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Neutralisation Reactions Tutorial

Key Concepts

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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 basewater+salt
    word equation for this reaction: hydrochloric acid+sodium hydroxidewater+sodium chloride

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

    nameformula
    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 hydroxidewater+sodium hydroxide
    skeletal chemical equation: HCl(aq)+NaOH(aq)H2O(l)+NaCl(aq)

  5. Balance the chemical equation:

    word equation: hydrochloric acid+sodium hydroxidewater+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)

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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 basewater+salt
    word equation for this reaction: sulfuric acid+potassium hydroxidewater+potassium sulfate

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

    nameformula
    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 hydroxidewater+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 (H2SO4(aq)) dissociates in water 3 producing 2H+(aq) + SO42-
    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 (K2SO4(aq)) is soluble in water because the salts of Group 1 metals are soluble.
    Potassium sulfate (K2SO4(aq)) dissociates in water producing 2K+(aq) + SO42-(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 hydroxidewater+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 hydroxidewater+potassium sulfate
    skeletal ionic equation:2H+(aq) + SO42-(aq)+K+ + OH-(aq)H2O(l)+2K+(aq) + SO42-(aq)
    number of H atoms: 2+         12
    balance the H atoms2H+(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): 12
    balance the K atoms2H+(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)

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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.
    SO42-(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)

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1. Since we are using the Arrhenius definition of an acid it is quite acceptable to use H+ (or H+(aq)) to represent the hydrogen ion.

2. These equations assume that the salt formed is soluble in water.
It is possible that the salt formed is not soluble in water, in which case the equation would be written as for a precipitation reaction.

3. The assumption here is that the neutralisation reaction will go to completion, driving the sulfuric acid to complete dissociation.