PRECIPITATION REACTIONS Chapter 17 Part 2. Insoluble Chlorides. All salts formed in this experiment are said to be INSOLUBLE and form precipitates when mixing moderately concentrated solutions of the metal ion with chloride ions. Insoluble Chlorides.
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All salts formed in this experiment are said to be INSOLUBLE and form precipitates when mixing moderately concentrated solutions of the metal ion with chloride ions.
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.
AgCl(s) <> Ag+(aq) + Cl(aq)
When the solution is SATURATED, experiment shows that [Ag+] = 1.34 x 105 M.
This is equivalent to the SOLUBILITYof AgCl.
What is [Cl]?
This is also equivalent to the AgCl solubility.
Make a chart.
AgCl(s) <> Ag+(aq) + Cl(aq)
some 0 0
 1.34 x 105
1.34 x 105
1.34 x 105
some  1.34 x 105
1.34 x 105
1.34 x 105
Ksp = [Ag+] [Cl]
= (1.34 x 105)(1.34 x 105)
= 1.80 x 1010
Ksp = solubility product constant
See Table 18.2 and Appendix J 18A & 18B
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I(aq)
Calculate Ksp if solubility =0.00130 M
Solution
Solubility = [Pb2+] = 1.30 x 103 M
[I] = _____________ ?
2(1.30 x 103 M)
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I(aq)
Calculate Ksp if solubility =0.00130 M
Solution
1. Solubility = [Pb2+] = 1.30 x 103 M
[I] = 2 x [Pb2+] = 2.60 x 103 M
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I(aq)
Calculate Ksp if solubility = 0.00130 M
Solution
1. Solubility = [Pb2+] = 1.30 x 103 M
[I] = 2 x [Pb2+] = 2.60 x 103 M
2. Ksp = [Pb2+] [I]2
= [Pb2+] {2 • [Pb2+]}2
= 4 [Pb2+]3
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I(aq)
Calculate Ksp if solubility = 0.00130 M
Solution
2. Ksp = 4 [Pb2+]3 = 4 (solubility)3
Ksp = 4 (1.30 x 103)3 = 8.8 x 109
Sample Problems
Hg2Cl2(s) <> Hg22+(aq) + 2 Cl(aq)
Ksp = 1.1 x 1018 = [Hg22+] [Cl] 2
If [Hg22+] = 0.010 M, what [Cl] is required to just begin the precipitation of Hg2Cl2?
What is the maximum [Cl] that can be in solution with 0.010 M Hg22+ without forming Hg2Cl2?
Hg2Cl2(s) Hg22+(aq) + 2 Cl(aq)
Ksp = 1.1 x 1018 = [Hg22+] [Cl]2
Recognize that
Ksp = product of maximum ion concentrations.
Precipitation begins when product of ion concentrations EXCEEDS the Ksp.
sp

18
[
Cl
]
=
=
1.1 x 10
M
4(0.010)
Precipitating an Insoluble SaltHg2Cl2(s) Hg22+(aq) + 2 Cl(aq)
Ksp = 1.1 x 1018 = [Hg22+] [2Cl]2
Solution
[Cl] that can exist when [Hg22+] = 0.010 M,
If this concentration of Cl is just exceeded, Hg2Cl2 begins to precipitate.
Hg2Cl2(s) Hg22+(aq) + 2 Cl(aq)
Ksp = 1.1 x 1018 = [Hg22+] [Cl]2
Now raise [Cl] to 1.0 M.
What is the value of [Hg22+] at this point?
Solution
[Hg22+] = Ksp / [Cl]2
= Ksp / (1.0)2 = 1.1 x 1018 M
The concentration of Hg22+ has been reduced by 1016 !
Sample Problems
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 1012
Ksp for PbCrO4 = 1.8 x 1014
Solution
The substance whose Ksp is first exceeded will precipitate first.
The ion requiring the lesser amount of CrO42 precipitate first.
19
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 1012
Ksp for PbCrO4 = 1.8 x 1014
Solution
[CrO42] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x 1014 / 0.020 = 9.0 x 1013 M
[CrO42] to ppt. Ag2CrO4 = Ksp / [Ag+]2
= 9.0 x 1012 / (0.020)2 = 2.3 x 108 M
PbCrO4 precipitates first.
Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42 to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 precipitates first.
