# 4.5: Writing Net Ionic Equations


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 actually 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 actually 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 actually 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 actually 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) S2–(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)

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