Chapter Fifteen - Kerman Medical University

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Transcript Chapter Fifteen - Kerman Medical University

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Chapter Fifteen
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Solubility products
AgCl(s) ⇄ Ag+(aq) + Cl-(aq).
Kc

Ag



(aq ) Cl  (aq )
AgCl(s)

  
K sp  K c  [AgCl]  Ag  Cl  .
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Slide 2 of 31
Solubility products
• Ca(OH)2(s)  Ca2+(aq) + 2OH-(aq)
•Ksp = [Ca2+][OH-]2
• Ag2CrO4(s)  2Ag+(aq) + CrO42-(aq)
•Ksp = [Ag+]2[CrO42 -]
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Slide 3 of 31
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Solubility or precipitation
AgCl(s) ⇄ Ag+(aq) + Cl-(aq).
• Q sp =[Ag+]×[Cl-]
– Q sp < K sp
More solubility
– Q sp = K sp
Saturation
•
– Q sp > K sp
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precipitation
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Solubility & Solubility products
CaF2(s) ⇄ Ca+2(aq) + 2F-(aq)
x(2 x)  4.0 10
2
Ksp = 4.0 × 10-11
11
•x = 2.2 × 10-4 M
•0.017 g CaF2/liter.
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Common ion effect & solubility
•
Solubility of AgCl in H2O
• AgCl(s) ⇄ Ag+(aq) + Cl-(aq).
•
Ksp=1.8 × 10-10
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x = 1.3 × 10-5
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Common ion effect & solubility
•
Solubility of AgCl in NaCl(0.1M)
• AgCl(s) ⇄ Ag+(aq) + Cl-(aq).
•X
•X+0.1
x( x  0.100)  1.8 10
x(0.100)  1.8 10
10
10
•x = 1.8 × 10-9 << 1.3 × 10-5.
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Common ion effect & solubility
•x = 1.8 × 10-9 0.1M NaCl<< 1.3 × 10-5 H2O
•
Solubility of AgCl in NaCl(M)
• AgCl(s) ⇄ Ag+(aq) + Cl-(aq).
•More Conc
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Soluble [Ag(Cl)2]-
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Salt effect & solubility
•
Solubility of AgCl in NaNO3
• AgCl(s) ⇄ Ag+(aq) + Cl-(aq).
• AgCl(s) ⇄ Ag+ + Cl-
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What Concentration of Ag+ is needed for precipitation
of Ag2CrO4 in 0.001M solution of CrO42-?
Ag2CrO4(s)  2Ag+ + CrO42KSP = [Ag+]2[CrO42-] = 1.1 x 10-12
[CrO42-] = 0.00100 M
[Ag+]2 = 1.1 x 10-12 / 0.00100 = 1.1 x 10-9
[Ag+] >3.3 x 10-5 M
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Solubility & pH
• If Anion=weak base
• Cation=weak acid
• CaF2Ca2+ +2F-
• H+
• AgClCl- + Ag+
• OH-
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• HF
AgOH
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Solubility & pH
• What Conc. Of NH4+ is necessary to prevent
the formation of Mg(OH)2 precipitate in a sol.
of 0.050M in Mg2+ & 0.050M in NH3. Ksp
Mg(OH)2=8.9×10-12,
• Kb NH3=1.8×10-5.
• [NH4]+>6.9 ×10-2M
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Precipitation of sulfides
• Pb2+ +H2S Pb(HS)2 PbS(s) + H2S
• PbS(s)  Pb2+ + S2Ksp=7.0× 10-29
• H2S(s)  2H+ + S2-
Ksp=1.1× 10-22
Fix
HCl
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Precipitation of sulfides
• What must be the H+ Conc. Of a Sol. That is
0.050M in Ni2+ to prevent the precipitation of
NiS when the sol. Is saturated with H2S.
• Ksp NiS=3×10-21
• [H+]> 0.04 M
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Precipitation of sulfides
• A sol. That is 0.050M in Cd2+ and 0.10M in H+
is saturated with H2S.What Conc. Of Cd2+
remains in sol. After CdS has precipitated. Ksp
CdS=1.0×10-28 .
• [Cd2+]= CdS=3.6×10-8 .
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Equilibria involving complex ions

• Zn2+ + 6H2O 
[Zn(H2O)6]2+
• Ligand=complexor




Cu 2  : NH 3 
[CuNH 3]2 
[Cu ( NH 3)2 ]2 
[Cu ( NH 3)3 ]2 
[Cu ( NH 3)4 ]2
• Coordination No.
• Chelate
• Ks =Kf =1/Kins=1/Kd
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Equilibria involving complex ions
•
•
•
•
What is the Sol. Of AgCl in 0.1M NH3 ?
Ksp AgCl=1.7×10-9
Kins [Ag(NH3)2]+ =6.0×10-8
S=4.8×10-3
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Equilibria involving complex ions
• A 0.010M Sol. Of AgNo3 is made in 0.50M NH3
and The complex [Ag(NH3)2]+ forms.
1) What is the Conc. Of Ag+ in the solution?
2) What percentage of total Conc. Of silver is in
the form of Ag+ ?
1)[Ag+]=2.6×10-9
2)2.6×10-5%
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Complex Ions with S2• CuS + S2-×××
• AS2S3 + S2- 2[AsS3]3• Al3+ + …..NH3  Al(OH)3(s)
• Zn2+ +…..NH3  [Zn(NH3)4]2+
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Amphoterism
• Mg2+ +…..OH- Mg(OH)2(s)
• Zn2+ + …...OH- [Zn(OH)4]2-
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