essay

Chemical Equilibrium

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• Paper mill on the Potomac River near Westernport, Maryland, neutralizes acid mine drainage in the water. Upstream of the mill, the river is acidic and lifeless; a victim of acid drainage from abandoned coal mines.

• Below the mill, the river teems with life. [Photo courtesy C. Dalpra, Potomac River Basin Commission

• CO2 has lowered the average pH of the ocean from its preindustrial value of 8.16 to 8.04 today.3 Without changes in our activities, the pH could be 7.7 by 2100. 2

• Professor Marc Edwards of Virginia Tech, a leading researcher on municipal drinking water, has confirmed that "there is nine times more chloride, which is the key ingredient in the corrosive water. . . in the Flint River than in Lake Huron water." Dr. Edwards suspects that road salt is at least partially to blame for this. Similarly, the Unites States Geological Survey has determined that concentrations of chloride ions have increased dramatically in urban rivers and streams in recent years. The USGS study found that "de-icing activity was the primary source of environmental chloride in urban areas of the northern US. . . ." As Dr. Edwards notes, these high concentrations of chloride ions are simply not present in Lake Huron water.

• https://www.youtube.com/watch?v=xob2_6qhy9c

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What Makes Flint River Water So Corrosive?

Flint drinking water pipes, showing different kinds of iron

corrosion and rust (Photo: Min Tang and Kelsey Pieper)

Introduction

1.) Equilibria govern diverse phenomena  Equilibria govern diverse phenomena from the folding of proteins to the action of

acid rain on minerals to the aqueous reactions used in analytical chemistry.

 This chapter introduces equilibria for the solubility of ionic compounds, complex formation, and acid-base reactions. Chemical equilibrium provides a foundation not only for chemical analysis, but also for other subjects such as biochemistry, geology, and oceanography

2.) Chemical equilibrium applies to reactions that can occur in both directions:

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First, system reaches equilibrium

Introduction

1.) Equilibria govern diverse phenomena  Protein folding, acid rain action on minerals to aqueous reactions

2.) Chemical equilibrium applies to reactions that can occur in both directions:

 reactants are constantly forming products and vice-versa  At the beginning of the reaction, the rate that the reactants are changing into the

products is higher than the rate that the products are changing into the reactants.  When the net change of the products and reactants is zero the reaction has

reached equilibrium.

At equilibrium the amount of reactants and products are constant, but not necessarily equal

Then, system continually exchanges products and reactants, while maintaining equilibrium distribution.

Reactants Product

5

Equilibrium Constant

1.) The relative concentration of products and reactants at equilibrium is a constant.

2.) Equilibrium constant (K):  For a general chemical reaction

Equilibrium constant:

Where: - small superscript letters are the stoichiometry coefficients - [A] concentration chemical species A relative to standard state

ba

dc

][][ ][][

BA DC

K =

6

Notice

Equilibrium Constant

- Equilibrium constant (K):  A reaction is favored when K > 1  K has no units, dimensionless

- Concentration of solutes should be expressed as moles per liter (M). - Concentrations of gases should be expressed in bars.

- Concentrations of pure solids, pure liquids and solvents are omitted ► are unity ► standard state is the pure liquid or solid

- Manipulating Equilibrium Constants

][ ]][[

HA AH

K 1 −+

=

1 ' 1 K/1

AH HA

K == −+ ]][[

][

Consider the following reaction:

Reversing the reaction results in a reciprocal equilibrium reaction:

7

Notice

K1

Chemical Equilibrium Equilibrium Constant

- Manipulating Equilibrium Constants

K2

K3

If two reactions are added, the new K is the product of the two individual K values:

]][[ ]H][[

]][[ ][C

][ ]][[

CHA CA

CH H

HA AH

KKK 213 +−

+

+−+ =⋅==

][ ]][[

HA AH

K 1 −+

= ]][[

][ CH

CH K 2 +

+ =

]][[ ]H][[

CHA CA

K 3 +−

=

8

2-equaitions are added

But multiply K

- Manipulating Equilibrium Constants  Example:

Kw= 1.0 x 10-14 Given the reactions and equilibrium constants:

KNH3= 1.8 x 10 -5

Find the equilibrium constant for the reaction:

Solution:

K1= Kw

K2=1/KNH3 K3=Kw*1/KNH3=5.6x10

-109

Equilibrium and Thermodynamics

1.) Equilibrium constant derived from the thermodynamics of a chemical reaction.  deals with the relationships and conversions between heat and other forms of

energy

2.) Enthalpy ∆H  ∆H – is the heat absorbed or released when the reaction takes place under

constant applied pressure

∆H = Hproducts – Hreactants  Standard enthalpy change (∆Ho)

all reactants and products are in their standard state.

 ∆Ho – negative  heat released - Exothermic Solution gets hot

 ∆Ho – positive  heat absorbed - Endothermic - Solution gets cold

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Equilibrium and Thermodynamics

3.) Entropy ∆S  Measure of a substances “disorder”  Greater disorder  Greater Entropy

- Relative disorder: Gas > Liquid > solid

∆S = Sproducts – Sreactants  ∆So – change in entropy when all species are in standard state.

- positive product more disorder

- negative  product less disorder

∆So = +76.4 J/(K.mol) at 25oC

More disorder for aqueous ions than solid

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Equilibrium and Thermodynamics

4.) Free Energy  Systems at constant temperature and pressure have a tendency

toward lower enthalpy and higher entropy

 Chemical reaction is favored if: - ∆H is negative  heat given off

and - ∆S is positive  more disorder

 Chemical reaction is not favored if: - ∆H is positive and ∆S is negative

 Gibbs Free Energy (∆G): determines if a reaction is favored or not when both ∆H and ∆S are positive or negative - A reaction is favored if ∆G is negative

where T is temperature (Kelvin)

Free energy: ∆G = ∆H -T∆S

∆G

12Memorize

spontaneous

Example:

Is the following reaction favored at 25oC?

∆Ho = -74.85 x 103 J/mol ∆So = -130.4 J/K.mol

Free energy: ∆G = ∆H –T∆S = (-74.85x103 J/mol) – (298.15K)(-130.4 J/K.mol)

∆G = -35.97 kJ/mol  ∆G negative  reaction favored

Favorable influence of enthalpy is greater than unfavorable influence of entropy

Chemical reaction is not favored if:

- ∆H is positive and ∆S is negative

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Equilibrium and Thermodynamics

Free Energy and Equilibrium  Relate Equilibrium constant to the energetic (∆H & ∆S) of a reaction  Equilibrium constant depends on ∆G:

 where R (gas constant) = 8.314472 J/(K.mol) T = temperature in kelvins

 The more negative ∆G  larger equilibrium constant  Example:

RT G o

eK ∆−

=

∆G = -35.97 kJ/mol

6)K15.298)(molK/(J314472.8)(mol/J10x97.35(RT G

10x00.2eeK .3

o

=== −− −∆

Because K is very large, HCl is very soluble in water and nearly completely ionized

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Notice-units Memorize

- Free Energy and Equilibrium  If ∆Go is negative or K >1 the reaction is spontaneous

- Reaction occurs by just combining the reactants

 is If ∆Go is positive or K < 1, the reaction not spontaneous - Reaction requires external energy or process to proceed

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Le Châtelier’s Principal 1.) What Happens When a System at Equilibrium is Perturbed?

 Change concentration, temperature, pressure or add other chemicals

 Equilibrium is re-established - Reaction accommodates the change in products, reactants,

temperature, pressure, etc. - Rates of forward and reverse reactions re-equilibrate

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Le Châtelier’s Principal

1.) What Happens When a System at Equilibrium is Perturbed?  Le Châtelier’s Principal:

- the direction in which the system proceeds back to equilibrium is such that the change is partially offset.

