1 Weak Bases in weak bases, only a small fraction of molecules accept H’s weak electrolyte most...

$Page 1: 1 Weak Bases in weak bases, only a small fraction of molecules accept H’s weak electrolyte most of the weak base molecules do not take H + from water.$
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Weak BasesWeak Bases

in weak bases, only a small fraction of molecules accept H’s

weak electrolyte

most of the weak base molecules do not take H+ from water

much less than 1% ionization in water

[HO–] << [weak base]

finding the pH of a weak base solution is similar to finding the pH of a weak acid

in weak bases, only a small fraction of molecules accept H’s

weak electrolyte

most of the weak base molecules do not take H+ from water

much less than 1% ionization in water

[HO–] << [weak base]

finding the pH of a weak base solution is similar to finding the pH of a weak acid

NH3 + H2O NH4+ + OH-

2

3

[NH3] [NH4+] [OH]

initial

change

equilibrium

Write the reaction for the base with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [OH-] from water is ≈ 0

since no products initially, Qc = 0, and the reaction is proceeding forward

NH3 + H2O NH4+ + OH-

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change

equilibrium

Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5


4

[NH3] [NH4+] [OH]

initial 0.100 0 0

change

equilibrium

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+x-x0.100 -x x x



5



determine the value of Kb from Table 15.8

since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 x x0.100 -x

6

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 x x

check if the approximation is valid by seeing if x < 5% of [NH3]init

the approximation is valid

x = 1.33 x 10-3



7

substitute x into the equilibrium concentration definitions and solve

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium0.100 x x x

x = 1.33 x 10-3

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3



8

use the [OH-] to find the [H3O+] using Kw

substitute [H3O+] into the formula for pH and solve

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3



9

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb

though not exact, the answer is reasonably close



10

Practice – Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6


11



Write the reaction for the base with water



since no products initially, Qc = 0, and the reaction is proceeding forward

B + H2O BH+ + OH-

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change

equilibrium

12

[B] [BH+] [OH]

initial 0.0015 0 0

change

equilibrium




+x+x-x

0.0015 -x x x

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6


13



determine the value of Kb

since Kb is very small, approximate the [B]eq = [B]init and solve for x

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x


14

check if the approximation is valid by seeing if x < 5% of [B]init


x = 4.9 x 10-5

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x




15


x = 4.9 x 10-5

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x

equilibrium 0.0015 x

x x

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x

equilibrium 0.0015 4.9E-5 4.9E-5



16



[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x




17

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated

Kb to the given Kb

the answer matches the given Kb

[B] [BH+] [OH]

initial 0.0015 0 ≈ 0

change -x +x +x




18

Acid-Base Properties of SaltsAcid-Base Properties of Salts

salts are water soluble ionic compounds

salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic

Example: NaHCO3 solutions are basic Na+ is the cation of the strong base NaOH HCO3

− is the conjugate base of the weak acid H2CO3

Conversely salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic

Example: NH4Cl solutions are acidic NH4

+ is the conjugate acid of the weak base NH3

Cl− is the anion of the strong acid HCl

salts are water soluble ionic compounds

salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic

Example: NaHCO3 solutions are basic Na+ is the cation of the strong base NaOH HCO3

− is the conjugate base of the weak acid H2CO3

Conversely salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic

Example: NH4Cl solutions are acidic NH4

+ is the conjugate acid of the weak base NH3

Cl− is the anion of the strong acid HCl

19

Anions as Weak BasesAnions as Weak Bases

every anion can be thought of as the conjugate base of an acid

therefore, every anion can potentially be a base A−(aq) + H2O(l) HA(aq) + OH−(aq)

the stronger the acid HA is, the weaker the conjugate base A- is

an anion that is the conjugate base of a strong acid is pH neutral

Cl−(aq) + H2O(l) HCl(aq) + OH−(aq)

an anion that is the conjugate base of a weak acid is basic

F−(aq) + H2O(l) HF(aq) + OH−(aq)

since HF is a weak acid, the position of this equilibrium favors the right

every anion can be thought of as the conjugate base of an acid

therefore, every anion can potentially be a base A−(aq) + H2O(l) HA(aq) + OH−(aq)

the stronger the acid HA is, the weaker the conjugate base A- is

an anion that is the conjugate base of a strong acid is pH neutral

Cl−(aq) + H2O(l) HCl(aq) + OH−(aq)

an anion that is the conjugate base of a weak acid is basic

F−(aq) + H2O(l) HF(aq) + OH−(aq)

