4.5: Writing Net Ionic Equations

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Learning Objectives

Identify what species are really present in an aqueous solution.
Identify possible products: insoluble ionic compound, water, weak electrolyte
Write net ionic equations for reactions that occur in aqueous solution.

WRITING NET IONIC EQUATIONS FOR CHEM 101A

Chemical reactions that occur in solution are most concisely described by writing net ionic equations. A net ionic equation is the most accurate representation of the actual chemical process that occurs. Writing these equations requires a familiarity with solubility rules, acid-base reactivity, weak electrolytes and special reactions of carbonates and bicarbonates. The following is the strategy we suggest following for writing net ionic equations in Chem 101A.

Step 1: Identify the species that are actually present, accounting for the dissociation of any strong electrolytes. (Insoluble ionic compounds do not ionize, but you must consider the possibility that the ions in an insoluble compound might still be involved in the reaction.)

Step 2: Identify the products that will be formed when the reactants are combined. The most common products are insoluble ionic compounds and water. See the "reactivity of inorganic compounds" handout for more information.

Step 3: Write the balanced equation for the reaction you identified in step 2, being certain to show the major species in your equation. This is the net ionic equation for the reaction.

Example \(\PageIndex{1}\): Writing Net Ionic Equations

Write a net ionic equation to describe the reaction that occurs when 0.100 M K₃PO₄ solution is mixed with 0.100 M Ca(NO₃)₂ solution

Step 1: The species that are present are:

From the

K₃PO₄

From the

Ca(NO₃)₂

K⁺

Ca²⁺

PO₄³^-

NO₃^-

Step 2: There are two possible combinations of ions here: K⁺ + NO₃^- (forming KNO₃) and Ca²⁺ + PO₄³^- (forming Ca₃(PO₄)₂). We know from the general solubility rules that Ca₃(PO₄)₂ is an insoluble compound, so it will be formed. KNO₃ is water-soluble, so it will not form.

Step 3: The reaction is the combination of calcium and phosphate ions to form calcium phosphate. The balanced equation for this reaction is:

\[\ce{3Ca^2+ (aq) + 2PO4^{3-}(aq) \rightarrow Ca3(PO4)2(s)}\]

Example \(\PageIndex{2}\): Writing Net Ionic Equations

Write a net ionic equation to describe the reaction that occurs when 0.1 M HC₂H₃O₂ solution is mixed with 0.1 M KOH solution

Step 1: The species that are present are:

From the HC₂H₃O₂	From the KOH
H⁺	K⁺
C₂H₃O₂^-	OH^-

Acetic acid, HC₂H₃O₂, is a weak acid. Think of the acid molecules as “potential” H⁺ and C₂H₃O₂^– ions, however, these “potential” ions are held together by a covalent bond. A small percentage of the acid molecules do actually ionize (break apart into ions) when they dissolve in water, but most of the weak acid molecules do not ionize.

Step 2: Reaction of an acid (source of H⁺) and a base (source of OH^-) will form water. The H⁺ from the HC₂H₃O₂ can combine with the OH^– to form H₂O. Note that KC₂H₃O₂ is a water-soluble compound, so it will not form.

Step 3: In order to form water as a product, the covalent bond between the H⁺ and the C₂H₃O₂^– ions must break. The H⁺ and OH^– will form water. The acetate ion is released when the covalent bond breaks. Remember to show the major species that exist in solution when you write your equation. Most of the acid molecules are not ionized, so you must write out the complete formula of the acid in your equation. The balanced equation for this reaction is:

\[\ce{HC2H3O2(aq) + OH^- (aq) \rightarrow H2O (l) + C2H3O2^- (aq)}\]

Example \(\PageIndex{3}\): Writing Net Ionic Equations

Write a net ionic equation to describe the reaction that occurs when solid Mg(OH)₂ and excess 0.1 M HCl solution

Step 1: The species that are present are:

From the Mg(OH)₂ solid	From the HCl
Mg²⁺	H⁺
OH^-	Cl^-

Notice that the magnesium hydroxide is a solid; it is not water soluble. The magnesium ions and the hydroxide ions will remain held together by ionic bonds even if they are in the presence of polar water molecules. However, these individual ions must be considered as possible reactants. Think of the solid ionic compound as a possible source of Mg²⁺ and OH^– ions. If a chemical reaction is possible, the ionic bonds between Mg²⁺ and OH^– will break.

