Kesetimbangan_Asam_Basa.pdf

7/30/2019 Kesetimbangan_Asam_Basa.pdf

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UNIVERSITY INDONESIAOF

AqueousEquilibria:

AcidandBase


2/101


AcidBaseConcepts:

Arrhenius theor of acids and bases.

Acids substances that dissociate in water to producehydrogen ions.

HA (aq) H+ (aq) + A (aq)

Bases substances that dissociate in water to produceh droxide ions.

MOH (aq) M+ (aq) + O H (aq)

Limitations to theory.

restricted to aqueous solutions doesn't account for the basicity of substances that don't

contain OH groups


3/101


AcidBaseConcepts:

Acid any substance that can transfer a proton to.

Base any substance that can accept a protonroton acce tor .

Acidbase reaction protontransfer reactions


4/101


AcidBaseConcepts:

whose formulas differ only by one proton.

.

A is the conjugate base of HA

.

BH+ is the conjugate acid of B


5/101


AcidBaseConcepts:

.

acid transfers a proton to the solvent, which acts.

HA (aq) + H2O (l) H3O+ (aq) + A (aq)

.

Base accepts a proton from the solvent, which

. NH3 (aq) + H2O (l) OH

(aq) + NH4+ (aq)


6/101


AcidBaseConcepts:

lone pairs of electrons.

the proton.


7/101


BronstedLowry


8/101


Bronsted

Lowry

acid


9/101


Bronsted

Lowry

base


10/101


AcidBaseConcepts:

Write the proton transfer equilibria for the

and identify the conjugate acidbase pairs in

each one: C6H5NH3

+ (anilinium ion),

H2PO4 (acid reaction),

NH2NH2 (hydrazine, a base), and

PO43.


11/101


AcidBaseConcepts:

+ +

Conjugate acidbase pairs:

6 5 3 6 5 2 2 3(acid)

+2 4 2 4 3


2 4

ac 4

ase ; 2 ase 3(acid)


12/101


AcidBaseConcepts:

+

Conjugate acidbase pairs: 2 2 2 3 2

(base)

4 2 4


4

ase 4

ac ; 2 ac

ase


13/101


AcidStrengthand

,

bases, H2O and A, are competing for protons.

proton transfer from the stronger acid to the

stron er base.


14/101


AcidStrengthand

aqueous solution..

H2O is a stronger base than A

solution.

v g ug . A is a stronger base than H2O


15/101



16/101


Conjugate

acid

basepairs


17/101


AcidStrengthand

Determine the direction of reactions involving:

acet c ac , acetate on, y rosu ur c ac , an

acetic acid, acetate ion, hydrogen sulfate, and4 .


18/101


AcidStrengthand

HSO4(aq)+CH3CO2

(aq)SO42(aq)+ CH3COOH


19/101


Introduction

to

aqueousacids

Introductiontoaqueousacids


20/101


Introductiontoaqueous

bases

IntroductiontoaqueousBases


21/101


HydratedProtons

+ +

an

y ron um ons

,

to the oxygen atom of a solvent water molecule. +

produces higher hydrates through hydrogen bonding with

other water molecules genera ormu a 2 n

+ +3

proton hydrated by an unspecified number of water

molecules.


22/101


Dissociationof

a er

Dissociation of water one water molecule

H2O (l) + H2O (l) H3O+ (aq) + O H (aq).

Ion roduct constant for water Kw.Kw = [H3O

+] [OH ]

Forward and reverse reactions are rapid.

Position of the equilibrium lies far to the left. In an aqueous solution, [H3O

+][OH] = 1.0 x 1014


23/101


Dissociationof

Relative values of H O+ and OH concentrations can be

a er

used to determine if an aqueous solution is neutral,acidic or basic.14

[H3O+

] .

[OH- ]

1.0 1014

[H3O+] = [OH], neutral solution and [H3O

+] = 1.0 x 107 M.

[OH-]

[H3O+ ]

H3+

> H

, ac so ut on an H3+

> 1. x 1

M. [H3O

+] < [OH], basic solution and [H3O+] < 1.0 x 107M.


