# Wilkes University Ionic Strength and Activity Chemistry Questionnaire

Homework Chapter 8
1. Using activity coefficients, calculate the solubility of Ag2CrO4 (expressed as moles of
CrO42‐ per liter) in 0.050 M KClO4
2. Write the charge and mass balance equations for the solutions given below.
*** Be sure to include all possible species in aqueous solution. ***
a. a solution saturated with barium oxalate (BaC2O4).
b. a solution containing 0.300 moles of MgBr2 in 500mL of water.
(Note: once dissolved, Mg2+ and Br- ions can react in water to form small
amounts of MgBr+ and MgOH+ ions)
3. Consider a saturated solution of SrSO4 in which the following reactions can occur:
SrSO4(s) ↔ Sr2+ + SO42‐
Ksp = 3.2 × 10‐7
SO42‐ + H2O ↔ HSO4‐ + OH‐
Kb = 9.8 × 10‐13
a. Write the charge and mass balance equations in this solution (SrSO4 in water).
b. Now, consider that the pH of the solution is fixed at 2.50. Solve for the [H+] and
[OH-] in this solution.
c. If the pH is fixed, this means we have a buffer, which means there must be other
salts (conjugate acid/base pairs) in solution. So we can NOT use our charge
balance from part a. (pure water). Solve for [Sr2+] in this solution using the mass
balance and any other K expressions needed.
Chapter 8: Charge Balance, Mass Balance, and
Systematic Treatment of Equilibrium (STOE)
Partially adapted from: Harris, D. C. Quantitative Chemical Analysis; W. H. Freeman and
Company: New York, 2016; Ch. 8
1. 0.250 moles of solid sodium carbonate (Na2CO3) are dissolved in 1.000 L of water.
a. Write a balanced equation to represent the dissolution of Na2CO3
b. Do any of the ions from part a. above react with water? If so, write balanced
equations for them.
c. Are there any other reactions that occur in this solution? If so, write balanced
equations for them.
d. Write the charge balance equation for this solution.
e. Write the mass balance equation(s) for this solution.
1
2. When aqueous zinc ions (Zn+2) and aqueous sulfate ions (SO4-2) interact, they readily form
the aqueous compound zinc sulfate (ZnSO4). The equilibrium constant for this reaction is
called a Kf (i.e. the K of formation). The value of Kf for ZnSO4 is 2.2×102.
Calculate the concentration of Zn+2 ions in a 0.010 M solution of ZnSO4.
Step 1) Write the PERTINENT REACTIONS
Step 2) Write the CHARGE BALANCE equation
Step 3) Write the MASS BALANCE equation(s)
2
Step 4) Write the EQUILIBRIUM CONSTANT EXPRESSION for each chemical reaction
*** Appendix G in your textbook lists the following Ka values for H2SO4: ***
Ka1 N/A (strong acid), Ka2=1.03×10-2
Step 5) COUNT the number of equations and unknowns.
Step 6) SOLVE for the unknowns (and simplify … use your chemical knowledge and brain!)

Look at the values of K for the equilibrium expressions in Step 4 … what do you notice?

Do you think we can ignore any variables? In other words, is any chemical species
concentration so SMALL it can be considered negligible?
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Consider your mass balance equation – can you simplify it now?

Consider your mass balance equation again … how are the [Zn2+] and [SO42-] related?

Is there an equation relating all three variables present in your simplified mass balance?

Write all three variables present in this equation in terms of [Zn2+]

Solve for [Zn2+]
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Notice we did not use [H+] or [OH-] yet. Which equation(s) are they present in?

Do you expect this ZnSO4 solution to be acidic, basic, or neutral? Support your answer.
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If you had to approximate a [H+] and [OH-], what would you use?

Given your approximation, can you reasonably “ignore” [H+] and [OH-] compared to
[Zn2+] and [SO42-]?

Could you simplify your charge balance equation, too?

