Activity
Introduction
1.) Hydration


Ions do not act as independent particles in solvent (water)
Surrounded by a shell of solvent molecules
Oxygen has a partial negative charge
and hydrogen partial positive charge
Oxygen binds cations
Hydrogen binds anions
Activity
Introduction
2.) H2O exchanges rapidly between bulk solvent and ion-coordination sites
Activity
Introduction
3.) Size of Hydration


Size and charge of ion determines number of bound waters
Smaller, more highly charged ions bind more water molecules
Activity – is related to the size of the hydrated species
Small Ions bind more water and
behave as larger species in solution
Activity
Effect of Ionic Strength on Solubility
1.) Ionic Atmosphere


Similar in concept to hydration sphere
Cation surrounded by anions and anions are surrounded by cations
- Effective charge is decreased
- Shields the ions and decreases attraction

Net charge of ionic atmosphere is less than ion
- ions constantly moving in/out of ionic atmosphere
Each ion-plus-atmosphere contains less net
charge and there is less attraction between
any particular cation and anion
Each ion see less of the other ions
charge and decreases the attraction
Activity
Effect of Ionic Strength on Solubility
2.) Ionic Strength (m)

Addition of salt to solution increases ionic strength
- Added salt is inert  does not interact or react with other ions

In general, increasing ionic strength increases salt solubility
- Opposite of common ion effect
The greater the ionic strength of a solution, the
higher the charge in the ionic atmospheres
More ions added, more ions can be
present in ionic atmospheres
Activity
Effect of Ionic Strength on Solubility
2.) Ionic Strength (m)

Measure of the total concentration of ions in solution
- More highly charged an ion is the more it is counted
- Sum extends over all ions in solution
m


1
1
c1 z12  c2 z22     ci z i2
2
2 i
where: Ci is the concentration of the ith species and zi is its charge
Activity
Effect of Ionic Strength on Solubility
2.) Ionic Strength (m)

Example: What is the ionic strength of a 0.0087 M KOH and 0.0002 M
La(IO3)3 solution? Assume complete dissociation and no formation of LaOH2+
Activity
Effect of Ionic Strength on Solubility
3.) Equilibria Involving Ionic Compounds are Affected by the Presence of All
Ionic Compounds in the Solution


Knowing the ionic strength is important in determining solubility
Example:
Ksp = 1.3x10-18
If Hg2(IO3)2 is placed in pure water, up to 6.9x10-7M will dissolve.
If 0.050 M KNO3 is added, up to 1.0x10-6M Hg2(IO3)2 will dissolve.
Occurs Due to Changes in the Ionic Strength & Activity Coefficients
Activity
Equilibrium Constant and Activity
1.) Typical Form of Equilibrium Constant
K



[C] c [D] d
[A] a [B] b
However, this is not strictly correct
Ratio of concentrations is not constant under all conditions
Does not account for ionic strength differences
2.) Activities, instead of concentrations should be used

Yields an equation for K that is truly constant
K
d
ACc AD
a b
AA
AB
where: AA, AB, AC, AD is activities of A through D
Activity
Equilibrium Constant and Activity
3.) Activities account for ionic strength effects

Concentrations are related to activities by an activity coefficient (g)
AC  C g C
where:
AC is activity of C
[C] is concentration of C
gC is activity coefficient of C
4.) “Real” Equilibrium Constant Using Activity Coefficients
K
d
ACc AD
a b
AA
AB

d
[C ]c g Cc [D ]d g D
a
b
[ A]a g A
[B ] b g B
Activity
Equilibrium Constant and Activity
4.) “Real” Equilibrium Constant Using Activity Coefficients

g is always ≤ 1

Activity coefficient measures the deviation from ideal behavior
- If g =1, the behavior is ideal and typical form of equilibrium constant is used

Activity coefficient depends on ionic strength
- Activity coefficient decrease with increasing ionic strength
- Approaches one at low ionic strength
Activity depends on hydrated radius (a) of the
ion. This includes the ion itself and any water
closely associated with it.
Activity
Equilibrium Constant and Activity
5.) Activity Coefficients of Ions

Extended Debye-Hϋckel Equation
 0.51z 2 m
log g 


m

1   a
305 


where:
g is the activity coefficient
a is ion size (pm)
z is the ion charge
m is the ionic strength

