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Precipitation Reaction Equations

Key Concepts

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Molecular Equation

Consider the reaction between aqueous solutions of sodium chloride, NaCl(aq), and silver nitrate, AgNO3(aq).

We can work out the possible products of the reaction by looking at the species present in each solution:

The possible products are salts made up of a cation (positively charged ion) and an anion (negatively charged ion):

  aqueous sodium chloride
(NaCl(aq))
  sodium ions
(Na+(aq)
chloride ions
(Cl-(aq))
aqueous silver nitrate
(AgNO3(aq))
silver ions
(Ag+(aq))
no salt
(both cations)
silver chloride
(AgCl)
nitrate ions
(NO3-(aq))
sodium nitrate
(NaNO3)
no salt
(both anions)

The possible products of the reaction are sodium nitrate, NaNO3, and silver chloride, AgCl.

From the solubility rules we find that:

We can write a word equation for this precipitation reaction:

  reactants (soluble) soluble
product
+ precipitate
word equation: sodium chloride + silver nitrate sodium nitrate + silver chloride

We can treat each reactant and each product as if it were a molecule and substitute the chemical formula for each "molecule" into the word equation.

Remember, use (aq) for "molecules" that are soluble and (s) for the precipitate (the insoluble salt)

  reactants (soluble) soluble
product
+ precipitate
word equation: sodium chloride + silver nitrate sodium nitrate + silver chloride
chemical equation: NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)

Check that the chemical equation is balanced!

chemical equation: NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)  
number Na "atoms" 1     = 1     balanced
number Cl "atoms" 1     =     1 balanced
number Ag "atoms"     1 =     1 balanced
number N atoms     1 = 1     balanced
number O atoms     3 = 3     balanced

For the precipitation reaction occuring between aqueous solutions of sodium chloride and silver nitrate the balanced molecular chemical equation is:

NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)

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Ionic Equation

An ionic equation is a balanced chemical equation that does NOT assume all reactants and products exist in solution as "molecules".

In the example above we found that mixing aqueous solutions of sodium chloride, NaCl(aq), and silver nitrate, AgNO3(aq), produced a precipitate of silver chloride, AgCl(s), and an aqueous solution of sodium nitrate, NaNO3(aq), for which we wrote the following balanced molecular chemical equation:

  reactants (soluble) soluble
product
+ precipitate
molecular equation: NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)

If a salt is soluble in water, it dissolves in water by breaking up into its ions which are completely surrounded by water, so the species in water are not "salt molecules" but cations and anions surrounded by water molecules:

The precipitate DOES exist as a "molecular" species, the water does not break the crystal lattice apart, so the silver chloride, AgCl(s), exists as a solid NOT as ions surrounded by water molecules.

So we can re-write our balanced molecular equation to show all the ionic species in the aqueous solution (that is, we write the chemical equation as an ionic equation):

  reactants (soluble) soluble
product
+ precipitate
molecular equation: NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)
ionic equation: Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgCl(s)

Check that the ionic equation is balanced!

ionic equation: Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgCl(s)  
number Na "atoms" 1             = 1         balanced
number Cl "atoms"     1         =         1 balanced
number Ag "atoms"         1     =         1 balanced
number N atoms             1 =     1     balanced
number O atoms             3 =     3     balanced

For the precipitation reaction that occurs when aqueous solutions of sodium chloride and silver nitrate are mixed, the balanced ionic equation for the reaction is:

Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

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Net Ionic Equation

A net ionic equation shows only the species that react to form the precipitate, and, the precipitate that is produced.

We need to identify the ions that are reacting and the ions that do not react to produce the precipitate (the spectator ions).