24/101



25/101


Dissociationofwater


26/101


Dissociationof

a er

Calculate the molarity of OH in a solution

3 . .

solution acidic, basic or neutral?

Is a so ut on w t an OH concentrat on o 8.5

105 M acidic, basic or neutral?


27/101


Dissociationof

+

a er

,

calculate the other from the relationship+

3 .

OH-

1 1014

2.9 1014

Since [H3O+] > 1.0 107 this solution is acidic.

.


28/101


Dissociationof

a er

14

H3O+

.

8.5 10 5 1.2 10

n s case, 3 < . ;

therefore, the solution is basic.


29/101


the hydronium ion concentration.+ 3 .

[H3O+] = antilog(pH) = 10pH.

the right of the decimal point.

number related to the integral power of 10 in the

exponential expression for [H3O+]


30/101


+

.

Acidic solutions, pH < 7.

, .

Neutral solutions, pH = 7.


31/101



32/101


change color in a specific pH range..

Have different colors in their acid (HIn) and .

Change color over a range of 2 pH units.

pH unit.


33/101


.

An electronic instrument that measures a pH.

Determines pH values more accurately than acid

base indicators.


34/101


Acidbaseindicators


35/101


Naturalindicators


36/101


Calculate the pH of a 1.25 x 103 M HNO3 solution.

HNO3 is a strong acid which completely dissociates

in water. Therefore, [H3O

+] = initial concentration of theundissociated acid.

[H3O+

] = 1.25 x 103

M pH = log [H3O

+] = log (1.25 x 103) = 2.903


37/101


ThepHofstrongacids

an

strong

ases

solution (HClO4, HCl, HNO3).+

3 .

[undissociated HA] = 0.

, = 3


38/101


ThepHofstrongacids

an

strong

ases

.

Alkali metal hydroxides, MOH.

exist in aqueous solution as alkalimetal cations and

hydroxide anions calculate pH from [OH]


39/101


ThepHofstrongacids

an

strong

ases

.

Alkalineearth metal hydroxides, M(OH)2 (M = Mg,, , .

less soluble than alkali hydroxides, therefore lower [OH]

most important strong base lime, CaO produces Ca(OH)2 when mixed with water

used in steel making, water purification, and chemical



40/101


Equilibria inSolutions

Weak acids artiall dissociated can write an acid

o

ea

c s

dissociation equilibrium equation.

HA (aq) + H2O (l) H3O+ (aq) + A (aq)

Pos t on o ac ssoc at on equ r um c aracter zeby the aciddissociation constant, Ka.

-

The lar er the value ofKa the stron er the acid.

Ka 3

[HA]

Ka values can be determined from pH measurements.



41/101


Aciddissociationconstant



42/101




43/101




44/101



o

ea

c s

The pH of a 0.200 M solution of nicotinic acid6 4 2 . .

Determine the value ofKa for this acid.



45/101



o

ea

c s

To determine Ka first write the balanced

the equilibrium equation that defines Ka.

HC6H4NO2 aq +H2O H3O+ aq +C6H4NO2 aq

+ -

Ka 3 6 4 2

[HC6H4NO2 ]



46/101



o

ea

c s

concentrations of the species in the.

concentration of H3O+ from the pH.

3 . .

x 103 M.



47/101



o

ea

c s

produces one H3O+ and C6H4NO2

ion; the H3O+

6 4 2 .

[H3O+] = [C6H4NO2

] = 1.66 x 103 M

T e HC6H4NO2 concentrat on at equ r umis equal to the initial concentration minus the

amount o 6 4 2 t at ssoc ates.



48/101

U S O SO


o

ea

c s

is equal to the initial concentration minus the6 4 2 .

[HC6H4NO2] = 0.200 (1.66 x 103) = 0.198 M

[H3O+ ][C6H4NO2

- ] 1.66 103 1.66 10 3 5

a

[HC6H4NO] 0.198

.

UNIVERSITY INDONESIAOFCalculating Equilibrium


49/101

CalculatingEquilibriumConcentrationsinWeak

AcidSolutions

solution of a weakacid can be calculated from.