Does your simplified charge balance agree with the relationship between [Zn2+] and
[SO42-] from your simplified mass balance?
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Chapter 8: Ionic Strength and Activity
Adapted from: Lantz, J., Langhus, D., Fischer-Drowos, S. The Importance of Ionic Strength.
and Lantz, J., Langhus, D. Activity and Activity Coefficients. In Analytical
Chemistry: A Guided Inquiry Approach; Lantz, J., Cole, R.; Wiley, 2014
Ionic strength is a measure of the total ion concentration of a solution. It is given the
symbol µ and has units of molarity.
µ = ½ ∑[C]izi2
Where z is the charge on an ion and [C] is its concentration in units of molarity. The total
number of ions in a solution changes the behavior of the ions in equilibrium
1. a. Calculate the ionic strength of 1.0×10-2 M KNO3 solution.
b. Calculate the ionic strength of 1.0×10-2 M Mg(NO3)2 solution.
A saturated solution of AgBr(s) in deionized water contains tiny concentrations of Ag+ and Br-.
AgBr (s) ↔ Ag+ (aq) + Br- (aq)
Ksp = 5.0×10-13
When an electrolyte such as KNO3 is added to such a saturated solution, the ionic atmospheres of
Ag+ and Br- accumulate some electrolyte ions in proportion to how many of the electrolyte ion
are present. This causes the charges on Ag+ and Br-, which are largely responsible for how these
ions interact in solution, to appear somewhat diminished. This diminishing of the charges on
Ag+ and Br- ions by their ionic atmospheres could be thought of as diminishing their “effective
concentrations.”
The activity of an ion, or its “effective concentration” due to the presence of an electrolyte is:
Activity = Ai = γi[i]
Where A is the activity of an ion i, [i] is its molar concentration, and γi is the so-called activity
coefficient, a proportionality constant between molar concentration and activity that quantifies
how much the effective concentration of ion i has been diminished. Note that γ values are
dependent upon the particular ion in question and the ionic strength of the solution.
1
Table 1.
Solution
Electrolyte
µ
Concentration (Ionic
Strength)
γAg+
γBr-
(Activity
Coefficient Ag+)
(Activity
Coefficient Br-)
1
1
7.07×10-7 M
[Ag+] = [Br-]
DI water
0
KNO3
1.0×10-5 M
1.0×10-4 M
1.0×10-3 M
1.0×10-2 M
1.0×10-1 M
1.0×10-5 M
1.0×10-4 M
1.0×10-3 M
1.0×10-2 M
1.0×10-1 M
0.996
0.988
0.964
0.897
0.745
0.996
0.988
0.964
0.897
0.745
7.1×10-7 M
7.2×10-7 M
7.3×10-7 M
7.9×10-7 M
9.5×10-7 M
Mg(NO3)2 1.0×10-5 M
1.0×10-4 M
1.0×10-3 M
1.0×10-2 M
1.0×10-1 M
3.0×10-5 M
3.0×10-4 M
3.0×10-3 M
3.0×10-2 M
3.0×10-1 M
0.994
0.980
0.940
0.837
0.642
0.994
0.980
0.941
0.840
0.658
7.1×10-7 M
7.2×10-7 M
7.5×10-7 M
8.4×10-7 M
1.1×10-6 M
2. a. Examine the activity coefficients for Ag+ (γAg+) and Br- (γBr-) in Table 1. What happens to
the magnitude of the activity coefficient, γ, as the concentration of KNO3 increases?
b. Compare the changes in magnitudes of γAg+ and γBr- in KNO3 to those in Mg(NO3)2.
Which electrolyte is having a bigger impact on these activity coefficients?
c. Why does Mg(NO3)2 have a bigger impact on activity coefficients than KNO3?
2
3. a. In low ionic strength solutions:
→ µ becomes small, as shown in Table 1. What value will γ approach?
→ Under these conditions, would activity (Ai) be ________________ concentration?
(less than)
(greater than)
b. In high ionic strength solutions:
→ µ becomes large, as shown in Table 1. What value will γ approach?
