acid-base equilibria and application

8/9/2019 Acid-base equilibria and application

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Acid-Base Equilibria

and

Neutralization Methods

Acid-Base Equilibria and

Neutralization MethodsReview units of concentration (calculations)

Acid Base theoryRelative strength of acids and bases

Chemical equilibrium

Electrolyte effects on chemical equilibria

Distribution of acid base as a function of pH

Systematic method of equilibrium calculation PBE, MBE, CBE

Calculation of pH

Buffer solutions

Acid base titrations/Applications

REVIEW: Units of Concentration

solutionofamount

soluteofamountionconcentrat

Analytical and Equilibrium Concentrations

Equilibrium Molarity, [X] = concentration of a given

dissolved form of the substance

Analytical Molarity, Cx = sum of allspecies of the

substance in solution

Molarity (M) - also called Equilibrium,

or Species Molarity

M = moles solute

liters of solution

Formality (F)or Analytical molarity

F = moles solute

liter of solution


Example:

What is the concentration (M and F) of a

solution prepared by dissolving exactly 1mol of acetic acid in 1 liter of solution?(The acid is 0.42% ionized)


Normality (N)

N = # of equivalents of soluteliter of solution

Molality (m)- defined as the number ofmoles of a substance per kilogram of solvent(not solution)

m = moles of solutekg of solvent



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Calculate the N, m and F of 5.700g H2SO4in1L solution (H2SO4=98.08 g/mole)

TRY THIS!

Show relationship between N and M.

Parts per thousand:

Cppt

= weight of substance x 1000weight of solution

Cppm= weight of substance x 106

weight of solution

Cppb= weight of substance x 109

weight of solution

Parts per million (ppm) or parts per billion (ppb)


Percent concentration (or parts per hundred)

Weight percent (w/w):C(w/w)= weight solute x 100%

weight solution

Volume percent (v/v):C(v/v)= volume solute x 100%

volume solution

Weight/Volume percent (w/v):

C(w/v)= weight solute(g) x 100%volume solution(mL)


Indicate

the unit!!!

Density and Specific Gravity

A. Density = mass/volume

B. Specific Gravity = Density of substanceDensity of water


Sample Calculations:

a.Calculate the molar concentration

and molality of commercially availableconcentrated acids and bases below:

Reagent

%w/w

Specific gravity

CH3COOHNH3

HCl

HF

HNO3

HClO4

H3PO4

H2SO4

99.7

29.0

37.2

49.5

70.5

71

86

96.5

1.05

0.90

1.19

1.15

1.42

1.67

1.71

1.84

Describe the preparation of the followingsolutiona. 2.00 L of 0.150 M HClO4from a 12.0 M

HClO4

b. 2.00 L of 0.150 M HClO4from aconcentrated solution that has a specific

gravity of 1.66%c. 100 mL of 0.1500 M of Na2SO4from

Na2SO4crystals.d. 250.0 ml of 100.0 ppm of Na from Na2SO4

crystals.

PREPARATION OF SOLUTION


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ACID BASE THEORY

Arrhenius theory - Svante Arrenhius(1857-1927)

Acids a substance that increases theconcentration of H3O

+when added to water

Bases- a substance that decreases theconcentration of H3O

+when added to H2Oor produces OH-

Bronsted and Lowry

Acidsproton donorsBases- proton acceptorsThe Bronsted and Lowry definition does notrequire that H3O

+be formed

A BrnstedLowry acid

must have a removable (acidic) proton.

HCl, H2O, H2SO4

Bronsted and Lowry

A BrnstedLowry base

must have a pair of nonbonding electrons.

NH3, H2O

Bronsted and Lowry

Assignment:

Samples of Bronsted and Lowry acid

and base but not Arrhenius

1. Amphiprotic species (or amphoteric)

species that have both acidic and basic properties

Fundamental ideas from Bronsted and Lowry
http://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.htmlhttp://www.kcsnet.or.kr/education/fame/arrhenius.html


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2. Conjugate Acids and Bases:

From the Latin word conjugare, meaning to join

together.

Reactions between acids and bases always yield

their conjugate bases and acids.

Fundamental ideas from Bronsted and Lowry Practice: Determine the conjugateacid base pair

OH-

3. Autoionization

Fundamental ideas from Bronsted and LowryLewis Acid-Base Concept

Lewis Acids

electron-pair acceptors.

Atoms with an empty valence orbital can be Lewisacids.

A compound with no Hs can be a Lewis acid.

Lewis Bases

electron-pair donors.

Anything that is a BrnstedLowry base is also a Lewisbase.

Lewis Acid-Base ConceptSTRENGTH OF ACIDS AND BASES

Strong acid- dissociates more or less

completely when it dissolves in water.- Their conjugate bases are quite weak.

Weak acid- dissociates only sl ightly when itdissolves in water

- Their conjugate bases are weak bases.

Substances with negligible acidity do not dissociatein water.

Their conjugate bases are exceedingly strong.


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Significant figure in pH calculations:

The number of digits in the mantissa would be thenumber of significant digits in x of log x.

Or

The number of significant figure in the x of log xwould be the number of digits in the mantissa.