To solve acidbase equilibrium problems, think

.



50/101


AcidSolutions

Calculate the concentration of all species

3 , 4 7 2 , 4 7 2,

the pH in 0.250 M butyric acid (HC4H7O2).



51/101


The s ecies resent initiall are HC H O acid

AcidSolutions

and H2O (acid or base)

The possible protontransfer reactions areHC4H7O2(aq) + H2O (l) H3O

+ (aq) + C4H7O2 (aq)

Ka = 1.5 x 105

H2O (l) + H2O (l) H3O+ (aq) + OH (aq)

Kw = 1.0 x 1014

Since Ka >> Kw, the principal reaction isdissociation of HC4H7O2.



52/101


HC H O =0.250x

AcidSolutions

[H3O+

]

=

x [C4H7O2 ]=x

Ka 1.5 10

5

[H3O+][C4H7O2

- ]

x x

Assume that x is negligible compared with the

4 7 2 .

. x 0.250.

Usin this value in the denominator solve for x.

x 1.9 103



53/101


AcidSolutions

Ka 1.5 105

[H3O+][C4H7O2- ]

HC H O x x

0.250 x

x 1.9 103

Thebigequilibriumconcentrationsare[HC4H7O2]=0.250 0.0019=

0.248M;[H3O+]=[C4H7O2

]=1.9x103 M

K 1.0 10 14[OH

-] w

[H3O+]

1.9 103 5.3 10

3 . .


P t Di i ti i


54/101

PercentDissociationin

ea

c

o u ons

[HA] dissociatedercen ssoc a on =

[HA] undissociated

Value depends on the acid.Increases with increasing value ofKa.


P t


55/101

Percent

dissociation


id di i ti


56/101

aciddissociation

constant



57/101

Pol rotic acids

one dissociable proton..

Each dissociation step has its own Ka.

Stepwise dissociation constants decrease inthe order Ka1 > Ka2 > Ka3.

More difficult to remove a positively chargedproton from a negative ion.



58/101

Pol rotic acids

acids, H2A, HA

, and H2O. 2 .

principal reaction dissociation of H2A

3



59/101



60/101


Equilibria in Solutions


61/101


o

ea

ases

basedissociation equilibrium equation+

2

Position of basedissociation equilibrium

constant, Kb.BH+ OH-

Kb

[B]


Equilibria in Solutions


62/101


C. Amines or anic com ounds that are weak

o

ea

ases

bases.

Derivatives of ammonia.

One or more hydrogen atoms are replaced by anothergroup.

. can be used for bonding to a proton

To solve for e uilibrium concentrations and

solution pH, use the same procedure for solvingweak acid problems.



63/101



64/101



65/101


RelationBetweenKa


66/101

For a con u ate acidbase air, can calculate

an

either Ka or Kb from the other.

Sum of an aciddissociation reaction and a base.

Equilibrium constant for the net reaction (dissociation

of water) = the product of the equilibrium constantsfor the individual reactions.

Ka Kb = Kw. ,

conjugate base must decrease as the strength of theacid increases.


RelationBetweenKa


67/101

EXAMPLE:

an

CalculateKb forlacticacid(HC3H5O3)givenKa =1.4

104.

b.

Kw =Ka Kb;

14

Kb w

Ka

Kb 1.4 10 4

7.14 1011

3 5 3

aq +2

3 5 3

+


AcidBasePropertiesof


68/101

a ts

base properties of the constituent cations andanions.

In an acidbase reaction, the influence of thestronger partner is dominant.

strong acid + strong base neutral solution

strong acid + weak base acidic solution




69/101

a ts

.

Derived from a strong base and a strong acid.

to produce H3O+ or OH ions.




70/101

a ts

.

Derivedfromaweakbaseandastrongacid. .

Cation isaweakacid.

+ 3

small,highlychargedcations arealsoacidic




71/101

a ts

.

Derived from a strong base and a weak acid. .

Anion is a weak base.




72/101

a ts

anions.

Derived from a weak acid and a weak base.

Both the cation and the anion can undergoprotontransfer reactions.

pH depends on the relative acid strength of thecation and the base strength of the anion.