Ksp (Ag2CrO4)= 9.0 x 1012
Ksp (PbCrO4) = 1.8 x 1014
How much Pb2+ remains in solution when Ag+ begins to precipitate?
Solution
We know that [CrO42] = 2.3 x 108 M to begin to precipitates Ag2CrO4.
What is the Pb2+ concentration at this point?
Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42 to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 precipitates first.
Ksp (Ag2CrO4)= 9.0 x 1012
Ksp (PbCrO4) = 1.8 x 1014
How much Pb2+ remains in solution when Ag+ begins to precipitate?
Solution
[Pb2+] = Ksp / [CrO42] = 1.8 x 1014 / 2.3 x 108 M
= 7.8 x 107 M
Lead ion has dropped from 0.020 M to < 106 M
Calculate the solubility of BaSO4 in: (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
BaSO4(s) Ba2+(aq) + SO42(aq)
Solution (a)
Solubility in pure water = [Ba2+] = [SO42] = s
Ksp = [Ba2+] [SO42] = s2
s = (Ksp)1/2 = 1.1 x 105 M
Calculate the solubility of BaSO4 in: (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
BaSO4(s) Ba2+(aq) + SO42(aq)
Solution (b)
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?___
Left
Less
Calculate the solubility of BaSO4 in: (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
BaSO4(s) Ba2+(aq) + SO42(aq)
Solution (b)
[Ba2+] [SO42]
initial
change
equilib.
0.010 0
+ s + s
0.010 + s s
Calculate the solubility of BaSO4 in:
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
BaSO4(s) Ba2+(aq) + SO42(aq)
Solution
Ksp = [Ba2+] [SO42] = (0.010 + s) (s)
s < 1.1 x 105 M (solubility in pure water), this means 0.010 + s is about equal to 0.010. Therefore, Ksp = 1.1 x 1010 = (0.010)(s)
s = 1.1 x 108 M = solubility in presence of added Ba2+ ion.
Calculate the solubility of BaSO4 in: (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 1010
BaSO4(s) Ba2+(aq) + SO42(aq)
Solution
Solubility in pure water = s = 1.1 x 105 M
Solubility in presence of added Ba2+ = 1.1 x 108 M
Le Chatelier’s Principle is followed!
Sample Problems
water.
0.0025 M calcium nitrate.
0.080 M sodium fluoride.
Write appropriate netionic equations.
CaC2O4(s) + 2 H+ <======> H2C2O4(aq) + Ca+2
Knet = Ksp. ( 1/Ka 1 ) . ( 1/Ka 2 )
AgBr(s) + 2 NH3(aq) <=====> Ag(NH3)2+(aq) + Br
Knet = Ksp. Kf =
( 3.3 x 1013 ) ( 1.6 x 107) = 5.3 x 106
1. If you add sufficient chromate ion to an aqueous suspension of PbCl2, can PbCl2 be converted to PbCrO4?
PbCl2 <> Pb2+ + 2 Cl
Pb2+ + CrO42 <> PbCrO4
1.7 x 105
1/1.8 x 1014
9.4 x 108
PbCl2 + CrO42 <> PbCrO4 + 2 Cl
Yes!
2. Can AgCl be dissolved by adding a solution of NH3?
Write the overall equation and determine the K value.
AgCl <> Ag+ + Cl
Ag+ + 2 NH3 <> Ag(NH3)2+
1.8 x 1010
1.6 x 107
2.9 x 103
AgCl + 2 NH3 <> Ag(NH3)2++ Cl
No, unless very high [NH3]
3. Can CaC2O4 be dissolved by adding a solution of HCl? Write the overall equation and determine the K value.
CaC2O4 <> Ca2+ + C2O42
H+ + C2O42 <> HC2O4
H+ + HC2O4 <> H2C2O4
2.3 x 109
1/6.4 x 105
1/5.9 x 102
6.1 x 104
CaC2O4 + 2 H+ <> H2C2O4 + Ca2+
No, unless very high [H+]
Ksp AgI = 1.5 x 10 16 Ksp PbI2 = 8.7 x 10 9.
1. A saturated solution of lead chloride contains 4.50 g of lead chloride per liter. Calculate the Ksp for lead chloride.
