16.18: Net Ionic Equations

At sports events around the world, we can see a small number of athletes fiercely competing on the field. They get tired, dirty, and sometimes hurt as they try to win the game. Surrounding them are thousands of spectators watching and cheering. Would the game be different without the spectators? Yes, it definitely would. They provide encouragement to the team and generate enthusiasm. The spectators are not really playing the game, but they are certainly a part of the process.

Net Ionic Equations

We can write a molecular equation for the formation of silver chloride precipitate:

$\ce{NaCl} + \ce{AgNO_3} \rightarrow \ce{NaNO_3} + \ce{AgCl}$

The corresponding ionic equation is:

$\ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) + \ce{AgCl} \left( s \right)$

If you look carefully at the ionic equation, you will notice that the sodium ion and the nitrate ion appear unchanged on both sides of the equation. When the two solutions are mixed, neither the $$\ce{Na^+}$$ nor the $$\ce{NO_3^-}$$ ions participate in the reaction. They can be eliminated from the reaction.

$\cancel{\ce{Na^+} \left( aq \right)} + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \cancel{\ce{NO_3^-} \left( aq \right)} \rightarrow \cancel{\ce{Na^+} \left( aq \right)} + \cancel{\ce{NO_3^-} \left( aq \right)} + \ce{AgCl} \left( s \right)$

A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction. In the above reaction, the sodium ion and the nitrate ion are both spectator ions. The equation can now be written without the spectator ions.

$\ce{Ag^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) \rightarrow \ce{AgCl} \left( s \right)$

The net ionic equation is the chemical equation that shows only those elements, compounds, and ions that are directly involved in the chemical reaction. Notice that in writing the net ionic equation, the positively-charged silver cation was written first on the reactant side, followed by the negatively-charged chloride anion. This is somewhat customary because that is the order in which the ions must be written in the silver chloride product. However, it is not absolutely necessary to order the reactants in this way.

Net ionic equations must be balanced by both mass and charge. Balancing by mass means making sure that there are equal numbers of each element. Balancing by charge means making sure that the overall charge is the same on both sides of the equation. In the above equation, the overall charge is zero, or neutral, on both sides of the equation. As a general rule, if you balance the molecular equation properly, the net ionic equation will end up being balanced by both mass and charge.

Example $$\PageIndex{1}$$

When aqueous solutions of copper (II) chloride and potassium phosphate are mixed, a precipitate of copper (II) phosphate is formed. Write a balanced net ionic equation for this reaction.

Solution:

Step 1: Plan the problem.

Write and balance the molecular equation first, making sure that all formulas are correct. Then write the ionic equation, showing all aqueous substances as ions. Carry through any coefficients. Finally, eliminate spectator ions and write the net ionic equation.

Step 2: Solve.

Molecular equation:

$3 \ce{CuCl_2} \left( aq \right) + 2 \ce{K_3PO_4} \left( aq \right) \rightarrow 6 \ce{KCl} \left( aq \right) + \ce{Cu_3(PO_4)_2} \left( s \right)$

Ionic equation:

$3 \ce{Cu^{2+}} \left( aq \right) + 6 \ce{Cl^-} \left( aq \right) + 6 \ce{K^+} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \rightarrow 6 \ce{K^+} \left( aq \right) + 6 \ce{Cl^-} \left( aq \right) + \ce{Cu_3(PO_4)_2} \left( s \right)$

Notice that the balancing is carried through when writing the dissociated ions. For example, there are six chloride ions on the reactant side because the coefficient of 3 is multiplied by the subscript of 2 on the copper (II) chloride formula. The spectator ions are $$\ce{K^+}$$ and $$\ce{Cl^-}$$ and can be eliminated.

Net ionic equation:

$3 \ce{Cu^{2+}} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \rightarrow \ce{Cu_3(PO_4)_2} \left( s \right)$