precipitation reactions solubility of salts section 18.4
DESCRIPTION
PRECIPITATION REACTIONS Solubility of Salts Section 18.4. Metal Chloride Salts. These products are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions. Analysis of Silver Group. - PowerPoint PPT PresentationTRANSCRIPT
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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
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
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.
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)
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..
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
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
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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