Consider this reaction:

At equilibrium:

Add excess CO(g):

To return to equilibrium (balance), some (not all) CO and H2 are converted to CH3OH

If all added CO was converted to CH3OH, then reaction would be unbalanced by the amount of product 17

Le Châtelier’s Principal

2.) Example:

Consider this reaction:

C25at101 CB

HCB K o11×==

+

+

23- 3

8-2 72

-

]r][rO[ ]][Or][r[

At one equilibrium state:

M 0.043][BrOM 1.0][Br

M 0.0030][CrM 0.10]O[CrM 5.0][H - 3

3-2 72

==

=== −

++

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Le Châtelier’s Principal

Example:

What happens when:

M 0.20M 0.10]O[Cr -272 tofromincreased

According to Le Châtelier’s Principal, reaction should go back to left to off-set dichormate on right:

Use reaction quotient (Q), Same form of equilibrium equation, but not at equilibrium:

( )( )( ) ( )( )

K102 0030.0043.0

0.520.00.1 CB

HCB Q 11

2

8 >×===

+

+

23- 3

8-2 72

-

]r][rO[ ]][Or][r[

C25at101 CB

HCB K o11×==

+

+

23- 3

8-2 72

-

]r][rO[ ]][Or][r[

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Because Q > K, the reaction must go to the left to decrease numerator and increase denominator.

Continues until Q = K:

1. If the reaction is at equilibrium and products are added (or reactants removed), the reaction goes to the left

2. If the reaction is at equilibrium and reactants are added ( or products removed), the reaction goes to the right

Example:

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Le Châtelier’s Principal

Affect of Temperature on Equilibrium

( )

( )

R S

RT H

R S

RT H

RTSTHRT G

oo

o

oo

ee

e

eeK

∆∆

∆∆

∆∆∆

⋅=

=

==

−

+−

−−−

Combine Gibbs free energy and Equilibrium Equations:

Only Enthalpy term is temperature dependent:

RT H o

e)T(K ∆−

∝ 21

1. Equilibrium constant of an endothermic reaction (∆Ho = +) increases if the temperature is raised.

2. Equilibrium constant of an exothermic reaction (∆Ho = -)decreases if the temperature is raised.

Le Châtelier’s Principal

Affect of Temperature on Equilibrium

∆H = +

∆H = -

∆

∆

1. The equilibrium constant of an endothermic reaction (ΔH° = +) increases if the temperature is raised. 2. The equilibrium constant of an exothermic reaction (ΔH° = −) decreases if the temperature is raised. 22

4.) Thermodynamics vs. Kinetics  Thermodynamics predicts if a reaction will occur

- determines the state at equilibrium

 Thermodynamics does not determine the rate of a reaction

- Will the reaction occur instantly, in minutes, hours, days or years? - In dealing with equilibrium problems, we are making thermodynamic

predictions, not kinetic predictions. - We are calculating what must happen for a system to reach equilibrium, but

not how long it will take.

-

While reaction is spontaneous, it takes millions of years to occur For example, dynamite remains unchanged indefinitely, until a spark sets off the

spontaneous, explosive decomposition. The size of an equilibrium constant tells us nothing about the rate (the kinetics) of the reaction. A large equilibrium constant does not imply that a reaction is fast.

∆G = -

spontaneous

Diamonds Graphite

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Solubility Product 1.) Equilibrium constant for the reaction which a solid salt dissolves to give

its constituent ions in solution ( for compounds with a very low solubility)

In chemical analysis, we encounter solubility in precipitation titrations, electrochemical reference cells, and gravimetric analysis.

The effect of acid on the solubility of minerals and the effect of atmospheric CO2 on the solubility (and death) of coral reefs are important in environmental science.

 Solid omitted from equilibrium constant because it is in a standard state 18

sp 102.1CHK −+ ×== 2-22 ]l][g[

We most commonly use the solubility product to find the concentration of one ion when the concentration of the other is known or fixed by some means

AgCl ↔ Ag+ + Cl- Ksp = [Ag+][Cl-] = 1.82 X 10-10

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EXAMPLE: what is the concentration of Hg22+in equilibrium with 0.10 M Cl− in a solution

of KCl containing excess, undissolved Hg2Cl2(s)?

