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Separating Metal Ions
Cu2+, Ag+, Pb2+
Ksp Values
AgCl
1.8 x 10-10
PbCl2
1.7 x 10-5
PbCrO4 1.8 x 10-14
Separating Salts by Differences in Ksp
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 CrO42ppts. 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. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
Calculate [CrO42-] required by each ion.
[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M
[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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1/30/2014
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 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 Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to
precipitate red Ag2CrO4 and yellow PbCrO4.
How much Pb2+ remains in solution when Ag+ begins to
precipitate?
Solution
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 < 7.8 x 10-7 M
Separating Salts by
Differences in Ksp
• 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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Separating Salts by
Differences in Ksp
• PbCl2(s) + CrO42- ↔ PbCrO4 + 2 ClSalt
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 productproduct-favored
Solubility and Complex Ions
Section 18.6
The combination of metal ions (Lewis
acids) with Lewis bases such as H2O and
NH3
------>
-----> COMPLEX IONS
Dissolving Precipitates
by forming Complex Ions
Formation of complex ions explains why
you can dissolve a ppt. by forming a
complex ion.
AgCl(s) + 2 NH3 ↔ Ag(NH3)2+ + Cl-
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Dissolving Precipitates
by forming Complex Ions
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.
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