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
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
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
Activity
Each ion see less of the other ions charge and decreases the attraction
Each ion-plus-atmosphere contains less net charge and there is less attraction between any particular cation and anion
Activity
Effect of Ionic Strength on Solubility
2.) Ionic Strength () 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 () 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
i
2ii
222
211 zc
2
1zczc
2
1
where: Ci is the concentration of the ith species and zi is its charge
Activity
Effect of Ionic Strength on Solubility
2.) Ionic Strength () 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
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
ba
dc
[B][A]
[D][C]K
bB
aA
dD
cC
AA
AAK
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 ()
4.) “Real” Equilibrium Constant Using Activity Coefficients
CC CA where:
AC is activity of C [C] is concentration of C
C is activity coefficient of C
bB
baA
a
dD
dcC
c
bB
aA
dD
cC
BA
DC
AA
AAK
][][
][][
Activity Equilibrium Constant and Activity
4.) “Real” Equilibrium Constant Using Activity Coefficients is always ≤ 1
Activity coefficient measures the deviation from ideal behavior- If =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 () 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
Only valid for concentrations ≤ 0.1M
In theory, is the diameter of hydrated ion
3051
z51.0log
2
where: is the activity coefficient
is ion size (pm)z is the ion charge is the ionic strength
Activity Equilibrium Constant and Activity
5.) Activity Coefficients of Ions In practiceis 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
: Li+ > Na+ > K+ > Rb+
Ideal behavior when = 1- low ionic strength- low concentration- low charge/large
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
M10.0
2
1
2
1
zc2
1
i
2ii
22
23
222
10.033M22M0.033(1)
1][NO2][Hg
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.
= 0.355 at = 0.10 M
Activity Equilibrium Constant and Activity
6.) Example 2:
What is the activity coefficient for H+ at = 0.025 M?
Note: Values for at = 0.025 are not listed in the table.
There are two possible ways to obtain in this case:
a.) Direct Calculation (Debye-Hϋckel)
88.0
05498.0log
)305/025.0900(1
025.0)1(51.0
3051
z51.0log
H
H
22
H
zH+
for H+ from table
Activity Equilibrium Constant and Activity
6.) Example 2:
What is the activity coefficient for H+ at = 0.025 M?
b.) Interpolation
Use values for H+ given at = 0.01 and 0.05 from table and assume linear change in with
To solve for H+ at = 0.025:
86.0914.001.005.0
01.0025.0914.0
H
Diff. in values at 0.01 and 0.05
at = 0.01
Fract. Of IntervalBetween 0.01 and 0.05
Activity Equilibrium Constant and Activity
6.) Example 2:
What is the activity coefficient for H+ at = 0.025 M?
b.) Interpolation
Use values for H+ given at = 0.01 and 0.05 from table and assume linear change in with
89.0
86.0914.001.005.0
01.0025.0914.0
H
H
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
- ≈ 1 for ≤ 0.1 M- or Ac = [C]
For gases, - ≈1 for pressures ≤ 1 atm- or A ≈ P, where P is pressure in atm
8.) Limitation of Debye-Hϋckel Equation Debye-Hϋckel predicts decreases as increases
- true up to = 0.10 M At higher , the equation is no longer accurate
- at ≥ 0.5 M, most ions actually show an increase in
with an increase in - at higher , 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
2.) Affect of pH with the Addition of a Salt Changes ionic strength Changes H+ and OH- activity
HH]Hlog[AlogpH
141001 .]OH[]H[KOHHw
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
and KCl acts as an “inert salt”?
Ksp = 5.6x10-23
Activity Using Activity Coefficients
1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants Example #2:
Note: KBr is not an inert salt, since Br- is also present in the Ksp reaction of Hg2Br2
What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KBr?
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
Initial Concentration solid 0 0
Final Concentration solid x x
FLi2
FLi
FLiFLisp
)x(
)x()x(
FLAAK
][]i[
Solution: Set-up the equilibrium equation in terms of activities
Activity Using Activity Coefficients
1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants Example #3:
Note: Both x and Li+,F- depend on the final amount of LiF dissolved in solution
To solve, use the method of successive of approximation
Solution: Assume Li+ = F- = 1. Solve for x.
041.0FLix
)x()x(107.1K 2FLi
23sp
][][
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 Values.
830.0
851.0
M041.0
F
Li
Obtained by using =0.041 and interpolating data in table
851.0
835.0907.001.005.0
01.0041.0907.0
Li
Li
830.0
810.0900.001.005.0
01.0041.0900.0
F
F
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 F and Li to re-estimate [Li+] and [F-].