since HF is a weak acid, the position of this equilibrium favors the right

20

Use the Table to Determine if the Given Anion Is Basic or Neutral

Use the Table to Determine if the Given Anion Is Basic or Neutral

a) NO3−

the conjugate base of a strong acid, therefore neutral

b) HCO3−

the conjugate base of a weak acid, therefore basic

c) PO43−


a) NO3−

the conjugate base of a strong acid, therefore neutral

b) HCO3−


c) PO43−


21

Relationship between Ka of an Acid and Kb of its Conjugate Base

Relationship between Ka of an Acid and Kb of its Conjugate Base

many reference books only give tables of Ka values because Kb values can be found from them

many reference books only give tables of Ka values because Kb values can be found from them

when you add equations, you multiply the K’s

22

Find the pH of 0.100 M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH


Na+ is the cation of a strong base – pH neutral. The CHO2

− is the anion of a weak acid – pH basic

Write the reaction for the anion with water



HCO2− + H2O HCHO2 + OH-

[CHO2−] [HCHO2] [OH-]

initial 0.100 0 ≈ 0

change

equilibrium


initial 0.100 0 ≈ 0

change

equilibrium

23

0.100 -x



Calculate the value of Kb from the value of Ka =1.8x10-4


+x+x-x

x x



24

since Kb is very small, approximate the [CHO2

−]eq = [CHO2

−]init and solve for x

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x




25

Find the pH of 0.100 M NaCHO2(aq) solutionFind the pH of 0.100 M NaCHO2(aq) solution

check if the approximation is valid by seeing if x < 5% of [CHO2

−]init


x = 2.4 x 10-6

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x


26



x = 2.4 x 10-6

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x

equilibrium0.100 −x x x


initial 0.100 0 ≈ 0

change -x +x +x


27




Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x


28


check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb

though not exact, the answer is reasonably close

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x


29

Polyatomic Cations as Weak AcidsPolyatomic Cations as Weak Acids

some polyatomic cations can be thought of as the conjugate acid of a base

therefore, some cations can potentially be an acid MH+(aq) + H2O(l) MOH(aq) + H3O+(aq)

the stronger the base MOH is, the weaker the conjugate acid MH+ is

a cation that is the counterion of a strong base (Na+, K+ etc) is pH neutral

a cation that is the conjugate acid of a weak base is acidicNH4

+(aq) + H2O(l) NH3(aq) + H3O+(aq)

since NH3 is a weak base, the position of this equilibrium favors the right

some polyatomic cations can be thought of as the conjugate acid of a base

therefore, some cations can potentially be an acid MH+(aq) + H2O(l) MOH(aq) + H3O+(aq)

the stronger the base MOH is, the weaker the conjugate acid MH+ is

a cation that is the counterion of a strong base (Na+, K+ etc) is pH neutral

a cation that is the conjugate acid of a weak base is acidicNH4

+(aq) + H2O(l) NH3(aq) + H3O+(aq)

since NH3 is a weak base, the position of this equilibrium favors the right

30

Metal Cations as Weak AcidsMetal Cations as Weak Acids

cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated

Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated

Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

31

Determine if the Given Cation is Acidic or NeutralDetermine if the Given Cation is Acidic or Neutral

a) C2H5NH3+ (Kb of C2H5NH2 = 5.6 x 10-4)

the conjugate acid of a weak base, therefore acidic

a) Ca2+ (Kb of Ca(OH)2 = 3.74 x 10-3)

the counterion of a (not so) strong base, therefore neutral

a) Cr3+

a highly charged metal ion, therefore acidic

a) C2H5NH3+ (Kb of C2H5NH2 = 5.6 x 10-4)

the conjugate acid of a weak base, therefore acidic

a) Ca2+ (Kb of Ca(OH)2 = 3.74 x 10-3)

the counterion of a (not so) strong base, therefore neutral

a) Cr3+

a highly charged metal ion, therefore acidic

32

Classifying Salt Solutions as Acidic, Basic, or NeutralClassifying Salt Solutions as Acidic, Basic, or Neutral

if the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO3)2 KBr

if the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C2H3O2)2 KNO2

if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH4Cl

if the salt cation is a highly charged metal ion and the anion is the

conjugate base of a strong acid, it will form an acidic solution Al(NO3)3

if the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO3)2 KBr

if the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C2H3O2)2 KNO2

if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH4Cl

if the salt cation is a highly charged metal ion and the anion is the

conjugate base of a strong acid, it will form an acidic solution Al(NO3)3

Tro, Chemistry: A Molecular Approach

33

if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base NH4F

HF is a stronger acid than NH4+

Ka of NH4+ is larger than Kb of the F−; therefore the solution will

be acidic

if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base NH4F

HF is a stronger acid than NH4+

Ka of NH4+ is larger than Kb of the F−; therefore the solution will

be acidic

Classifying Salt Solutions as Acidic, Basic, or NeutralClassifying Salt Solutions as Acidic, Basic, or Neutral

