Acid-base equilibria and application

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Acid-Base Equilibria and Neutralization Methods Acid-Base Equilibria and Neutralization Methods

Review units of concentration concentration (calculations) Acid Base theory Relative strength strength of acids and bases Chemical equilibrium Electrolyte 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/Applicat titrations/Applications ions

REVIEW:

REVIEW: Units of Concentration Concentration concentrat ion 

amount of solute

Molarity

(M)

amount of soluti on

Units of Concentration Concentration

- also called called Equilibrium, Equilibrium, or Species Molarity

M=

moles solute liters of solution

Analytical and Equilibrium Concentrations Equilibrium Molarity, [X] = concentration of a given dissolved form of the substance Analytical Molarity, C x = sum of all species of the substance in solution

REVIEW:

Formality (F) or Analytical Analytical molarity F=


Example: What is the concentration (M and F) of a solution prepared by dissolving exactly 1 mol of acetic acid in 1 liter of solution? (The acid is 0.42% ionized)

moles solute liter of solution

REVIEW:


Normality (N) N

=

# of equivalents of solute liter of solution

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

moles of solute kg of solvent

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REVIEW:


REVIEW:


Parts per thousand: thousand :

TRY THIS!

Calculate the N, m and F of 5.700g H 2SO4 in 1L solution (H 2SO4=98.08 g/mole)

Show relationship between N and M.

REVIEW:


Percent concentration (or parts per hundred) •

•

•

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

Cppt = weight of substance x 1000 weight of solution Parts per million (ppm) or parts per billion (ppb) Cppm = weight of substance weight of solution

x 106

Cppb = weight of substance weight of solution

x 10 9

REVIEW:


Density and Specific Gravity A. Density = mass/volume B. Specific Specific Gravity = Density of substance Density of water

Volume percent (v/v): C(v/v) = volume solute x 100% volume solution Indicate the unit!!! Weight/Volume percent (w/v): C(w/v) = weight solute(g) x 100% volume solution(mL)

Sample Calculations: a.Calculate the molar concentration and molality of commercially available concentrated acids and bases below: Reagent CH3COOH NH3 HCl HF HNO3 HClO4 H3PO4 H2SO4

%w/w

Specific gravity

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

PREPARATION OF SOLUTION Describe the preparation of the following solution a. 2.00 L of 0.150 M HClO 4 from a 12.0 M HClO4 b. 2.00 L of 0.150 M HClO 4 from a concentrated solution that has a specific gravity of 1.66% c. 100 mL of 0.1500 M of Na 2SO4 from Na2SO4 crystals. d. 250.0 ml of 100.0 ppm of Na from Na 2SO4 crystals.

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REVIEW:


REVIEW:


Parts per thousand: thousand :

TRY THIS!

Calculate the N, m and F of 5.700g H 2SO4 in 1L solution (H 2SO4=98.08 g/mole)

Show relationship between N and M.

REVIEW:


Percent concentration (or parts per hundred) •

•

•

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

Cppt = weight of substance x 1000 weight of solution Parts per million (ppm) or parts per billion (ppb) Cppm = weight of substance weight of solution

x 106

Cppb = weight of substance weight of solution

x 10 9

REVIEW:


Density and Specific Gravity A. Density = mass/volume B. Specific Specific Gravity = Density of substance Density of water

Volume percent (v/v): C(v/v) = volume solute x 100% volume solution Indicate the unit!!! Weight/Volume percent (w/v): C(w/v) = weight solute(g) x 100% volume solution(mL)

Sample Calculations: a.Calculate the molar concentration and molality of commercially available concentrated acids and bases below: Reagent CH3COOH NH3 HCl HF HNO3 HClO4 H3PO4 H2SO4

%w/w

Specific gravity

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

PREPARATION OF SOLUTION Describe the preparation of the following solution a. 2.00 L of 0.150 M HClO 4 from a 12.0 M HClO4 b. 2.00 L of 0.150 M HClO 4 from a concentrated solution that has a specific gravity of 1.66% c. 100 mL of 0.1500 M of Na 2SO4 from Na2SO4 crystals. d. 250.0 ml of 100.0 ppm of Na from Na 2SO4 crystals.

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ACID BASE THEORY

Bronsted Bronsted and Lowry

Arrhenius theory - Svante Arrenhius (1857-1927) Acids – a substance that increases the concentration of H 3O+ when added to water

Bases - a substance that decreases the concentration of H 3O+ when added to H2O or produces OH-

Acids – proton donors Bases - proton acceptors The Bronsted and Lowry definition does not require that H3O+ be formed



A Brønsted –Lowry acid … …must have a removable (acidic) proton .

HCl, H2O, H2SO4

A Brønsted –Lowry base… …must have a pair of nonbonding electrons.

