1. No category

Chapter 11:
Solving Equilibrium Problems for
Complex Systems
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 For simultaneous equilibria in aqueous solutions, BaSO4(s) in
water for example, there are three equilibria:
BaSO4(s)  Ba2+ + SO42(1)
SO42- + H3O+  HSO4- +H2O
(2)
2H2O  H3O+ + OH(3)
The addition of H3O+ causes:
(2) shift right and (1) shift right.
since
Ba2+ + OAc-  BaOAc+
(4)
The addition of OAc- causes:
(1) shift right
*The introduction of a new equilibrium system into a solution
does not change the equilibrium constants for any existing
equilibria.
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11 A Solving multiple-equilibrium problems using A systematic
method
Three types of algebraic equations are used to solve multipleequilibrium problems:
(1) equilibrium-constant expressions
(2) mass-balance equations
(3) a single charge-balance equation
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11 A-1 Mass-Balance Equations
 Mass-balance equations: The expression that relate the
equilibrium concentrations of various species in a solution to
one another and to the analytical concentrations of the
various solutes. These equations are a direct result of the
conservation of mass and moles.
A weak acid HA dissolved in water for example:
HA+ H2O  H3O+ + A(1)
2H2O  H3O+ + OH(2)
mass equation 1: cHA = [HA] + [A-]
cHA is analytical concentration, [HA] and [A-] are equilibrium
concentration.
mass equation 2: [H3O+] = [A-] + [OH-]
since [H3O+] = [H3O+]from HA + [H3O+]from H2O ,
where [H3O+]from HA = [A-] , [H3O+]from H2O = [OH-]
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*
*
: conc. of H3O+ at equilibrium
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*
*
*
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11 A-2 Charge-Balance Equation
Charge-Balance Equation: An expression relating the
concentrations of anions and cations based on charge neutrality
in any solution.
Charge balance equation:
n1[C1+n1] + n2[C2+n2] + ..... = m1[A1-m1] + m2[A2-m2] + .....
Example: A solution contains H+, OH–, K+, H2PO4–, HPO42–，, and
PO43–，what is the charge balance equation?
Solution:
[H+] + [K+] = [H2PO4–] + 2[HPO42–] + 3[PO43–] + [OH–]
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+ [Cl-]
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11A-3 Steps for solving problems with several equilibria
Figure 11-1 A systematic method for solving multiple-equilibrium problems.
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11A-4 Using Approximations to Solve Equilibrium Calculations
 Approximations can be made only in charge-balance and
mass-balance equations, never in equilibrium-constant
expressions.
 If the assumption leads to an intolerable error, recalculate
without the faulty approximation to arrive at a tentative
answer.
11A-5 Use of Computer Programs to Solve MultipleEquilibrium Problems
 Several software packages are available for solving multiple
nonlinear simultaneous equations include Mathcad,
Mathematica, Solver, MATLAB, TK, and Excel.
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 A simple example of systematic calculations
Q Calculate [H3O+] and [OH-] in pure water
A Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Step 8:
2H2O  H3O+ + OH[H3O+]=? and [OH-]=?
2 unknowns
[H3O+][OH-] = 1x10-14
(1)
mass-balance equation:
[H3O+]=[OH-]
(2)
charge-balance equation:
[H3O+]=[OH-]
(3)
equations (2) and (3) are identical, omit equation (3)