34

Determine whether a solution of the following salts is acidic, basic, or neutral


a) SrCl2Sr2+ is the counterion of a strong base, pH neutralCl− is the conjugate base of a strong acid, pH neutralsolution will be pH neutral

b) AlBr3

Al3+ is a small, highly charged metal ion, weak acidBr− is the conjugate base of a strong acid, pH neutralsolution will be acidic

b) CH3NH3NO3

CH3NH3+ is the conjugate acid of a weak base, acidic

NO3− is the conjugate base of a strong acid, pH neutral

solution will be acidic

a) SrCl2Sr2+ is the counterion of a strong base, pH neutralCl− is the conjugate base of a strong acid, pH neutralsolution will be pH neutral

b) AlBr3

Al3+ is a small, highly charged metal ion, weak acidBr− is the conjugate base of a strong acid, pH neutralsolution will be acidic

b) CH3NH3NO3

CH3NH3+ is the conjugate acid of a weak base, acidic

NO3− is the conjugate base of a strong acid, pH neutral

solution will be acidic

35



d) NaCHO2

Na+ is the counterion of a strong base, pH neutral

CHO2− is the conjugate base of a weak acid, basic

solution will be basic

e) NH4F

NH4+ is the conjugate acid of a weak base, acidic

F− is the conjugate base of a weak acid, basic

Ka(NH4+) > Kb(F−); solution will be acidic

d) NaCHO2

Na+ is the counterion of a strong base, pH neutral

CHO2− is the conjugate base of a weak acid, basic

solution will be basic

e) NH4F

NH4+ is the conjugate acid of a weak base, acidic

F− is the conjugate base of a weak acid, basic

Ka(NH4+) > Kb(F−); solution will be acidic

36

Polyprotic AcidsPolyprotic Acids

since polyprotic acids ionize in steps, each H has a separate Ka

Ka1 > Ka2 > Ka3

generally, the difference in Ka values is great enough so that the second ionization does not happen to a large enough extent to affect the pH most pH problems just do first ionization except H2SO4 use [H2SO4]o = [H3O+] for the second ionization

[A2-] = Ka2 as long as the second ionization is negligible

since polyprotic acids ionize in steps, each H has a separate Ka

Ka1 > Ka2 > Ka3

generally, the difference in Ka values is great enough so that the second ionization does not happen to a large enough extent to affect the pH most pH problems just do first ionization except H2SO4 use [H2SO4]o = [H3O+] for the second ionization

[A2-] = Ka2 as long as the second ionization is negligible

37

38

[HSO4 -] [SO4

2 -] [H3O+]

initial 0.0100 0 0.0100

change

equilibrium

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°CFind the pH of 0.0100 M H2SO4(aq) solution @ 25°C

Write the reactions for the acid with water


Enter the initial concentrations – assuming the [HSO4

−] and [H3O+] is ≈ [H2SO4]

HSO4- + H2O SO4

2- + H3O+ Ka2= 0.012

H2SO4 + H2O HSO4- + H3O+ Ka1= big!

39




[HSO4 -] [SO4

2 -] [H3O+]

initial 0.0100 0 0.0100

change −x +x +x

equilibrium 0.0100 −x x 0.0100 −x


40

Ka for HSO4− = 0.012expand and solve for x

using the quadratic formula


41

x = 0.0045


[HSO4 -] [SO4

2 -] [H3O+]

initial 0.0100 0 0.0100

change −x +x +x

equilibrium 0.0100 −x x 0.0100 −x

[HSO4 -] [SO4

2 -] [H3O+]

initial 0.0100 0 0.0100

change −x +x +x

equilibrium 0.0055 0.0045 0.0145


42


[HSO4 -] [SO4

2 -] [H3O+]

initial 0.0100 0 0.0100

change −x +x +x

equilibrium 0.0055 0.0045 0.0145


43

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the answer matches

Ka for HSO4− = 0.012

[HSO4 -] [SO4

2 -] [H3O+]