2. The Ksp for Al(OH)3 is 1.9 x 1033. Calculate the molar solubility of Al(OH)3 and determine [Al3+] and [OH1].
3. What is the molar solubility of BaSO4 in a solution that contains 0.100 M Na2SO4? (Ksp for BaSO4 = 1.1 x 1010)
4. Will precipitation occur when 50.0 ml of 0.030 M Pb(NO3)2 is added to 50.0 ml of 0.0020 M KBr? (Ksp for lead bromide = 6.3 x 106)
5. Would it be possible to separate a solution containing 0.0020 M Pb2+ and 0.030 M Ag+ by adding drops of Na2CO3 solution? (Ksp for lead carbonate = 1.5 x 1013 and
Ksp for silver carbonate = 8.2 x 1012)
6. Can CuBr be dissolved by adding a solution of NaCl? Write the overall equation and determine the K value
1. 1.7 x 105
2. 2.9 x 109 M, 2.9 x 109 M, 8.7 x 109 M
3. 1.1 x 109 M
4. no
5. yes
6. No, unless [Cl] is very large,
K = 5.3 x 104
The End
Hg2Cl2(s) <> Hg2+(aq) + 2 Cl(aq)
Ksp = 1.1 x 1018
Lead(II) Chloride
PbCl2(s) <> Pb2+(aq) + 2 Cl(aq)
Ksp = 1.9 x 105
Silver Chloride
AgCl(s) <> Ag+(aq) + Cl(aq)
Ksp = 1.8 x 1010
1. A saturated solution of CuCl has a gram solubility of 0.05643 g/L. Calculate the Ksp.
(0.05643g/L)(1 mole/99.0g) = 0.000570 M
CuCl(s) <> Cu+(aq) + Cl(aq)
Ksp = [Cu+] [Cl]
= (0.000570)(0.000570)
= 3.25 x 107
Solid
 0.000570
0.000570
0.000570
0.000570
0.000570
Solid
2. A saturated solution of PbBr2 has [Pb2+] = 1.05 x 101 M. Calculate the Ksp.
PbBr2(s) <> Pb2+(aq) + 2 Cl(aq)
Ksp = [Pb2+] [Cl]2
= (0.0105)(0.0210)2
= 4.63 x 103
Solid
 0.0105
0.0105
0.0210
Solid
0.0105
0.0210
3. A saturated solution of Ag2CrO4 has [Ag+] = 1.6 x 104 M. Calculate the Ksp.
Ag2CrO4(s) <> 2 Ag+(aq) + CrO42(aq)
Ksp = [Ag+]2 [CrO42]
= (1.6 x 104)2(8.0 x 105)
= 2.0 x 1012
Solid
 8.0 x 105
1.6 x 104
8.0 x 105
Solid
1.6 x 104
8.0 x 105
1. The Ksp of SrCO3 is 7.0 x1010. Calculate the molar solubility of SrCO3.
SrCO3(s) <> Sr2+(aq) + CO32(aq)
Ksp = [Sr2+] [CO32]
= (s)(s)= s2 = 7.0 x 1010
s = 2.6 x 105 M
Solid
 s
s
s
Solid
s
s
2.The Ksp of Ca(OH)2 is 7.9 x106. Calculate the molar solubility of Ca(OH)2.
Ca(OH)2(s) <> Ca2+(aq) + 2 OH(aq)
Ksp = [Ca2+] [OH]2
= (s)(2s)2 = 4s3 = 7.9 x 106
s = 1.3 x 102 M
Solid
 s
s
2s
Solid
s
2s
3.The Ksp of Al(OH)3 is 2.0 x 1033. Calculate the molar solubility of Al(OH)3.
Al(OH)3(s) <> Al3+(aq) + 3 OH(aq)
Ksp = [Al3+] [OH]3
= (s)(3s)3 = 27s4 = 2.0 x 1033
s = 2.9 x 109 M
Solid
 s
s
3s
Solid
s
3s
Will mixing 200. mL 5.0 x 106 M mercury(I) nitrate and 100. mL 5.0 x 108 M sodium chloride cause a precipitate to form? Hg2Cl2(s) <> Hg22+(aq) + 2 Cl(aq)
Q = [Hg22+] [Cl] 2
[Hg22+] = 5.0 x 106 (200./300.) = 3.3 x 106 M
[Cl] = 5.0 x 108 (100./300.) = 1.7 x 108 M
Q = (3.3 x 106)(1.7 x 108) 2 = 9.5 x 1022
Q < Ksp No ppt
Will mixing 100. mL 0.20 M magnesium nitrate and 300. mL 0.40 M sodium oxalate cause a precipitate to form?