182-2 2 102.1]l][g[

−+ ×== CHK sp

Because Hg2Cl2 is so slightly soluble, additional Cl− obtained from Hg2Cl2 is negligible compared with 0.10 M Cl−.

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18102.1 −×=spK

 Solid omitted from equilibrium constant because it is in a standard state

Solubility Product 1.) Saturated Solution – contains excess, undissolved solid

 Solution contains all the solid capable of dissolving under the current conditions

2.) Physical meaning of solubility :if an aqueous solution is left in contact with excess solid, the solid will dissolve until the condition of Ksp is satisfied

 Amount of undissolved solid remains constant  Excess solid is required to guarantee ion concentration is

consistent with Ksp

3.) If ions are mixed together such that the concentrations exceed Ksp, the solid will precipitate.

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–

e product [Ca2+][ SO4] is constant at equilibrium in the

presence of excess solid CaSO4. If the concentration of Ca2+ were increased by adding another source of Ca2+, such as CaCl2, then the concentration must decrease so that the product [Ca2+][SO4] remains constant. In other words, less CaSO4(s) will dissolve if Ca2+ or is already present from some other source.

Decrease in the solubility of MgF2 by the addition of NaF

Common Ion Effect

Effect of Adding a Second Source of an Ion on Salt Solubility  Equilibrium re-obtained following Le Châtelier’s Principal  Reaction moves away from the added ion

A salt will be less soluble if one of its constituent ions is already present in the solution

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Selective Separation by Precipitation "A useful application of the solubility product is the separation of one substance from

another by precipitating one from solution." This is the basis of Qualitative Analysis.

Precipitations can sometimes be used to separate ions from each other. For example, consider a solution containing lead(II) (Pb2+) and mercuryHg22+ ions, each at a concentration of 0.010 M. Each forms an insoluble iodide , but the mercury (I)iodide is considerably less soluble, as indicated by the smaller value of Ksp:

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Is it possible to lower the concentration of [Hg22+ ] by 99.990% by selective precipitation with I−, without precipitating Pb2+ ? If each of [Hg22+ ], [Pb2+ ]at a concentration of 0.010 M) We are asking whether we can lower [Hg22+ ] by 99.990% to 0.010% of 0.010 M= = 1.0 × l0−6M without precipitating Pb2+. Here is the experiment:

1-We add enough I− to precipitate 99.990% of [Hg22+ ] if all the I− reacts with [Hg22+ ] and none reacts with Pb2+.

2-To see if any Pb2+ should precipitate, we need to know the concentration of I− in equilibrium with precipitated Hg2I2(s) plus the remaining 1.0 × 10−6M .

01.0 100

01.0 x

Pb2+ Hg22+ both 0.010 M

-Separation by Precipitation

Is it possible to lower the concentration of[Hg22+ ] by 99.990% by selective precipitation with I−, without precipitating Pb2+ ? (each of [Hg22+ ], [Pb2+ ]at a concentration of 0.010 M)

• 1-We add enough I− to precipitate 99.990% (1.0 × 10−6M left)of [Hg22+ ] if all the I− reacts with [Hg22+ ] and none reacts with Pb2+.

• 2-To see if any Pb2+ should precipitate, we need to know the concentration of I− in equilibrium with precipitated Hg2I2(s) plus the remaining 1.0 × 10−6M .

3-Will this concentration of I− cause 0.010 M Pb2+ to precipitate? We answer this question by seeing if the solubility product of PbI2 is exceeded Q vs Ksp

4-The reaction quotient, Q, is 4.6 × 10 −25, which is less than Ksp(= 7.9 × 10−9) for PbI2. Therefore, Pb2+ will not precipitate and separation of Pb2+ and [Hg22+ ] is feasible. We predict

that adding I− to a solution of Pb2+ and [Hg22+ ] will precipitate virtually all the mercury(I) before any lead(II) precipitates. 30

