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?

3

•

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+]

4

•

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.

5

•

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?

6

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)

(about equal to)

Support your answer.

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)

(about equal to)

Support your answer.

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?

3

5. Summarize the effect of ionic strength, activity coefficients, and activity on the solubility of

a sparingly soluble salt such as AgBr(s).

4

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

5

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

6

9/16/2021

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

ions is already present

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

4

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

5

6

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9/16/2021

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

7

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

9

10

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

12

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9/16/2021

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

13

14

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

15

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

18

3

9/16/2021

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

20

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

21

22

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,

we must include ALL products in mass balance ***

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

23

24

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9/16/2021

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

25

26

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

27

28

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!)

29

30

5

9/16/2021

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

31

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:

35

36

6

9/16/2021

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

41

42

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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