Neutralisation Reactions Chemistry Tutorial

Key Concepts

• Neutralisation (or neutralization) was defined by Arrhenius as a reaction in which hydrogen ions (H⁺) combine with hydroxide ions (OH^-) to form water (H₂O)¹:

  H⁺ + OH^- → H₂O

• The neutralisation reaction (or neutralization reaction) takes place when an Arrhenius acid (HX) reacts with an Arrhenius base (MOH).

  HX_(aq) + MOH_(aq) → H₂O + MX_(aq)

• When an Arrhenius acid is added to an Arrhenius base, we say that the acid will neutralise the base.
  When an Arrhenius base is added to an Arrhenius acid, we say that the base will neutralise the acid.
• A salt (MX) is a compound whose ions are left after an acid has neutralised a base:

  HX_(aq) + MOH_(aq) → H₂O + MX_(aq)

  
  The salt formed is an ionic compound and is named by placing the name of the metal first, followed by the the name of the non-metal with an appropriate ending:

  acid	anion	salt example
  hydrochloric acid (HCl)	chloride (Cl-)	sodium chloride (NaCl)
  sulfuric acid (H2SO4)	sulfate (SO42-)	sodium sulfate (Na2SO4)
  nitric acid (HNO3)	nitrate (NO3-)	sodium nitrate (NaNO3)
  phosphoric acid (H3PO4)	phosphate (PO43-)	sodium phosphate (Na3PO4)
  acetic acid (ethanoic acid) (CH3COO-)	acetate (ethanoate) (CH3COO-)	sodium acetate (sodium ethanoate) (CH3COONa)
  oxalic acid (ethanedioc acid) (HOOCCOOH)	oxalate (ethanedioate) (-OOCCOO-)	sodium oxalate (sodium ethanedioate) (NaOOCCOONa)

• A neutralisation reaction can be summarized as:

  general word equation:	Arrhenius acid	+	Arrhenius base	→	water	+	salt
  general molecular chemical equation: (all species shown as molecules)	HX(aq)	+	MOH(aq)	→	H2O(l)	+	MX(aq)
  general ionic chemical equation: (showing which species are ions and which are molecules)	H+(aq) + X-(aq)	+	M+(aq) + OH-(aq)	→	H2O(l)	+	M+(aq) + X-(aq)
  general net ionic chemical equation2: (showing only the species that react)	H+(aq)	+	OH-(aq)	→	H2O(l)

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Example: Writing a Molecular Equation for a Neutralisation Reaction

In a molecular equation, all the species are represented as molecules, even the compounds that exist only as ions in the solution.

Example: Write a balanced molecular equation for the reaction in which sodium hydroxide is neutralised by hydrochloric acid.

1. Determine which species is the Arrhenius acid and which is the Arrhenius base:

   An Arrhenius acid contains hydrogen: hydrochloric acid
   An Arrhenius base contains hydroxide ions: sodium hydroxide

2. Write the word equation for the neutralisation reaction:

   general word equation:	Arrhenius acid	+	Arrhenius base	→	water	+	salt
   word equation for this reaction:	hydrochloric acid	+	sodium hydroxide	→	water	+	sodium chloride

3. Write the chemical formula for each species and include its state:

   name	formula
   hydrochloric acid	HCl(aq)
   sodium hydroxide	NaOH(aq)
   water	H2O(l)
   sodium chloride	NaCl(aq)
   (The salts of Group 1 metals are soluble in water.)

4. Write the skeletal chemical equation by substituting the names of each species with its chemical formula:

   word equation:	hydrochloric acid	+	sodium hydroxide	→	water	+	sodium hydroxide
   skeletal chemical equation:	HCl(aq)	+	NaOH(aq)	→	H2O(l)	+	NaCl(aq)

5. Balance the chemical equation:

   word equation:	hydrochloric acid	+	sodium hydroxide	→	water	+	sodium hydroxide
   skeletal chemical equation:	HCl(aq)	+	NaOH(aq)	→	H2O(l)	+	NaCl(aq)
   number of H atoms:	1	+	1	=	2
   number of O atoms:			1	=	1
   number of chlorine atoms (Cl):	1			=			1
   number of sodium atoms (Na):			1	=			1
   balanced molecular equation:	HCl(aq)	+	NaOH(aq)	→	H2O(l)	+	NaCl(aq)

Example: Writing an Ionic Equation for a Neutralisation Reaction

In an ionic equation, species that exist as ions in solution are shown as ions, and species that exist as molecules in the solution are shown as molecules.

Example: Write a balanced ionic equation for the reaction in which sulfuric acid is neutralised by potassium hydroxide.