Example:

pH = 2.460[H+] = 0.00347

CHEMICAL EQUILIBRIUM

"[ ]"means..

The equilibrium, molar concentration of a solute

The partial pressure (in atmospheres) for a gas.

Is assumed to be "1" if the species is..o A pure liquid in excess.

o A pure solid.

o The solvent in a dilute solution.

o The solvent, water.

BA

DC

BA

DCKeq

aA + bB cC + dD

When a stress is imposed on a system at equilibrium

the system will respond in such a way as to relieve

the stress.

Addition of a reactant

Change in pressure or volume of a gas

Change in temperature


Le ChteliersPrinciple Types of equilibria

1. Dissociation constant of waterKw

2. Ionization constant for acid - Ka

3. Ionization constant for base - Kb


Types of equilibria

1. Dissociation constant of water - Kw


OHOHKw 3

At 25C, Kw= 1.0 10-14

2

2

3

][

]][[

OH

OHOHKequil

]][[][ 32

2

OHOHOHKequil

Calculate pH of pure water at 25C and 50C.

Practice :


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2. Ionization constant for acid - Ka

][

]][[3

HAc

AcOHKa

Kacan be used to distinguish between

strong acids and weak acids.

Types of equilibriaIonization constant for acid - Ka

The greater the value of Ka, the stronger the acid.

3. Ionization constant for base - Kb

Types of equilibria

NH3 + H2O NH4+ + H3O

+

][

]][[

3

34

NH

OHNHKb

Assignment: Why is it that H2O does not appear in

equilibrium constant expression for aqueoussolutions?

Relationship between Ka and Kb for

conjugate acid - base pair

][

]][[3

HAc

AcOHKa

][

]][[

Ac

OHHAcK

b

HAc + H2O Ac- + H3O

+

Ac- + H2O HAc + OH-

Show that: Kw= kakb

Practice:

What is the Kb for the equilibrium whereKa is 6.2 X 10-10

CN- + H2O HCN + OH-

Assign:For a diprotic acid, derive relationship betweeneach of two acids and their conjugate base


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ELECTROLYTE EFFECTS ON

CHEMICAL EQUILIBRIA

ELECTROLYTE EFFECTS

Ideal Solutions

Equilibrium constants are independent of

the presence and concentration of electrolytes.

Real Solutions

Constants vary based on electrolyte concentration.This results in deviation from ideal behavior.

BA

DC

BA

DCKeq '

aA + bB cC + dD

Electrolyte Effects

Consider the effect of added NaCl to increase

the size of the Kspfor barium sulfate:

1. At 0 M NaCl, Ksp= 1.1 x 10-10.

2. At 1 x 10-3M NaCl, Ksp 1.8 x 10-10.

3. At 1 x 10-2M NaCl, Ksp 2.85 x 10-10.

Consider the effect of added NaCl to increase the size

of the Kafor acetic acid:

1. At 0 M NaCl, Ka= 1.75 x 10-5.

2. At 1 x 10-2

M NaCl, Ka 2.1 x 10-5

.3. At 1 x 10-1M NaCl, Ka 2.7 x 10

-5.

Electrolyte Effects

The electrolyte effect is dependent upon its ionic

strength.

- a measure of the total concentration of

ions in solution

2222

1CBA

ZCZBZA

CBA ,,are the molar concentrations

of each ionic species

and ZA, ZB, ZC.... are thecharges on each species

Ionic strength,

Sample calculation:

Calculate the ionic strength () of a 0.1 M

NaNO3solution.

Calculate the ionic strength () of a 0.1 M

Mg(NO3)2solution.

Calculate the ionic strength () of 0.020 M KBr

plus 0.010 M Na2SO4

Ionic strength,


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The higher the charge in the ionic atmosphere the

greater the ionic strength of a solution.

The stoichiometry of the electrolyte determines the

ionic strength.

Type of

electrolyte

Example

Compound

Ionic

strength

1:1

1:2 or 2:1

1:3 or 3:1

2:2

NaCl

MgCl2,Na2SO4

Al(NO3)3

MgSO4

C

3C

6C

4C

Ionic strength, Activity and Activity Coefficients

Activity,ax

, is the "effective" concentration of a

chemical species.

- Term used to account for the effects of

electrolytes on chemical equilibria.

Activity coefficient,x, is the numerical factor

necessary to convert the molar concentration

of the chemical species to activity.

XaXX

Where: [X] = molar concn

of species X

X = activity coefficient

Dimensionless factor that measure the

deviation of behavior from ideality.

Assumed to be "1" for unchargedmolecules.

Calculation of Activity Coefficients

Debye-Huckel Equation

Extended Debye-Huckel equation

Davies equation

The activity coefficient X

Activity and Activity Coefficients

- permits the calculation of activity coefficients of ion

from their charge and their average size.

where Zx= charge of the ion x

= ionic strength

x = effective diameter of the hydrated ion in

nanometers

x

Xx

uZ

3.31

512.0log

2

Debye-Huckel equation


The constants 0.51 and 3.3 are applicable to

aqueous solutions at 25 oC

The value of is approximately 0.3 nm for most

singly charged ions; for ions with higher charge,

may be as large as 1.0 nm

when 0.01, then the equation becomes

Xx Z2512.0log

x

Xx

uZ

3.31

512.0log

2

Debye-Huckel equation

This equation is referred to as the Debye

Huckel Limiting Law (DHLL)

DHLL can be used to calculate/approximate

activity coefficients in solutions of very low

ionic strength.

