Precipitation Reactions
What are Precipitation Reactions ?
** One common
type of reaction that occurs in aqueous solution is the precipitation reaction,
which results in the formation of an insoluble product, or precipitate.
** A precipitate
is an insoluble solid that separates from the solution. Precipitation reactions
usually involve ionic compounds.
** For example,
when an aqueous solution of lead(II) nitrate [Pb(NO3)2]
is added to an aqueous solution of potassium iodide (KI), a yellow precipitate
of lead(II) iodide (PbI2) is formed:
Potassium
nitrate remains in solution. Figure blow shows this reaction in progress.
** The
preceding reaction is an example of a metathesis reaction (also called a
double displacement reaction), a reaction that involves the exchange of parts
between the two compounds.
(In this case,
the cations in the two compounds exchange anions, so Pb2+ ends up
with I2 as PbI2 and K+ ends up with NO3-
as KNO3 .)
** The precipitation reactions are examples of
metathesis reactions.
Solubility
** How can we
predict whether a precipitate will form when a compound is added to a solution
or when two solutions are mixed?
** It depends
on the solubility of the solute, which is defined as the maximum amount of
solute that will dissolve in a given quantity of solvent at a specific
temperature.
** Chemists
refer to substances as soluble, slightly soluble, or insoluble in a qualitative
sense.
** A substance
is said to be soluble if a fair amount of it visibly dissolves when added to
water. If not, the substance is described as slightly soluble or insoluble.
** All ionic
compounds are strong electrolytes, but they are not equally soluble.
** Table (1)
classifies a number of common ionic compounds as soluble or insoluble. Keep in
mind, however, that even insoluble compounds dissolve to a certain extent.
** Figure blow
shows Appearance of several
precipitates. From left to right: CdS, PbS, Ni(OH)2, and Al(OH)3.
Example (1)
Classify the
following ionic compounds as soluble or insoluble:
(a) silver
sulfate (Ag2 SO4 )
(b) calcium
carbonate (CaCO3 )
(c) sodium
phosphate (Na3 PO4 )
Strategy:
Although it is not necessary to memorize the
solubilities of compounds, you should keep in mind the following useful rules:
all ionic compounds containing alkali metal cations; the ammonium ion; and the
nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, we
need to refer to Table (1).
Solution:
(a)
According to Table (1) , Ag2 SO4 is insoluble.
(b)
This is a carbonate and Ca is a Group 2A metal. Therefore, CaCO3 is
insoluble.
(c)
Sodium is an alkali metal (Group 1A) so Na3 PO4 is
soluble.
Molecular Equations, Ionic Equations, and Net Ionic Equations
(a) Molecular Equations
** The equation
describing the precipitation of lead(II) iodide is called a molecular equation
because the formulas of the compounds are written as though all species existed
as molecules or whole units.
** A molecular
equation is useful because it identifies the reagents [that is, lead(II)
nitrate and potassium iodide]. If we wanted to bring about this reaction in the
laboratory, we would use the molecular equation. However, a molecular equation
does not describe in detail what actually is happening in solution.
(b) Ionic Equations
** when ionic
compounds dissolve in water, they break apart into their component cations and anions.
** To be more
realistic, the equations should show the dissociation of dissolved ionic compounds
into ions. Therefore, returning to the reaction between potassium iodide and lead(II)
nitrate, we would write:
** The
preceding equation is an example of an ionic equation, which shows dissolved species
as free ions.
** To see
whether a precipitate might form from this solution, we first combine the cation
and anion from different compounds; that is, PbI2 and KNO3
.
** Referring to
Table (1) , we see that PbI2 is an insoluble compound and KNO3
is soluble. Therefore, the dissolved KNO3 remains in solution as
separate K+ and NO3- ions, which are called
spectator ions, or ions that are not involved in the overall reaction. Because
spectator ions appear on both sides of an equation, they can be eliminated from
the ionic equation.
(c) Net Ionic Equations
** Finally, we
end up with the net ionic equation, which shows only the species that actually take
part in the reaction.
conclusion
** Looking at
another example, we find that when an aqueous solution of barium chloride (BaCl2)
is added to an aqueous solution of sodium sulfate (Na2SO4
), a white precipitate is formed.
** Treating this as a metathesis reaction, the
products are BaSO4 and NaCl. From Table (1) we see that only BaSO4
is insoluble. Therefore, we write the molecular equation as:
The ionic
equation for the reaction is:
Canceling the
spectator ions (Na+ and Cl-) on both sides of the
equation gives us the net ionic equation:
The following
four steps summarize the procedure for writing ionic and net ionic equations:
(1)
Write a balanced molecular equation for the reaction, using the correct
formulas for the reactant and product ionic compounds. Refer to Table (1) to
decide which of the products is insoluble and therefore will appear as a
precipitate.
(2)
Write the ionic equation for the reaction. The compound that does not appear as
the precipitate should be shown as free ions.
(3)
Identify and cancel the spectator ions on both sides of the equation. Write the
net ionic equation for the reaction.
(4)
Check that the charges and number of atoms balance in the net ionic equation. These
steps are applied in Example 2
Example (2)
Predict what
happens when a potassium phosphate (K3 PO4 ) solution is
mixed with a calcium nitrate [Ca(NO3)2 ] solution. Write
a net ionic equation for the reaction?
Strategy:
From the given information, it is useful to first
write the unbalanced equation:
What happens when ionic compounds
dissolve in water? What ions are formed from the dissociation of K3
PO4 and Ca(NO3)2 ? What happens when the
cations encounter the anions in solution?
Solution:
In solution, K3PO4
dissociates into K+ and PO43- ions and Ca(NO3)2
dissociates into Ca2+ and NO3- ions. According
to Table (1) , calcium ions (Ca2+) and phosphate ions (PO43-)
will form an insoluble compound, calcium phosphate [Ca3(PO4
)2], while the other product, KNO3 , is soluble and
remains in solution. Therefore, this is a precipitation reaction.
We follow the
stepwise procedure just outlined.
Step 1: The
balanced molecular equation for this reaction is:
Step 2: To
write the ionic equation, the soluble compounds are shown as dissociated ions:
Step 3:
Canceling the spectator ions (K+ and NO3-) on
each side of the equation, we obtain the net ionic equation:
Step 4:
Note that because we balanced the molecular equation fi rst, the net ionic
equation is balanced as to the number of atoms on each side and the number of
positive (+6) and negative (-6) charges on the left-hand side is the same.
Reference: Chemistry / Raymond Chang ,Williams College /(10th edition).
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