MgC2O4(s) <> Mg2+(aq) + C2O42(aq)
Q = [Mg2+][C2O42]
[Mg2+] = 0.20 (100./400.) = 0.050 M
[C2O42] = 0.40 (300./400.) = 0.30 M
Q = (0.050)(0.30) = 1.5 x 102
Ksp = 8.6 x 105 Q > Ksp ppt
Will mixing 1.0 L 0.00010 M sodium chloride and 2.0 L 0.0090 M silver nitrate cause a precipitate to form?
AgCl(s) <> Ag+(aq) + Cl(aq)
Q = [Ag+][Cl]
[Ag+] = 0.0090 (2.0/3.0) = 0.0060 M
[Cl] = 0.00010 (1.0/3.0) = 0.000033 M
Q = (0.0060)(0.000033) = 2.0 x 107
(Ksp = 1.8 x 1010) Q > Ksp ppt
What [Sr2+] is required to ppt SrSO4 in a 0.20 M Na2SO4 solution? (Ksp = 2.8 x 107)
SrSO4(s) <> Sr2+(aq) + SO42(aq)
Ksp = [Sr2+] [SO42]
2.8 x 107 = (x)(0.20)
x = 1.4 x 106 M = [Sr2+]
For ppt [Sr2+] > 1.4 x 106 M
x
0.20
How many moles of HCl are required to ppt AgCl from 100. mL 0.10 M AgNO3? (Ksp = 1.8 x 1010)
AgCl(s) <> Ag+(aq) + Cl(aq)
Ksp = [Ag+] [Cl] 1.8 x 1010 = (0.10)(x)
x = 1.8 x 109 M = [Cl]
1.8 x 109 mole/L)(0.100 L) = 1.8 x 1010 mole
For ppt mole HCl > 1.8 x 1010
0.10
x
Calculate [Cl] required to ppt PbCl2 from 0.100 M Pb(NO3)2. (Ksp = 1.7 x 105)
PbCl2(s) <> Pb2+(aq) + 2 Cl(aq)
Ksp = [Pb2+] [Cl]2
1.7 x 105 = (0.100)(x)2
x = 1.3 x 102 M = [Cl]
For ppt [Cl] > 1.3 x 102 M
0.100
x
If [Cl] is raised to 0.10 M, calculate [Pb2+]
PbCl2(s) <> Pb2+(aq) + 2 Cl(aq)
Ksp = [Pb2+] [Cl]2
1.7 x 105 = (x)(0.10)2
x = 1.7 x 103 M = [Pb2+]
x
0.10
200.
Precipitating an Insoluble Salt100. mL 0.200 M silver nitrate is mixed with 100. mL 0.100 M hydrochloric acid. Calculate [Ag+] and [Cl]. (Ksp = 1.8 x 1010)
AgCl(s) < Ag+(aq) + Cl(aq)
20.0 10.0
10.0
10.0
10.0
0
[Ag+] =
= 0.0500 M
Precipitating an Insoluble Salt
AgCl(s) <> Ag+(aq) + Cl(aq)
Ksp = [Ag+] [Cl]
1.8 x 1010 = (0.0500)(x)
x = 3.6 x 109 M = [Cl]
Solid
0.0500
 x
x
x
Solid
0.0500
x
Remember in H2O: s= 2.6 x 105M
Common Ions1. The Ksp of SrCO3 is 7.0 x1010. Calculate the molar solubility of SrCO3 in 0.10 M Na2CO3.
SrCO3(s) <> Sr2+(aq) + CO32(aq)
Ksp = [Sr2+] [CO32]
= (s)(.10)= s2 = 7.0 x 1010
s = 7.0 x 109 M
0.10
Solid
 s
s
s
Solid
s
0.10
Remember in H2O: s= 1.3 x 102M
Common Ions2. The Ksp of Ca(OH)2 is 7.9 x106. Calculate the molar solubility of Ca(OH)2 in 0.50 M NaOH.
Ca(OH)2(s) <> Ca2+(aq) + 2 OH(aq)
Ksp = [Ca2+] [OH]2
7.9 x 106 = (s)(0.50)2
s = 3.2 x 105 M
Solid
0.50
 s
s
2s
Solid
s
0.50
Remember in H2O: s= 2.9 x 109 M
Common Ions3. The Ksp of Al(OH)3 is 2.0 x 1033. Calculate the molar solubility of Al(OH)3 in 1.0 M KOH.