Xx Z

2512.0log


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Sample calculation: Calculate the activity

coefficient for Hg2+in a solution that has an

ionic strength of 0.085. = 0.5 nm


For the general equilibrium

aA + bB cC + dD

ba

dc

BA

DCKeq

][][

][]['

ConcentrationEquilibrium

Constant

b

B

a

A

d

D

c

C

aa

aaKeq

ThermodynamicEquilibrium

Constant

XaXX

Since:

B

B

A

A

D

D

C

C

BA

DCKeq

BA

BA

DC

DC

BA

DC

'KeqKeqBA

DC

Then:

b

B

a

A

d

D

c

C

aa

aaKeq

DISTRIBUTION OF ACID BASE AS

A FUNCTION OF PH

Objective:

To find an expression for the fraction of an

acids and bases ()as a function of pH.

An fraction is the ratio of the equilibrium

concentration of one specific form of a solute

divided by the total concentration of all forms

of that solute in an equilibrium mixture.

DISTRIBUTION OF ACID BASE AS A

FUNCTION OF PH


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Monoprotic system

Consider 0.10 M HAc Ka = 1.8 x 10 -5

HAc (aq)+ H2O H+

(aq)+ Ac(aq)

MBE: CHAc= [HAc ] + [Ac-]

PBE: [H+ ] = [OH- ] + [Ac-]

][

]][[

HAc

AcHKa


FUNCTION OF PH

o= the fraction of HAc

1 = the fraction of Ac-

cHA

oC

HAc

AcHAc

HAc ][

][][

][

HAc

Ac

AcHAc

Ac ][

][][

][1


FUNCTION OF PH

HAc ][ AcRearranging Ka to get and

][

][][

H

HAcKaAc

aK

AcHHAc

]][[][


FUNCTION OF PH

][

]][[

HAc

AcHKa

Substituting the above equations into o

expression


FUNCTION OF PH

][][

][

AcHAc

HAco

][

][][

H

HAcKaAc

aHA

oKH

H

c

HAc

][

][][

Substituting the above equations into 1

expression


FUNCTION OF PH

][][

][1

AcHAc

Ac

aK

AcHHAc

]][[][

When: [HAc] = [Ac-]

1

o

Ka = [H+]

pKa = pH


FUNCTION OF PH

Assign: Derive this


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Example: Calculate values:

1. at pH 2.00

2. at pH 4.74

3. at pH 6.50


FUNCTION OF PHDistribution Diagram

Plot of the fraction of Ac-present in each of

the 2 forms as a function of pH.

CH3COOH (aq) H++ CH3COO

(aq)

Practice: 0.20 MHCN Ka= 4.91010

HCN + H2O CN+ H3O

+

Distribution Diagram

1. Draw the distribution

diagram

2. Calculate the

concentration at pH 3.5

of HCN and CN-

3. What species will

predominate at pH 2.5

and pH 12

4. At what pH will

[HCN]=[CN-]

Derivation of Expressions for

Polyprotic Weak AcidsFor H2CO3

fraction of H2CO3

fraction of HCO3-

fraction of CO32-

211

2

2

][][

][

aaa

oKKKHH

H

211

2

1

1

][][

][

aaa

a

KKKHH

KH

211

2

21

2

][][aaa

aa

KKKHH

KK

Distribution diagram: Carbonate

systemKa1=4.45 x 10

-7

Ka2=4.45 x 10-11

General form for the polyprotic acid HnA

D

HAHn

D

HKAH a

n

1

1

Where D = [H+]n + K1[H+]n-1+ K1K2[H

+]n-2+ . + K1K2K3.Kn


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Assignment:

Derive the alpha expressions for all PO4-

bearing species in a phosphoric acid or

phosphate solution. Draw the distribution

diagram

SYSTEMATIC METHOD OF

EQUILIBRIUM CALCULATION

SYSTEMATIC METHOD OF EQUILIBRIUM

CALCULATION

Four types of algebraic expressions commonly

derived/used in solving multiple equilibria.

Equilibrium constant expressions (Ka, Kb,

Ksp, Kf, etc.).

Mass-balance equations (MBE)

Charge-balance equations (PBE)

Proton Balance equations (PBE)

Mass balance equations (MBE )

Equations that relate the equilibrium

concentrations of species in a solution to each

other and the formal (analytical)

concentrations of the solutes.

It states that , that is analyticalconcentration of an acid or base is equal to

the sum of concentrations of the protonated

and unprotonated species.