Only valid for concentrations ≤ 0.1M

In theory, a is the diameter of hydrated ion
Activity
Equilibrium Constant and Activity
5.) Activity Coefficients of Ions

In practice, a is an empirical value, provide agreement between activity and
ionic strength
- sizes can not be taken literally
- trends are sensible  small, highly charged ions have larger effective sizes
a: Li+ > Na+ > K+ > Rb+
Ideal behavior when g = 1
- low ionic strength
- low concentration
- low charge/large a
Activity
Activity Coefficients
from Debye-Hϋckel
Equation
Activity
Equilibrium Constant and Activity
6.) Example 1:
What is the activity coefficient of Hg22+ in a solution of 0.033 M Hg2(NO3)2?
Solution: Step 1 – Determine m
m
1
ci z i2

2 i



1
    [Hg 2  2 ] 22  [NO 3 ] 12
2
1
    (1)0.033 M 22  20.033M  12  0.10 M
2

Activity
Equilibrium Constant and Activity
6.) Example 1:
What is the activity coefficient of Hg22+ in a solution of 0.033 M Hg2(NO3)2?
Solution: Step 2 – Identify Activity Coefficient from table at corresponding
ionic strength.
g = 0.355 at m = 0.10 M
Activity
Equilibrium Constant and Activity
6.) Example 2:
What is the activity coefficient for H+ at m = 0.025 M?
Note: Values for g at m = 0.025 are not listed in the table.
There are two possible ways to obtain g in this case:
a.) Direct Calculation (Debye-Hϋckel)
zH+
 0.51z 2 m
 0.51( 1 )2 0.025
log g H  


 1  ( 900 0.025 / 305 )
m

1   a
305 


 log g H   0.05498
 g H   0.88
a for H+ from table
m
Activity
Equilibrium Constant and Activity
6.) Example 2:
What is the activity coefficient for H+ at m = 0.025 M?
b.) Interpolation
Use values for gH+ given at m = 0.01 and 0.05 m from table and assume
linear change in g with m.
To solve for gH+ at m = 0.025:
Fract. Of Interval
Between 0.01 and
0.05
 0.025  0.01 
g H   0.914  
0.914  0.86 
0
.
05

0
.
01

 Diff. in g values at
gH at m = 0.01
0.01 and 0.05
Activity
Equilibrium Constant and Activity
6.) Example 2:
What is the activity coefficient for H+ at m = 0.025 M?
b.) Interpolation
Use values for gH+ given at m = 0.01 and 0.05 m from table and assume
linear change in g with m.
 0.025  0.01 
0.914  0.86 
0
.
05

0
.
01


g H   0.914  
 g H   0.89
Note: This value is slightly different from the calculated
value (0.88) since it is only an estimate.
Activity
Equilibrium Constant and Activity
7.)
Activity Coefficients of Gasses and Neutral Molecules

For nonionic, neutral molecules
-

g ≈ 1 for m ≤ 0.1 M
or Ac = [C]
For gases,
-
g ≈1 for pressures ≤ 1 atm
or A ≈ P, where P is pressure in atm
8.) Limitation of Debye-Hϋckel Equation


Debye-Hϋckel predicts g decreases as m increases
true up to m = 0.10 M
At higher m, the equation is no longer accurate
at m ≥ 0.5 M, most ions actually show an increase in g
-
with an increase in m
at higher m, solvent is actually a mixture instead of just
water
Hydration sphere is mixture of water
and salt at high concentration
Activity
pH
1.)
When we measure pH with a pH meter, we are measuring the negative
logarithm of the hydrogen ion activity

Not measuring concentration
pH   log AH    log[ H  ] g H 
2.) Affect of pH with the Addition of a Salt

Changes ionic strength  Changes H+ and OH- activity
K w  [ H  ] g H  [ OH  ] g OH   1.0  10 14
Activity
pH
2.) Affect of pH with the Addition of a Salt

