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Net Ionic Equations

Objective: In this lesson, the student will learn how to identify and write net ionic equations.

Net ionic equations result from double displacement reactions in which free ions in aqueous solution react to form either a precipitate, a gas, a weak electrolyte, or water.

Strategy for Writing Net Ionic Equations

If one reactant is an acid, check for gaseous products (i.e., CO2, H2S, SO2, or NH3 gases).If one reactant is a base, check for gaseous products (i.e., NH3 or CO2 gases).An acid + a base produces a salt + water.Check for precipitates using the activity series of metals and a solubility table.

Activity Series of Metals & Solubility Table



Remember!

All chemical equations must be balanced.In double displacement reactions, the sum of ion charges on both sides of the equation must be balanced.

Example

Write a balanced net ionic equation for the reaction of aqueous solutions of sodium sulfide and hydrochloric acid.

Na2S (aq) + HCl (aq) → ?

If one reactant is an acid (HCl), check for gaseous products.

Na2S (aq) + HCl (aq) → H2S (g) + NaCl (aq)

Balance.

Na2S (aq) + 2 HCl (aq) → H2S (g) + 2 NaCl (aq)

The above is called a molecular equation: all reactants and products are expressed as compounds.

From the molecular equation, identify the ions in solution.  Note that the H2S is a gas and is not in aqueous solution.

2 Na+ (aq)  + S2- (aq) + 2 H+ (aq) + 2  Cl- (aq)  →   H2S (g) + 2 Na+ (aq) + 2 Cl- (aq)

The aqueous 2 H+ and S2-  ions react to form H2S (g).  The remaining aqueous 2 Na+ and 2 Cl- ions appear on both sides of the equation and are spectators.  That is, they do not participate in the reaction and can be eliminated.

2 Na+ (aq)  + S2- (aq) + 2 H+ (aq) + 2  Cl- (aq)  →   H2S (g) + 2 Na+ (aq) + 2 Cl- (aq)

After eliminating the spectator ions from the reaction, the net ionic equation is

S2- (aq) + 2 H+ (aq)  → H2S (g)






Randall K
Experienced Chemistry Tutor
Montclair State University
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