→ Under these conditions, would activity (Ai) be ________________ concentration?
(less than)
(greater than)
4. Consider Table 1 again. As the concentration of either electrolyte (KNO3 or Mg(NO3)2)
increases, what happens to [Ag+] and [Br-]? For which electrolyte is there a more
dramatic change? Why?
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5. Summarize the effect of ionic strength, activity coefficients, and activity on the solubility of
a sparingly soluble salt such as AgBr(s).
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5. Including activity coefficients, calculate the solubility of AgBr(s) in a solution of 5×10-3M
KNO3. Compare your results to Table 1 – do your results seem sensible?
***Hint*** Table 8-1 in this packet (and your textbook) is useful for activity coefficients
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6. Including activity coefficients, calculate the pH of a 0.0100M KNO3 solution. Think about
your calculated value – does this seem reasonable?
***Hint*** Table 8-1 in this packet (and your textbook) is useful for activity coefficients
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Chapter 8: Activity/Equilibrium
Ionic Strength (µ)
Recall: Ch. 6 Common Ion Effect
Consider the following:
Activity Coefficients (ɣ)
• Interpolating
• Using Activity Coefficients
Why does KCl precipitate?
Common Ion Effect:
A salt will be less soluble
if one of its constituent
in the solution
Systematic Treatment of
Equilibrium (STOE)
• Charge Balance
• Mass Balance
• STOE process
1
2
Solubility of Salts
Ionic Strength
IONIC STRENGTH (µ): Measure of the total concentration of
ions in solution
Solubility ~ 0.015M
1
μ = (c1 z12 + c2 z22 + c3 z32 +
2
BUT – when 0.050M KNO3 is added to the solution,
Solubility of CaSO4 INCREASES by ~30%
KNO3 is an “inert salt,” (neither K+ or NO3- are in the CaSO4 solute)
It’s NOT the common ion effect
What’s going on?
Add salt (KNO3) to the solution, the IONIC STRENGTH increases
3
)=
1
 ci zi2
2
c = concentration of an ion
z = charge on the ion
1:1 Electrolytes
• (cation and anion both z=1)
• Ionic Strength = molarity
• ex: NaNO3
Electrolytes w/ other Stoichiometry
• (cation and anion different z)
• Ionic Strength > molarity
• ex: Na2SO4
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Ionic Strength
Chapter 8: Activity/Equilibrium
Why does solubility increase with IONIC STRENGTH?
Ionic Strength (µ)
(-) anions surround a dissolved (+) cation
Activity Coefficients (ɣ)
• Interpolating
• Using Activity Coefficients
(+) cations surround a dissolved (-) anion
Systematic Treatment of
Equilibrium (STOE)
• Charge Balance
• Mass Balance
• STOE process
Ionic atmosphere decreases attraction between ions
Greater the ionic strength, greater the charge in the ionic atmosphere
= LESS net charge = LESS attraction between anion/cation
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Activity Coefficients
Activity Coefficients
To account for µ, concentration replaced by Activities
To account for µ, concentration replaced by Activities
“Correct” Form of Equilibrium Constant:
aA + bB
cC + dD
Activity coefficient ()
• Measures the deviation of behavior from ideality
• Numeric value used to adjust concentration for
ionic strength
• If  = 1, behavior would be ideal
[C]c [D]d
Kc =
[A]a [B]b
K=
[C]c γ Cc [D]d γ dD
[A]a γ aA [B]b γ bB
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8
Activity Coefficients
Table 8-1
Activity coefficients (ɣ) and ionic strength (µ) are
