Chapter Sixteen

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Chapter Sixteen

• More Equilibria In Aqueous Solutions
• Slightly Soluble Salts And
• Complex Ions

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Contents
1. Solubility Product Constant, Ksp and Molar
Solubility 2. Common Ion Effect In Solubility
Equilibria 3. Will Precipitation Occur? Is it
Complete? 4. Effect Of pH On Solubility 5. Equilib
ria Involving Complex Ions 6. Qualitative
Inorganic Analysis
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The Solubility Product Constant, Ksp and Molar
Solubility

• Solubility The number of grams of solute in one
  liter of a saturated solution (g/L).
• Molar solubility (s) The number of moles of
  solute in one liter of a saturated solution (M).
• Solubility product constant (Ksp) The product of
  the molar concentrations of the constituent ions
• AxBy(s) ? xAm(aq) yBn(aq) Ksp
  AmxBny
• s xs ys
• Solubility equilibria calculations
• Determining a value of Ksp from experimental
  data
• Calculating equilibrium concentrations with
  known Ksp

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Some Ksp At 25 oC
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• Example 16.1
• Write a solubility constant expression for
  equilibrium in a saturated aqueous solution of
  each of the following slightly soluble salts.
• iron(III) phosphate, FePO4
• chromium(III) hydroxide, Cr(OH)3

Solution
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Example 16.2 At 20 oC, a saturated aqueous
solution of silver carbonate contains 32 mg of
Ag2CO3 per liter of solution. Calculate Ksp for
Ag2CO3 at 20oC. (FW Ag2CO3 275.8)
Solution
Ksp Ag2CO32 (2.3 x 104)2(1.2 x 104)
6.3 x 1012
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Example 16.3 Ksp for Ag2SO4 is 1.4x105 at 25 oC.
Calculate molar solubility of silver sulfate at
25oC.
Solution
2s
s
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Example 16.4 Show the trend of increasing for the
solubility of AgI (Ksp8.5x1017),
AgBr(Ksp5.0x1013), AgCl(Ksp1.8x1010).
Solution
Increasing Ksp, increasing molar
solubilities Thus, AgI lt AgBr lt AgCl
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2. Common Ion Effect In Solubility Equilibria

• Effect in solubility by common ion
• Followed Le Châteliers principle, i.e., the
  solubility of a slightly soluble ionic compound
  is lowered when a second solute that contains a
  common ion.

• Example
• For equilibrated Ag2SO4(s) ? 2Ag(aq) SO42(aq)
• What is happened once adding Na2SO4.
• What is happened once adding AgNO3.
• What is happened once adding Ag2SO4(s)

• Ans
• Shift left
• Shift left
• No effect

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Example 16.5 Calculate the molar solubility of
Ag2SO4 in 1.00 M Na2SO4(aq). (Ksp of Ag2SO4
1.4x105)
Solution
Ksp Ag2SO42, 1.4 x 105 (2s)2(1.00
s) (2s)2(1.00) 1.4 x 105 4s2 1.4 x 105 s
molar solubility 1.9 x 103 mol Ag2SO4/L
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2) Solubility and Activities

• Inert ions, i.e., that are not common to the
  precipitate can affect solubility.
• e.g., CaF2 is more soluble in 0.010 M Na2SO4
  than it is in water because of interionic
  attractions, i.e., each Ca2 and F is surrounded
  by ions of opposite charge, which interfere the
  interaction of Ca2 with F.
• The effective concentrations, called activities,
  of Ca2 and F are lower than their actual
  concentrations.

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1) Whether Precipitation Occurs
3. Will Precipitation Occur? Is it Complete?

• Calculating Qip, the ion product reaction
  quotient
• Qip gt Ksp precipitation occur
• Qip Ksp saturated, i.e., at equilibrium
• Qip lt Ksp precipitation cannot occur
• The effect of dilution when solutions are mixed
  must be considered.

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Example 16.6 If 1.00 mg of Na2CrO4 is added to
225 mL of 0.00015 M Ag(NO3)(aq), will a
precipitate form? (FW Na2CrO4 162.0, Ksp Ag2CrO4
1.1x1012)
Solution
Qip Aginitial2CrO42initial (1.5 x
104)2(2.74 x 105) 6.2 x 1013
Qip lt Ksp, no precipitation occurs.
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Example 16.8 If 0.100 L of 0.0015 M MgCl2 and
0.200 L of 0.025 M NaF are mixed, should a
precipitate of MgF2 form? (Ksp MgF2 3.7x108)
Solution
Qip Mg2F2 (5.0 x 104)(1.7 x 102)2
1.4 x 107 Qip gt Ksp, precipitation should occur
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2) Whether Precipitation Is Complete

• A slightly soluble solid never totally
  precipitates from solution.
• Generally, complete precipitation is considered
  as 99.9 of the target ion is precipitated, i.e.,
  0.1 left in solution.
• Complete precipitation is favored when
• A very small value of Ksp.
• A high initial concentration of the target ion.
• A concentration of common ion (precipitating
  reagent) that greatly exceeds that of the target
  ion.

