Matter and Measurement

1
The Ksp of chromium (III) iodate in water is 5.0
x 10-6. Estimate the molar solubility of the
compound. Cr(IO3)3(s) ? Cr3(aq) 3 IO3-(aq) 1
mole Cr(IO3)3 produces 1 mole Cr3 and 3 moles
IO3- Ksp Cr3(aq) IO3-(aq)3 (s) (3s)3
5.0 x 10-6 where s is the solubility of
Cr(IO3)3 s 0.021 Molar solubility Cr(IO3)3 of
0.021 M
2
Precipitation from Solution

• If equal volumes of aqueous solutions of 0.2 M
  Pb(NO3)2 and KI are mixed will PbI2(s)
  precipitate out? Ksp of PbI2 is 1.4 x 10-8
• Use the reaction quotient, Q, to predict whether
  precipitation will occur
• Pb(NO3)2 (aq) 2 KI(aq) -gt PbI2 (s) 2 KNO3
  (aq)
• Net ionic equation Pb2 (aq) 2I- (aq) -gt PbI2
  (s)
• The reverse of this reaction defines Ksp
• PbI2 (s) ? Pb2 (aq) 2I- (aq)

3

• Ksp Pb2 (aq) I- (aq)2
• If Q gt Ksp precipitation if Q lt Ksp no
  precipitation
• Equal volumes of Pb(NO3)2 and KI are mixed
• On mixing, volume of mixed solution is twice
  initial volume
• Pb2 (aq) 0.2M / 2 0.1 M
• I- (aq) 0.1 M
• Q Pb2(aq) I- (aq)2 (0.1)(0.1)2 0.001
  M
• Q gt Ksp PbI2(s) precipitates

4
Common Ion Effect

• Adding NaCl to a saturated solution of AgCl
  lowers the solubility of AgCl, reducing the
  amount of Ag(aq) and Cl- (aq)
• AgCl(s) ? Ag(aq) Cl-(aq)
• The common ion effect is the reduction in the
  solubility of a sparingly soluble salt by the
  addition of a soluble salt that has an ion in
  common with it.
• Example of LeChateliers principle.

5

• AgCl(s) ? Ag(aq) Cl- (aq) Ksp 1.6 x 10-10
  at 25oC
• Ag (aq) Cl- (aq) 1.3 x 10-5 M
• concentration of dissolved AgCl 1.3 x 10-5 M
• Dissolve AgCl in a solution of 0.10 M NaCl .
  What is the solubility of AgCl in the NaCl
  solution?
• Cl-(aq) 0.10 M
• Since Ksp at 25oC is a constant,
• Ag(aq) Ksp / Cl- (aq) 1.6 x 10-9 M
• Concentration of dissolved AgCl 1.6 x 10-9 M

6
Selective Precipitation

• A mixture of cations in solution can be separated
  by adding anions with which they form salts with
  different solubilities.
• A few drops of a solution
• of Pb(NO3)2(aq) is
• added to a solution of
• KI(aq), yellow PbI2(s)
• is formed

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8

• Ca(OH)2(s) (Ksp 5.5 x 10-6 ) and Mg(OH)2(s)
  (Ksp 1.1 x 10-11 ). A sample of sea water
  contains, among other solutes, the following
  concentrations of soluble cations 0.050 M
  Mg2(aq) and 0.010 M Ca2(aq). Determine the
  order in which each ion precipitates as solid
  NaOH is added, and give the concentration of OH-
  when precipitation of each begins. Assume no
  volume change on addition of solid NaOH.
• M(OH)2(s) ? M2(aq) 2 OH-(aq) (M Ca or Mg)
• OH- (aq) (Ksp / Ca2(aq))0.5 0.023 M
• OH- (aq) (Ksp / Mg2(aq))0.5 1.5 x 10-5
  M
• Mg(OH)2(s) will precipitate at a OH- (aq) 1.5
  x 10-5 M
• Ca(OH)2(s) will precipitate at a OH- (aq)
  0.023 M
• Mg(OH)2(s) precipitates first

9
Dissolving Precipitates

• The solubility of insoluble compounds can often
  be increased by addition of acids.
• ZnCO3 (s) ? Zn2 (aq) CO32- (aq)
• Adding acid like HNO3(aq)
• CO32- (aq) 2 HNO3(aq) -gt CO2 (g) H2O(l) 2
  NO3- (aq)
• Addition of acids reacts with the anions in
  solution lowering the concentration of the anion.
  