related through the EXTENDED DEBYE-HUCKLE Equation
log γ =
−0.51z 2 μ
1 + α μ/305
 = activity coefficient
 = ionic strength
z = ion charge
 = ion size (in picometers, pm)
Extended Debye-Huckel equation Works well for µ ≤ 0.1M
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Table 8-1 Activity Coefficients
Table 8-1 Activity Coefficients
What if the  you need is not on the table?
All ions of the same size and charge appear in
the same group and have the same .
Option 1: log γ =
11
−0.51z 2 μ
1 + α μ/305
Option 2: Interpolate
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Interpolating Activity Coefficients
Linear interpolation
Interpolating Activity Coefficients
Linear interpolation is a simple ratio:
Assume the values between two
neighboring entries of a table can be
approximated by a straight line.
Unknown y interval known x interval
=
Δy
Δx
Unknown y interval known x interval
=
Δy
Δx
0.83 − y
20 − 16
=
 y = 0.766
0.83 − 0.67 20 − 10
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Activity Coefficients
In GENERAL, for µ 0 to 0.1M:
• As µ increases, ɣ decreases
• ɣ → 1 as µ → 0
• As magnitude of charge on ion
increases, ɣ deviates from 1
• Activity corrections more
important for z ± 3 than z ± 1
• Smaller the size (α), more
important activity effect
becomes
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Chapter 8: Activity/Equilibrium
Ionic Strength (µ)
Activity Coefficients (ɣ)
• Interpolating
• Using Activity Coefficients
Systematic Treatment of
Equilibrium (STOE)
• Charge Balance
• Mass Balance
• STOE process
16
Using Activity Coefficients
Calculate the pH of water containing 0.10M KCl at 25°C
using activity coefficients
17
Using Activity Coefficients
Calculate the pH of water containing 0.10M KCl at 25°C
using activity coefficients
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Using Activity Coefficients
Calculate the pH of water containing 0.10M KCl at 25°C
using activity coefficients
19
Using Activity Coefficients
Calculate the pH of water containing 0.10M KCl at 25°C
using activity coefficients
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Chapter 8: Activity/Equilibrium
Ionic Strength (µ)
Charge Balance
CHARGE BALANCE: Algebraic statement of electroneutrality
Activity Coefficients (ɣ)
• Interpolating
• Using Activity Coefficients
• Sum (+) charges in solution = sum (-) charges in solution
• Solutions must have zero (0) total charge
Systematic Treatment of
Equilibrium (STOE)
• Charge Balance
• Mass Balance
• STOE process
[C] = concentration of cation
n = charge on cation
[A] = concentration of anion
m = charge on anion
There can ONLY be ONE charge balance in a system
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Mass Balance
Mass Balance
There may be MORE THAN ONE mass balance in a system
MASS BALANCE: statement of conservation of matter
• Quantity of all species containing a particular atom (or
group of atoms) must be equal to the amount of that
atom (or group) initially delivered to solution
CH 3CO 2 H
CH 3CO 2− + H +
K a = 1.75×10−5
Mass balance at equilibrium:
*** When a compound dissociates in different ways,
Ex: 0.025 M solution of of phosphoric acid (H3PO4):
0.025 M = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
Formal concentration =  all relevant species after reaching equilibrium
i
There may be MORE THAN ONE mass balance in a system
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Mass Balance
La(IO3 )3 (s)
3+