initial 0.0100 0 0.0100

change −x +x +x

equilibrium 0.0055 0.0045 0.0145


44

Strengths of Binary AcidsStrengths of Binary Acids

the more δ+ H-X δ- polarized the bond, the more acidic the bond

the stronger the H-X bond, the weaker the acid

binary acid strength increases to the right across a period

H-C < H-N < H-O < H-F

binary acid strength increases down the column

H-F < H-Cl < H-Br < H-I

the more δ+ H-X δ- polarized the bond, the more acidic the bond

the stronger the H-X bond, the weaker the acid

binary acid strength increases to the right across a period

H-C < H-N < H-O < H-F

binary acid strength increases down the column

H-F < H-Cl < H-Br < H-I

45

Strengths of Oxyacids, H-O-YStrengths of Oxyacids, H-O-Y

the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond

the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond

the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond

the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond


46

Lewis Acid - Base Theory

electron sharing electron donor = Lewis Base = nucleophile

must have a lone pair of electrons electron acceptor = Lewis Acid = electrophile

electron deficient when Lewis Base gives electrons from lone pair to Lewis Acid, a

covalent bond forms between the moleculesNucleophile: + Electrophile Nucleophile:Electrophile product called an adduct other acid-base reactions also Lewis


47

Example - Complete the Following Lewis Acid-Base ReactionsLabel the Nucleophile and

ElectrophileOH

H C H

+ OH-1

OH

H C H

OH

OH

H C H

+ OH-1 Electrophile

Nucleophile

••

••••


48

Practice - Complete the Following Lewis Acid-Base ReactionsLabel the Nucleophile and

Electrophile BF3 + HF

CaO + SO3

KI + I2

49

Practice - Complete the Following Lewis Acid-Base Reactions

Label the Nucleophile and Electrophile

BF3 + HF H+1BF4-1

CaO + SO3 Ca+2SO4-2

KI + I2

KI3

F

B F

F

H F••

••

•• +

F

B F

F

F-1

H+1

NucElec

O

S O

O

••

••Ca+2 O -2

•• •• +

O

S O

O

O

-2

Ca+2

ElecNuc

I I K+1 I -1••

••

•• •• +ElecNuc K+1 I I I -1

50

What is Acid Rain?What is Acid Rain?

natural rain water has a pH of 5.6 naturally slightly acidic due mainly to CO2

rain water with a pH lower than 5.6 is called acid rain

acid rain is linked to damage in ecosystems and structures

natural rain water has a pH of 5.6 naturally slightly acidic due mainly to CO2

rain water with a pH lower than 5.6 is called acid rain

acid rain is linked to damage in ecosystems and structures

51

What Causes Acid Rain?What Causes Acid Rain? many natural and pollutant gases dissolved in the air are nonmetal

oxides CO2, SO2, NO2

nonmetal oxides are acidic

CO2 + H2O H2CO3

2 SO2 + O2 + 2 H2O 2 H2SO4

processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural – volcanoes and some bacterial action man-made – combustion of fuel

weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced

many natural and pollutant gases dissolved in the air are nonmetal oxides CO2, SO2, NO2

nonmetal oxides are acidic

CO2 + H2O H2CO3

2 SO2 + O2 + 2 H2O 2 H2SO4

processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural – volcanoes and some bacterial action man-made – combustion of fuel

weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced

52

Damage from Acid RainDamage from Acid Rain

pH of Rain in Different RegionspH of Rain in Different Regions

Sources of SO2 from UtilitiesSources of SO2 from Utilities

55

Damage from Acid RainDamage from Acid Rain

acids react with metals, and materials that contain carbonates

acid rain damages bridges, cars, and other metallic structures

acid rain damages buildings and other structures made of limestone or cement

acidifying lakes affecting aquatic life

dissolving and leaching more minerals from soil

making it difficult for trees

acids react with metals, and materials that contain carbonates

acid rain damages bridges, cars, and other metallic structures

acid rain damages buildings and other structures made of limestone or cement

acidifying lakes affecting aquatic life

dissolving and leaching more minerals from soil

making it difficult for trees

56

Acid Rain LegislationAcid Rain Legislation

1990 Clean Air Act attacks acid rain force utilities to reduce SO2

result is acid rain in northeast stabilized and beginning to be reduced

1990 Clean Air Act attacks acid rain force utilities to reduce SO2

result is acid rain in northeast stabilized and beginning to be reduced

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1 Weak Bases in weak bases, only a small fraction of molecules accept H’s weak electrolyte most...

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