two unknowns two different equations (1) and (2), OK
Approximation, omit
equation (2) substitute into equation (1)
[H3O+] [OH-] = [H3O+]2 = 1x10-14
∴ [H3O+] = 1x10-7 and [OH-] = 1x10-7
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11B Calculating solubilities by the systematic method
11B-1 Solubility of metal hydroxides
 for High Ksp value, pH controlled by the solubility
Example 11-5
Calculate the molar solubility of Mg(OH)2 in water.
Solution
Step 1. Balanced equations
2+
Mg (aq)
+2OH -(aq)
Mg(OH)2(s)
2H 2 O(l)
+
H 3O(aq)
+OH (aq)
Step 2. Unknown(s)?
s=[Mg 2+ ]=? [H 3O + ]=? [OH - ]=?
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Step 3. Equilibrium-constant expressions
K sp =[Mg 2+ ][OH - ]2 =7.1x10-12
(1)
K w =[H 3O + ][OH - ]=1x10 14
(2)
Step 4. Mass-balance equation(s)
[OH - ]=2[Mg 2+ ]  [H 3O + ]
(3)
Step 5. Charge-balance equation
[OH - ]=2[Mg 2+ ]  [H 3O + ]
(4)
identical to equation (3)
Step 6. Independent equations and unknowns
3 unknowns and 3 independent equations, OK
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Step 7. Make approximations
assume [H3O+ ] <<2[Mg 2+ ], equation (3) simplifies to
[OH - ]  2[Mg 2+ ]
Step 8. Solve the equations
substitute (5) into (1)
(5)
K sp =[Mg 2+ ][OH - ]2 =[Mg 2+ ](2[Mg 2+ ])2 =7.1x10-12
[Mg 2+ ]  s=1.21x10-4 M
substitute into (5) [OH - ]=2.42x10-4 M
substitute into (2) [H 3O + ]=4.1x10-11 M
Step 9. Check
[H3O+ ] << 2[Mg 2+ ]
OK
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 for Low Ksp value, pH ≈7, controlled by autoprotolysis of water
Example 11-6
Calculate the molar solubility of Fe(OH)3 in water.
Solution
Step 1. Balanced equations
+3OH
Fe3+
(aq)
(aq)
Fe(OH)3(s)
2H 2O (l)
+
+OH (aq)
H 3O (aq)
Step 2. Unknown(s)?
s=[Fe3+ ]=? [H 3O + ]=? [OH - ]=?
Step 3. Equilibrium-constant expressions
K sp =[Fe3+ ][OH - ]3 =2x10-39
(1)
K w =[H 3O + ][OH - ]=1x10 14
(2)
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Step 4. Mass-balance equation(s)
[OH - ]=3[Fe3+ ]  [H 3O + ]
(3)
Step 5. Charge-balance equation
[OH - ]=3[Fe3+ ]  [H 3O + ]
(4)
identical to equation (3)
Step 6. Independent equations and unknowns
3 unknowns and 3 independent equations, OK
Step 7. Make approximations
assume 3[Fe3+ ]<<[H 3O + ], equation (3) simplifies to
[OH - ]  [H 3O + ]
(5)
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Step 8. Solve the equations
substitute (5) into (2)
[OH - ]=[H 3O + ]=1.0x10-7 M
substitute into (1)
K sp =[Fe3+ ][OH - ]3 =[Fe3+ ](1.0x10-7 )3 =2x10-39
[Fe3+ ]  s=2x10-18 M
Step 9. Check
3[Fe3+ ]  [H 3O + ]
OK
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11B-2 The Effect of pH on Solubility
*The solubility of precipitates containing an anion with basic
properties, a cation with acidic properties, will depend on pH.
(simultaneous equilibria)
 Solubility Calculations When the pH Is Constant ([OH-] and
[H3O+] are known)
Example 11-7
Calculate the molar solubility of calcium oxalate in a solution
that has been buffered so that its pH is constant and equal to
4.00.
Solution
Step 1. Balanced equations
CaC2O4(s)
2+
2Ca (aq)
+C2O4(aq)
H 2C2O 4(aq) +H 2O
HC2O-4(aq) +H 2O
+
H 3O(aq)
+HC 2O-4(aq)
+
H 3O(aq)
+C2O 2-4(aq)
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Step 2. Unknown(s)
s=[Ca 2+ ]=? [C 2O 2-4 ]=? [HC 2O -4 ]=? [H 2 C 2O 4 ]=?
Step 3. Equilibrium-constant expressions
[Ca 2 ][C 2O 24 ]  K sp  1.7 10 19
(1)
[H 3O  ][HC2 O 4 ]
 K1  5.60 102
[H 2 C2 O 4 ]
(2)
[H 3O  ][C2 O 42 ]
5