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Example 16.9 To a solution with Ca2 0.0050
M, we add sufficient solid ammonium oxalate,
(NH4)2C2O4(s), to make the initial C2O42
0.0051 M. Will precipitation of Ca2 as
CaC2O4(s) be complete? CaC2O4(s) ? Ca2(aq)
C2O42(aq) Ksp 2.7 x 109
Solution
Step 1 whether precipitation occured Ca2
0.0050 M, and C2O42 0.0051 M. Qip
Ca2C2O42 (5.0 x 103)(5.1 x 103) 2.6 x
105 Qip gt Ksp, precipitation occur
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Step 2 The solubility equilibrium
Excess C2O42 0.0051M 0.0050 M 0.0001 M

• Ksp Ca2C2O42 s(0.0001 s) ? s x 0.0001
  2.7 x 109
• s Ca2 3 x 105 M
• Recalled initial Ca2 5x103 M

Ans Left Ca2 gt 0.1, precipitation is
incomplete
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3) Selective Precipitation
Example AgCl(s) ? Ag(aq) Cl(aq) Ksp 1.8 x
1010 AgI(s) ? Ag(aq) I(aq) Ksp 8.5 x 1017

• When AgNO3(aq) is added to an aqueous solution
  containing Cl and I, the first precipitate to
  form yellow AgI(s).
• (b) Essentially all the I has precipitated
  before the precipitation of white AgCl(s) begins.

(a)
(b)
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Example 16.10 An aqueous solution that is 2.00 M
in AgNO3 is slowly added from a buret to an
aqueous solution that is 0.0100 M in Cl and
0.0100 M I. AgCl(s) ? Ag(aq) Cl(aq) Ksp
1.8 x 1010 AgI(s) ? Ag(aq) I(aq) Ksp 8.5 x
1017

• Which ion, Cl or I, is the first to precipitate
  from solution?
• When the second ion begins to precipitate, what
  is the remaining concentration of the first ion?
• Is separation of the two ions by selective
  precipitation feasible?

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• Solution
• To precipitate Cl

To precipitate I
Ans I is the first ion to precipitate
(b) When Cl begins to precipitate, the remaining
I
Ans 4.7 x 109 M
(c)
Ans I remaining is far below the 0.1,
separation of the two anions is feasible.
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4. Effect Of pH On Solubility
Example 1 AgCl(s) ? Ag(aq) Cl(aq) pH
decrease (H increase), no effect on
solubility Example 2 CaF2(s) ? Ca2(aq)
2F(aq) Dissolution reaction 2H3O(aq) 2F(aq)
? 2HF (aq) Acid-base reaction CaF2(s)
2H3O(aq) ? Ca2(aq) 2HF(aq) 2H2O(l) pH
decrease (H increase), solubility
increase. Example 3 Mg(OH)2(s) ? Mg2(aq)
2OH(aq) Dissolution reaction 2H3O(aq)
2OH(aq) ? 4H2O(l) Acid-base reaction Mg(OH)2(s)
2H3O(aq) ? Mg2(aq) 4H2O(l) pH decrease
(H3O increase), solubility increase.
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Example 4 CaCO3(s) ? Ca2(aq)
CO32(aq) 2HCl 2H2O ? 2H3O(aq)
Cl(aq) CO32(aq) H3O(aq) ? HCO3(aq)
H2O(l) HCO32(aq) H3O(aq) ? H2CO3 (aq)
H2O(l) H2CO3 (aq) ? CO2(g) H2O(l) CaCO3(s)
2HCl ? Ca2(aq) 2Cl(aq) H2O(l) CO2 (g) pH
decrease (H3O increase), solubility increase.
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Example 16.11 What is the molar solubility of
Mg(OH)2(s) in a buffer solution having OH
1.0x105 M, that is, pH 9.00? Mg(OH)2(s) ?
Mg2(aq) 2OH(aq) Ksp 1.8 x 1011
Solution
Ans s 0.18 M
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1) Complex-Ion
5. Equilibria Involving Complex Ions

• Consists of a central metal atom or ion bonded
  with ligands.
• Metal center, Lewis acid, accepts electron pairs
• Ligands, Lewis bases, donate electron pairs.
• At least one lone pair of electron in the Lewis
  structures of ligands (Lewis base).
• The ligands and metal center are jointed by
  coordinate covalent (dative) bonds.
• The constant of formation reaction of a complex
  ion is called a formation constant (or stability
  constant).

