Transcript Chapter 8 cont.(Systematic Equilibrium)

Help !
I’m
dissolving
SYSTEMATIC
TREATMENT OF
EQUILIBRIUM AND ACID
BASE EQUILIBRIUM
OR
Using complex algebra to calculate
concentrations when multiple equilibria are
in play

We can look for two more pieces of info:
Charge Balance: Electroneutrality of the solution; the
sum of the positive charge in solution equals the
sum of negative charges in the solution.
Mass Balance: Conservation of matter; the quantity of
all species in a solution containing a particular atom
(or group of atoms) must equal to the amount of that
atom (or group) delivered to the solution.
SYSTEMATIC TREATMENT OF EQUILIBRIUM

We can look for two more pieces of info:
Charge Balance: Electroneutrality of the solution; the
sum of the positive charge in solution equals the
sum of negative charges in the solution.
[H+] + 2[Ca2+] = [OH-] + [F-]
(4)
Mass Balance: Conservation of matter; the quantity of
all species in a solution containing a particular atom
(or group of atoms) must equal to the amount of that
atom (or group) delivered to the solution.
2[Ca2+] = [F-] + [HF]
(5)
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SYSTEMATIC TREATMENT OF EQUILIBRIUM
How does the solubility of CaF2 depend on pH?
CaF2(s)  Ca2+ + 2F-
Ksp = [Ca2+][F-]2 = 3.910-11
F-+H2O  HF + OH-
[HF][OH  ]
11
Kb 

1
.
5

10
[F  ]
H2O  H+ + OH-
Kw = [H+][OH-] = 1.010-14
(1)
(2)
(3)
Five unknowns: [Ca2+], [F-], [HF], [H+], and [OH-]
Three equations mean we need more equations…
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SYSTEMATIC TREATMENT OF EQUILIBRIUM

We can look for two more pieces of info:
Charge Balance: Electroneutrality of the solution; the
sum of the positive charge in solution equals the
sum of negative charges in the solution.
[H+] + 2[Ca2+] = [OH-] + [F-]
(4)
Mass Balance: Conservation of matter; the quantity of
all species in a solution containing a particular atom
(or group of atoms) must equal to the amount of that
atom (or group) delivered to the solution.
2[Ca2+] = [F-] + [HF]
(5)
Ksp = [Ca2+][F-]2 = 3.910-11
[HF][OH  ]
11
Kb 

1
.
5

10
[F  ]
(1)
(2)
Kw = [H+][OH-] = 1.010-14
(3)
[H+] + 2[Ca2+] = [OH-] + [F-]
(4)
2[Ca2+] = [F-] + [HF]
(5)
We need to substitute to get things in terms of Ca2+ or H+
SOLVING THE EQUATIONS
[HF][OH  ]
11
Kb 

1
.
5

10
[F  ]
Kw = [H+][OH-] = 1.010-14
2[Ca2+] = [F-] + [HF]
Ksp = [Ca2+][F-]2 = 3.910-11
Combine eq. 2 and eq. 5, we have…
K b [F  ]
2
[F ] 

2
[
Ca
]

[OH ]

2
2
[
Ca
]

[F ] 
Kb
1
[OH  ]
(B)
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SOLVING THE EQUATIONS
(B)
Combine (B) and eq. 1, we have…
2
2[Ca ]
[F ] 
Kb
1
[OH  ]

(C)
 K sp 
Kb 
1 

[Ca ]  
 
 4  [OH ] 
2
K sp




2
2[Ca ] 
 [Ca2 ]

Kb 
1

 

[
OH
]

2
It’s much simpler if we
can consider the pH
fixed (how would we do
that?)
2



1
3
If we can fix pH…
[H+] = 1.0×10-3 M
pH = 3.00
Kw
[OH-] = 1.0×10-11 M
Kb
[HF-] =1.5[F-]
Mass
[Ca2+] = 3.9×10-4 M
[F-] =0.80[Ca2+]
Ksp
Applications of coupled equilibria in the modeling of environmental problems
The [Ca] removed from marble
stone (largely dissolution of
CaCO3) increases as the [H+] of
acid rain increases.
CaCO3(s) + 2H+(aq) 
Ca2+(aq) + CO2(g) + H2O(l)
SO2(g) + H2O(l)  H2SO3(aq)
oxidation
H2SO4(aq)
Deposits include CaSO4•2H2O
(gypsum), which accumulates
creating a black residue.
www.chem.wits.ac.za/chem212-213-280
http://pubs.usgs.gov/gip/acidrain/5.html
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Acid rain also releases Al, Hg, and Pb into the environment.
Total [Al] as a function of pH in
1000 Norwegian lakes.
www.chem.wits.ac.za/chem212-213-280
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