FLi
23sp )x(107.1K
830.0851.0 FLi substitute
M049.0FLix
)830.0)(851.0()x(107.1K 23sp
][][
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
F, Li
[F-], [Li+]
Use to calculate new concentrations.
Use concentrations to calculate new and
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.
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
][][][][n 2111 AmAmCnC 2122
(positive charge) (negative charge)
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
-,HPO42- and
PO43-, the charge balance would be:
The coefficient in front of each species always equals the magnitude of the charge on the ion.
][][][][][][ 34
2442 PO3HPO2POHOHKH
For a solution composed of 0.0250 mol of KH2PO4 and 0.0300 mol of KOH in 1.00L:
[H+] = 5.1x10-12M [H2PO4-] = 1.3x10-6 M
[K+] = 0.0550 M [HPO42-] = 0.0220M
[OH-] = 0.0020M [PO43-] = 0.0030M
M 0.0550 M 0.0550
3(0.0030)2(0.0220)101.30.0550105.1
][][][][][][612-
34
2442 PO3HPO2POHOHKHCharge balance:
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
][][ 2323 COCHHCOCHM050.0
Mass balance for 0.050 M in water:
Include ALL products in mass balance: H3PO4 H2PO4-,HPO4
2-, PO43-
][][][][ -34
-24
-4243 POHPOPOHPOHM025.0
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 PO4
3- and Ag+ when it dissolves.
Solution: If phosphate remained as PO43-, then
but, PO43- reacts with water
][][ 34PO3Ag
][][][][][ -34
-24
-4243 POHPOPOHPOH3Ag
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
-.
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
Equilibrium Applying the Systematic Treatment of Equilibrium
1.) Example #1: Ionization of water
Kw Kw = 1.0x10-14 at 25oC
Step 1: Pertinent reactions:
][][ OHH
][][ OHH
][][][ OHHOH2 :[H2O], [H+], [OH-] determined by KwNot True!
Step 2: Charge Balance:
Step 3: Mass Balance
Equilibrium Applying the Systematic Treatment of Equilibrium
1.) Example #1: Ionization of water
Step 4: Equilibrium constant expression:
14OHHw 100.1OHHK ][][
Step 5: Count equations and unknowns:
Two equations: ][][ OHH
14OHHw 100.1OHHK ][][
(1)
(2)
(1)
(2)
][ H
][ OH
Two unknowns:
Equilibrium Applying the Systematic Treatment of Equilibrium
1.) Example #1: Ionization of water
Step 6: Solve:
Ionic strength () of pure water is very low, H+ and OH- ~ 1
14OHHw 100.1OHHK ][][
][][ OHH
substitute
14100.1H1H 1][][
00.7)100.1log(HlogAlogpH100.1OHH 7H
7
H
][][][
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:
Kw
Kw = 1.0x10-14
Kbase
Kacid
Kion pair
Ksp
Kbase = 9.8x10-13
Kacid = 2.0x10-13
Kion pair = 5.0x10-3
Ksp = 2.4x10-5
This information is generally given:
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[ -4
24
2 OHHSOSO2HCaOHCa
Step 3: Mass Balance:
sulfate] [Totalcalcium] [Total
Doesn’t matter what else happens to these ions!
][][][][][][ (aq)CaSOHSOSOCaOH(aq)CaSOCa 44244
2
Step 4: Equilibrium constant expression (one for each reaction):
14OHHw 100.1OHHK ][][
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
5SO
24Ca
2sp 104.2SOCaK 2
4
][][
34pairion 100.5(aq)CaSOK ][
13
Ca
HCaOHacid 100.2
Ca
HCaOHK
][
][][
13
SO2-
4
OHHSO-4
base 108.9SO
OHHSOK
2-4
4
][
][][
Step 5: Count equations and unknowns:
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
Seven Equations:
][][][][][]2[ -4
24
2 OHHSOSO2HCaOHCa
][][][][][][ (aq)CaSOHSOSOCaOH(aq)CaSOCa 44244
2
5SO
24Ca
2sp 104.2SOCaK 2
42
][][ 34pairion 100.5(aq)CaSOK ][
14OHHw 100.1OHHK ][][
13
Ca2
HCaOHacid 100.2
Ca
HCaOHK
2
][
][][13
SO2-
4
OHHSO-4
base 108.9SO
OHHSOK
2-4
4
][
][][
Seven Unknowns:
(aq)][CaSO],[OH],[HSO],[SO],[H],[CaOH],[Ca 4-
424
2
(1)
(2)
(3) (4)
