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Chem 142: Lecture 9
Precipitation Reactions (4.4, 4.5)
Ions in Solution (4.6)
Selective Precipitation and
Stoichiometry (4.7, 4.8)
Reaction Classes
• Precipitation: synthesis of an ionic solid
– a solid precipitate forms when aqueous solutions of
certain ions are mixed
• Acid‐Base: proton transfer reactions
– acid donates a proton to a base, forming a molecule (water
or another weak acid) and an aqueous salt
– Acid: proton‐donor; Base: proton‐acceptor
• Oxidation‐Reduction: electron transfer reactions
– electron transfer from one species to another, causing a
change in the oxidation state of the two species
– OIL RIG: Oxidation Is Loss (of e‐), Reduction Is Gain (of e‐)
– includes combustion, the reaction of a substance with
oxygen
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Precipitation Reactions
• Sometimes when we mix two solutions, an insoluble
solid will form:
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + KNO
2 3(aq)
• The solid, called a precipitate (or insoluble salt) is
insoluble in water.
• It is so insoluble that when its component ions find each
other in solution, they very rapidly get locked together in
large clumps, driving the rxn towards products.
P R (cont.)
• The main challenge in precipitation reaction is predicting
what (if anything) will form.
• Example: Addition of potassium chromate to barium
nitrate.
K2CrO4 (aq) + Ba(NO3)2
What is the precipitate?
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P R (cont.)
• One approach to understanding the chemistry is to
consider the ions that are present in solution:
– K2CrO4 (aq) consists of K+ (aq) and CrO42‐ (aq)
– Ba(NO3)2 (aq) consists of Ba2+ (aq) and NO3‐ (aq)
P R (cont.)
• When combining ions for see which precipitate could be
formed, the precipitate must have a net charge of zero.
• Possible choices are BaCrO4 and KNO3.
• Answer: BaCrO4.
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P R (cont.)
BaCl2(aq) AgNO3(aq) BaCl2(aq) + AgNO3(aq)
Cl- NO3- Cl- NO3-
Ba2+ Ag+ Ag+
Cl- Cl-
Cl- Ba2+ Ag+ -
Ag+
NO3
Cl- NO3- Ba2+ Cl- Ba2+
Cl-
BaCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Ba(NO3)2(aq)
Barium and nitrate
Solid silver chloride
ions are left in
precipitates out of
aqueous solution.
solution.
P R (cont.)
• Notice in the previous example we know that Ba(NO3)2
isn’t the precipitate since in our first example this
compound was soluble in water.
• Therefore, one way to determine if a precipitate will for
is to simply study combinations of reactions where you
know that one the potential precipitates is soluble.
• That’s exactly how precipitation or “solubility” rules have
been determined.
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P R (cont.)
Concept Quiz
Two electrodes connected to a light bulb are immersed in a 0.1 M solution of KNO3.
Which of the following pictures best represents the brightness of the bulb?
“bright” “dim” “dark”
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P R (cont.)
• Predict the outcome of the following reactions:
• KNO3 (aq) + BaCl2 (aq) KCl and Ba(NO3)2
Both are soluble
• Na2SO4 (aq) + Pb(NO3)2 (aq) PbSO4 and NaNO3
PbSO4 precipitates
• KOH (aq) + Fe(NO3)3 (aq) Fe(OH)3 and KNO3
Fe(OH)3 precipitates
Ions in Solution
• Conventional Equation: a bookkeeping of all species
present, and arranged for charge neutrality.
Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2NaNO3(aq)
• Complete Ionic Equation: all aqueous species are split up into
their component ions.
Ba2+(aq) + 2NO3-(aq) + BaSO4(s) + 2Na+(aq) + 2NO3-(aq)
2Na+(aq) + SO42-(aq)
• Net Ionic Equation: indicates exactly the chemical change that
occurs, and nothing more.
Ba2+(aq) + SO42-(aq) BaSO4(s)
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IiS (cont.)
• Notice in the previous example that some ions appears as
solvated species on both the reactant and product sides of
the chemical equation:
Ba2+(aq) + 2NO3-(aq) + BaSO4(s) + 2Na+(aq) + 2NO3-(aq)
2Na+(aq) + SO42-(aq)
• Ions that appear on both the reactant and product side are
referred to as spectator ions.
• Identifying spectator ions allows for one to focus on the
chemistry of interest.
• Consistent with this focus, precipitation reactions are generally
written as the Net Ionic Equation.
IiS (cont.)
• Example: Write the balanced chemical equation, complete,
and net ionic equation for the reaction of potassium chloride
and silver nitrate.
• First: Remember nomenclature and write the chemical
reaction.
KCl (aq) AgNO3 (aq)
AgCl (s) KNO3 (aq)
• Next: Using solubility rules write the ionic equation.
K (aq) Cl (aq) Ag (aq) + NO3 (aq)
AgCl (s) K (aq) NO3 (aq)
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IiS (cont.)
• Last: Cancel spectator ions for either side of the complete
ionic equation to obtained the net ionic equation
K (aq) Cl (aq) Ag (aq) + NO3 (aq)
AgCl (s) K (aq) NO3 (aq)
Cl (aq) Ag (aq)
AgCl (s)
Selective Precipitation and Stoich.
• We’ve seen that some ionic
compounds are soluble, and other
are not.
• Can use this behavior to remove
species selectively.
• Example: Separating Ag+ from Ba2+
from Fe3+
• Notice that selective precipitation is
nothing more than an application of
the solubility rules.
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S P & S (cont.)
• We can apply the mathematic of stoichiometry to
precipitation reactions.
• Example: What mass of NaCl must be added to 1.50 L of a
0.100 M AgNO3 solution to precipitate all the Ag+?
• First step: Determine the moles of Ag+ in solution.
0.100 mol AgNO3
1.50 L solution 0.150 mol AgNO3
1 L of solution
1 mol Ag
0.150 mol AgNO3 0.150 mol Ag
1 mol AgNO3
S P & S (cont.)
• Next: Look at the stoichiometry of the precipitation reaction
to determine the moles of Cl‐ needed.
Ag + (aq) Cl- (aq)
AgCl (s)
1 mol Cl
0.150 mol Ag
0.150 mol Cl
1 mol Ag
• Finally, determine the mass of NaCl:
1 mol NaCl
0.150 mol Cl 0.150 mol NaCl
1 mol Cl
58.4 g NaCl
0.150 mol NaCl 8.76 g NaCl
1 mol NaCl
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S P & S (cont.)
• Of course, we can also think about limiting reactants in
precipitation reactions.
• Example: Calculate the mass of BaSO4 formed with 0.500 L of
1.20 M Na2SO4 is mixed with 0.250 L of 0.350 M Ba(NO3)2.
• First, determine the moles of Ba2+ and SO42‐
1.20 mol Na 2SO 4
0.500 L solution 0.600 mol Na 2SO 4
1 L of solution
0.600 mol SO 24
0.350 mol Ba NO3 2
0.250 L solution 0.0875 mol Ba NO3 2
1 L of solution
0.0875 mol Ba 2 limiting
S P & S (cont.)
• Once the limiting reagent is identified, you know the moles of
precipitate that will be formed. All that’s left is to determine
the mass:
233.43 g BaSO 4
0.0875 mol BaSO 4 20.4 g BaSO 4
1 mol BaSO 4
• One can use this limiting reagent approach to determine the
amount of one ion is a sample by adding the other reactant in
the precipitation reaction in excess.
• For example, you could figure out the amount of Ag+ in a
sample by adding a NaCl solution until precipitation was
complete. The amount of AgCl formed is entirely dependent
on the amount of Ag+ initially present.
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