1. Determine which species is the Arrhenius acid and which is the Arrhenius base:

   An Arrhenius acid contains hydrogen: sulfuric acid
   An Arrhenius base contains hydroxide ions: potassium hydroxide

2. Write the word equation for the neutralisation reaction:

   general word equation:	Arrhenius acid	+	Arrhenius base	→	water	+	salt
   word equation for this reaction:	sulfuric acid	+	potassium hydroxide	→	water	+	potassium sulfate

3. Write the chemical formula for each species and include its state:

   name	formula
   sulfuric acid	H2SO4(aq)
   potassium hydroxide	KOH(aq)
   water	H2O(l)
   potassium sulfate	K2SO4(aq)
   (The salts of Group 1 metals are soluble in water.)

4. Write the skeletal chemical molecular equation by substituting the names of each species with its chemical formula:

   word equation:	sulfuric acid	+	potassium hydroxide	→	water	+	potassium sulfate
   skeletal molecular equation:	H2SO4(aq)	+	KOH(aq)	→	H2O(l)	+	K2SO4(aq)

5. Determine which of these "molecular" species exists as ions in the aqueous solution:

   Sulfuric acid (H₂SO_4(aq)) dissociates in water ³ producing 2H⁺_(aq) + SO₄^2-
   Potassium hydroxide (KOH_(aq)) dissociates in water producing K⁺_(aq) + OH^-_(aq)
   Water: H₂O_(l) (water dissociates only very, very slightly, so we can treat it as a molecule)
   Potassium sulfate (K₂SO_4(aq)) is soluble in water because the salts of Group 1 metals are soluble.
   Potassium sulfate (K₂SO_4(aq)) dissociates in water producing 2K⁺_(aq) + SO₄^2-_(aq)

6. Write the skeletal ionic equation by substituting the molecular formula of each soluble ionic species with the formula of its ions:

   word equation:	sulfuric acid	+	potassium hydroxide	→	water	+	potassium sulfate
   skeletal molecular equation:	H2SO4(aq)	+	KOH(aq)	→	H2O(l)	+	K2SO4(aq)
   skeletal ionic equation:	2H+(aq) + SO42-(aq)	+	K+ + OH-(aq)	→	H2O(l)	+	2K+(aq) + SO42-(aq)

7. Balance the chemical equation:

   word equation:	sulfuric acid	+	potassium hydroxide	→	water	+	potassium sulfate
   skeletal ionic equation:	2H+(aq) + SO42-(aq)	+	K+ + OH-(aq)	→	H2O(l)	+	2K+(aq) + SO42-(aq)
   number of H atoms:	2	+	1	≠	2
   balance the H atoms	2H+(aq) + SO42-(aq)	+	K+ + 2OH-(aq)	→	2H2O(l)	+	2K+(aq) + SO42-(aq)
   check the number of H atoms:	2	+	2	=	4
   number of O atoms:	4	+	2	=	2	+	4
   number of sulfur atoms (S):	1			=			1
   number of potassium atoms (K):			1	≠			2
   balance the K atoms	2H+(aq) + SO42-(aq)	+	2K+ + 2OH-(aq)	→	2H2O(l)	+	2K+(aq) + SO42-(aq)
   check number of potassium atoms (K):			2	=			2
   balanced ionic equation:	2H+(aq) + SO42-(aq)	+	2K+ + 2OH-(aq)	→	2H2O(l)	+	2K+(aq) + SO42-(aq)

Example: Writing a Net Ionic Equation for a Neutralisation Reaction

In a net ionic equation, species that are ions in solution and that do not take part in the reaction are not shown.
Ions that do not take part in the reaction are referred to as spectator ions.

Example: Wtite a balanced net ionic equation for the reaction in which sulfuric acid is neutralised by potassium hydroxide.

1. Write the balanced ionic equation for the reaction (as shown above):

   balanced ionic equation:

   2H+(aq) + SO42-(aq)

   +

   2K+ + 2OH-(aq)

   →

   2H2O(l)

   +

   2K+(aq) + SO42-(aq)

2. Determine which ionic species do not take part in the reaction:

   K⁺_(aq) does not take part in the reaction because it is seen to be both a reactant and a product.
   SO₄^2-_(aq) does not take part in the reaction because it is seen to be both a reactant and a product.

3. Remove the non-participating (spectator) ions from the ionic equation and balance the net ionic equation:

   balanced ionic equation:	2H+(aq) + SO42-(aq)	+	2K+ + 2OH-(aq)	→	2H2O(l)	+	2K+(aq) + SO42-(aq)
   remove non-participating ions:	2H+(aq) + SO42-(aq)	+	2K+ + 2OH-(aq)	→	2H2O(l)	+	2K+(aq) + SO42-(aq)
   rewrite the ionic equation:	2H+(aq)	+	2OH-(aq)	→	2H2O(l)
   balance the equation by dividing throughout by 2:	2/2H+(aq)	+	2/2OH-(aq)	→	2/2H2O(l)
   balanced net ionic equation:	H+(aq)	+	OH-(aq)	→	H2O(l)