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Precipitation Reactions
Graphic: Wikimedia Commons User Tubifex
Double Replacement Reactions
The ions of two compounds exchange places in an
aqueous solution to form two new compounds.
AX + BY  AY + BX
One of the compounds formed is usually a
precipitate (an insoluble solid), an insoluble gas
that bubbles out of solution, or a molecular
compound, usually water.
Precipitates
•
•
•
Soluble – solid dissolves in solution; (aq) is used
in reaction.
Insoluble – solid does not dissolve in solution;
(s) is used in reaction.
Insoluble and slightly soluble are often used
interchangeably.
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Solubility Rules – AP Chemistry
All sodium, potassium, ammonium, and nitrate
salts are soluble in water.
Solubility Rules – Mostly Soluble
Ion
NO3-
Solubility
Soluble
Exceptions
None
ClO4-
Soluble
None
Na+
Soluble
None
K+
Soluble
None
NH4+
Soluble
None
Cl-, I-
Soluble
Pb2+, Ag+, Hg22+
SO42-
Soluble
Ca2+, Ba2+, Sr2+, Pb2+, Ag+, Hg2+
Solubility Rules – Mostly Insoluble
Ion
CO32-
Solubility
Insoluble
Exceptions
Group IA and NH4+
PO43-
Insoluble
Group IA and NH4+
OH-
Insoluble
Group IA and Ca2+, Ba2+, Sr2+
S2-
Insoluble
Groups IA, IIA, and NH4+
Double replacement forming a precipitate…
Lead(II) nitrate + potassium iodide  lead(II) iodide + potassium nitrate
Double replacement (ionic) equation
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)
Complete ionic equation shows compounds as aqueous ions
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  PbI2(s) + 2K+(aq) + 2 NO3-(aq)
Net ionic equation eliminates the spectator ions
Pb2+(aq) + 2 I-(aq)  PbI2(s)
The Reaction of K2CrO4(aq) and
Ba(NO3)2(aq)
Ba(NO3)2 (aq) + K2CrO4(aq) → BaCrO4(s) + KNO3 (aq)
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Formula Equation (Molecular Equation)
•
•
•
Gives the overall reaction stoichiometry but not
necessarily the actual forms of the reactants and
products in solution.
Reactants and products generally shown as
compounds.
Use solubility rules to determine which
compounds are aqueous and which compounds
are solids.
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
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Complete Ionic Equation
•
Represents as ions all reactants and products that are strong
electrolytes.
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq) 
AgCl(s) + Na+(aq) + NO3(aq)
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Net Ionic Equation
•
Includes only those solution components
undergoing a change.
 Show only components that actually react.
Ag+(aq) + Cl(aq)  AgCl(s)
•
Spectator ions are not included (ions that do not
participate directly in the reaction).
 Na+ and NO3 are spectator ions.
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11
Solving Stoichiometry Problems for Reactions in
Solution
Identify the species present in the combined solution, and
determine what reaction if any occurs.
Write balanced net ionic equation for the reaction.
Calculate moles of reactants.
Determine which reactant is limiting.
Calculate moles of product(s), as required.
Convert to grams or other units, as required.
Example:
10.0 mL of 0.25M Barium chloride is reacted with 10.0mL of
0.35M sodium sulfate.
Write the balanced reaction
How many grams of ppt will form?
Example:
50.0mL of 0.10M sodium phosphate is mixed with 50.0mL of
0.10M copper II chloride.
Write the balanced reaction
How many grams of ppt will form?
Example
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
 What precipitate will form?
 What mass of precipitate will form?
1.1 g Pb3(PO4)2
20
Let’s Think About It
•
Where are we going?

•
To find the mass of solid Pb3(PO4)2 formed.
How do we get there?






What are the ions present in the combined solution?
What is the balanced net ionic equation for the reaction?
What are the moles of reactants present in the solution?
Which reactant is limiting?
What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?
21
Example 5
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
 What is the concentration of nitrate ions left
in solution after the reaction is complete?
0.27 M
22
Let’s Think About It
•
•
Where are we going?
 To find the concentration of nitrate ions left in
solution after the reaction is complete.
How do we get there?
 What are the moles of nitrate ions present in the
combined solution?
 What is the total volume of the combined solution?
23
Example 6
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
 What is the concentration of phosphate ions
left in solution after the reaction is complete?
0.011 M
24
Let’s Think About It
•
•
Where are we going?
 To find the concentration of phosphate ions left in
solution after the reaction is complete.
How do we get there?
 What are the moles of phosphate ions present in
the solution at the start of the reaction?
 How many moles of phosphate ions were used up
in the reaction to make the solid Pb3(PO4)2?
 How many moles of phosphate ions are left over
after the reaction is complete?
 What is the total volume of the combined solution?
25
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change). What is the
concentration of nitrate ions left in solution after
the reaction is complete?