3
La (aq) + 3IO (aq)
• Extent of dissociation is not known.
• Ratio of dissociated species is known.
Charge and Mass Balance Example
Example: Write the charge and mass balance equations for a saturated
aqueous solution of the slightly soluble salt Ag3PO4.
3+
[IO−
3 ] must be 3 × greater than [La ]
[Total iodate] = 3[total lanthanum]
[IO3− ] = 3[La3+ ]
If solution also contains other ions LaIO32+ and LaOH2+, then
2+
3+
2+
[IO3− ] + [LaIO2+
3 ] = 3([La ] + [LaIO3 ] + [LaOH ])
Total IO3- species
Total La3+ species
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Charge and Mass Balance Example
Example: Write the charge and mass balance equations for a saturated
aqueous solution of the slightly soluble salt Ag3PO4.
Chapter 8: Activity/Equilibrium
Ionic Strength (µ)
Activity Coefficients (ɣ)
• Interpolating
• Using Activity Coefficients
Systematic Treatment of
Equilibrium (STOE)
• Charge Balance
• Mass Balance
• STOE process
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Systematic Treatment of Equilibrium
This is simply a way to deal with all types of chemical equilibria
The basic idea: write as many independent algebraic equations as
we have unknowns (chemical species)
Systematic Treatment of Equilibrium
Step – By – Step Instructions:
1) Write the PERTINENT REACTIONS
2) Write the CHARGE BALANCE equation … there can ONLY be 1!
• Write all the chemical equilibria plus ….
3) Write the MASS BALANCE equations … there may be > 1
• CHARGE BALANCE: Sum of the + charges in solution =
sum of (-) charges in solution
4) Write the EQUILIBRIUM CONSTANT EXPRESSION for each
chemical reaction (Activity coefficients will appear here!)
• MASS BALANCE:
Quantity of all species of an atom in solution = the amount of
that atom that went into solution
5) Count the equations and unknowns
# of unknowns must = # of equations
6) SOLVE for the unknowns (and simplify!)
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Systematic Treatment of Equilibrium
Systematic Treatment of Equilibrium
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
1) Write the PERTINENT REACTIONS
2) Write the CHARGE BALANCE equation
This will be the second equilibrium in EVERY aqueous solution
Sum of + species must = sum of – species
3) Write the MASS BALANCE equations
All of the ammonia delivered to the solution is either in the form NH3 or NH4+
So … now we know our goal is to find the [NH3], [NH4+], [H+], and [OH-]
F = “formal concentration” = all species of a substance in solution
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32
Systematic Treatment of Equilibrium
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
4) Write the EQUILIBRIUM CONSTANT EXPRESSION for each
chemical reaction (Activity coefficients will appear here!)
33
Systematic Treatment of Equilibrium
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
5) Count the equations and unknowns
# of unknowns must = # of equations
34
Systematic Treatment of Equilibrium
Systematic Treatment of Equilibrium
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
6) SOLVE for the unknowns (and simplify!)
6) SOLVE for the unknowns (and simplify!)
Mass balance tells us:
Haha – even “simple” equilibrium problems can be tough … let’s simplify