Example:
What is the pH of a solution containing 0.010M HCl plus 0.040 M
KClO4?
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #1:
What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KCl, where
Ksp = 5.6x10-23
and KCl acts as an “inert salt”?
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #2:
What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KBr?
Note: KBr is not an inert salt, since Br- is also present in the Ksp reaction of Hg2Br2
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #3:
What is the true concentration of Li+ and F- in a saturated solution of LiF in water?
Note: Only LiF is present in solution. Ionic strength is only determined by the
amount of LiF that dissolves
Solution: Set-up the equilibrium equation in terms of activities
Initial Concentration
solid
0
0
Final Concentration
solid
x
x
 K sp  ALi  AF   [Li ]g Li  [F  ]g F 
 ( x )g Li  ( x )g F 
 ( x )2 g Li  g F 
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #3:
Note: Both x and gLi+,gF- depend on the final amount of LiF dissolved in solution
To solve, use the method of successive of approximation
Solution: Assume gLi+ = gF- = 1. Solve for x.
K sp  1.7  10 3  ( x )2 g Li  g F   ( x )2
 x  [Li  ]  [F  ]  0.041
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #3:
Solution: Step 2 use the First Calculated Value of [Li+] and [F-] to Estimate
the Ionic Strength and g Values.
 m  0.041M
 g Li  0.851
g F  0.830
 0.041  0.01 
g Li  0.907  
0.907  0.835 
0
.
05

0
.
01


g Li  0.851
Obtained by using m=0.041
and interpolating data in
table
 0.041  0.01 
0.900  0.810 
0
.
05

0
.
01


g F  0.900  
g F  0.830
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #3:
Solution: Step 3 use the calculated values for gF and gLi to re-estimate [Li+]
and [F-].
K sp  1.7  10 3  ( x )2 g Li  g F 
substitute
g Li  0.851 g F  0.830
K sp  1.7  10 3  ( x )2 ( 0.851 )( 0.830 )
 x  [Li  ]  [F  ]  0.049 M
Activity
Using Activity Coefficients
1.)
Activity Coefficients Need to be Considered for Accurate Answers
Involving Equilibrium Constants

Example #3:
Solution: Repeat Steps 2-3 Until a Constant Value for x is obtained
For this example, this occurs after 3-4 cycles, where x = 0.050M
gF, gLi
Use concentrations to
calculate new m and g.
[F-], [Li+]
Use g to calculate
new concentrations.
Equilibrium
Systematic Treatment of Equilibrium
1.) Help Deal with Complex Chemical Equilibria



Set-up general equations
Simplify using approximations
Introduce specific conditions  number of equations = number of unknowns
2.) Charge Balance

The sum of the positive charges in solution equals the sum of the negative
charges in solution.
n1[C1]  n2 [C2 ]    m1 [ A1]  m2 [ A2 ]  
(positive charge)
(negative charge)
where
[C] is the concentration of a cation
n is the charge of the cation
[A] is the concentration of an anion
m is the charge of the anion
A solution will not have a net charge!
Equilibrium
Systematic Treatment of Equilibrium
2.) Charge Balance

If a solution contains the following ionic species: H+, OH-,K+,H2PO4-,HPO42and PO43-, the charge balance would be:
[H  ]  [K  ]  [OH  ]  [H2 PO 4 ]  2[HPO 42  ]  3[PO 43  ]
The coefficient in front of each species always equals the magnitude of the
charge on the ion.
For a solution composed of 0.0250 mol of KH2PO4 and 0.0300 mol of KOH in 1.00L:
[H+] = 5.1x10-12M
[K+] = 0.0550 M
[OH-] = 0.0020M
Charge balance:
[H2PO4-] = 1.3x10-6 M
[HPO42-] = 0.0220M
[PO43-] = 0.0030M
[H  ]  [K  ]  [OH  ]  [H2 PO4 ]  2[HPO42  ]  3[PO43  ]
5.1  10 -12  0.0550  1.3  10  6  2(0.0220)  3(0.0030)
0.0550 M  0.0550 M
Equilibrium
Systematic Treatment of Equilibrium
3.) Mass Balance



Also called material balance
Statement of the conservation of matter
The quantity of all species in a solution containing a particular atom must equal
the amount of that atom delivered to the solution
Acetic acid
Acetate
Mass balance for 0.050 M in water:
0.050 M  [CH 3CO2 H ]  [CH 3CO2 ]
Include ALL products in mass balance: H3PO4 H2PO4-,HPO42-, PO43-
0.025 M  [H3 PO 4 ]  [H2 PO 4- ]  [HPO 42 - ]  [PO 43 - ]
Equilibrium
Systematic Treatment of Equilibrium
3.) Mass Balance

Example #1:
Write the mass balance for a saturated solution of the slightly soluble salt Ag3PO4, which
produces PO43- and Ag+ when it dissolves.
Solution: If phosphate remained as PO43-, then
[ Ag  ]  3[PO 43  ]
but, PO43- reacts with water

[ Ag  ]  3  [H3 PO 4 ]  [H2 PO 4- ]  [HPO 42 - ]  [PO 43 - ]

Equilibrium
Systematic Treatment of Equilibrium
3.) Mass Balance

Example #2:
Write a mass balance for a solution of Fe2(SO4)3, if the species are Fe3+, Fe(OH)2+,
Fe(OH)2+, Fe2(OH)24+, FeSO4+, SO42- and HSO4-.
Equilibrium
Systematic Treatment of Equilibrium
1.) Write all pertinent reactions.
2.) Write the charge balance equation.