• The insoluble compound then dissolves further to
  increase the concentration of the anion in
  solution.
• Another example of LeChateliers principle in
  action.

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• The solubility of a solid can be increased by
  removing an ion from solution.
• Acids can be used to dissolve hydroxides,
  sulfides, sulfites, or carbonate precipitates.
• Mg(OH)2(s) ? Mg2(aq) 2 OH-(aq) Ksp 1.1 x
  10-11
• In acidic pH concentration of OH- is lowered
  increases solubility of the metal hydroxide

11
Estimate the solubility of Fe(OH)3 at 25oC in a
solution buffered to a pH of 2.9. Ksp (Fe(OH)3)
1.1 x 10-36 pOH 11.1 OH-(aq) 7.94 x 10-12
M Fe(OH)3(s) ? Fe3(aq) 3 OH-(aq) Ksp 1.1 x
10-36 Fe3(aq) OH-(aq)3 Fe3(aq) Ksp /
OH-(aq)3 1.1 x 10-36 / (7.94 x 10-12 )3
2.2 x 10-3 M molar solubility of Fe(OH)3 In
pure water, molar solubility of Fe(OH)3 is 4.5
x 10-10 M
12
Complex Ion Formation

• The formation of a complex can remove an ion,
  affecting the solubility equilibrium.
• Example reaction between a Lewis acid such as a
  metal cation and a Lewis base such as NH3.
• Ag(aq) 2 NH3(aq) ? Ag(NH3)2(aq)
• If NH3 is added to a saturated solution of AgCl,
  the Ag complexes with the NH3, removing the Ag
  from solution, increasing the solubility of AgCl
• If enough NH3 is added, all the AgCl will
  dissolve.

13

• Both dissolution and complex formation are at
  equilibrium
• AgCl(s) ? Ag(aq) Cl-(aq) Ksp Ag(aq)
  Cl-(aq)
• Ag(aq) 2 NH3(aq) ? Ag(NH3)2(aq)
• Formation constant, Kf equilibrium constant for
  complex formation
• Kf Ag(NH3)2(aq) / (Ag(aq) NH3(aq)
  2)
• 1.6 x 107 at 25oC

14

• Calculate the molar solubility of AgCl in 0.10 M
  NH3(aq) given that Ksp 1.6 x 10-10 for AgCl and
  Kf 1.6 x 107 for Ag(NH3)2.
• AgCl(s) ? Ag(aq) Cl-(aq)
• Ag(aq) 2 NH3(aq) ? Ag(NH3)2(aq)
• Overall AgCl(s) 2 NH3(aq) ? Ag(NH3)2(aq)
  Cl-(aq)
• K Ksp Kf
• Molar solubility of AgCl Cl-(aq)
• Also, Ag(NH3)2(aq) Cl-(aq)

15
NH3(aq) Ag(NH3)2(aq)
Cl-(aq) Initial 0.10 0
0 Change -2x x
x Equilibrium 0.10 - 2x
x x K
Ag(NH3)2(aq) Cl-(aq) / NH3(aq)2 Ksp Kf
2.6 x 10-3 x 4.6 x 10-3 Molar solubility of
AgCl is 4.6 x 10-3 M Compare with 1.3 x 10-5 M in
pure water
16
Qualitative Analysis

• Qualitative Analysis involves the separation and
  identification of ions by techniques such as
  complex formation, selective precipitation, and
  control of the pH of a solution.
• A solution of Pb2(aq), Hg22(aq), Ag (aq),
  Cu2(aq), Zn2(aq)

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18
(1) (2) (3) (1) Add HCl. Precipitate
Hg2Cl2, AgCl, PbCl2 (2) Add H2S. Precipitate
CuS (3) Make solution basic (add NH3),
precipitates ZnS
19
1) Precipitate of Hg2Cl2, AgCl, PbCl2 Rinse the
precipitate in hot water PbCl2 dissolves Add
CrO42- to precipitate Pb2 as PbCrO4(s) To the
Hg2Cl2, AgCl precipitates add NH3 to form
Ag(NH3)2 complex which dissolves. Ag(NH3)2(aq)
Cl-(aq) 2 H3O(aq) ? AgCl(s) 2 NH4(aq)
2 H2O(l)