NH3, H2O

Fundamental ideas from Bronsted and Lowry

Assignment: Samples of Bronsted and Lowry acid and base but not Arrhenius

1. Amph Amphipr iprotic otic speci species es (or amph amphoter oteric) ic) – species that have both acidic and basic properties

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Practice: Determine the conjugate acid base pair

2. Conjugate Acids and Bases: •

•

From the Latin word conjugare, meaning “to join together.”

OH-

Reactions between acids and bases always yield their conjugate bases and acids.


3. Autoionization

Lewis Acid-Base Concept

Lewis Acids

• •

•

Lewis Acid-Base Concept

Lewis Bases

electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids. A compound with no H’s can be a Lewis acid.

STRENGTH 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 slightly when it dissolves in water - Their conjugate bases are weak bases. •

electron-pair donors.

•

Anything that is a Brønsted –Lowry base is also a Lewis base.

Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong.

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STRENGTH OF ACIDS AND BASES


In any acid-base reaction, the equilibrium favors the reaction that moves the proton to the stronger base.

Acetate is a stronger base than H2O, so the equilibrium favors the backward reaction (K< 1).

HCl(aq) + H2O(l )  H3O+(aq) + Cl –(aq) acid

base

The stronger base “wins” the proton.

c. base

c. acid

H3O+(aq) + C2H3O2 –(aq)

HC2H3O2(aq) + H2O


STRENGTH OF ACIDS AND BASES Leveling solvent

Assign: Explain why HF behave as a weak acid while HI, HCl, HBr are all strong acid

– two or more acids give the same acid strength –

one in which all acids are dissociated to the same degree and have equal acid or base strength

Example: HCl(aq) + H 2O(l )  H3O+(aq) + Cl –(aq) HClO4(aq) + H 2O(l )  H 3O+(aq) + HClO4- (aq)

STRENGTH OF ACIDS AND BASES Differentiating solvent – allows relative strength to be measured — a solvent where various acids dissociate to different degrees and have different strengths.

e.g. Which is a stronger acid HCl or HClO 4? HClO4

+ HAc

 H2Ac+

+

ClO4-

HCl

+ HAc

 H2Ac+

+

Cl-

HClO4 dissociate 5000 times greater in Hac than HCl

p- Functions The negative logarithmn (to the base 10) of the number pX = -log X pH = - log H+ pOH =- log OHpKw = - log Kw pH - is a measure of the acidity or alk alinity of a substance Assign: Show that: pH

+

pOH = 14

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Significant figure in pH calculations:

CHEMICAL EQUILIBRIUM

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

aA +

bB

Or

Keq

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

cC 

+ dD

C   D   A  B  C

D

A

B

"[ ]" means..  The equilibrium, molar concentration of a solute  The partial pressure (in atmospheres) for a gas.

Example:

 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.

pH = 2.460 [H+] = 0.00347

CHEMICAL EQUILIBRIUM Le Châtelier’s Principle

CHEMICAL EQUILIBRIUM Types of equilibria

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

1. Dissociation constant of water – Kw 2. Ionization constant for acid - Ka

3. Ionization constant for base - Kb

Change in pressure or volume of a gas Change in temperature

CHEMICAL EQUILIBRIUM Types of equilibria 1. Dissociation constant of water - Kw



K equil 



[ H 3O ][OH ] 2

[ H 2O ]

K equil [ H 2O ]2  [ H 3O  ][OH  ]

Practice :









K w  H 3O OH



•

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

At 25°C, K w = 1.0  10-14

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Ionization constant for acid - Ka

Types of equilibria

The greater the value of K a, the stronger the acid.

2. Ionization constant for acid - Ka

K a 

[ H 3O  ][ Ac  ] [ HAc ]

can be used to distinguish between strong acids and weak acids. K a

Types of equilibria 3. Ionization constant for base - K b

NH3 + H2O

NH4+ 

K b 

H3O+

+

Relationship between Ka and Kb for conjugate acid - base pair Ac-

HAc + H2O

+

H3O+



[ NH 4 ][ H 3O ] [ NH 3]

Ac- + H2O

HAc 

Assignment: Why is it that H 2O does not appear in equilibrium constant expression for aqueous solutions?

K a 



[ H 3O ][ Ac ] [ HAc ]

+

OH-

[ HAc ][OH ] 

K b



[ Ac ] 

Show that: Kw= kakb

Practice: What is the Kb for the equilibrium where Ka is 6.2 X 10-10 CN- + H2O

HCN +

OH-

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

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ELECTROLYTE EFFECTS aA +

ELECTROLYTE EFFECTS ON CHEMICAL EQUILIBRIA

bB

cC Keq

Ideal Solutions

' 

+ dD C C  D  D  A A  B  B

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.

Electrolyte Effects

Electrolyte Effects

Consider the effect of added NaCl to increase the size of the K sp for barium sulfate: 1. At 0 M NaCl, K sp = 1.1 x 10

-10

2. At 1 x 10 M NaCl, K sp ≈ 1.8 x 10 3. At 1 x

10 -2 M

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

.