Q < Ksp

Pb2+ Hg22+ both 0.010 M

Acids and Bases Arrhenius An acid is any substance that ionizes in water to give hydrogen ions, thus hydronium ions HA + H2O  H3O+ + A- A base is a substance which ionizes in water to give hydroxyl ions MOH  M+ + OH- (strong) B + H2O  BH+ + OH- (weak)

Protic Acids and Bases – transfer of H+ (proton) from one molecule to another

 Hydronium ion (H3O+) – combination of H+ with water (H2O)

 Acid – is a substance that increases the concentration of H3O+

 Base – is a substance that decreases the concentration of H3O+ - base also causes an increase in the concentration of OH- in aqueous solutions

Brønsted-Lowry – definition does not require the formation of H3O+  Extended to non-aqueous solutions or gas phase

 Acid – proton donor

 Base – proton acceptor

acid base salt

acid

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Acids and Bases Lewis Theory

An acid is any substance which can accept a lone pair of electrons(electron poor)

HCl + :NH3  H:NH3+ + Cl-

A base is a substance which can donate a lone pair of electrons (electron rich)

HCl + :NH3  H:NH3+ + Cl-

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Acids and Bases

Salts – product of an acid-base reaction  Any ionic solid  Acid and base neutralize each other and form a salt  Most salts with a single positive and negative charge dissociate completely

into ions in water

Conjugate Acids and Bases

Products of acid-base reaction are also acids and bases

A conjugate acid and its base or a conjugate base and its acid in an aqueous system are related to each other by the gain or loss of H+ 33

Acids and Bases

Autoprotolysis – Water undergoes self-ionization, called autoprotolysis, in which it acts as both an

acid and a base:

 Extent of these reactions are very small

water

14 w 100.1OHHK

−+ ×== ]][[ -

Acetic acid 15105.3K −×=

- H3O+ is the conjugate acid of water - OH- is the conjugate base of water - Kw is the equilibrium constant for the dissociation of water

Protic solvents have a reactive H+, and all protic solvents undergo autoprotolysis. An example is acetic acid:

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Acids and Bases

pH – negative logarithm of H+ concentration

 A solution is acidic if [H+] > [OH-]  A solution is basic if [H+] < [OH-]

 An aqueous solution has a neutral pH if [H+]=[OH-] - This occurs when [H+] = [OH-] = 10-7M or pH = 7

]Hlog[pH +−≈

CatpKKpOHpH oww 2500.14]log[ ==−=+

14 w 100.1OHHK

−+ ×== ]][[ -

35

pH  pH values for some common samples

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pH  Example:

What is the pH of a solution containing 1x10-6 M H+?

What is [OH-] of a solution containing 1x10-6 M H+?

Please try more examples on page 107 very important

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Strengths of Acids and Bases  Depends on whether the compound react nearly completely or partially to

produce H+ or OH-

 strong acid or base completely dissociate in aqueous solution - equilibrium constants are large - everything else termed weak

Strong  no undissociated HCl or KOH

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Strengths of Acids and Bases

 weak acids react with water by donating a proton - only partially dissociated in water - equilibrium constants are called Ka – acid dissociation constant - Ka is small

 weak bases react with water by removing a proton - only partially dissociated in water - equilibrium constants are called Kb – base dissociation constant - Kb is small

][ ]][[

HA AH

K a −+

= Ka

Kb ][

]][[ B

OHBH K b

−+ =

Equivalent Ka

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Strengths of Acids and Bases

Ka Kb

Amines are weak bases, and ammonium ions are weak acids. The “parent” of all amines is ammonia, NH3. When a base such as methylamine reacts with water, the product is the conjugate acid.