Kb (for the anion) > Ka (for the cation); basic solutionKa (for the cation) = Kb (for the anion); neutral solution




73/101

a ts

Determine whether aqueous solutions of the, .

Write the hydrolysis reaction for those

.NH4Br, K3PO4, CsI, [(CH3)3NH]F.




74/101

a ts

aqueous solution of a salt, we must determine

1) a strong acid/strong base reaction,

, 3) strong acid/weak base reaction, or

w w .




75/101

a ts

acid:3 2,

derived from NaOH and HCl and from Ca(OH)2 and

HNO res ectivel . Neither cation nor anion hydrolyzes.




76/101

a ts

acid:

conjugate base.

2 3 2 2. The anion hydrolyzes to produce OH(aq) ions.

.

The solution has a pH above 7.




77/101

a ts

acid:

conjugate acid.

4 3 3. The cation hydrolyzes to produce H+(aq) ions.

.

The solution has a pH below 7.




78/101

a ts

Examples are NH4C2H3O2, NH4CN, and FeCO3. .

The pH of the solution depends on the extent towhich each ion hydrolyzes.

The pH of a solution of NH4CN is greater than 7because CN (Kb = 2.0 10

5) is more basic than

NH4+ Ka = 5. 1

s ac c.

Consequently, CN hydrolyzes to a greater extent than+

4 .




79/101

a ts




80/101

a ts

and the strong acid HBr (aq). Acidic (NH4

+ isthe weak conjugate acid of NH ).

NH4+ (aq) + H2O NH3 (aq) + H3O

+ (aq)

K PO : derived from the stron base KOH aand the weak acid HPO42 (aq).

Basic PO 3 is the weak con u ate base of

HPO42

.) PO 3 a + H O HPO 2 a + OH




81/101

a ts

and the strong acid HI (aq). Neutral.


Acidbase


82/101

propertiesof

salts


Acidbase


83/101

propertiesof

salts


FactorsthatAffectAcid


84/101

trengt

often determined by the strength and polarity .

The weaker the HA bond, the stronger the acid.

,acid.


Factorsthataffect


85/101

acidstrength




86/101

trengt

.

HAbondstrengthdecreasesdownagroupinthe.

increaseinthesizeofAdownagroupproducespoorerorbitaloverlapandaweakerbond

HAbondstrengthisthemostimportantfactorfordeterminingacidstrengthforbinaryacidsof

elementsinthesamecolumn.




87/101

trengt

.

HA bond polarity is the most important factor for

elements in the same row.

increaseinHAbondpolarityastheelectronegativity ofAincreases

ingeneral,electronegativities increasefromlefttoright


Factorsthataffect

id h


88/101

acidstrength


Factorsthataffect

id h


89/101

acidstrength



t t


90/101

=

trengt

, ,

and m are integers).

groups.

can also be bonded to one or more ox en atoms Dissociation of oxoacid involves breaking an OH

bond.



t t


91/101

=

trengt

and m are integers).

Stren th of the acid increases as the OH bond isweakened or becomes more polar.

For oxoacids that contain the same number of OH

,increases with increasing electronegativity of Y.

For oxoacids that contain the same atom Y but differentnum ers o oxygen atoms, ac strengt ncreases w t

increasing oxidation number of Y, which increases, inturn, with an increasing number of oxygen atoms.


Factorsthataffect

id t th


92/101

acidstrength


Factorsthataffect

acid strength


93/101

acidstrength


Factorsthataffect

acid strength


94/101

acidstrength


LewisAcidsandBases


95/101

.

All Lewis bases are BrnstedLowry bases.

.


LewisAcidsandBases


96/101

.

More general definition than BrnstedLowry.

vacant valence orbitals that can accept a share in

Cationic Lewis acids metal ions.

and oxides of nonmetals.


Lewisacidbase

theory


97/101

theory

Lewisacidbasetheory


Lewisacidbase

theory


98/101

theory


Lewisacidbase

theory


99/101

theory

UNIVERSITY INDONESIAOF Keyconcept


100/101


101/101

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