Al(OH)3(s) <> Al3+(aq) + 3 OH(aq)
Ksp = [Al3+] [OH]3
2.0 x 1033 = (s)(1.0)3
s = 2.0 x 1033 M
1.0
Solid
 s
s
3s
Solid
s
1.0
4. Calculate the solubility of calcium chromate in 0.0050 M calcium chloride. (Ksp = 7.1 x 104)
CaCrO4(s) <> Ca2+(aq) + CrO42(aq)
Ksp = [Ca2+] [CrO42]
7.1 x 104 = (0.0050 + s)(s)
s = 2.4 x 102 M
Solid
0.0050
 s
s
s
Solid
0.0050 + s
s
1. Separation of .10 M Ag+ and .10 M Pb2+
AgBr Ksp=3.3 x 1013
PbBr2 Ksp=6.3 x 106
Plan: Add Br until all AgBr is ppt,
but no PbBr2 is ppt.
a. Calculate [Br] required to ppt.
b. Calculate [Ag+] left in solution.
The substance whose Ksp is first exceeded will precipitate first.
The ion requiring the lesser amount of Br precipitate first.
Separating Salts by Differences in Ksp
AgBr(s) <> Ag+(aq) + Br(aq)
Ksp = [Ag+] [Br]
3.3 x 1013 = (.10)(x)
x = 3.3 x 1012 = [Br]
For ppt [Br] > 3.3 x 1012
.10
x
Separating Salts by Differences in Ksp
PbBr2(s) <> Pb2+(aq) + 2 Br(aq)
Ksp = [Pb2+] [Br]2
6.3 x 106 = (.10)(x)2
x = 7.9 x 103 = [Br]
For ppt [Br] > 7.9 x 103
.10
x
Separating Salts by Differences in Ksp
For ppt AgBr [Br] > 3.3 x 1012
For ppt PbBr2 [Br] > 7.9 x 103
Therefore;
Start ppt of AgBr [Br] > 3.3 x 1012
Max. ppt of AgBr [Br] > 7.9 x 103
At max. ppt of AgBr, what is the [Ag+] left in solution?
Separating Salts by Differences in Ksp
AgBr(s) <> Ag+(aq) + Br(aq)
Ksp = [Ag+] [Br]
3.3 x 1018 = (x)(7.9 x 103)
x = 4.2 x 1011 = [Ag+]
Good Separation: [Ag+] < 105
x
7.9 x 103
2. Separation of .10 M CO32 and .10 M C2O42
BaCO3 Ksp=8.1 x 109
BaC2O4 Ksp=1.1 x 107
Plan: Add Ba2+ until all BaCO3 is ppt,
but no BaC2O4 is ppt.
a. Calculate [Ba2+] required to ppt.
b. Calculate [CO32] left in solution.
The substance whose Ksp is first exceeded will precipitate first.
The ion requiring the lesser amount of Ba2+ precipitate first.
Separating Salts by Differences in Ksp
BaCO3(s) <> Ba2+(aq) + CO32(aq)
Ksp = [Ba2+][CO32]
8.1 x 109 = (x)(.10)
x = 8.1 x 108 = [Ba2+]
For ppt [Ba2+] > 8.1 x 108
x
.10
Separating Salts by Differences in Ksp
BaC2O4(s) <> Ba2+(aq) + C2O42(aq)
Ksp = [Ba2+] [C2O42]
1.1 x 107 = (x)(.10)
x = 1.1 x 106 = [Ba2+]
For ppt [Ba2+] > 1.1 x 106
x
.10
Separating Salts by Differences in Ksp
For ppt BaCO3 [Ba2+] > 8.1 x 108
For ppt BaC2O4 [Ba2+] > 1.1 x 106
Therefore;
Start ppt of BaCO3 [Ba2+] > 8.1 x 108
Max. ppt of BaCO3 [Ba2+] > 1.1 x 106
At max. ppt of BaCO3, what is the
[CO32] left in solution?
Separating Salts by Differences in Ksp
BaCO3(s) <> Ba2+(aq) + CO32(aq)
Ksp = [Ba2+][CO32]
8.1 x 109 = (1.1 x 106)(x)
x = 7.4 x 103 = [CO32]
Poor Separation: [CO32] > 105
1.1 x 106
x