-3

Consider the effect of added NaCl to increase the size of the K a for acetic acid:

-10

.

NaCl, K sp ≈ 2.85 x 10 -10.

2. At 1 x 10 -2 M NaCl, Ka ≈ 2.1 x 10 -5. 3. At 1 x 10 -1 M NaCl, Ka ≈ 2.7 x 10 -5. The electrolyte effect is dependent upon its ionic strength.

Ionic strength, µ -

 

Ionic strength, µ

a measure of the total concentration of ions in solution

1 2

 A Z   B Z  C Z 2

A

2

B

2

C

 

 A,  B ,C  are the molar concentrations of each ionic species and ZA, ZB, ZC .... are the charges on each species

Sample calculation: Calculate the ionic strength ( µ) of a 0.1 M NaNO3 solution. Calculate the ionic strength ( µ) of a 0.1 M Mg(NO3)2 solution. Calculate the ionic strength ( µ) of 0.020 M KBr plus 0.010 M Na 2SO4

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Activity and Activity Coefficients

Ionic strength, µ 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.

Activity,a x , is the "effective" concentration of a chemical species. - Term used to account for the effects of electrolytes on chemical equilibria.

Activity coefficient, ɣ , is the numerical factor necessary to convert the molar concentration of the chemical species to activity. x

Type of electrolyte 1:1 1:2 or 2:1 1:3 or 3:1 2:2

Example Compound NaCl MgCl2,Na2SO4 Al(NO3)3 MgSO4

Ionic strength C 3C 6C 4C

Where:

a X   X  X 

 X = activity coefficient


Activity and Activity Coefficients Debye-Huckel equation

The activity coefficient  X •

- permits the calculation of activity coefficients of ion from their charge and their average size.

Dimensionless factor that measure the deviation of behavior from ideality.

2

•

Assumed to be "1" for uncharged molecules.

•

Calculation of Activity Coefficients –

Debye-Huckel Equation

–

Extended Debye-Huckel equation

–

Davies equation

 log  x 

where

0.512  Z X  1  3.3 x

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 •

 log  x  0.512Z 2 X

1  3.3 x

u 

Z x = charge of the ion x α x

 log  x 

0.512  Z X 

µ = ionic strength

2

Debye-Huckel equation

[X] = molar conc’n of species X

= effective diameter of the hydrated ion in nanometers

u 

 log  x  0.512Z 2 X



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.



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Activity and Activity Coefficients Sample calculation: Calculate the activity coefficient for Hg 2+ in a solution that has an ionic strength of 0.085. α = 0.5 nm

For the general equilibrium Since: •

aA +

bB

Concentration Equilibrium Constant Thermodynamic Equilibrium Constant

cC

+

Keq 

Keq 

dD c

Keq ' 

a X   X  X 

d

[C ] [ D ] a

Then:

c

d

a

b

a A a B

C

D

D

A

b

[ A] [ B ]

aC a D

C    D Keq    A   B  C

A

B



B

Keq 

 C  D

c

d

a

b

a A a B

   D   A  B 

 C  D C  A B

aC a D

C

D

A

B

Keq '

 A B

DISTRIBUTION OF ACID BASE AS A FUNCTION OF PH Objective:

DISTRIBUTION OF ACID BASE AS A FUNCTION OF PH

•

To find an expression for the fraction of an acids and bases (α) as a function of pH.

•

An α 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.

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Monoprotic system •

Consider 0.10 M HAc

Ka = 1.8 x 10 -5

HAc (aq) + H2O

H+ (aq) + Ac – (aq)

•

αo= the fraction of HAc

•

α1 = the fraction of Ac  o



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

PBE:



Ka 

 1

Ka [ HAc ] 

[ H ]

•

 o

[ HAc ] 

[ HAc ] 



 1



[ HAc ]  [ Ac ]

[ Ac  ]  [ HAc ]  [ Ac ]

[ HAc ] 



[ Ac  ] c HA

[ Ac ]  

[ HAc ]  [ Ac  ]

 o 

K a

[ HAc] c HA

Ka [ HAc ] 

[ H ]





[ H ] 

[ H ]  K a


When: 

[ H ][ Ac ] K a



[ HAc ]



[ H ][ Ac ]





c

Substituting the above equations into αo expression



[ H ][ Ac ]

Substituting the above equations into α1 expression [ Ac ]

C HA


DISTRIBUTION OF ACID BASE AS A FUNCTION OF PH •



[ HAc ]

[ HAc ]





[ HAc ]  [ Ac ]



[ H ][ Ac ]

Rearranging Ka to get  HAc  and [ Ac  ]

[ Ac ] 






Ka 

[ HAc ]

[HAc] = [Ac-]  o

  1

Assign: Derive this

Ka = [H+] pKa = pH

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Distribution Diagram •

Example: Calculate α values: 1. at pH 2.00

Plot of the fraction of Ac - present in each of the 2 forms as a function of pH. H+ + CH3COO – (aq)

