Post on 13-Jan-2016
description
1
PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts
Section 18.4Section 18.4
PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts
Section 18.4Section 18.4
2
Analysis of Silver Analysis of Silver GroupGroup
Analysis of Silver Analysis of Silver GroupGroup
All salts formed in All salts formed in this experiment are this experiment are said to be said to be INSOLUBLEINSOLUBLE and and form when mixing form when mixing moderately moderately concentrated concentrated solutions of the solutions of the metal ion with metal ion with chloride ions.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
Although all salts formed in this experiment
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
5Analysis Analysis of Silver of Silver GroupGroup
Analysis Analysis of Silver of Silver GroupGroup
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
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
6Analysis Analysis of Silver of Silver GroupGroup
Analysis Analysis of Silver of Silver GroupGroup
AgCl(s) Ag+(aq) + Cl-(aq)
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
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
7
8
Lead(II) ChlorideLead(II) ChloridePbClPbCl22(s) (s) Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)
KKspsp = 1.9 x 10 = 1.9 x 10-5-5
9
SolutionSolution
1. 1. Solubility = [Pb2+] = 1.30 x 10-3 M
[I-] = ?
[I-] = 2 x [Pb2+] = 2.60 x 10-3 M
Solubility of Lead(II) IodideSolubility of Lead(II) IodideSolubility of Lead(II) IodideSolubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 0.00130 M
10
SolutionSolution
2. Ksp = [Pb2+] [I-]2
= [Pb2+] [2Pb2+]2
Ksp = 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
PbI2(s) Pb2+(aq) + 2 I-(aq)
11
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
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?
12
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 = maximum ion concs.
Precip. begins when product of .
EXCEEDS the Ksp.
13
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble SaltHg2Cl2(s) Hg2
2+(aq) + 2 Cl-(aq)
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.
14
Precipitating an Insoluble SaltPrecipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18
Now raise [Cl-] to 1.0 M. What is the value of [Hg2
2+]
[Hg22+] = Ksp / [Cl-]2
= Ksp / (1.0)2 = 1.1 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016 !
15
The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes the
equilibrium to shift back to reactant.
16
Common Ion EffectCommon Ion Effect
PbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5
17
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [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 mol/L
The Common Ion EffectThe Common Ion Effect
18
SolutionSolution
Solubility in pure water = 1.1 x 10-5 mol/L.
Now dissolve BaSO4 in water already containing 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?___
The Common Ion EffectThe Common Ion Effect
BaSO4(s) Ba2+(aq) + SO42-(aq)
LEFT
LESS
19
Solution
[Ba2+] [SO42-]
initial
change
equilib.
The Common Ion EffectThe Common Ion Effect
+ y
0.010 00
+ y
0.010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
20
SolutionSolution
Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)
Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means
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
The Common Ion EffectThe Common Ion Effect
BaSO4(s) Ba2+(aq) + SO42-(aq)
21
SUMMARY
Solubility in pure water =
SO42-(aq) = x = 1.1 x 10-5 M
Solubility in presence of added Ba2+ SO4
2-(aq) = 1.1 x 10-8 M
Le Chatelier’s Principle is followed!
The Common Ion EffectThe Common Ion Effect
BaSO4(s) Ba2+(aq) + SO42-(aq)
22
23
Separating Metal Ions Cu2+, Ag+, Pb2++
Separating Metal Ions Cu2+, Ag+, Pb2++
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
24Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO4
2-. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
SolutionSolution
The substance whose Ksp is first exceeded precipitates first.
The ion requiring the lesser amount of CrO4
2- ppts. first.
25Separating Salts by Differences in Separating Salts by Differences in KKspsp
[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-1
Calculate [CrO42-] 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
26
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
27Separating 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
28
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
29
Separating Salts by Separating Salts by Differences in KDifferences in Kspsp
• PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 1.7 x 10-5
PbCrO4 1.8 x 10-14
PbClPbCl22(s) (s) Pb Pb2+2+ + 2 Cl + 2 Cl-- KK11 = K = Kspsp
PbPb2+2+ + CrO + CrO442-2- PbCrO PbCrO44 KK22 = 1/K = 1/Kspsp
KKnetnet = K = K11 • K • K22 = 9.4 x 10 = 9.4 x 1088
Net reaction is product-favoredNet reaction is product-favored
30
31
Formation of complex ions explains why you can dissolve a ppt. by forming a complex
ion.
Dissolving Precipitates Dissolving Precipitates by forming Complex Ionsby forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
32
Examine the solubility of AgCl in ammonia.
AgCl(s) Ag+ + Cl- Ksp = 1.8 x 10-10
Ag+ + 2 NH3 Ag(NH3)2+ Kform = 1.6 x 107
-------------------------------------
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
Knet = Ksp • Kform = 2.9 x 10-3
By adding excess NH3, the equilibrium shifts to the right.
Dissolving Precipitates Dissolving Precipitates by forming Complex Ionsby forming Complex Ions
33
34
35