For now … IGNORE activities. We can come back to them later
Eliminate one variable at a time … let’s tackle [OH-] in charge balance first
We can sub in our new equation:
and get:
We know [OH-] = Kw/[H+] …. So, we get:
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Systematic Treatment of Equilibrium
Example: Find the concentrations of species in a aqueous solution
containing 0.0100 mol NH3 in 1.000L
Systematic Treatment of Equilibrium
TA-DA! One variable. Now … how do we solve????
6) SOLVE for the unknowns (and simplify!)
Now, we have [NH4+] and [NH3] in terms of H+
Plug these, along with [OH-] = Kw/[H+] into the Kb … ignoring ɣ
37
Method 0:
Guess and check for values of H+
Then solve for OH-, and NH3 and NH4+
THEN … from these concentrations, calculate the ionic
strength
Find the activity coefficients from Table 8-1
Plug coefficients in, and RESOLVE for all the concentrations
Repeat until the concentration values agree within 1% with
new coefficients
38
Systematic Treatment of Equilibrium
TA-DA! One variable. Now … how do we solve????
Method 1:
Use Excel Goal
Seek to solve
equation with
one unknown
39
Systematic Treatment of Equilibrium
TA-DA! One variable. Now … how do we solve????
Method 2:
SIMPLIFY MORE using your chemical intuition
40
Systematic Treatment of Equilibrium
TA-DA! One variable. Now … how do we solve????
Systematic Treatment of Equilibrium
Example: Find [Zn2+] in 0.010 M ZnSO4 (aq). Ignore the activity.
1) Write the PERTINENT REACTIONS
Method 2: (aka, use your brain!)
• We know NH3 is a base.
So … we know [OH-] >> [H+]
i.e. … at pH 9, [OH-] = 10-5M, while [H+] = 10-9M
That’s 10,000 units different!!!!
we can NEGLECT [H+] in comparison to Kw/[H+] in our equation!
2) Write the CHARGE BALANCE equation
3) Write the MASS BALANCE equations
4) Write the EQUILIBRIUM CONSTANT EXPRESSION for each
chemical reaction
5) Count the equations and unknowns
# of unknowns must = # of equations
6) SOLVE for the unknowns (and simplify!)
BAM. Quadratic equation – totally solve-able.
Now, solve for [OH-], then other species.
Continue to IGNORE activities unless otherwise specified
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7
pk.
кр
Formula
pK.
кар
Chromates: L = Cro
Formula
Azides: L = N;
Cul.
Agl.
Hg2L2
TIL
PdLa(Q)
8.31
8.56
9.15
3.66
8.57
4.9 x 10
28 x 10-9
-10
22 x 10+
2.7 x 10
9
7.1 x 10
Bal
Cul
Agal
HgL
TIL
9.67
5.44
11.92
8.70
12.01
2.1 x 10
-10
3.6 X 106
1.2 x 10-12
20 x 10″
9.8 x 10
Bromates: L = Broz
Bal H0 (1)
Agl.
25.41
36.72
5.11
4.26
3.78
5.10
3.9 X 10-26
1.9 X 10
-37
TIL
PbL.
Cobalticyanides: L = C(CN)
7.8 x 10 Ag L
5.5 x 10 5
(Hg.)3L2
1.7 X 104
7.9 x 10
Cyanides: L = CN
Agl.
HgL2
5 x 10
ZnL2 (h)
5.0 x 10-13
5.6 X 10 Ferrocyanides: L = Fe(CN)
Ag L
1.3 X 10-19
Zn L.
2.1 x 10
CdL
Pb,L
15.66
39.3
15.5
22 x 10-16
5 x 1040
3 x 10-16
Bromides: L = Br
Cul.
Agl.
HgL2
TIL
HEL(f)
8.3
12.30
22.25
5.44
18.9
5.68
3.6 X 106
Pbl.2
44.07
15.68
17.38
18.02
8.5 X 1065
21 x 10-16
4.2 x 10-18
9.5 x 10-19
Cal
Sch2
BaL
Carbonates: L = co
MgL
Cal (calcite)
Cal (aragonite)
SOL.
Bal
YL
LaL
MAL
Fel
COL
NiL
Cul
AgL
HgL
ZnL
Cdl
Ры.
2.77
8.13
10.50
8.58
5.82
18.7
28.3
7.44
واها
7.46
8.35
8.22
9.03
8.30
30.6
33.4
9.30
10.68
9.98
6.87
9.63
11.09
16.05
10.00
13.74
13.13
1.7 x 10-3
7.4 x 10
3.2 x 10-11
26 X 10′
1.5 x 10-6
2 x 10-19