CH3COOH (aq)

2. at pH 4.74 3. at pH 6.50

Distribution Diagram Practice: 0.20 M HCN Ka = 4.9×10 –10

HCN + H2O

CN – + H3O+ 1. Draw the distribution diagram 2. Calculate the concentration at pH 3.5 of HCN and CN3. What species will predominate at pH 2.5 and pH 12 4. At what pH will [HCN]=[CN-]

Distribution diagram: Carbonate system

Derivation of Expressions for Polyprotic Weak Acids For H2CO3

fraction of H2CO3

 2

 o



[ H ]  2

fraction of HCO3-

fraction of CO32-



[ H ]  [ H ] K a1  K a1 K a 2 

 1



 2



[ H ] K a1  2



[ H ]  [ H ] K a1  K a1 K a 2

K a1 K a 2  2



[ H ]  [ H ] K a1  K a1 K a 2

General form for the polyprotic acid HnA

Ka1=4.45 x 10 -7 Ka2=4.45 x 10 -11

 H  H A  



n

H n1 A 



D

K a1  H





D

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 PO 4bearing species in a phosphoric acid or phosphate solution. Draw the distribution diagram

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 )

SYSTEMATIC METHOD OF EQUILIBRIUM CALCULATION

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 C   HA  A  , that is analytical HA

concentration of an acid or base is equal to the sum of concentrations of the protonated and unprotonated species.

Mass balance equations (MBE )

Example:

Write a mass balance expressions for the ff: a) HA, H2A, H3A b) 0.50 M CH3COOH c) 0.50 M HCN

ASSIGN: Write the MBE for the following solutions:

1) 0.10 F H2SO4

d) 0.50 M H2CO3

2) 0.20 F Na2S

e) NaHCO3

3) 0.10 F NH4Cl

f) Na2CO3

4) 0.20 F Na2H2Y

g) H3PO4 h) Sat’d solution of Ag3PO4

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Proton Balance Equation (PBE)

Proton Balance Equation (PBE) Matches the concentration of species which have released protons with those which have consumed.

Practice: Write a proton balance expression for the following: proton rich = proton poor

a) H2O b) strong acid - HBr

# protons consumed = # protons released

c) strong base - KOH d) weak acid(monoprotic) - HCN

proton rich = proton poor

e) weak base(monobasic) NH 3 f) Salt - KCN

Proton Balance Equation (PBE) Try this: Write a proton balance expression for the following: •

ASSIGN: Write PBE for the following acids and bases:

PBE for weak base(monobasic)

1. H2CO3 2. Na2CO3 3. H3PO4 4. Na3PO4

NH3 CH3COONa •

PBE for polyprotic acids and bases H2S

5. NaH2PO4 6. Na2HPO4 7. Na2H2Y

NaHS Na2S

Charge Balance Equation (CBE) Equations that express the electrical neutrality of a solution by equating the molar concentrations of the positive and negative charges.

Charge Balance Equation (CBE) Example: Write CBE for 0.1 M solution of NaHCO3

NaHCO3(s)

(2)

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

sum (+) charge = sum (-) charge

Neutral species are not included

Na+(aq) + HCO3-(aq)

(1)

(3)

2H2O(l)

H3O+(aq) + CO32-(aq) OH-(aq) + H2CO3(aq)

H3O+(aq) + OH-(aq)

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Steps in solving problems of multiple equilibria

TRY THIS! Write the charge balance expression 1) HBr

Write balanced reactions for all equilibria.

2) Na2HPO4

Write all equations (PBE, MBE, CBE) Write out all equilibrium constants and obtain their values

3) H3PO4

Steps in solving problems of multiple equilibria Count the number of unknown species in the equilibrium system and count the number of independent algebraic relationships available in the MBE, CBE, and Keq expressions. If the number of independent equations > the number of unknowns, a solution is possible. If the number of independent equations < the number of unknowns, you must either find more equilibria or else a solution is not possible.

Steps in solving problems of multiple equilibria Make suitable approximations to simplify the algebra. Solve the algebraic equations. Check the validity of the approximation.

Calculation of pH Strong acids and bases

Strong acids and bases

Calculate the pH of 0.50 M HX Write balanced reactions

Write out all equilibrium constants and obtain their values

H+

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

1. HX 2. H2O

+

X-

H+

+ OH-

Write all equations (PBE, MBE, CBE) PBE: [H+] = [OH-] + [X-] MBE: 0.50 M =[HX] + [X-] CBE: [H+] = [OH -]

+ [X-]

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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Strong acids and bases Strong acids and bases

Make suitable approximations to simplify the algebra

Calculate pH a.) 0.025 M KOH

Solve the algebraic equations Check the validity of the approximation.

b.) 0.15 M Ba(OH) 2 c.) 1.0 x 10 -8 M KOH d.) 1.0 x 10 -6 M HCl

Strong acids and bases

Weak acids and bases

General formula for strong acid would be:

 H  

C HX  C HX  4 Kw 2

This formula is applicable for strong acid with concentration  HX   10 6 M 

For strong acid with  HX   10 M 6

 H    X   C 

Then



HX

Consider a weak acid HA (0.50 M) Write balanced reactions 1. HA H+ + A2. H2O H+ + OHWrite all equations (PBE, MBE, C BE) PBE: [H+] = [OH-] + [A-] MBE: CHA = 0.50 M = [HA] + [A-] CBE: [H+] = [OH-] + [A-]

pH   log C HX

Weak acids and bases Write out all equilibrium constants and obtain their values Kw = 1.0 x 10-14 = [H+] [OH-]

 H  A 

Weak acids and bases Make suitable approximations to simplify the algebra Two assumptions to remove the unknown terms:



Ka 



 HA

Count the number of species: [H+] [OH-] [A-] and [HA] Count the number of independent algebraic relationships:

1. Most of the H+ come from the dissociation of the acid and contribution from water is negligible. PBE: [H+] = [OH -] + [A-] 2. Most of the acid HA remains in the undissociated form such that [HA] >>[A -] MBE: CHA = 0.50 M = [HA] + [A -]

PBE, MBE, Kw Solvable

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pH Calculations

Weak acids and bases

Exercise Calculate the pH of 0.10 M CH 3COOH, Ka = 1.8 x 10 -5

Substituting this to Ka expression

 H  A  Ka  



 HA

 H 

 2

 H  H   



 H 

 2



C HA

pertinent equilibria



C HA

CH3COOH H2O

 KaC HA

 

H+ + CH3COOH+ + OH-



H  KaC HA 4 species: H+, OH-, CH3COOH, CH3COO-

Check the validity of the approximation.

pH Calculations

pH Calculations

4 independent equations

Since the solution is acidic, we can assume that [OH-] is very small





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

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

MBE 0.10 = [CH 3COOH] + [CH3COO-]

K a 

[CH 3COO  ][H  ] [CH 3COOH ]



MBE can be rewritten as



0.10 - [CH3COO-] = [CH3COOH]



K w  [H ][OH ]

0.10 - [H+] = [CH3COOH]

pH Calculations Substituting and simplifying equations



5 K a  1.8  10  

[H  ][H  ] 0.10  [H  ]

  0  [H ]2  Ka[H ]  0.10 Ka

Method of Successive Approximations Exercise Calculate the pH of 0.10 M CH 3COOH, Ka = 1.8 x 10 -5 Steps are the same



For the last step, approximations are used to solve for the unknown



Using the quadratic equation





3

[H ]  1.33  10 M pH  2.88

0.10 = [CH3COOH] + [CH3COO-]

5 K a  1.8  10  

[H  ][H  ] 0.10  [H  ]

assume that [H+] is negligible

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Method of Successive Approximations [H  ]I  

Method of Successive Approximations

5 3 K a  0.10  (1.8  10  )(0.10)  1.34  10  M

Substitute the answer to the Ka expression, then solve for [H+]II K a  1.8  10

5



[H  ]IV  1.33  10 3 M

[H  ][H  ] 0.10  [H  ] 

K a  1.8  10  5 

III 3 [H  ]  1.33  10  M

Iterations have converged

[H  ]V  1.33  10 3 M



[H ][H ] 0.10  1.34  10  3

[H  ]II  1.33  10 3 M

Method of Successive Approximations Exercise Calculate the pH of 5.0 x 10 -7 M NaOH solution

Method of Successive Approximations 3 independent equations



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

pertinent equilibria



NaOH H2O

 

Na+ + OHH+ + OH-

MBE

F NaOH  5.0  10 7 M  [Na  ]

[H+] may be derived from the K w expression 3 species: H+, OH-, Na+


[H  ] 

K w [OH  ]


Substituting and simplifying equations



K w 

[OH ]

 F NaOH  [OH  ]

Assume water has very small contribution in [OH -]



K w 

[OH ]

 F NaOH  [OH  ]

K w [OH  ]I

1.0  10 14 5.0  10

7

 F NaOH  [OH  ]II

 5.0  10  7  [OH  ]II

II 7 [OH  ]  5.2  10  M

[OH  ]I  F NaOH  5.0  10 7 M

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Method of Successive Approximations K w  II

[OH ]

pH of Polyprotic Acids/Bases

 F NaOH  [OH  ]III

1.0  10 14 5.2  10  7 [OH ]

H2B - diprotic acids



 5.0  10  7  [OH  ]III

[OH  ]III  5.19  10  7 M  IV

7

 5.19  10 M

 V

[OH ]  5.19  10  7 M

pH Calculations

[OH  ]  5.19  10 7 M

H2B + H2O

pH Calculations

Ka1 Ka2



 Ka2

HL + H2O 2H2O





L- + H3O+

H3O+ + OH-

K a 2 

[H 3O  ][ B 2  ] [HB  ]

K a 1 

K b 2

K a 2

HL



L-

K b 1

Assume that any solution containing an appreciable quantity of H 2L+ will contain essentially no L- (Since Ka1  Ka2)



pertinent equilibria HL + H 3O+

+

B2-

pH Calculations





+ H2O

 H3O+

H2L+

pH Calculations

H2L+ + H2O

[H 2B ]

Example Alanine Hydrochloride is a salt consisting of th e diprotic weak acid H 2L+ and Cl -. Calculate the pH of 0.10 M H 2L+ solution (Ka1 = 4.487 x 10 -3, Ka2 = 1.358 x 10-10).

