1 chapter 16 precipitation equilibrium solubility product principle
TRANSCRIPT
![Page 1: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/1.jpg)
1
Chapter 16
Precipitation EquilibriumSolubility Product Principle
![Page 2: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/2.jpg)
2
Solubility Product Constants
Silver chloride, AgCl,is rather insoluble in water.Careful experiments show that if solid AgCl is
placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.
aqaq ClAgClAg
![Page 3: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/3.jpg)
3
Solubility Product Constants
The equilibrium constant expression for this dissolution is called a solubility product constant. Ksp=solubility product constant
Molar concentration of ions raised to their stoichiometric powers at equilibrium 10sp 101.8ClAgK
![Page 4: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/4.jpg)
4
Solubility Product Constants
Solubility product constant for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound.
Consider the dissolution of silver sulfide in water.
Ag S 2 Ag + S2+ 2-
H O+ 2-2 100%
![Page 5: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/5.jpg)
5
Solubility Product Constants
Its solubility product expression is
K Ag Ssp
2 2 10 10 49.
![Page 6: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/6.jpg)
6
Solubility Product Constants
The dissolution of solid calcium phosphate in water is represented as
Ca PO 3 Ca 2 PO3
2+43
2 s
H O2
43
100%
2
![Page 7: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/7.jpg)
7
Solubility Product Constants
Its solubility product constant expression is
K Ca POsp2 3
43 2
10 10 25.
![Page 8: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/8.jpg)
8
Solubility Product Constants
In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as
Ksp has a fixed value for a given system at a given temperature
M Y r M s Y
K M Y
r s s
H Os r
100%
sps r r s
2
![Page 9: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/9.jpg)
9
Solubility Product Constants
The same rules apply for compounds that have more than two kinds of ions.
An example is calcium ammonium phosphate.
CaNH PO Ca NH PO
K Ca NH PO
4 4 s2
4 43
sp2
4 43
![Page 10: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/10.jpg)
10
Determination of Solubility Product Constants
Example: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl.
Molar solubility can be calculated from the data:
? .
.
mol AgClL
g AgClL
mol AgCl143 g AgCl
mol AgClL
0 00192 1
134 10 5
![Page 11: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/11.jpg)
11
Determination of Solubility Product Constants
The equation for the dissociation of silver chloride and its solubility product expression are
AgCl Ag Cl
1.34 10 1.34 10
K Ag Cl
s
-5 -5
sp
M M
![Page 12: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/12.jpg)
12
Determination of Solubility Product Constants
Substitution into the solubility product expression gives
K Ag Clsp
134 10 134 10
18 10
5 5
10
. .
.
![Page 13: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/13.jpg)
13
Uses of Solubility Product Constants
We can use the solubility product constant to calculate the solubility of a compound at 25oC.
Example: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC.
Ksp= 1.1 x 10-10.
![Page 14: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/14.jpg)
14
Uses of Solubility Product Constants
Example: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. Ksp= 1.1 x 10-10.
BaSO Ba SO
Ba SO
4 s aq2+
4 aq
242
2
1011 10
xM xM xM
Ksp .
![Page 15: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/15.jpg)
15
Uses of Solubility Product Constants
Substitute into solubility product expression and solve for x, giving the ion concentrations.
x x
x M
11 10
10 10
10
5
.
.
Ba SO242
![Page 16: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/16.jpg)
16
Uses of Solubility Product Constants
Now we can calculate the mass of BaSO4 in 1.00 L of saturated solution.
?
.
g BaSOL
1.0 10 molL
234 gmol
g
45
BaSOL
4
2 3 10 3
![Page 17: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/17.jpg)
17
The Reaction Quotient in Precipitation Reactions
Use solubility product constants to calculate the concentration of ions in a solution and whether or not a precipitate will form.
Example: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?
![Page 18: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/18.jpg)
18
The Reaction Quotient in Precipitation Reactions
Example: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?
K SO 2 K SO
Pb NO Pb NO
Will PbSO precipitate?
2 4H O 100% +
42
3 2H O 100% 2+
3-
4
2
2
2
![Page 19: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/19.jpg)
19
The Reaction Quotient in Precipitation Reactions
Calculate the Qsp for PbSO4. Solution volumes are additive. Concentrations of the important ions are:
MM
M
MM
M
Pb2+
SO 42-
2+
42-
mL 0.10200 mL
Pb
mL 0.010200 mL
SO
1000 050
1000 0050
.
.
![Page 20: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/20.jpg)
20
The Reaction Quotient in Precipitation Reactions
Finally, we calculate Qsp for PbSO4.
Q Pb SO
K for PbSO
Q K therefore solid forms
sp2
42
sp 4
sp sp
0 050 0 0050
2 5 10
18 10
4
8
. .
.
.
![Page 21: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/21.jpg)
21
Fractional Precipitation
Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. Look at a solution that contains Cu+, Ag+, and Au+ We could precipitate them as chlorides
![Page 22: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/22.jpg)
22
Fractional Precipitation
Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. Look at a solution that contains Cu+, Ag+, and Au+ We could precipitate them as chlorides
13
sp
10sp
7sp
100.2ClAuKCl AuAuCl
108.1ClAgKCl AgAgCl
109.1ClCuKClCuCuCl
![Page 23: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/23.jpg)
23
Fractional Precipitation
Example: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal I chlorides.
![Page 24: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/24.jpg)
24
Fractional Precipitation
Example: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal I chlorides.