Sum of positive charges equals the sum of negative charges in solution
3.) Write the mass balance equations. There may be more than one.


Conservation of matter
Quantity of all species in a solution containing a particular atom must equal
the amount of atom delivered to the solution
4.) Write the equilibrium constant expression for each chemical reaction.

Only step where activity coefficients appear
5.) Count the equations and unknowns

Number of unknowns must equal the number of equations
6.) Solve for all unknowns
7.) Verify any assumptions
Equilibrium
Applying the Systematic Treatment of Equilibrium
1.) Example #1:

Ionization of water
Kw
Kw = 1.0x10-14 at 25oC
Step 1: Pertinent reactions:
Step 2: Charge Balance:
[H  ]  [OH  ]
Step 3: Mass Balance
[H  ]  [OH  ]


Not True! [H2O ]  [H ]  [OH ] :[H2O], [H+], [OH-] determined by Kw
Equilibrium
Applying the Systematic Treatment of Equilibrium
1.) Example #1:

Ionization of water
Step 4: Equilibrium constant expression:
Kw  [H  ]g H  [OH  ]g OH   1.0  10 14
Step 5: Count equations and unknowns:
Two equations:
Two unknowns:
(1)
[H  ]  [OH  ]
(2)
Kw  [H  ]g H  [OH  ]g OH   1.0  10 14
(1)
[H  ]
(2)
[OH  ]
Equilibrium
Applying the Systematic Treatment of Equilibrium
1.) Example #1:

Ionization of water
Step 6: Solve:
Ionic strength (m) of pure water is very low, gH+ and gOH- ~ 1
substitute
Kw  [H  ]g H  [OH  ]g OH   1.0  10 14
[H  ]  [OH  ]
[H  ]  1  [H  ]  1  1.0  10 14
[H  ]  [OH  ]  1.0  10 7  pH   log AH    log[H  ]g H   log( 1.0  10 7 )  7.00
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 1: Pertinent reactions:
This information is
generally given:
Ksp
Kion pair
Kacid
Kbase
Kw
Ksp = 2.4x10-5
Kion pair = 5.0x10-3
Kacid = 2.0x10-13
Kbase = 9.8x10-13
Kw = 1.0x10-14
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 2: Charge Balance:
2[Ca 2  ]  [CaOH  ]  [H  ]  2[SO 42  ]  [HSO 4 ]  [OH - ]
Step 3: Mass Balance:
[Total calcium]  [Total sulfate]
Doesn’t matter what
else happens to
these ions!
[Ca 2  ]  [CaSO 4 (aq)]  [CaOH  ]  [SO 42  ]  [HSO 4 ]  [CaSO 4 (aq)]
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 4: Equilibrium constant expression (one for each reaction):
K sp  [Ca2  ]g Ca [SO42  ]g SO2   2.4  10 5
4
K ion pair  [CaSO4 (aq)]  5.0  10 3
K acid 
K base 
[CaOH  ]g CaOH  [H  ]g H 
[Ca  ]g Ca
[HSO4- ]g HSO  [OH  ]g OH 
4
[SO4- 2 ]g SO - 2
4
 2.0  10 13
 9.8  10 13
Kw  [H  ]g H  [OH  ]g OH   1.0  10 14
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 5: Count equations and unknowns:
Seven Equations:
(1) 2[Ca 2  ]  [CaOH  ]  [H  ]  2[SO 42  ]  [HSO 4 ]  [OH - ] (CB)
(2) [Ca 2  ]  [CaSO 4 (aq)]  [CaOH  ]  [SO 42  ]  [HSO 4 ]  [CaSO 4 (aq)] (MB)
(3) K base 
[HSO4- ]g HSO  [OH  ]g OH 
4
[SO4- 2 ]g SO - 2
 9.8  10
13
4
(5)K sp  [Ca2  ]g Ca2  [SO42  ]g SO2   2.4  10 5
4
(4) K acid 
[CaOH  ]g CaOH  [H  ]g H 
[Ca
2
]g Ca2 
 2.0  10 13
(6) Kion pair  [CaSO4 (aq)]  5.0  10 3
(7) Kw  [H  ]g H  [OH  ]g OH   1.0  10 14
Seven Unknowns:
[Ca 2  ], [CaOH  ], [H  ], [SO 42  ], [HSO 4 ], [OH - ], [CaSO 4 (aq)]
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 6: Solve (Not Easy!):
- don’t know ionic strength  don’t know activity coefficients
- where to start with seven unknowns
Make Some Initial Assumptions:
 At first, set all activities to one to calculate ionic strength
 At first, ignore equations with small equilibrium constants
K base 
[HSO4- ]g HSO  [OH  ]g OH 
4
[SO4- 2 ]g SO - 2
 9.8  10
13
4