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Common ligands examples

• Complex ion formation example

Ag(aq) 2 Cl(aq) AgCl2(aq)
AgCl2 Kf 1.2 x 108
AgCl2
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Complex Ion Formation Example
Cu2(aq) 4NH3(aq) ? Cu(NH3)42(aq) Blue
Deep blue-violet
Cu(NH3)42
SO42
Dilute CuSO4(aq)
After adding NH3(aq)
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Some Formation Constants
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2) Solubility of AgCl(s) in NaCl(aq)
Ag(aq) Cl(aq) ? AgCl(s) AgCl(s) Cl(aq) ?
AgCl2(aq) Ag(aq) 2Cl(aq) ? AgCl2(aq)
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3) Complexation effect on Solubility

• AgCl(s) in NH3(aq) solution for example
• AgCl(s) ? Ag(aq) Cl(aq) Ksp 1.8 x 1010
• Ag(aq) 2NH3(aq) ? Ag(NH3)2(aq) Kf 1.6 x
  107
• AgCl(s) 2NH3(aq) ? Ag(NH3)2(aq)
  Cl(aq) Kc Ksp x Kf 2.9 x 103
• In the presence of a high concentration of NH3, a
  significant amount of AgCl dissolves.

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Example 16.13 Calculate the concentration of Ag,
in an aqueous solution prepared as 0.10 M AgNO3
and 3.0 M NH3. Ag(aq) 2NH3(aq) ?
Ag(NH3)2(aq) Kf 1.6 x 107
Assume all shift right firstly
Solution
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Complex Ions in AcidBase Reactions

• H2O is a Lewis base act as ligands in complex
  ions
• Al(H2O)63 Fe(H2O)63
• For those complexes consisted with the small,
  highly central charged metal ion weakens an OH
  bond, may give up its proton to another water
  molecule.
• Those complex ion acts as an acid caused by the
  hydrolysis of the hydrated ion.

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Complex Ion Ionization Example
Fe(H2O)63 H2O Fe(H2O)5OH2
H3O
Ka 1 x 107
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5) Amphoteric hydoxides and oxides

• Amphoteric hidroxides Al(OH)3, Zn(OH)2, and
  Cr(OH)3 etc.
• Example
• Al(OH)3(s) 3H3O(aq) ? Al(H2O)63(aq)
• Al(OH)3(s) OH(aq) ? Al(OH)4(aq)
• Amphoteric oxides Al2O3, ZnO, and Cr2O3 etc.
• Example
• Al2O3(s) 6H3O(aq) 3H2O(l) ?
  2Al(H2O)63(aq)
• Al2O3(s) 2OH(aq) 3H2O(l) ? 2Al(OH)4(aq)

Dissolution by forming complex
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6. Qualitative Inorganic Analysis

• Classical qualitative inorganic analysis
• Identify inorganic ions by acid-base chemistry,
  precipitation reactions, oxidation-reduction, and
  complex-ion formation etc.
• Qualitative interest in what is present, not
  how much is present.
• Classical qualitative analysis relevant to all
  the basic concepts of equilibria in aqueous
  solutions.

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2) Outline of qualitative analysis for some
cations
The group numbers used in the qualitative
analysis scheme are unrelated to group numbers in
the periodic table
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3) Group 1 Cations (Chloride group Pb2, Hg22,
Ag)

• Adding HCl resulted in PbCl2(s), Hg2Cl2(s),
  AgCl(s) if the sample contains those cations.
• Separation by filtration or centrifugation
• Supernatant for other groups cations analysis.
• Precipitates for Pb2, Hg22, Ag.
• Group 1 cations analysis
• i) Analyzing for Pb2(aq)
• Treat the precipitate with hot water for
  dissolving PbCl2(s), i.e., releasing Pb2(aq).
• Treat the washings with K2CrO4 (aq). If Pb2(aq)
  was present, forming a yellow precipitate,
  PbCrO4(s).

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• ii) Analyzing for Ag
• Treat the undissolved precipitate (in step i))
  with NH3(aq) for dissolving AgCl(s), i.e.,
  releasing Ag(aq).
• (Save precipitate for iii) Hg22 analysis).
• Treat the supernatant with HNO3(aq). If Ag(aq)
  was present, reforming a white, AgCl(s).
• iii) Analyzing for Hg22
• Treat the undissolved precipitate (in step ii))
  with NH3(aq). If Hg22 present, Hg22 was
  present, forming dark gray mixture, Hgo(s) and
  HgNH2Cl(s).

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End of Chapter 16
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