AuCl. eprecipitat torequired Cl ofion concentrat theCalculate
last. eprecipitat willCuCl reasoning, same By the
first. esprecipitatit so Ksmallest thehas AuCl
-
sp
![Page 25: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/25.jpg)
25
Fractional Precipitation
Example: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal I chlorides.
M11
1313
sp13
-
sp
100.2010.0
100.2
Au
100.2Cl
K100.2ClAu
AuCl. eprecipitat torequired Cl ofion concentrat theCalculate
last. eprecipitat willCuCl reasoning, same By the
first. esprecipitatit so Ksmallest thehas AuCl
![Page 26: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/26.jpg)
26
Fractional Precipitation
Repeat the calculation for silver chloride.
K Ag Cl
ClAg
sp
18 10
18 10 18 100 010
18 10
10
10 10
8
.
. ..
. M
![Page 27: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/27.jpg)
27
Fractional Precipitation
For copper (I) chloride to precipitate.
K Cu Cl
ClCu
sp+
+
19 10
19 10 19 100 010
19 10
7
7 7
5
.
. ..
. M
![Page 28: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/28.jpg)
28
Fractional Precipitation
We have calculated the [Cl-] required to precipitate AuCl, [Cl-] >2.0 x 10-11 Mto precipitate AgCl, [Cl-] >1.8 x 10-8 Mto precipitate CuCl, [Cl-] >1.9 x 10-5 M
We can calculate the amount of Au+ precipitated before Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before Cu+ begins to precipitate.
![Page 29: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/29.jpg)
29
Fractional Precipitation Example: Calculate the percent of Au+ ions that
precipitate before AgCl begins to precipitate. Use the [Cl-] from before to determine the [Au+]
remaining in solution just before AgCl begins to precipitate.
![Page 30: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/30.jpg)
30
Fractional Precipitation Example: Calculate the percent of Au+ ions that
precipitate before AgCl begins to precipitate. Use the [Cl-] from before to determine the [Au+]
remaining in solution just before AgCl begins to precipitate.
Au Cl
AuCl
Au Au unprecipitated
2 0 10
2 0 10 2 0 10
18 10
11 10
13
13 13
8
5
.
. .
.
.
![Page 31: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/31.jpg)
31
Fractional Precipitation
The percent of Au+ ions unprecipitated just before AgCl precipitates is
% Au
Au
Auunprecipitated+ unprecipitated
original
unprecipitated
100%
11 100 010
100 0 00011%8.
..
![Page 32: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/32.jpg)
32
Fractional PrecipitationThe percent of Au+ ions unprecipitated just before AgCl precipitates is
Therefore, 99.99989% of the Au+ ions precipitates before AgCl begins to precipitate.
% Au
Au
Auunprecipitated+ unprecipitated
original
unprecipitated
100%
11 100 010
100 0 00011%8.
..
![Page 33: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/33.jpg)
33
Fractional PrecipitationSimilar calculations for the concentration of Ag+
ions unprecipitated before CuCl begins to precipitate gives atedunprecipit Ag105.9Ag
109.1
108.1
Cl
108.1Ag
108.1ClAg
6
5
1010
10
![Page 34: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/34.jpg)
34
Fractional Precipitation
The percent of Au+ ions unprecipitated just before AgCl precipitates is
atedunprecipit
6
original
atedunprecipit+atedunprecipit
%095.0100010.0
105.9
%100Ag
AgAg %
![Page 35: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/35.jpg)
35
Fractional PrecipitationThe percent of Au+ ions unprecipitated just before AgCl precipitates is
Thus, 99.905% of the Ag+ ions precipitates before CuCl begins to precipitate.
atedunprecipit
6
original
atedunprecipit+atedunprecipit
%095.0100010.0
105.9
%100Ag
AgAg %
![Page 36: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/36.jpg)
36
Factors that Affect Solubility
CaF2 (s) Ca+2 (aq)+ 2F- (aq)
Addition of a Common ion (F- from NaF) Solubility decreases Equilibrium shifts to left
Changes in pH (H+ reacts with F-) Solubility increases (with increasing pH) Equilibrium shifts to right
![Page 37: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/37.jpg)
37
Factors that Affect SolubilityAg+(aq) + 2NH3(aq) Ag(NH3)2
+ (aq)
complex ion
Formation of a complex ion Lewis Acid base chemistry
Calculate Kf Formation Constant
![Page 38: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/38.jpg)
38
Factors that Affect SolubilityAg+(aq) + 2NH3(aq) Ag(NH3)2
+ (aq)
complex ionAgCl(s) Ag+(aq) + Cl-(aq)
In formation of complex ion Removed Ag+ from the equilibrium Equilibrium shifts to right Favors dissolving AgCl
![Page 39: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/39.jpg)
39
Synthesis Question
Most kidney stones are made of calcium oxalate, Ca(O2CCO2). Patients who have their first kidney stones are given an extremely simple solution to stop further stone formation. They are told to drink six to eight glasses of water a day. How does this stop kidney stone formation?
![Page 40: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/40.jpg)
40
Synthesis Question
stone. solida as than
rather solution in ions towardsright, theto
mequilibriu theshifts water more Drinking
COOCaOHCa(COO)
oxalate calciumfor
expressionproduct Solubility
2aq2
2aq2s2
![Page 41: 1 Chapter 16 Precipitation Equilibrium Solubility Product Principle](https://reader036.vdocuments.site/reader036/viewer/2022081516/56649dab5503460f94a9ac5b/html5/thumbnails/41.jpg)
41
Group Question
The cavities that we get in our teeth are a result of the dissolving of the material our teeth are made of, calcium hydroxy apatite. How does using a fluoride based toothpaste decrease the occurrence of cavities?