AHACHA

Example:

Write a mass balance expressions for the ff:

a) HA, H2A, H3A

b) 0.50 M CH3COOH

c) 0.50 M HCN

d) 0.50 M H2CO3

e) NaHCO3

f) Na2CO3

g) H3PO4

h) Satdsolution of Ag3PO4


ASSIGN:

Write the MBE for the following solutions:

1) 0.10 F H2SO4

2) 0.20 F Na2S

3) 0.10 F NH4Cl

4) 0.20 F Na2H2Y



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Proton Balance Equation (PBE)

Matches the concentration of species which

have released protons with those which have

consumed.

# protons consumed = # protons released

proton rich = proton poor

Practice:

Write a proton balance expression for the following:

a) H2O

b) strong acid - HBr

c) strong base - KOH

d) weak acid(monoprotic) - HCN

e) weak base(monobasic) NH3

f) Salt - KCN

proton rich = proton poor


PBE for weak base(monobasic)

NH3

CH3

COONa

PBE for polyprotic acids and bases

H2S

NaHS

Na2S

Try this:Write a proton balance expression for the following:


ASSIGN:

Write PBE for the following acids and bases:

1. H2CO3

2. Na2CO3

3. H3PO44. Na3PO4

5. NaH2PO4

6. Na2HPO4

7. Na2H2Y

Charge Balance Equation (CBE)

Equations that express the electrical neutrality

of a solution by equating the molar

concentrations of the positive and negative

charges.

Neutral species are not included

sum (+) charge = sum (-) charge

Example: Write CBE for 0.1 M solution of NaHCO3

(1) NaHCO3(s) Na+(aq) + HCO3

-(aq)

(2) HCO3-(aq) + H2O(l) H3O

+(aq) + CO3

2-(aq)

HCO3-(aq) + H2O(l) OH

-(aq) + H2CO3(aq)

(3) 2H2O(l) H3O+

(aq) + OH-(aq)

Charge Balance Equation (CBE)


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Write the charge balance expression

1) HBr

2) Na2HPO4

3) H3PO4

TRY THIS!Steps in solving problems of multiple

equilibria

Write balanced reactions for all equilibria.

Write all equations (PBE, MBE, CBE)

Write out all equilibrium constants and obtain

their values

Count the number of unknown species in theequilibrium system and count the number ofindependent algebraic relationships availablein the MBE, CBE, and Keq expressions.

If the number of independent equations > thenumber of unknowns, a solution ispossible.

If the number of independent equations < thenumber of unknowns, you must either findmore equilibria or else a solution isnotpossible.

Steps in solving problems of multiple

equilibria

Make suitable approximations to simplify the

algebra.

Solve the algebraic equations.

Check the validity of the approximation.

Steps in solving problems of multiple

equilibria

Calculation of pH

Calculate the pH of 0.50 M HX

Write balanced reactions

1. HX H+ + X-

2. H2O H+ + OH-


PBE: [H+] = [OH-] + [X-]

MBE: 0.50 M =[HX] + [X-]

CBE: [H+] = [OH-] + [X-]

Strong acids and bases

Write out all equilibrium constants and obtain their

values

Kw = 1.0 x 10-14 = [H+] [OH-]

Count the number of species [H+] [OH-] [X-] [HA]

Count the number of independent algebraic

relationships

PBE, MBE, Kw

solution is possible



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Make suitable approximations to simplify the

algebra

Solve the algebraic equations



Calculate pH

a.) 0.025 M KOH

b.) 0.15 M Ba(OH)2

c.) 1.0 x 10-8 M KOH

d.) 1.0 x 10-6 M HCl


General formula for strong acidwould be:

2

4KwCCH

HXHX

This formula is applicable for strongacid with concentration

For strong acid with

MHX 610

MHX 610

HXCXH

HXCpH logThen

Strong acids and bases Weak acids and bases

Consider a weak acid HA (0.50 M)

Write balanced reactions

1. HA H+ + A-

2. H2O H+ + OH-


PBE: [H+] = [OH-] + [A-]

MBE: CHA= 0.50 M = [HA] + [A-]

CBE: [H+] = [OH-] + [A-]

Write out all equilibrium constants and obtain their

values

Kw = 1.0 x 10-14 = [H+] [OH-]

Count the number of species: [H+] [OH-] [A-] and

[HA]

Count the number of independent algebraic

relationships:

PBE, MBE, Kw

Solvable

HA

AHKa

Weak acids and bases

Make suitable approximations to simplify the algebra

Two assumptions to remove the unknown terms:

1. Most of the H+come from the dissociation of theacid and contribution from water is negligible.

PBE: [H+] = [OH-] + [A-]

2. Most of the acid HA remains in the undissociatedform such that [HA] >>[A-]

MBE: CHA= 0.50 M = [HA] + [A-]

Weak acids and bases


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Substituting this to Ka expression

HAHA C

H

C

HH

HA

AHKa

2

HA

HA

KaCH

KaCH

2


Weak acids and bases pH Calculations

Exercise

Calculate the pH of 0.10 M CH3COOH, Ka= 1.8 x 10-5

pertinent equilibria

CH3COOH H+ + CH3COO

-

H2O H+ + OH-

4 species: H+, OH-, CH3COOH, CH3COO-

pH Calculations

4 independent equations

PBE

[H+] = [OH-] + [CH3COO-]