K acid 
[CaOH  ]g CaOH  [H  ]g H 
[Ca
2
]g Ca2 
 2.0  10 13
[H+]=[OH-]=1x10-7, remaining chemical reactions are independent of water
K sp  [Ca2  ]g Ca2  [SO42  ]g SO2   2.4  10 5
4
K ion pair  [CaSO4 (aq)]  5.0  10 3
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 6: Solve (Not Easy!):
Assumptions Reduce Number of Equations and Unknowns:
 Three unknowns:
[Ca 2  ], [SO 42  ], [CaSO 4 (aq)]

Three equations
K sp  [Ca2  ]g Ca2  [SO42  ]g SO2   2.4  10 5
4
K ion pair  [CaSO4 (aq)]  5.0  10 3
Mass balance and charge
2 

Charge
balance: 2[Ca
22[SO 2  ] 2 [HSO  ]  [OH - ]
2  2  ]  [CaOH
Mass
balance:
]

[
H
]

balance
reduces to:
[
Ca
]

[
SO
4
[Ca ]  [CaSO 4 (aq)]  [CaOH4 ]  4[SO 4 ]  [HSO
4 ]  [CaSO 4 (aq)]
[H+
] =small
[OH
]equilibrium
Low concentrations
 -small
equilibrium
constant
Low concentrations
constant
Simple
Cancellation
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 6: Solve (Not Easy!):
K ion pair  [CaSO4 (aq)]  5.0  10 3
So, [CaSO4] is known
Therefore, only two equations and two unknowns:
K sp  [Ca  ]g Ca [SO42  ]g SO2   2.4  10 5
substitute
4
[Ca 2  ]  [SO 42  ] and g Ca2   1;
g SO2   1
4
[Ca 2  ]  1  [Ca 2  ]  1  2.4  10 5  [Ca 2  ]1  4 .9  10 3 M
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 6: Solve (Not Easy!):
Given:
[SO 42  ]1  [Ca 2  ]1  4 .9  10 3 M
Determine Ionic Strength:
m
1
ci z i2

2 i


Determine Activity Coefficients:

1
    [Ca 2 ] 22  [SO24- ] 22
2
1
    (4.9  10 - 3 ) 22  (4.9  10 - 3 ) 22
2
 0.020 M
From table
g Ca2   0.628 ; g SO2   0.606
4

Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 6: Solve (Not Easy!):
Use activity coefficients and Ksp equation to calculate new concentrations:
K sp  [Ca2  ]g Ca [SO42  ]g SO2   2.4  10 5
4
g Ca2   0.628 ; g SO2   0.606
4
[Ca 2  ]  0.628  [Ca 2  ]  0.606  2.4  10 5  [Ca 2  ]2  7 .9  10 3 M
Use new concentrations to calculate new ionic strength and activity coefficients:
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:


Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4
Step 6: Solve (Not Easy!):
Repeat process until calculated numbers converge to a constant value:
Iteration
gCa2+
gSO42-
[Ca2+] (M)
m(M)
1
1
1
0.0049
0.020
2
0.628
0.606
0.0079
0.032
3
0.570
0.542
0.0088
0.035
4
0.556
0.526
0.0091
0.036
5
0.551
0.520
0.0092
0.037
6
0.547
0.515
0.0092
0.037
Stop, concentrations
converge
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #2:
Solubility of Calcium Sulfate
Find concentrations of the major species in a saturated solution of CaSO4