K

5.42

10
2
[HC2 O 4 ]
(3)
Step 4. Mass-balance equation(s)
[Ca 2+ ] = s = [C 2O 42- ]+[HC 2O -4 ]+[H 2C 2 O 4 ] (4)
Step 5. Charge-balance equation
buffer is not specified, omit
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Step 6. Independent equations and unknowns
4 unknowns and 4 independent equations, OK
Step 7. Make approximations
omit
Step 8. Solve the equations
from equation (3)
[ H 3O  ][C2O42- ] 1.00 10-4 [C2O42- ]
2[ HC2O ] 


1.85[
C
O
2 4 ] (5)
-5
K2
5.42 10
4
substitute (5) into (2)
[ H 3O  ][ HC2O4- ] 1.00 10-4 1.85[C2O42- ]
[ H 2C2O4 ] 

K1
5.60 10-2
 3.30 10 -3[C2O42- ]
(6)
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substitute (5) and (6) into (4)
[Ca 2 ]  [C2O42- ]  1.85[C2O42- ]  3.30  10-3[C2O42- ]  2.85[C2O42- ]
[Ca 2 ]
[Ca ][C2O ]  [Ca ] 
 K sp =1.7  109
2.85
[Ca 2 ]  s  7.0  105 M
2
24
2
Also
[C2O 42- ]=2.46x10 5 M [HC 2O -4 ]=4.54x105 M
[H 2C2O 4 ]=8.11x10 8 M
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 Solubility Calculations When the pH Is Variable ([OH-] and
[H3O+] are unknown)
 for High Ksp value (pH controlled by the solubility)
omit
 for Low Ksp value (pH≈7, controlled by autoprotolysis of
water)
omit
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11B-3 The Effect of Undissociated Solutes on Precipitation
Calculations
For example, a saturated solution of AgCl(s) contains significant
amounts of undissociated silver chloride molecules, AgCl(aq)
complexs:
AgCl(s)
AgCl(aq)
+
Ag(aq)
+ Cl -(aq)
AgCl(aq)
AgCl(s)
+
Ag(aq)
+ Cl (aq)
K s =3.6x10-7
K d =5.0x10-4
K sp =K s K d  1.8x10-10
Example 11-8
Calculate the solubility of AgCl in distilled water.
Solution
+
solubility = s = [AgCl(aq) ] + [Ag(aq)
]
= 3.6x10-7 
K sp = = 3.6x10-7  1.34x105 = 1.38x105 M
neglecting [AgCl(aq) ] leads to -3% error.
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11B-4 The Solubility of Precipitates in the Presence of
Complexing Agents
The solubility increase in the presence of reagents that form
complexes with the anion or the cation of the precipitate.
Ex. F- prevent the precipitation of Al(OH)3
Al(OH)3(s)
Al3+
+
6F
(aq)
(aq)
3+
Al(aq)
+ 3OH (aq)
3AlF6(aq)
 for High stability constant
omit
 for Low High stability constant
omit
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• Complex formation with a common ion to the precipitate
may increase in solubility by large excesses of a common ion.
Example
Given:
AgCl(s)
AgCl(aq)
AgCl(s)
AgCl(aq)
+
Ag(aq)
+ Cl(aq)
+
Ag(aq)
+ Cl(aq)
AgCl(s) + Cl(aq)

AgCl 2(aq)
+ Cl -(aq)

AgCl 2(aq)
2
AgCl3(aq)
What is the the concentration of KCl at which the solubility of
AgCl is a minimum?
Solution
omit
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Figure 11-2 The effect of chloride ion concentration on the solubility of AgCl.
The solid curve sows the total concentration of dissolved AgCl.
The broken lines show the concentrations of the various silver-containing
species.
CKCl = 0.003 M
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11C Separation of ions by control of the concentration of the
precipitating agent
11C-1 Calculation of the feasibility of separations
Generally, complete precipitation is considered as 99.9%
of the target ion is precipitated, i.e., 0.1% left.
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[Fe3+][OH-]3 = 2x10-39
[Mg2+][OH-]2 = 7.1x10-12
Fe3+會先沈澱
設剩餘 0.1% 之Fe3+為 complete precipitation for Fe(OH)3，
則 [Fe3+] = 0.1x0.1% = 1x 10-4 M
1 x 10-4 x [OH-]3 = 2x10-39
[OH-] = 3 x 10-12 M 完全沈澱Fe3+所需之[OH-]
Mg(OH)2 開始沉澱之 [OH-]：
0.1 x [OH-]2 = 7.1x10-12
[OH-] = 8.4 x 10-6 M Mg2+開始沈澱之[OH-]
控制水溶液之 [OH-] = 3 x 10-12 ~ 8.4 x 10-6 M，可將 0.1 M
的 Fe3+ 與 0.1 M 的 Mg2+ 分離。
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11C-2 Sulfide Separations
 Saturated H2S, [H2S](aq) = 0.1 M
H 2S  H 2O

HS  H 2O
(1)  (2)
[H 3O  ][HS ]
H 3O + HS K1 
 9.6  108 (1)
[H 2S]



H 3O + S
[H 3O  ][S2 ]
14
K2 

1.3

10
[HS- ]
2
(2)
[H 3O  ]2 [S2 ]
=K1K 2
[H 2S]
[H 3O  ]2 [S2 ]  [H 2S]K1K 2  (0.10)(9.6  10 8 )(1.3  10 14 )  1.2 x1022
1.2 x1022
[S ] 
[H 3O  ]2
2
[S2-] in H2S saturated solution depend on the pH
 Metal sulfide solubility for (M2+S2-) in saturated H2S
MS(s)
2
K sp =[M 2 ][S2 ]
2+
M (aq)
+ S(2-aq)
[M ] = solubility =
K sp
2-
[S ]
=
K sp  [H 3O  ]2
1.2  1022
* The solubility of MS in H2S saturated solution depend on the pH
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Homework (Due 2014/11/27)
Skoog 9th edition, Chapter 11, Questions and Problems
11-5 (e) (g)
11-6 (e) (g)
11-7 (a)
11-14
End of Chapter 11
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