5 x 1023
3.6 X 10%
ThL
3.5 x 10-5
Fluorides: L=F
4.5 x 10
LiL
6.0 x 10
MgL2
9.3 x 10
-30
5.0 x 10
25 x 10-31
4.0 X 1024
5.0 x 10
-10
2.1 x 10-11
P PbL
1.0 x 10-10
1.3 x 10-7
23 x 10-20 Hydroxides: L = OH
MgL2 (amorphous)
8.1 X 10-12
MgL2 (brucite crystal)
8.9 X 10-17
Cal
1.0 X 10-10
1.8 X 10-14
BaL2.8H,0
7.4 x 10
-14
YL
LaL
Cela
UO (U4+ + 40H)
1.9 X 10-7
UO2L2 (UO3+ + 2OH)
1.8 X 100
MAL
1.2 x 10-18
Fel
1.8 X 10
-4
COL
1.7 X 105
NiL
9.2
11.15
5.19
3.6
23.2
20.7
21.2
56.2
22.4
12.8
15.1
14.9
15.2
6 X 10-10
7.1 x 10-12
6.5 X 106
3 x 10-
6 X 1024
2 x 10-21
6 x 10-22
6 X 1057
4 x 1023
1.6 X 10-13
7.9 x 10-16
1.3 x 10-15
6 X 10-16
Chlorides: L = CI
Cul
AgL
HgL2
TIL
РЫ,
6.73
9.74
17.91
3.74
4.78
PKF
K
PR
K
19.32
34.4
29.8
38.8
44.5
23.5
28.5
15.52
14.35
25.44
29.4
15.42
Formula
Phosphates: L = PO
MgHL-3H,0 (Mg2+ + HL?)
CaHL.2H20 (Ca? + HL.)
SHL (S2+ + HL?-) (b)
BaHL (Ba²+ + HL.) (b)
Lal (1)
FeL2 – 8,0
Fel-2H-0
(VO), L2 (3V02+ + 21)
Agal
Hg,HL (Hg3+ + HL?)
ZnL2- 4H20
Pb L2 (c)
4.8 x 10-20
4.0 X 1035
1.6 X 10
1.6 X 10″
3 X 104
3 X 1024
3 x 10-2
3.0 x 10-16
45 x 10-15
3.6 X 10
-26
4 x 10-30
3.8 x 1016
3 X 106
3 x 1034
10-37
13 x 10-5
6x 10-27
8 x 10-16
5 X 10-16
5.78
6.58
6.92
7.40
22.43
36.0
26.4
25.1
17.55
12.40
35.3
43.53
21.0
21.63
1.7 X 106
2.6 x 107
1.2 x 10
-7
4.0 x 10
3.7 x 10-23
1 x 10 %
4 x 10-27
8 x 10-26
28 x 10-18
4.0 x 10-13
5 x 103
3.0 x 10-4
IX 10-21
2.3 x 10-22
5.5
Formula
Hydroxides: L-OH-(Continued)
Cul
VLE
CrL. (d)
FeL
COL, (a)
VOLVO2+ + 2OH)
POL
Znly (amorphous)
COL (B)
HgO (red) (Hg* + 20H)
Cu 0 (-20* + 20H)
Ag20(2Ag + 20H)
AULA
AIL: (0)
Gal (amorphous)
InL3
SnO(Sn2+ + 20H)
PbO (yellow) (Pb2+ + 2OH)
PbO (red) (Pb2+ + 20H)
lodates: L = 103
Cal.
SOL
Bal.
YL
Lal
Cel
Thl (f)
UOL (VOS+ 2103 e)
CL (1)
Agl.
HEL2
TIL
Galg)
InL )
33.5
37
36.9
26.2.
15.1
15.3
4.62
6.50
9.96
10.37
4.83
6.13
6.20
2.4 x 10
3
3.2 x 10-7
1.1 X 10-10
43 x 10-11
1.5 x 10-5
7.4 X 10′
6.3 x 107
6.15
6.48
8.81
10.15
10.99
10.86
14.62
7.01
5.3
7.51
17.89
5.51
5.41
7.64
12.61
7.1 x 107
3.3 x 10-7
15 x 109
7.1 x 10-11
1.0 x 10-11
1.4 x 10-11
2.4 x 10-15
9.8 x 10
5 x 10-6
3.1 x 10
1.3 X 10
-18
3.1 x 10
-6
3.9 x 10-6
23 x 10-5
25 x 10-13
Sulfates: L= so
Cal
Srl
Bal
Ral (b)
AgL
HAL
Phl.
Sulfides: L=52
Mol (pink)
MnL (green)
Fel.
CoL (a)
CoL (B)
Nil(a)
Nil (B)
1.3 x 10
NiL (1)
جام
10.5
13.5
18.1
21.3
25.6
19.4
24.9
26.6
36.1
48.5
50.1
21.2
24.7
22.5
27.0
52.7
53.3
25.9
27.5
CdL
PbL
lodides: L=1
Cul
Agl.
CH HgL (CHHg + 1) (bg)
CHỊCH HgL = CHCH H’+1)
TIL
HgL2
SnL (1)
PL
Oxalates: L = co
Cal. (b, d)
Srl (b, d)
Bal. (b, d)
La Lg (b, d)
ThL2 ()
UOL (UO’ + Co (b,d)
3 x 10-11
3 X 10-14
8 x 10-19
5x 10-22
3 x 10 26
4 x 10-20
-25
3 x 10-27
8 x 1037
3 x 10-49
8 x 10-51
6 x 10-22
2 x 10-25
3 x 10-23
1 X 10-27
2 x 1053
5 x 10-54
26
3 x 10-2
-70
Cul.
Cu L
AgaL
TIL
Zal (a)
ZAL. (B)
CAL
Hyl. (black)
HEL (red)
Snl
POL
In L
Thiocyanates: L = SCN
12.0
16.08
11.46
4.11
7.23
28.34
5.08
8.10
Ix 10-12
83 x 10-17
35 x 10
-12
7.8 X 105
5.9 x 10 %
4.6 x 10
8.3 X 106
7.9 X 109
1.3 x 10
69.4
4 x 10
Culo
Agl.
Hg,L2
TIL
7.9
6.4
6.0
25.0
21.38
8.66
13.40
11.97
19.52
3.79
19.56
13 x 10
4 x 10-7
1 X 106
IX 105
4.2 x 10-22
2.2 x 10
9
-4
4.0 x 10-14
1.1 X 10-12
3.0 x 10-20
1.6 X 10
2.8 x 10-20
HEL

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