MBE0.10 = [CH3COOH] + [CH3COO

-]

][

]][[

3

3

COOHCH

HCOOCHKa

]][[ OHHKw

pH Calculations

Since the solution is acidic, we can assume that

[OH-] is very small

PBE

[H+] = [OH-] + [CH3COO-]

MBE can be rewritten as

0.10 - [CH3COO-] = [CH3COOH]

0.10 - [H+] = [CH3COOH]

0.10 = [CH3COOH] + [CH3COO-]

pH Calculations

Substituting and simplifying equations

][10.0

]][[108.1 5

H

HHKa

Using the quadratic equation

88.21033.1][

3

pHMH

KaHKaH 10.0][][0 2

Method of Successive Approximations

Steps are the same

Exercise

Calculate the pH of 0.10 M CH3COOH, Ka= 1.8 x 10-5

For the last step, approximations are used to

solve for the unknown

][10.0

]][[

108.1

5

H

HH

Ka assume that[H+] is negligible


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MKH aI 35 1034.1)10.0)(108.1(10.0][

Substitute the answer to the Kaexpression, then

solve for [H+]II

][10.0

]][[108.1 5

H

HHKa

3

5

1034.110.0

]][[108.1

HH

Ka

MH II 31033.1][


MH III 3

1033.1][

MH IV 31033.1][

MH V 31033.1][

Iterations have

converged


Exercise

Calculate the pH of 5.0 x 10-7M NaOH solution


NaOH Na+ + OH-

H2O H+ + OH-

3 species: H+, OH-, Na+


3 independent equations

PBE

[H+] + [Na+] = [OH-]

MBE ][100.5 7 NaMFNaOH

[H+] may be derived from the Kwexpression

][][

OH

KH w


][][

OHF

OH

KNaOH

w


Assume water has very small contribution in [OH-]

][][

OHF

OH

KNaOH

w

MFOH NaOHI 7100.5][


II

NaOHI

w OHFOH

K][

][

MOH

OH

II

II

7

7

7

14

102.5][

][100.5100.5

100.1


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III

NaOHII

w OHF

OH

K][

][

MOH

MOH

MOH

OH

V

IV

III

III

7

7

7

7

7

14

1019.5][

1019.5][

1019.5][

][100.5102.5

100.1

Iterations have

converged72.7

1093.1][

1019.5][

8

7

pH

MH

MOH

pH Calculations

pH of Polyprotic Acids/Bases

H2B - diprotic acids

H2B + H2O H3O+ + HB-

HB- + H2O H3O+ + B2-

][

]][[

2

31

BH

HBOHKa

][

]][[ 232

HB

BOHKa

pH Calculations

Usually Ka1Ka2

Ka1 H3O+from neutral species

Ka2 H3O+from a charged species

pH of Polyprotic Acids/Bases

pH Calculations

Example

Alanine Hydrochloride is a salt consisting of the

diprotic weak acid H2L+and Cl-. Calculate the pH of

0.10 M H2L+solution (Ka1= 4.487 x 10

-3, Ka2= 1.358 x

10-10).

H2L+ HL L-

1aK 2aK

1bK2bK

pH Calculations


H2L+ + H2O HL + H3O

+

HL + H2O L- + H3O

+

2H2O H3O+ + OH-

][

]][[

2

31

LH

OHHLKa

][

]][[ 32

HL

OHLKa

]][[ 3 OHOHKw

pH Calculations

Assume that any solution containing an

appreciable quantity of H2L+will contain

essentially no L- (Since Ka1Ka2)

PBE

[H3O+] = [OH-] + [HL]

MBE

][][2

2

HLLHCLH


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pH Calculations

Since H2L+is a weak acid

PBE

[H3O+] = [OH-] + [HL]

MBE

][][2

2

HLLHCLH

][][2

2 HLCLHLH

][][32

2

OHCLHLH

pH Calculations


][

]][[

2

31

LH

OHHLKa

][

]][[

3

33

1

2

OHC

OHOHK

LH

a

Solving the quadratic equation or using MSA

72.1

1091.1][ 23

pH

MOH

Example

Calculate the pH of 0.10 M L-solution (Ka1= 4.487 x

10-3, Ka2= 1.358 x 10-10).

pH Calculations


L-

+ H2O

HL + OH-

HL + H2O H2L+ + OH-

2H2O H3O+ + OH-

5

10

14

21 10364.710358.1

100.1

a

w

b K

K

K

12

3

14

1

2 10229.210487.4

100.1

a

wb

K

KK

]][[ 3 OHOHKw

pH Calculations

Assume that any solution containing an

appreciable quantity of L-will contain

essentially no H2L+ (Since Kb1Kb2)

PBE [H3O+] + [HL] = [OH-]

MBE ][][ HLLFL

][][ HLFL L

Equilibrium constant expression

][

]][[1

L

OHHLKb


pH Calculations

][

]][[1

L

OHHLKb

][

][ 2

1

OHF

OHK

L

b

Solving the quadratic equation or using MSA

43.11

10736.3][

10677.2][

12

3

3

pH

MOHMOH

pH Calculations

Example

Calculate the pH of 0.10 M HL solution (Ka1= 4.487 x10-3, Ka2= 1.358 x 10

-10).


