PRECIPITATION REACTIONS Solubility of Salts Section 18.4
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PRECIPITATION REACTIONSSolubility of
SaltsSection 18.4
PRECIPITATION REACTIONSSolubility of
SaltsSection 18.4
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Metal Chloride SaltsMetal Chloride Salts
These products are said to be
INSOLUBLE and form when mixing
moderately concentrated
solutions of the metal ion with
chloride ions.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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3Analysis Analysis of Silver of Silver GroupGroup
Analysis Analysis of Silver of Silver GroupGroup
The products are said to be insoluble, they
do dissolve to some SLIGHT extent.
AgCl(s) = Ag+(aq) + Cl-(aq)
When equilibrium has been established, no
more AgCl dissolves and the solution is
SATURATED.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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4Analysis Analysis of Silver of Silver GroupGroup
Analysis Analysis of Silver of Silver GroupGroup
AgCl(s) = Ag+(aq) + Cl-(aq)
When solution is SATURATED, expt. shows that
[Ag+] = 1.67 x 10-5 M.
This is equivalent to the SOLUBILITY of AgCl.
What is [Cl-]?
[Cl-] is equivalent to the AgCl solubility.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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AgCl(s) = Ag+(aq) + Cl-(aq)
Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (1.67 x 10-5)(1.67 x 10-5)
= 2.79 x 10-10
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Kc = [Ag+] [Cl-] = 2.79 x 10-10
Because this is the product of “solubilities”, we call it
Ksp = solubility product constant
See Table 18.2 and Appendix J
Solubility Product Constant
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Lead(II) ChlorideLead(II) Chloride
PbCl2(s) = Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5
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SolutionSolution
Solubility = [Pb2+] = 1.30 x 10-3 M
[I-] = ?
[I-] = 2 x [Pb2+] = 2.60 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) = Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 0.00130 M
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Calculate Ksp
Ksp = [Pb2+] [I-]2
= X{2 X}2
Ksp = 4 X3 = 4[Pb2+]3
Solubility of Lead(II) IodideSolubility of Lead(II) IodideSolubility of Lead(II) IodideSolubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9
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Precipitating an Insoluble Precipitating an Insoluble SaltSalt
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
Hg2Cl2(s) = Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0.010 M, what [Cl-] is req’d to just
begin the precipitation of Hg2Cl2?
That is, what is the maximum [Cl-] that can be
in solution with 0.010 M Hg22+ without
forming Hg2Cl2?
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Precipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble Salt
Hg2Cl2(s) = Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of
maximum ion concs..
Precip. begins when product of
ion concs. EXCEEDS the Ksp.
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Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0.010
M,
[Cl ] = Ksp
0.010 = 1.1 x 10-8 M[Cl ] =
Ksp
0.010 = 1.1 x 10-8 M
If this conc. of ClIf this conc. of Cl-- is just exceeded, Hg is just exceeded, Hg22ClCl22
begins to precipitate.begins to precipitate.
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Hg2Cl2(s) = Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18
Now use [Cl-] = 1.0 M. What is the value of [Hg2
2+] at this point?
Solution
[Hg22+] = Ksp / [Cl-]2
= Ksp / (1.0)2 = 1.1 x 10-18 M
The concentration of Hg22+ has been
reduced by 1016 !
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The Common Ion EffectThe Common Ion Effect
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The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes the
equilibrium to shift back to reactant.
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17Common Ion EffectCommon Ion Effect
PbCl2(s) = Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5
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Calculate the solubility of BaSO4
BaSO4(s) = Ba2+(aq) + SO42-(aq)
(a) In pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
The Common Ion EffectThe Common Ion Effect
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Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) = Ba2+(aq) + SO42-(aq)
Solution
Solubility = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)1/2 = 1.1 x 10-5 M
Solubility in pure water = 1.1 x 10-5 M
BaSO4 in pure water
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Solution
Solubility in pure water = 1.1 x 10-5 mol/L.
Now starting with 0.010 M Ba2+.
Which way will the “common ion” shift the equilibrium? ___ Will
solubility of BaSO4 be less than or greater than in pure water?___
BaSO4 in in 0.010 M Ba(NO3)2.
BaSO4(s) = Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initial
change
equilib.
The Common Ion EffectThe Common Ion Effect
+ y
0.010 0
+ y
0.010 + y y
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Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)
Because y < 1.1 x 10-5 M (pure),
0.010 + y is about equal to 0.010. Therefore,
Ksp = 1.1 x 10-10 = (0.010)(y)
y = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion.
The Common Ion EffectThe Common Ion EffectSolutionSolution
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SUMMARY
Solubility in pure water = x = 1.1 x 10-5 M
Solubility in presence of added Ba2+ = 1.1 x 10-8 M
Le Chatelier’s Principle is followed!Add to the right: equilibrium goes to the left
The Common Ion EffectThe Common Ion Effect
BaSO4(s) = Ba2+(aq) + SO42-(aq)
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24Separating Metal Separating Metal Ions Ions
CuCu2+2+, Ag, Ag++, Pb, Pb2+2+
Separating Metal Separating Metal Ions Ions
CuCu2+2+, Ag, Ag++, Pb, Pb2+2+
Ksp Values
AgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14
Ksp Values
AgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14
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26Separating Salts by Separating Salts by Differences in KDifferences in Kspsp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to
precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
The substance whose Ksp is first exceeded precipitates first.
The ion requiring the lesser amount of CrO4
2- ppts. first..
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27Separating Salts by Differences Separating Salts by Differences in Kin Kspsp
[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M
SolutionSolutionCalculate [CrO4
2-] required by each ion.
[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2
= 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M
PbCrO4 precipitates first
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A solution contains 0.020 M Ag+ and Pb2+. Add CrO4
2- to precipitate red Ag2CrO4 and yellow PbCrO4.
PbCrO4 ppts. first.
Ksp (Ag2CrO4)= 9.0 x 10-12
Ksp (PbCrO4) = 1.8 x 10-14
How much Pb2+ remains in solution when Ag+ begins to precipitate?
Separating Salts by Differences in Separating Salts by Differences in KKspsp
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29Separating Salts by Differences in Separating Salts by Differences in KKspsp
We know that [CrO42-] = 2.3 x 10-8 M to begin to
ppt. Ag2CrO4.
What is the Pb2+ conc. at this point?
[Pb2+] = Ksp / [CrO42-]
= 1.8 x 10-14 / 2.3 x 10-8 M
= 7.8 x 10-7 M
Lead ion has dropped from
0.020 M to < 10-6 M
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Separating Salts by Separating Salts by Differences in KDifferences in Kspsp
Separating Salts by Separating Salts by Differences in KDifferences in Kspsp
• Add CrO42- to solid PbCl2. The less
soluble salt, PbCrO4, precipitates
• PbCl2(s) + CrO42- = PbCrO4 + 2 Cl-
• Salt Ksp
PbCl2 1.7 x 10-5
PbCrO4 1.8 x 10-14
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31Separating by KSeparating by Kspsp
• PbCl2(s) + CrO42- = PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 1.7 x 10-5
PbCrO4 1.8 x 10-14
PbCl2(s) = Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- = PbCrO4 K2 =
1/Ksp
Knet = K1 • K2 = 9.4 x 108
Net reaction is product-favored
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