H3O+ from neutral species H3O+ from a charged species



+

[H 3O  ][HB  ]

K a1 

pH Calculations

pH of Polyprotic Acids/Bases

Usually Ka1

HB-

  Iterations [H ]  1.have 93  10 8 M converged pH  7.72

HB-



 H3O+

K a1  K a 2 

[HL ][H 3O  ] [H 2L ] [L ][H 3O  ]

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

[HL ]

  K w  [H 3O ][OH ]

MBE

C H L  [ H 2 L ]  [ HL ] 2

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pH Calculations

pH Calculations Substituting and simplifying equations



Since H2L+ is a weak acid



PBE

K a1 

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

[HL ][H 3O  ] [H 2L ] 

K a1 

MBE 

C H L  [ H 2 L ]  [ HL ]



[ H 3O ][ H 3O ] 

C H L  [ H 3O ] 2

2

Solving the quadratic equation or using MSA





[ H 2 L ]  C H L  [ HL ]

[H 3O  ]  1.91  10 2 M

2





[ H 2 L ]  C H L  [ H 3O ]

pH  1.72

2

pH Calculations

pH Calculations

Example Calculate the pH of 0.10 M L - solution (Ka1 = 4.487 x 10-3, Ka2 = 1.358 x 10 -10). pertinent equilibria



L- + H2O



HL + H2O

2H2O





HL + OH -

K b1 

H2L+ + OH-

K b2 

K w K a 2

K w K a1



1.0  10

14

1.358  10 10



1.0  10 14 4.487  10  3



Assume that any solution containing an appreciable quantity of L - will contain essentially no H2L+ (Since Kb1  Kb2)



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

PBE  7 .364  10 5

F L  [L ]  [HL ]

MBE  2.229  10 12

Equilibrium constant expression K b 1 



K w  [H 3O ][OH ]

H3O+ + OH-

pH Calculations Substituting and simplifying equations [HL ][OH  ] [L ]

F L  [OH  ]

Solving the quadratic equation or using M SA



  [OH ]  2 .677  10 3 M 

pH  11 .43

Example Calculate the pH of 0.10 M HL solution (K a1 = 4.487 x 10-3, Ka2 = 1.358 x 10 -10). pertinent equilibria

[OH ]

[H 3O ]  3 .736  10

[L ]



 2

K b 1 

[HL ][OH  ]

pH Calculations



K b 1 

[L ]  F L  [HL]

 12

M

HL + H2O



L- + H3O+

K a 2 

HL + H2O



H2L+ + OH-

K b 2 

2H2O



H3O+ + OH-

[L ][H 3O  ] [HL ]

K w K a1



[H 2L ][OH  ] [HL ]

  K w  [H 3O ][OH ]

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pH Calculations CBE

[H3O+] + [H2L+] = [OH-] + [L -]

MBE

FHL = [H2L+] + [HL] + [L -]

pH Calculations Next we solve Kb2 for [H2L+]



K b 2 

Assume that [H2L+] and [L-] are significantly lower than [HL]

[H 2L ][OH  ] [HL ]



MBE

[H 2L ] 

FHL = [H2L+] + [HL] + [L -] [H 2 L ] 

pH Calculations Next we solve Ka2 for [L-]

K w [HL ] [OH  ]K a1 [H 3O  ]F HL K a1

pH Calculations

[H3O+] + [H2L+] = [OH-] + [L-]

CBE

[HL ]



[L ] 



[L ] 



[H 3O ] 

K a 2 [HL ]

[H 3O  ] 

F HL [H 3O ] K a 1

[H 3O  ]



K a 2F HL

[H 3O  ] 1 





[H 3O ]

pH Calculations [H 3O  ]2 

K a 2 F HL  K w

 F HL     1  K a1 

K a1K a 2F HL  K a1K w

[H 3O  ] 

K a1



[L ][H 3O  ]



K w

Substituting equations into the charge balance equation



K a 2 





F HL 

K w 

[H 3O ]



K a 2 F HL [H 3O  ]

1

 K w  K a 2 F HL  K a1  [H 3O  ]

pH Calculations For a solution of 0.10 M HL



[H 3O  ] 

(4.487  10 3)(1.358  10 10 )  7.807  10 7 M

pH  6.11

F HL  K a1

K a1 K a 2

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pH Calculations Exercise Calculate pH 1.) 0.10 M H2SO3 solution (Ka1 = 1.2 x 10 -2, Ka2 = 6.6 x 10 -8). 2.) 0.10 M H2C2O4 solution (Ka1 = 5.6 x 10 -2, Ka2 = 5.42 x 10 -5).