HL + H2O L- + H3O

+

HL + H2O H2L+ + OH-

2H2O H3O+ + OH-

][

]][[ 32

HL

OHLKa

][

]][[ 2

1

2HL

OHLH

K

KK

a

wb

]][[ 3 OHOHKw


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pH Calculations

Exercise Calculate pH

1.) 0.10 M H2SO3solution(Ka1= 1.2 x 10

-2, Ka2= 6.6 x 10-8).

2.) 0.10 M H2C2O4solution

(Ka1= 5.6 x 10-2, Ka2= 5.42 x 10

-5).

3.) 0.04 M CH3NH2solution

(Kb= 4.35 x 10-4)

4.) 0.200 M Na2CO3

CHEM 32

Calculations required to establish titration curve

Initial- Solution of only strong acid (solution ofH3O

+)

Pre equivalence- Excess moles of strong acid +limiting moles of strong base (solution of H3O

+)

Equivalence Point- [H3O+] = [OH-]

Post equivalence- Excess moles of strong base+ limiting moles of strong acid (solution of OH -)

Sample problem:

Derive a curve for the titration of 50.00 mL0.0500M HCl with 0.1000 M NaOH

1. Initially, before any base is added to the acidsample, the [H3O

+]total= CHA + [H3O+]water.

If the CHAis greater than 10-6 M, the [H3O

+]watercan be ignored.

2, Preequivalence - As strong base is added but

prior to equivalence, [H3O+] is consumed

3. At equivalence:

The acid and base have reacted at the

stoichiometric ratio.

[H3O+] = [OH1-]

All the acid is

consumed; only base is

present.

The amount of base is

calculated from the

excess added beyond

equivalence.

Figure 1: Titrat ion curv e of a strong base

t i t rat ing a strong acid

3. At equivalence:


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4. Beyond equivalence:

All the acid is consumed; only base is

present.The amount of base is calculated from the

excess added beyond equivalence.

Note that...

If the CAcid> 10-6 M, we have assumed that the

water contribution to the hydronium ionconcentration can be ignored.

If the CAcid< 10-8 M, you can also assume that the

water is primarily responsible for the hydroniumion concentration, and that the added acid isinsignificant.

Only when the CAcidis between 10-8- 10-6M must

the water contribution to the hydronium ionconcentration be considered

Region Major constituents Comments

1.Initial

2. Pre-equivalence3. At eq pt.4. After eq pt.

HCl

HCl + NaOHNaClNaCl + NaOH

SUMMARY:

Sample problem:

Derive a curve for the titration of 50.00 mL 0.0500M

HCl with 0.1000 M NaOHPractice:

Fill up the table for the titration of HBr with

NaOH

SAMPLE: 50.00 mL of a 0.100 M HBr(aq)

solution.

TITRANT: stepwise addition of a 0.200 M

KOH(aq)solution

TITRANT ADDED

pH OF RESULTING

SOLUTION

0.00 mL

10.00 mL

24.90 mL

25.00 mL

25.10 mL

30.00 ml

40.00 ml

Calculate the pH during the titration of 50.00 mL

of 0.0500 M NaOH with 0.1000 M HCl after

the addition of the following volumes of

reagent (a). 24.50 mL (b). 25.00 mL (c). 25.50

mL


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Figure 2: Titration curve of a strong acid

titrating a strong base

Derive a curve for the

titration of 50.00 mL

of 0.1000 M acetic

acid(Ka = 1.75 x 10-5)

with 0.1000 M NaOH

Practice:

SAMPLE: 50.00 mL of a 0.100 M HF(aq)solution

TITRANT: 0.200 M KOH(aq)solution

titrant added

region classification

pH of resulting solution

0.00 mL

10.00 mL

24.90 mL

25.00 mL

25.10 mL

30.00 ml

40.00 ml

Titration of weak base with a strong acid

Titrant: HCl

Analyte: NH3

Region

Major constituents

Comment

1. Initial

2. Before E.P.3. At E.P.

4. After E.P.

NH3NH3 + NH4Cl

NH4ClNH4Cl + HCl

Summary:

Practice:

A 50.00 mL aliquot of 0.0500 M NaCN is

titrated with 0.1000 M HCl. Calculate thepH after the addition of (a). 0.00 (b).10.00 (c). 25.00 and (d). 26.00 mL ofacid.

Titration curve and effect of

concentration and Ka

As the acid becomes

dilute the EP become

shorter, become less

distinct, the more

limited the choice of

indicator.


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TITRATING POLYFUNCTIONAL

ACIDS AND BASES

Regions:

1. Initial - No titrant added.

2. Titrant added, but beforeEP1

3. At EP1

4. After EP1, but before theEP2

5. At EP2

6. After EP2

Sample problem

Consider 20.00mL of 0.100 M H2A, titrated with

0.100 M NaOH.

(Ka1= 1.00 x 10-4, Ka2= 1.00 x 10

-8)


ACIDS AND BASES

Try this!