Step 2: Reaction of an acid (source of H⁺) and a base (source of OH^-) will form water. The H⁺ from the HCl can combine with the OH^– from the solid Mg(OH)₂ to form H₂O. Note that MgCl₂ is a water-soluble compound, so it will not form.

Step 3: In order to form water as a product, the ionic bond between the magnesium and hydroxide ions must break. The OH^– and H⁺ will form water. The magnesium ion is released into solution when the ionic bond breaks. Remember to show the major species that exist in solution when you write your equation. The magnesium hydroxide is a solid reactant, so you must write out the complete formula in your equation. The balanced equation for this reaction is:

\[\ce{Mg(OH)2(s) + 2H^+ (aq) \rightarrow 2H2O(l) + Mg^2+ (aq)}\]

Example \(\PageIndex{4}\): Writing Net Ionic Equations

Write a net ionic equation to describe the reaction that occurs when 0.1 M KHCO₃ solution is mixed with excess 0.1 M HNO₃ solution

Step 1: The species that are present are:

From the KHCO₃	From the HNO₃
K⁺	H⁺
HCO₃^-	NO₃^-

Step 2: From the “reactivity of inorganic compounds” handout, we know that when carbonate or bicarbonate ions react with acids, carbon dioxide and water are the normal products. Be sure to refer to the handout for details of this process.

Step 3: The reaction is the combination of bicarbonate ions and hydrogen ions that will first form carbonic acid (H₂CO₃). However, carbonic acid can only exist at very low concentrations. Under normal circumstances, carbonic acid decomposes into CO₂ and H₂O. The balanced equation for this reaction is:

\[\ce{HCO3^- (aq) + H^+ (aq) \rightarrow H2O(l) + CO2(g)}\]

Supplemental Exercises: Writing Net Ionic Equations

For each of the following, write the net ionic equation for the reaction that will occur when the two substances are mixed. If no reaction occurs, write “no reaction.” Note: the reactions are grouped according to the difficulty that typical students have with them—our groupings may not match your own experience and ability. (Answers are available below.)

Easy reactions:

1) 0.1 M AgNO₃ and 0.1 M KBr

2) 0.1 M CaCl₂ and 0.1 M NaNO₃

3) 0.1 M Fe(NO₃)₃ and 0.1 M Na₂CO₃

4) 0.1 M KOH and 0.1 M CoBr₂

5) 0.1 M HNO₃ and 0.1 M Ba(OH)₂

6) 0.1 M Pb(NO₃)₂ and 0.1 M MgSO₄

7) 0.1 M Na₂S and 0.1 M MnI₂

8) 0.1 M K₃PO₄ and 0.1 M CuCl₂

9) 0.1 M HCl and 0.1 M NaC₂H₃O₂

10) 0.1 M NiSO₄ and 0.1 M FeCl₃

Harder reactions:

1) 0.1 M HC₂H₃O₂ and 0.1 M KOH

2) 0.1 M NH₃ and 0.1 M HBr

3) 1 M HCl and solid Mn(OH)₂

4) excess 1 M HNO₃ and solid AlPO₄

5) 0.1 M AgNO₃ and 0.1 M NaOH

6) 0.1 M HClO and 0.1 M Ba(OH)₂ (no precipitate forms)

Still harder reactions:

1) 0.1 M Na₂HPO₄ and 0.1 M HI (equal volumes)