Step 7: Check Assumptions:
Are [CaOH  ] and [HSO 4- ]  [Ca 2 ] and [SO 42 - ] ?
With:
K base 
[OH  ]  [H  ]  1  10 7 M
[SO 42  ]  [Ca 2  ]  0.0092 M
[HSO4- ]g HSO  [OH  ]g OH 
4
[SO4- 2 ]g SO - 2
 9.8  10 13  [HSO4- ] 
(Kbase )[SO4- 2 ]


[OH ]
(9 .8  10 -13 )[0 .0092 ]
[1.0  10
-7
 9  10  8 M
]
4
K acid 
[CaOH  ]g CaOH  [H  ]g H 
[Ca2  ]g Ca
 2.0  10
13

 [CaOH ] 
(K acid )[Ca2  ]
[H  ]

(2 .0  10 -13 )[0 .0092 ]
[1.0  10 7 ]
 2  10  8 M
Both [HSO4-] and [CaOH+] are ~ 5 times less than [Ca2+] and [SO42-]
 assumption is reasonable
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 1: Pertinent reactions:
Ksp
K1
Kw
Step 2: Charge Balance:
2[Mg 2  ]  [MgOH  ]  [H  ]  [OH - ]
Ksp = 7.1x10-12
K1 = 3.8x102
Kw = 1.0x10-14
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 3: Mass Balance (tricky):
[OH-] = 2[Mg2+]:
But, two sources of OH-, [OH-] = [H+]:
Account for both sources of OH-::
[OH - ]  [MgOH  ]  2 {[Mg 2  ]  [MgOH  ]}  [H  ]
Species containing
OH-
Species containing Mg+
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 4: Equilibrium constant expression (one for each reaction):
K sp  [Mg2  ]g Mg2  [OH  ]g OH   7.1  10 12
K1 
[MgOH  ]g MgOH 
[Mg 2  ]g Mg [OH ]g OH 
 3.8  10 2
Proper to write equilibrium equations
using activities, but complexity of
manipulating activity coefficients is a
nuisance.
Kw  [H  ]g H  [OH  ]g OH   1.0  10 14
Most of the time we will omit activity coefficients
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 5: Count equations and unknowns:
Four equations:
(1) 2[Mg 2  ]  [MgOH  ]  [H  ]  [OH - ] CB=MB
(2) K sp  [Mg2  ]g Mg2  [OH  ]2 g OH   7.1  10 12
(3)
K1 
[MgOH  ]g MgOH 
[Mg 2  ]g Mg [OH ]g OH 
 3.8  10 2
(4) Kw  [H  ]g  [OH  ]g   1.0  10 14
H
OH
Four unknowns:
[Mg 2  ], [MgOH  ], [H  ], [OH - ]
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 6: Solve (Not Easy!):
Assumption to Reduce Number of Equations and Unknowns:
 Solution is very basic [OH-] >> [H+], neglect [H+]
2[Mg 2  ]  [MgOH  ]  [OH - ] CB=MB
Rearrange K1 (ignore activity coefficients):
K1 
[MgOH  ]g MgOH 
[Mg 2  ]g Mg [OH  ]g OH 
 [MgOH  ]  K1 [Mg2  ][OH  ]
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 6: Solve (Not Easy!):
Substitute K1 into Mass or Charge Balance:
K1  [MgOH  ]  K1 [Mg 2  ][OH ]
2[Mg 2  ]  [MgOH  ]  [OH - ]
2[Mg 2  ]  K1 [Mg 2  ][OH  ]  [OH - ]
Solve for [Mg2+]:
[Mg
2
]
[OH  ]
2  K1 [OH  ]
Equilibrium
Applying the Systematic Treatment of Equilibrium
2.) Example #3:


Solubility of Magnesium Hydroxide
Find concentrations of the major species in a saturated solution of Mg(OH)2
Step 6: Solve (Not Easy!):
Substitute [Mg2+] into Ksp equation:
[Mg
2
]
[OH  ]
K sp  [Mg2  ][OH  ]2  7.1  10 12
2  K1 [OH  ]
Reduces to a single equation
with a single variable:
K sp 
[OH  ]3
2  K1 [OH  ]
Solve using spreadsheet, vary [OH-] until obtain correct value for Ksp (7.1x10-12)
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