CHEM 32

3.) 0.04 M CH 3NH2 solution (Kb = 4.35 x 10 -4) 4.) 0.200 M Na2CO3

Sample problem:

Calculations required to establish titration curve Initial - Solution of only strong acid (solution of H3O+) 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 -)

Derive a curve for the titration of 50.00 mL 0.0500M HCl with 0.1000 M NaOH 1. Initially, before any base is added to the acid sample, the [H3O+]total = CHA + [H3O+]water. If the CHA is greater than 10-6 M, the [H3O+]water can be ignored.

3. At equivalence:

2, Preequivalence - As strong base is added but prior to equivalence, [H 3O+] is consumed 3. At equivalence: The acid and base have reacted at the stoichiometric ratio.

All the acid is consumed; only base is present. The amount of base is calculated from the excess added beyond equivalence.

[H3O+] = [OH1-] Figure 1: Titration curv e of a strong base titrating a strong acid

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Note that...

4. Beyond equivalence: All the acid is consumed; only base is present.

If the CAcid > 10-6 M, we have assumed that the water contribution to the hydronium ion concentration can be ignored. If the CAcid < 10-8 M, you can also assume that the water is primarily responsible for the hydronium ion concentration, and that the added acid is insignificant. Only when the CAcid is between 10-8 - 10-6 M must the water contribution to the hydronium ion concentration be considered

The amount of base is calculated from the excess added beyond equivalence.

•

Sample problem: Derive a curve for the titration of 50.00 mL 0.0500M HCl with 0.1000 M NaOH

SUMMARY: Region

Major constituents

1.Initial 2. Pre-equivalence 3. At eq pt. 4. After eq pt.

HCl HCl + NaOH NaCl NaCl + NaOH

TITRANT ADDED

0.00 mL 10.00 mL 24.90 mL 25.00 mL 25.10 mL 30.00 ml 40.00 ml

Comments

pH OF RESULTING SOLUTION

Practice:

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

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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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 Figure 2: Titration curve of a strong aci d titrating a strong base

Titration of weak base with a strong acid

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

Analyte: NH3

TITRANT: 0.200 M KOH(aq) solution titrant added 0.00 mL 10.00 mL 24.90 mL 25.00 mL 25.10 mL 30.00 ml 40.00 ml

region classification

pH of resulting solution

Practice: A 50.00 mL aliquot of 0.0500 M NaCN is titrated with 0.1000 M HCl. Calculate the pH after the addition of (a). 0.00 (b). 10.00 (c). 25.00 and (d). 26.00 mL of acid.

Summary:

Region 1. Initial 2. Before E.P. 3. At E.P. 4. After E.P.

Major constituents NH3 NH3 + NH4Cl NH4Cl NH4Cl + HCl

Comment

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 before EP1 3. At EP 1 4. After EP1, but before the EP2 5. At EP 2 6. After EP2

TITRATING POLYFUNCTIONAL ACIDS AND BASES Sample problem

Consider 20.00mL of 0.100 M H 2A, titrated with 0.100 M NaOH. (Ka1 = 1.00 x 10 -4, Ka2 = 1.00 x 10 -8)

TITRATING POLYFUNCTIONAL ACIDS AND BASES

Acid-Base Indicators Acid-base indicators (pH indicators) are weak organic acids or weak organic bases that change color as a function of ionization state.

Try this! Find the pH of a 50 mL solution of a 0.10 M H2CO3 after addition of 0, 25, 50, 75, 100, and 150 mL of 0.10 M NaOH. K a1=4.3x10 -7 and ka2 = 4.8x10-11

Act as a second acid or base in the solution being titrated and must be weaker than the analyte so that it reacts last with the titrant.

Acid-Base Indicators

Acid-Base Indicators •

For a typical indicator

 In    10

we can see the In - color...

 In    10

we can see the HIn color

 HIn 

 H O  In  

Ka 



3

 HIn 

 H O   Ka  HIn   In  

3



 HIn 

 In  

pH  pKa  log

 HIn 

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Acid-Base Indicators Selected pH indicators Common name Methyl yellow Methyl orange Methyl red Chlorophenol red Bromthymol Blue Cresol Purple Phenolhpthalein Thymolphthalein


Ex. Phenolhpthalein (phe)

pKa Indicator Range 3.3 2.9-4.0 4.2 3.1-4.4 5.0 4.2-6.2 6.0 4.8-6.4 7.1 6.0-7.6 8.3 7.4-9.0 9.7 8.0-9.8 9.9 9.3-10.5

acid form

Acid-Base Indicators 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 HAc with NaOH A: 0.1000 M HAc with 0.1000 M NaOH B: 0.001000 M HAc with 0.00100 M NaOH

base form

Acid-Base Indicators Titration curves for HCl with NaOH A: 50.00 mL 0.0500 M HCl with 0.100 M NaOH B: 50.00 mL of 0.000500 M HCl with 0.00100 M NaOH