(5) (6)
(7)
(CB)
(MB)
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
13
Ca2
HCaOHacid 100.2
Ca
HCaOHK
2
][
][][13
SO2-
4
OHHSO-4
base 108.9SO
OHHSOK
2-4
4
][
][][
5SO
24Ca
2sp 104.2SOCaK 2
42
][][ 34pairion 100.5(aq)CaSOK ][
[H+]=[OH-]=1x10-7, remaining chemical reactions are independent of water
At first, ignore equations with small equilibrium constants
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:
Three equations
(aq)][CaSO],[SO],[Ca 424
2
5SO
24Ca
2sp 104.2SOCaK 2
42
][][
34pairion 100.5(aq)CaSOK ][
][][ 24
2 SOCaMass balance and charge balance reduces to: ][][][][][]2[ -
424
2 OHHSOSO2HCaOHCa
Low concentrations small equilibrium constant[H+] = [OH-]
Charge balance:Mass balance: ][][][][][][ (aq)CaSOHSOSOCaOH(aq)CaSOCa 44244
2
Simple CancellationLow concentrations small equilibrium constant
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!):
34pairion 100.5(aq)CaSOK ][ So, [CaSO4] is known
5SO
24Casp 104.2SOCaK 2
4
][][
substitute
][][ 24
2 SOCa 1;1 24
2 SOCa and
M109.4Ca104.21Ca1Ca 32522 1][][][
Therefore, only two equations and two unknowns:
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!):
M109.4CaSO 3224
11 ][][
M020.0
2
1
2
1
zc2
1
i
2ii
23-23-
2-24
22
2)10(4.92)10(4.9
2][SO2][Ca
606.0;628.0 24
2 SOCa
From table
Determine Ionic Strength:
Determine Activity Coefficients:
Given:
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:
5SO
24Ca
2sp 104.2SOCaK 2
4
][][ 606.0;628.0 24
2 SOCa
M109.7Ca104.2606.0Ca628.0Ca 3522 2 2][][][
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 Ca2+ SO42- [Ca2+] (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:
?][][][][ -24
2-4 SOandCaHSOandCaOHAre
M0092.0CaSO 224 ][][ M101HOH 7 ][][
M109100.1
0092.0108.9
OH
SOKHSO108.9
SO
OHHSOK 8
7-
13-2-4-
413
SO2-
4
OHHSO-4
base2-
4
4
][
])[(
][
])[(][
][
][][base
With:
M102100.1
0092.0100.2
H
CaKCaOH100.2
Ca
HCaOHK 8
7
13-213
Ca2
HCaOHacid
][
])[(
][
])[(][
][
][][acid
Both [HSO4-] and [CaOH+] are ~ 5 times less than [Ca2+] and [SO4
2-] 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:
KwKw = 1.0x10-14
K1
Ksp
K1 = 3.8x102
Ksp = 7.1x10-12
Step 2: Charge Balance:
][][][]2[ -2 OHHMgOHMg
[OH-] = 2[Mg2+]:
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):
But, two sources of OH-, [OH-] = [H+]:
Account for both sources of OH-::
][]}[]{[][][ HMgOHMg2MgOHOH 2-
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):
14OHHw 100.1OHHK ][][
12OHMg
2sp 101.7OHMK 2
][]g[
2
OHMg2
MgOH1 108.3
OMg
MgOHK
]H[][
][ Proper to write equilibrium equations using activities, but complexity of manipulating activity coefficients is a nuisance.
Most of the time we will omit activity coefficients
Step 5: Count equations and unknowns:
Four equations:
Four unknowns:
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
][OH],[H],[MgOH],[Mg -2
][][][]2[ -2 OHHMgOHMg CB=MB
12OH
2Mg
2sp 101.7OHMK 2
][]g[
2
OHMg2
MgOH1 108.3
OMg
MgOHK
]H[][
][
14OHHw 100.1OHHK ][][
(1)
(2)
(3)
(4)
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[ -2 OHMgOHMg CB=MB
Rearrange K1 (ignore activity coefficients):
]][[][][][
][
OHMgKMgOH
OHMg
MgOHK 2
1
OHMg2
MgOH1
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:
]H][[][ OMgKMgOHK 211][][]2[ -2 OHMgOHMg
][]][[]2[ -21
2 OHOHMgKMg
Solve for [Mg2+]:
][
][][
OHK2
OHMg
1
2
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:
][
][][
OHK2
OHMg
1
2 1222sp 101.7OHMK ]][g[
][
][ 3
OHK2
OHK
1sp
Reduces to a single equation with a single variable:
Solve using spreadsheet, vary [OH-] until obtain correct value for Ksp (7.1x10-12)