Find the pH of a 50 mL solution of a 0.10 M

H2CO3after addition of 0, 25, 50, 75, 100, and150 mL of 0.10 M NaOH. Ka1=4.3x10-7and ka2

= 4.8x10-11


ACIDS AND BASESAcid-Base Indicators

Acid-base indicators (pH indicators) are weak organic

acids or weak organic bases that change color as a

function of ionization state.

Act as a second acid or base in the solution being

titrated and must be weaker than the analyte so thatit reacts last with the titrant.

Acid-Base Indicators

HInInOH

Ka

3

In

HInKaOH

3

HIn

InpKapH

log

For a typical indicator

10

HIn

In

10

HIn

In


we can see the In-color...

we can see the HIn color


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Common name pKa Indicator Range

Methyl yellow

Methyl orange

Methyl red

Chlorophenol red

Bromthymol Blue

Cresol Purple

Phenolhpthalein

Thymolphthalein

3.3

4.2

5.0

6.0

7.1

8.3

9.7

9.9

2.9-4.0

3.1-4.4

4.2-6.2

4.8-6.4

6.0-7.6

7.4-9.0

8.0-9.8

9.3-10.5

Selected pH indicators


Ex. Phenolhpthalein (phe)

acid form base form


Selecting the Proper indicator

The transition range of the indicator should

overlap the steepest part of the transition

curve.

Indicator range = pKa 1


Titration curves for

HCl with NaOH

A:50.00 mL 0.0500

M HCl with 0.100

M NaOHB:50.00 mL of

0.000500 M HCl

with 0.00100 M

NaOH


Titration curves

for HAc with

NaOH

A: 0.1000 M HAc

with 0.1000 M

NaOH

B: 0.001000 M

HAc with

0.00100 M

NaOH


Indicator Error

Determinate error

difference between the endpoint and the equivalencepoint

pH at which the indicator completes its color change isnot the same as the pH of the EP

Indeterminate error

inability to decide if its end point or not



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Sample problem 1: For bromcresol green, the followinggeneral reaction exists.

Indicate the reaction that will take place when acid is added

base is added

What is the pH transition range of the indicator

What species and color will predominate in a solution of pH 8.0

ph 2.0

acidic endpoint basic

bromocresol green yellow green blue 4.0-5.6


Sample problem 2:

Consider the indicator, phenol red which has a

yellow HIn form, a red In-form, and a Ka of 5.0

x 10-8. What is the ratio of HIn/In- .

If a few drops of this indicator are added to a

solution of pH 2.3, what color would the

solution be?


APPLICATIONS OF ACID BASE

TITRATIONS

APPLICATIONS OF ACID BASE

TITRATIONS

Concept of equivalence (review)

Preparation of standard acid solutions

Standardization of acids

Primary standards for acids

Preparation of standard base solutions

Standardization of bases

Applications of neutralization titrationsElemental analysis

Determination of inorganic substances

Determination of organic functional groups


Describe the preparation of 5.00 L of 0.1000 N

Na2CO3(105.99 g/mol) from the primary-

standard solid, assuming the solution is to be

used for titrations in which the reaction is

CO32- + 2H+H2O + CO2

Exactly 50.00 mL of an HCI solution required

29.71 mL of 0.03926 N Ba(OH)2to give an end

point with bromocresol green indicator.

Calculate the normality of the HC!.



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Preparation of standard acid solutions

Strong acids are used like HCl, H2SO4, HClO4

HClwidely used for titration of bases

H2SO4and HClO4useful for titrations where

chloride ion interferes by forming precipitates

HNO3are seldom used because of their oxidizing

properties


Na2CO3available commercially or can beprepared by heating purified NaHCO3between2700C to 3000C for 1 hour

2NaHCO3 Na2CO3 + H2O + CO2

Na2CO3

CO32-+ H3O

+ HCO3- + H2O pH 8.3

HCO3- + H3O

+ H2CO3 + H2O pH 3.8

H2CO3CO2+ H2O


THAM or TRIS - tris-hyhydroxymethyl)aminomethane

(HOCH3)CNH2


sodium tetraborate - Na2B4O7

Ex. 27.65 mL of HCl were used to titrate

0.4916 g of Borax. Molarity of HCl is:

Standardization of acidsPreparation of standard base solutions

Effect of CO2absorption

Bases (e.g. NaOH ) react rapidly with atmospheric CO2

to produce carbonate

CO2(g)+ 2OH- CO3

2- + H2O

So that this carbonate will also be titrated

CO32- + 2H3O

+ H2CO3 + 2H2O

But no error is incurred because it cancels out by the

OH- consumed


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Effect of CO2absorption

But if an indicator in the basic transition range is usedthen each CO3

2 reacted with only one hydronium ion

when the color change is observed

CO32- + 2H3O

+ HCO3- + 2H2O

The effective concentration of the base is thus

diminished by absorption of CO2, and a systematic

error (called a carbonate error) results.


e.g. NaOH

Standardization of bases

Primary standards:

Potassium Hydrogen Phthalate,KHC8H4O4

Benzoic acid

Potassium hydrogen iodate,KH(IO3)2

Applications of neutralization titrations

Elemental analysis

Nitrogen

sulfur

Determination of inorganic substances

ammonium salts

nitrates and nitrites

carbonate and carbonate mixtures

Determination of organic functional groups

Determination of salts

1. Digestion

2. Distillation

3. Titration

Kjeldahl Method for Determining Nitrogen

Digestion in strong sulfuric acid in the presence

of a catalyst

Sample + H2SO4 (NH4)2SO4(aq)+ CO2(g)+ SO2(g)+ H2O(g)

Conversion of ammonium ions into ammonia

gas, heated and then distilled.