You should learn to recognize whether a compound is acidic or basic

H + + :NH3  H:NH3+

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ACID FORMULA Ka pKa ACID FORMULA Ka pKa

acetic acid H(C2H3O2)or CH3COOH 1.74 E-5 4.76 hydrocyanic acid HCN 6.17 E-10 9.21

ascorbic acid (1) H2(C6H6O6) 7.94 E-5 4.10 hydrofluoric acid HF 6.31 E-4 3.20

ascorbic acid (2) (HC6H6O6)- 1.62 E-12 11.79 lactic acid H(C3H5O3) 8.32 E-4 3.08

boric acid (1) H3BO3 5.37 E-10 9.27 nitrous acid HNO2 5.62 E-4 3.25

boric acid (2) (H2BO3)- 1.8 E-13 12.7 octanoic acid H(C8H15O2) 1.29 E-4 4.89

boric acid (3) (HBO3)= 1.6 E-14 13.8 oxalic acid (1) H2(C204) 5.89 E-2 1.23

butanoic acid H(C4H7O2) 1.48 E-5 4.83 oxalic acid (2) (HC2O4)- 6.46 E-5 4.19

carbonic acid (1) H2CO3 4.47 E-7 6.35 pentanoic acid H(C5H9O2) 3.31 E-5 4.84

carbonic acid (2) (HCO3)- 4.68 E-11 10.33 phosphoric acid (1) H3PO4 6.92 E-3 2.16

chromic acid (1) H2CrO4 1.82 E-1 0.74 phosphoric acid (2) (H2PO4)- 6.17 E-8 7.21

chromic acid (2) (HCrO4)- 3.24 E-7 6.49 phosphoric acid (3) (HPO4)= 2.09 E-12 12.32

citric acid (1) H3(C6H5O7) 7.24 E-4 3.14 propanoic acid H(C3H5O2) 1.38 E-5 4.86

citric acid (2) (H2C6H5O7)- 1.70 E-5 4.77 sulfuric acid (2) (HSO4)- 1.05 E-2 1.98

citric acid (3) (HC6H5O7)= 4.07 E-7 6.39 sulfurous acid (1) H2SO3 1.41 E-2 1.85

formic acid H(CHO2) 1.78 E-4 3.75 sulfurous acid (2) (HSO3)- 6.31 E-8 7.20

heptanoic acid H(C7H13O2) 1.29 E-5 4.89 uric acid H(C5H3N4O3) 1.29 E-4 3.89

hexanoic acid H(C6H11O2) 1.41 E-5 4.84

Some Common Weak Acids (carboxylic acids) Look also at Appendix G

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Some Common Weak Bases (amines)

BASE FORMULA Kb pKb

alanine C3H5O2NH2 7.41 E-5 4.13

Ammonia NH3 (NH4OH) 1.78 E-5 4.75

dimethylamine (CH3)2NH 4.79 E-4 3.32

ethylamine C2H5NH2 5.01 E-4 3.30

glycine C2H3O2NH2 6.03 E-5 4.22

hydrazine N2H4 1.26 E-6 5.90

methylamine CH3NH2 4.27 E-4 3.37

trimethylamine (CH3)3N 6.31 E-5 4.20

 The Ka or Kb of an acid or base may also be written in terms of “pKa” or “pKb”

 As Ka or Kb increase  pKa or pKb decrease

- a strong acid/base has a high Ka or Kb and a low pKa or pkb

)Klog(pK aa −= )Klog(pK bb −=

42

][ ]][[

HA AH

K 1 −+

=

- Relationship Between Ka and Kb For a conjugate acid and its base or a conjugate base and its acid

][ ]][[

HA AH

K a −+

=

][ ]][[

−

− =

A OHHA

K b

]][[ ][

]][[ ][

]][[ −+ −

−−+ =⋅=

⋅=

OHH A

OHHA HA

AH

KKK baw

baw KKK ⋅= 43Memorize

Relationship Between Ka and Kb Example:

The Ka for Acetic Acid is 1.75x10-5. What is Kb for acetate (its conjugate base)?

10 5

14 1075

10751 1001 −

−

− ×=

×

× =⇒

=⇒⋅=

. ).( ).(

K

K/KKKKK

b

awbbaw

44

Polyprotic Acids and Bases – can donate or accept more than one proton  Ka or Kb are sequentially numbered

- Ka1,Ka2,Ka3 Kb1,Kb2,Kb3

45

(6-36)

(6-37)

:

(6-38)

(6-39)

Memorize

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