We saw above that we can write an ionic equation to represent the reaction between aqueous solutions of sodium chloride, NaCl(aq), and silver nitrate, AgNO3(aq), to produce the precipitate (insoluble salt) silver chloride, AgCl(s):

  soluble reactant species soluble product
species
+ precipitate
(insoluble salt)
ionic equation: Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgCl(s)  

Let's highlight the aqueous species that occur as both reactants and products, that is, the species that DO NOT take part in the precipitation reaction (the spectator ions) in red:

  soluble reactant species soluble product
species
+ precipitate
(insoluble salt)
ionic equation: Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgCl(s)  

Since these spectator ions (Na+(aq) and NO3-(aq)) are present on both the reactant side and the product side of the equation and they do not take part in the reaction and we can effectively ignore them, so let's remove them from the ionic equation to produce a net ionic equation:

  soluble reactant species soluble product
species
+ precipitate
(insoluble salt)
ionic equation: Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgCl(s)  
net ionic equation:     Cl-(aq) + Ag+(aq)             AgCl(s)  

We end up with a net ionic equation to represent the reaction between aqueous solutions of sodium chloride, NaCl(aq), and silver nitrate, AgNO3(aq), to produce the precipitate (insoluble salt) silver chloride, AgCl(s):

Ag+(aq) + Cl-(aq) → AgCl(s)

And this is a very interesting equation!
This equation tells us that if we take ANY aqueous solution containing silver ions, Ag+(aq), and mix it with ANY aqueous solution containing chloride ions, Cl-(aq), a precipitate (insoluble salt) of silver chloride, AgCl(s), will be produced!

That is, if we mix a source of silver ions such as an aqueous solution of silver nitrate4, AgNO3(aq) (Ag+(aq) + NO3-(aq)) with a source of chloride ions, then solid silver chloride, AgCl(s), will precipitate:

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Worked Example

Question:
An aqueous solution of barium nitrate, Ba(NO3)2(aq), is mixed with dilute sulfuric acid, H2SO4(aq).
A white precipitate forms.

  1. Write a balanced molecular equation for this precipitation reaction.
  2. Write a balanced ionic equation for this precipitation reaction.
  3. Write a balanced net ionic equation for this reaction.

Solution:

  1. Write a balanced molecular equation for this precipitation reaction.

    Step 1: Use the solubility rules to predict the name and formula of the precipitate:

      aqueous barium nitrate
    (Ba(NO3)2(aq))
      barium ions
    (Ba2+(aq))
    nitrate ions
    (NO3-(aq))
    aqueous sulfuric acid
    (H2SO4(aq))
    hydrogen ions
    (H+(aq))
    no salt
    (both cations)
    hydrogen nitrate
    (nitric acid, HNO3(aq))
    sulfate ions
    (SO42-(aq))
    barium sulfate
    (BaSO4(s))
    no salt
    (both anions)

    Solubility rules: all nitrates are soluble so hydrogen nitrate (nitric acid, HNO3(aq)) is soluble.
    All sulfates are soluble EXCEPT those of silver, lead, mercury(I), barium, strontium and calcuim. So barium sulfate is insoluble (BaSO4(s)).

    Step 2: Write the word equation:

      reactants (soluble) soluble
    product
    + precipitate
    word equation: barium nitrate + sulfuric acid nitric acid + barium sulfate

    Step 3: Substitute each word for the corresponding chemical formula including each state:

      reactants (soluble) soluble
    product
    + precipitate
    word equation: barium nitrate + sulfuric acid nitric acid + barium sulfate
    chemical formula: Ba(NO3)2(aq) + H2SO4(aq) HNO3(aq) + BaSO4(s)

    Step 4: Balance the molecular equation:

      reactants (soluble) soluble
    product
    + precipitate
    unbalanced molecular equation: Ba(NO3)2(aq) + H2SO4(aq) HNO3(aq) + BaSO4(s)  
    number Ba "atoms": 1     =     1 balanced
    number N atoms: 2     1     unbalanced
    need 2 molecules HNO3: Ba(NO3)2(aq) + H2SO4(aq) 2HNO3(aq) + BaSO4(s)  
    number N atoms: 2     = 2     balanced
    number O atoms: 6 + 4 = 6 + 4 balanced
    number H atoms:     2 = 2     balanced
    number S atoms:     1 =     1 balanced
    check number Ba "atoms":
    (in new equation)
    1     =     1 balanced

