Activity Introduction 1.)Hydration Ions do not act as independent particles in solvent (water) Surrounded by a shell of solvent molecules Oxygen has

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

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


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


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


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


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


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

][][

][][


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.


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


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

Activity Coefficients from Debye-Hϋckel Equation


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


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


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


6.) Example 2:


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


6.) Example 2:


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.


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



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?



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



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

][][



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



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

][][



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!


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:


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


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


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



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:



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

][][][


2.) Example #2: Solubility of Calcium Sulfate Find concentrations of the major species in a saturated solution of CaSO4


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:




][][][][][]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 ][][



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

][

][][




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)



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



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




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:




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:




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:




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



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


2.) Example #3: Solubility of Magnesium Hydroxide Find concentrations of the major species in a saturated solution of Mg(OH)2


KwKw = 1.0x10-14

K1

Ksp

K1 = 3.8x102

Ksp = 7.1x10-12


][][][]2[ -2 OHHMgOHMg

[OH-] = 2[Mg2+]:



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+



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


Four equations:

Four unknowns:



][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)




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




Substitute K1 into Mass or Charge Balance:

]H][[][ OMgKMgOHK 211][][]2[ -2 OHMgOHMg

][]][[]2[ -21

2 OHOHMgKMg

Solve for [Mg2+]:

][

][][

OHK2

OHMg

1

2




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)

Excel Demo of Goal Seek

Documents

Activity Introduction 1.)Hydration Ions do not act as independent particles in solvent (water) Surrounded by a shell of solvent molecules Oxygen has