(NH4)2SO4 + 2NaOH 2NH3+ Na2SO4+ 2H2O


Titration

back titration - the ammonia is captured by a carefully

measured excess of a standardized acid solution in

the receiving flask

Receiver: 2NH3+ 2H2SO4 (NH4)2SO4+ H2SO4

Titration: H2SO4+ 2NaOH Na2SO4+ 2H2O



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direct titration Sample problem:

A 0.7121 g sample of wheat flour was analyzedby the kjeldahl method. The ammonia formedby addition of concentrated base afterdigestion withH2SO4was distilled into 25.00mL of 0.04977 M HCl. The excess HCl was thenback-titrated with 3.97 mL of 0.04012 MNaOH. Calculate the percent protein in theflour.


Kjeldahl Method for Determining Nitrogen Kjeldahl Method for Determining Nitrogen

Titration curves for NaOH

DOUBLE INDICATOR

Titration curves for NaHCO3 Titration curves for Na2CO3

DOUBLE INDICATOR


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Sample problem:

An impure mixture that may contain NaOH, Na2CO3,

and/or NaHCO3with other inert matter was analyzed.

0.5000 g of this sample was titrated to the

phenolphthalein endpoint with 0.1042 N NaOH,

requiring 16.33 mL. A second 0.5000 g sample was

titrated to the modified methyl orange endpoint,

requiring 48.15 mL. Determine (a) the component(s)

present from the three listed; and (b) the %

composition of each of the component(s) that are(is)

present.

Sample problem:

The alkalinity of natural waters is usually controlled by

OH, CO32, and HCO3, which may be present

singularly or in combination. Titrating a 100.0-mL

sample to a pH of 8.3 requires 18.67 mL of a 0.02812

M solution of HCl. A second 100.0-mL aliquot requires

48.12 mL of the same titrant to reach a pH of 4.5.

Identify the sources of alkalinity and their

oncentrations in parts per million.

A 1.200 g sample of mixture containing NaOH

and Na2CO3was dissolved and titrated with

0.5000 N HCl. With phenolphthalein as

indicator, the solution turns colorless after the

addition of 30.00 mL of the acid. Methylorange is then added, and 5.00 mL more of

the acid is required to reach the endpoint.

Wht is the percentage composition of the

sample?

Sample problem:

A 1.200 g sample of mixture containing

NaHCO3 and Na2CO3 was dissolved and

titrated with 0.5000 N HCl. With

phenolphthalein as indicator, the solution

turns colorless after the addition of 15.00 mLof the acid. Methyl orange is then added, and

22.00 mL more of the acid is required to reach

the endpoint. Wht is the percentage

composition of the sample?

Sample problem:

An impure mixture that may contain NaOH,Na2CO3, and/or NaHCO3with other inert matterwas analyzed. 1.000 g of this sample wasdissolved in 30.00 mL dH2O and titrated to thephenolphthalein endpoint with 0.1042 N NaOH,requiring 16.33 mL. A second 1.000 g g samplewas titrated to the methyl orange endpoint,requiring 21.17 mL. Determine (a) thecomponent(s) present from the three listed; and(b) the % composition of each of thecomponent(s) that are(is) present.

Sample problem:

A 1.200 g sample of mixture containing NaOH

and Na2CO3was dissolved and titrated with

0.5000 N HCl. With phenolphthalein as

indicator, the solution turns colorless after the

addition of 30.00 mL of the acid. Methyl

orange is then added, and 5.00 mL more of

the acid is required to reach the endpoint.

Wht is the percentage composition of the

sample?

Sample problem:


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A 1.200 g sample of mixture containing

NaHCO3 and Na2CO3 was dissolved and

titrated with 0.5000 N HCl. With

phenolphthalein as indicator, the solution

turns colorless after the addition of 15.00 mL

of the acid. Methyl orange is then added, and

22.00 mL more of the acid is required to reach

the endpoint. Wht is the percentage

composition of the sample?

Sample problem:

An impure mixture that may contain NaOH,

Na2CO3, and/or NaHCO3with other inert matterwas analyzed. 1.000 g of this sample wasdissolved in 30.00 mL dH2O and titrated to thephenolphthalein endpoint with 0.1042 N NaOH,requiring 16.33 mL. A second 1.000 g g samplewas titrated to the methyl orange endpoint,requiring 21.17 mL. Determine (a) thecomponent(s) present from the three listed; and(b) the % composition of each of thecomponent(s) that are(is) present.

Sample problem:

acid-base equilibria and application

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