    Step 5. Write balanced molecular equation:

    Ba(NO3)2(aq) + H2SO4(aq) → 2HNO3(aq) + BaSO4(s)

  2. Write a balanced ionic equation for this precipitation reaction.

    Step 1: Write the balanced molecular equation:

    Ba(NO3)2(aq) + H2SO4(aq) → 2HNO3(aq) + BaSO4(s)

    Step 2: Split any soluble ionic species up into its ions:

    Ba(NO3)2(aq) → Ba2+(aq) + 2NO3-(aq)
    H2SO4(aq) → 2H+(aq) + SO42-(aq)
    2HNO3(aq) → 2H+(aq) + 2NO3-(aq)

    NOTE: an insoluble salt is NOT spilt up into its ions!
    The formula for the precipitate is still BaSO4(s)

    Step 3: Substitute each soluble "molecule" in the molecular equation for the formula of its ions:

      soluble reactants soluble
    product
    + precipitate  
    balanced
    molecular
    equation:
    Ba(NO3)2(aq) + H2SO4(aq) 2HNO3(aq) + BaSO4(s)  
    ionic equation: Ba2+(aq) + 2NO3-(aq) + 2H+ + SO42-(aq) 2H+(aq) + 2NO3-(aq) + BaSO4(s)  

    Step 4: Balance the ionic equation:

    ionic equation: Ba2+(aq) + 2NO3-(aq) + 2H+ + SO42-(aq) 2H+(aq) + 2NO3-(aq) + BaSO4(s)  
    number "Ba" atoms: 1             =         1 balanced
    number N atoms:     2         =     2     balanced
    number O atoms:     6   +   4 =     6 + 4 balanced
    number H "atoms":         2     = 2         balanced
    number S atoms:             1 =         1 balanced

    Step 5: Write the balanced ionic equation for the precipitation reaction:

    Ba2+(aq) + 2NO3-(aq) + 2H+(aq) + SO42-(aq) → 2H+(aq) + 2NO3-(aq) + BaSO4(s)

  3. Write a balanced net ionic equation for this reaction.

    Step 1: Write the balanced ionic equation:

    Ba2+(aq) + 2NO3-(aq) + 2H+(aq) + SO42-(aq) → 2H+(aq) + 2NO3-(aq) + BaSO4(s)

    Step 2: Identify the spectator ions, those ions that appear on both left hand side and the right hand side of the equation as ions and not as part of the precipitate:

    Ba2+(aq) + 2NO3-(aq) + 2H+(aq) + SO42-(aq)2H+(aq) + 2NO3-(aq) + BaSO4(s)

    Spectator ions are NO3-(aq) and H+(aq)

    Step 3: Remove the spectator ions from the ionic equation to make the net ionic equation:

    ionic equation: Ba2+(aq) + 2NO3-(aq) + 2H+(aq) + SO42-(aq)2H+(aq) + 2NO3-(aq) + BaSO4(s)

    net ionic equation: Ba2+(aq) + SO42-(aq) → BaSO4(s)

    Step 4: Write the net ionic equation for the precipitation reaction:

    Ba2+(aq) + SO42-(aq) → BaSO4(s)

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1. "insoluble" refers to a solid with an extremely low solubility.

2. You can also use Ksp, solubility products, to predict precipitation.

3. Precipitation reactions can be thought of as equilibrium systems:
example: M+(aq) + X-(aq) MX(s)
but the equilibrium position lies very far to the right, that is, the formation of the precipitate is significantly favoured so we can approximate the description of the reaction to:
example: M+(aq) + X-(aq) → MX(s)

4. Are you wondering why silver nitrate is usually used as the source of silver ions in precipitation reactions?
If you check the solubility rules you will find that, generally speaking, silver salts are insoluble, or only slightly soluble, in water EXCEPT silver nitrate!