2) 0.1 M NH₃ and 0.1 M Fe(NO₃)₂

3) 0.1 M NaHCO₃ and 0.1 M HCl

4) 0.1 M K₂CO₃ and 0.1 M HNO₃ (equal volumes)

5) 0.1 M H₃PO₄ and 0.1 M NH₃ (equal volumes)

Very tricky reactions:

1) 0.1 M AgNO₃ and 0.1 M NH₃

2) solid BaCO₃ and excess 2 M HC₂H₃O₂

3) solid Cu(OH)₂ and 1 M H₂SO₄ (equal numbers of moles)

4) solid Ag₂O and excess 2 M HCl

5) 0.1 M H₃PO₄ and excess 1 M KOH

AnswerS TO NET IONIC EQUATIONS PRACTICE PROBLEMS

Easy reactions:

1) Ag⁺(aq) + Br^–(aq) --> AgBr(s)

2) no reaction

3) 2 Fe³⁺(aq) + 3 CO₃²^–(aq) --> Fe₂(CO₃)₃(s)

4) 2 OH^–(aq) + Co²⁺(aq) --> Co(OH)₂(s)

5) H⁺(aq) + OH^–(aq) --> H₂O(l)

6) Pb²⁺(aq) + SO₄²^–(aq) --> PbSO₄(s)

7) S^2–(aq) + Mn²⁺(aq) --> MnS(s)

8) 2 PO₄³^–(aq) + 3 Cu²⁺(aq) --> Cu₃(PO₄)₂(s)

9) H⁺(aq) + C₂H₃O₂^–(aq) --> HC₂H₃O₂(aq)

10) no reaction

Harder reactions:

1) HC₂H₃O₂(aq) + OH^–(aq) --> C₂H₃O₂^–(aq) + H₂O(l)

2) NH₃(aq) + H⁺(aq) --> NH₄⁺(aq)

3) 2 H⁺(aq) + Mn(OH)₂(s) --> Mn²⁺(aq) + 2 H₂O(l)

4) 3 H⁺(aq) + AlPO₄(s) --> Al³⁺(aq) + H₃PO₄(aq)

5) 2 Ag⁺(aq) + 2 OH^–(aq) --> Ag₂O(s) + H₂O(l)

6) HClO(aq) + OH^–(aq) --> ClO^–(aq) + H₂O(l)

Still harder reactions:

1) HPO₄²^–(aq) + H⁺(aq) --> H₂PO₄^–(aq)

2) Fe²⁺(aq) + 2 NH₃(aq) + 2 H₂O(l) --> Fe(OH)₂(s) + 2 NH₄⁺(aq)

3) HCO₃^–(aq) + H⁺(aq) --> H₂O(l) + CO₂(g)

4) CO₃²^–(aq) + H⁺(aq) --> HCO₃^–(aq)

5) H₃PO₄(aq) + NH₃(aq) --> H₂PO₄^–(aq) + NH₄⁺(aq)

Very tricky reactions:

1) 2 Ag⁺(aq) + 2 NH₃(aq) + H₂O(l) --> Ag₂O(s) + 2 NH₄⁺(aq)

2) BaCO₃(s) + 2 HC₂H₃O₂(aq) --> Ba²⁺(aq) + 2 C₂H₃O₂^–(aq) + H₂O(l) + CO₂(g)

3) Cu(OH)₂(s) + H⁺(aq) + HSO₄^–(aq) --> Cu²⁺(aq) + 2 H₂O(l) + SO₄²^–(aq)

4) Ag₂O(s) + 2 H⁺(aq) + 2 Cl^–(aq) --> 2 AgCl(s) + H₂O(l)

5) Three reactions will occur, one after the other:

H₃PO₄(aq) + OH^–(aq) --> H₂PO₄^–(aq) + H₂O(l)

H₂PO₄^–(aq) + OH^–(aq) --> HPO₄²^–(aq) + H₂O(l)

HPO₄²^–(aq) + OH^–(aq) --> PO₄³^–(aq) + H₂O(l)