Acid-Base Indicators Indicator Error Determinate error difference between the endpoint and the equivalence point pH at which the indicator completes its color change is not the same as the pH of the EP

Indeterminate error inability to decide if its end point or not

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Acid-Base Indicators Sample problem 1: For bromcresol green, the following general reaction exists.

acidic endpoint basic bromocresol green yellow green blue 4.0-5.6 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 – –

Acid-Base Indicators 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?

pH 8.0 ph 2.0

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 titrations Elemental analysis Determination of inorganic substances Determination of organic functional groups

Concept of equivalence (review) Describe the preparation of 5.00 L of 0.1000 N Na2CO3 (105.99 g/mol) from the primarystandard solid, assuming the solution is to be used for titrations in which the reaction is

Concept of equivalence (review) Exactly 50.00 mL of an HCI solution required 29.71 mL of 0.03926 N Ba(OH) 2 to give an end point with bromocresol green indicator. Calculate the normality of the HC!.

CO32- + 2H+  H2O + CO2

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Preparation of standard acid solutions

Standardization of acids

Strong acids are used like HCl, H 2SO4, HClO4 HCl – widely used for titration of bases H2SO4 and HClO4 – useful for titrations where chloride ion interferes by forming precipitates HNO3 are seldom used because of their oxidizing properties

Na2CO3 – available commercially or can be prepared by heating purified NaHCO3 between 2700C to 3000C for 1 hour 2NaHCO3




Na2CO3

+

H2O +

CO2

Standardization of acids THAM or TRIS - tris-hyhydroxymethyl)aminomethane

Na2CO3

(HOCH3)CNH2 2-

+

CO3 + H3O 

HCO3-

+ H2O

HCO3- + H3O+  H2CO3 +

pH 8.3

H2O pH 3.8

H2CO3  CO2 + H2O


Preparation of standard base solutions Effect of CO2 absorption

sodium tetraborate - Na2B4O7 Bases (e.g. NaOH ) react rapidly with atmospheric CO2 to produce carbonate CO2(g) + •

Ex. 27.65 mL of HCl were used to titrate 0.4916 g of Borax. Molarity of HCl is:

2OH-

 CO32- +

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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Preparation of standard base solutions

Preparation of standard base solutions

Effect of CO2 absorption

e.g. NaOH But if an indicator in the basic transition range is used then each CO32 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.

Standardization of bases Primary standards: Potassium Hydrogen Phthalate, KHC8H4O4 Benzoic acid Potassium hydrogen iodate, KH(IO3)2

Applications of neutralization titrations

Kjeldahl Method for Determining Nitrogen

Elemental analysis

1. Digestion

Nitrogen sulfur

2. Distillation

Determination of inorganic substances

3. Titration

ammonium salts nitrates and nitrites carbonate and carbonate mixtures

Determination of organic functional groups

Determination of salts

Kjeldahl Method for Determining Nitrogen Digestion in strong sulfuric acid in the presence of a catalyst Sample + H2SO4 → (NH4)2SO4(aq) + CO 2(g) + SO2(g) + H2O(g)

Conversion of ammonium ions into ammonia gas, heated and then distilled. (NH4)2SO4 +

Kjeldahl Method for Determining Nitrogen 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

2NaOH → 2NH3 + Na2SO4 + 2H2O

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Kjeldahl Method for Determining Nitrogen •

direct titration

Kjeldahl Method for Determining Nitrogen Sample problem:

A 0.7121 g sample of wheat flour was analyzed by the kjeldahl method. The ammonia formed by addition of concentrated base after digestion withH2SO4 was distilled into 25.00 mL of 0.04977 M HCl. The excess HCl was then back-titrated with 3.97 mL of 0.04012 M NaOH. Calculate the percent protein in the flour.



DOUBLE INDICATOR

DOUBLE INDICATOR

Titration curves for NaOH

Titration curves for NaHCO3

Titration curves for Na2CO3

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Sample problem:

Sample problem:

An impure mixture that may contain NaOH, Na 2CO3, and/or NaHCO3 with 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.

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.

Sample problem:

Sample problem:

•

A 1.200 g sample of mixture containing NaOH and Na2CO3 was 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: •

An impure mixture that may contain NaOH, Na2CO3, and/or NaHCO3 with other inert matter was analyzed. 1.000 g of this sample was dissolved in 30.00 mL dH2O and titrated to the phenolphthalein endpoint with 0.1042 N NaOH, requiring 16.33 mL. A second 1.000 g g sample was titrated to the methyl orange endpoint, requiring 21.17 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.

•

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: •

A 1.200 g sample of mixture containing NaOH and Na2CO3 was 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?

31

Acid-base equilibria and application

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