectator ions, which yields water.
H+(aq) + OH-(aq) ➜ H2O(l)

The spectator ions in KOH(aq) and HNO3(aq) are: K+(aq) and NO3-(aq)
How to Identify and List the Spectator Ions in a Reaction
Step I : Write the equation (with chemical formulas) for which spectator ions are to be determined.
Step II : Express each of the reactants and products in terms of its cations and anions.
Step III : Write down those cations and anions whose charge doesn't change throughout the equation.
Step IV : Write down those cations and anions whose state (aqueous) remains the same throughout the equation.

The ions which satisfy both the criteria (step 3 and step 4) are said to be spectator ions.

Step 1 :
Consider the complete ionic equation in which sodium chloride reacts with copper sulfate. The products are copper chloride and sodium sulfate.
2NaCl(aq) + CuSO4(aq) ➜ CuCl2(s) + Na2SO4(aq) ------- Equation I

Step 2 :
The complete ionic equation for equation I is as follows:
2Na+(aq) + 2Cl-(aq) + Cu2+(aq) + SO42-(aq) ➜ CuCl2(s) + 2Na+(aq) + SO42-(aq)

The spectator ions in this equation are: Na+ and SO42-. These ions remain unaltered over both sides of the equation besides watching over other ions.

The net ionic equation on canceling the spectator ions from both the sides is:
2Cl-(aq) + Cu2+(aq) ➜ CuCl2(s)

All the reactants and products are in aqueous form, and thus, the ions will cancel out each other due to the same charges.
Example of Spectator Ions
Na2CO3(aq) and MgSO4(aq)
Step 1 : MgSO4(aq) + Na2CO3(aq) ➜ MgCO3(s) + Na2SO4(aq)
Step 2 : Mg2+(aq) + SO42-(aq) + 2Na+(aq) + CO32-(aq) ➜ MgCO3(s) + 2Na+(aq) + SO42-(aq)

The spectator ions in Na2CO3(aq) and MgSO4(aq) are: Na+(aq) and SO42-(aq)
From the working of the spectator ions, it is clearly visible that these ions play a significant role when it comes to balancing the charges of a chemical equation. They carry out the entire charge-balancing act without disturbing the equilibrium of the equation.