Copyright 2011 Pearson Education, Inc. Chapter 15 Acids and Bases: Part B. Roy Kennedy Massachusetts…

Copyright 2011 Pearson Education, Inc.

Chapter 15Acids and Bases:

Part B.

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro


Polyprotic Acids• Acid molecules often have more than one ionizable H –

these are called polyprotic acids the ionizable H’s may have different acid strengths or be equal1 H = monoprotic, 2 H = diprotic, 3 H = triproticHCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic

• Polyprotic acids ionize in stepseach ionizable H is removed sequentially

• Removing of the first H automatically makes removal of the second H harder H2SO4 is a stronger acid than HSO4

2Tro, Chemistry: A Molecular Approach, 2/e


Percent Ionization• Another way to measure the strength of an acid

is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionizationthe higher the percent ionization, the stronger the acid

• Because [ionized acid]equil = [H3O+]equil



Example 15.9: What is the percent ionization of a 2.5 M HNO2 solution?

4

write the reaction for the acid with waterconstruct an ICE table for the reactionenter the Initial Concentrationsdefine the change in concentration in terms of xsum the columns to define the equilibrium concentrations

HNO2 + H2O NO2 + H3O+

[HNO2] [NO2−] [H3O+]

initial 2.5 0 ≈0

changeequilibrium

[HNO2] [NO2−] [H3O+]

initial 2.5 0 ≈ 0

changeequilibrium

+x+xx2.5 x x x

Tro, Chemistry: A Molecular Approach, 2/e



5

determine the value of Ka from Table 15.5

because Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x

Ka for HNO2 = 4.6 x 10−4

[HNO2] [NO2−] [H3O+]

initial 2.5 0 ≈ 0change −x +x +xequilibrium 2.5−x ≈2.5 x x




6

substitute x into the equilibrium concentration definitions and solve


[HNO2] [NO2−] [H3O+]

initial 2.5 0 ≈ 0change −x +x +xequilibrium 2.5 0.034 0.0342.5 x x x

x = 3.4 x 10−2




7

apply the definition and compute the percent ionization


[HNO2] [NO2−] [H3O+]

initial 2.5 0 ≈ 0change −x +x +xequilibrium 2.5 0.034 0.034

because the percent ionization is < 5%, the “x is small” approximation is valid



Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2?

(Ka = 1.4 x 10−5 @ 25 °C)




9

write the reaction for the acid with water

construct an ICE table for the reaction

enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HC6H4NO2 + H2O C6H4NO2 + H3O+

[HA] [A−] [H3O+]

initial 0.012 0 ≈ 0changeequilibrium



[HA] [A−] [H3O+]

initial 0.012 0 0changeequilibrium


10

represent the change in the concentrations in terms of xsum the columns to find the equilibrium concentrations in terms of xsubstitute into the equilibrium constant expression

+x+xx0.012 x x x




Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5

11

determine the value of Ka

because Ka is very small, approximate the [HA]eq = [HA]init and solve for x

[HA] [A−] [H3O+]

initial 0.012 0 ≈ 0change −x +x +xequilibrium 0.012 x x0.012 x





12


x = 4.1 x 10−4

[HA] [A−] [H3O+]

initial 0.012 0 ≈ 0change −x +x +xequilibrium 0.012−x x x




13

apply the definition and compute the percent ionization

because the percent ionization is < 5%, the “x is small” approximation is valid

[HA] [A−] [H3O+]

initial 0.012 0 ≈ 0change −x +x +xequilibrium 0.012 4.1E-04 4.1E-04



Relationship Between [H3O+]equilibrium & [HA]initial

• Increasing the initial concentration of acid results in increased [H3O+] at equilibrium

• Increasing the initial concentration of acid results in decreased percent ionization

• This means that the increase in [H3O+] concentration is slower than the increase in acid concentration



Why doesn’t the increase in H3O+ keep up with the increase in HA?

• The reaction for ionization of a weak acid isHA(aq) + H2O(l) A−

(aq) + H3O+(aq)

• According to Le Châtelier’s Principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particleswe can reduce the (aq) concentrations by using a more dilute initial

acid concentration

• The result will be a larger [H3O+] in the dilute solution compared to the initial acid concentration

• This will result in a larger percent ionization


Copyright 2011 Pearson Education, Inc.16

Finding the pH of Mixtures of Acids• Generally, you can ignore the contribution of the

weaker acid to the [H3O+]equil

• For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H3O+] is negligible

• For mixtures of weak acids, generally only need to consider the stronger for the same reasons as long as one is significantly stronger than the other,

and their concentrations are similar



Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

17

write the reactions for the acids with water and determine their Kas

if the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reactionenter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HF + H2O F + H3O+ Ka = 3.5 x 10−4

[HF] [F−] [H3O+]


HClO + H2O ClO + H3O+ Ka = 2.9 x 10−8

H2O + H2O OH + H3O+ Kw = 1.0 x 10−14




18

represent the change in the concentrations in term of xsum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

[HF] [F-] [H3O+]


+x+xx0.150 x x x




19

because Ka is very small, approximate the [HF]eq = [HF]init and solve for x

Ka for HF = 3.5 x 10−4

[HF] [F-] [H3O+]





20

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

Ka for HF = 3.5 x 10−4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0change −x +x +xequilibrium 0.150 x x

the approximation is valid

x = 7.2 x 10−3




21


Ka for HF = 3.5 x 10−4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0change −x +x +xequilibrium 0.150-x x x

x = 7.2 x 10−3

[HF] [F-] [H3O+]





22

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

Ka for HF = 3.5 x 10−4

[HF] [F-] [H3O+]





Ka for HF = 3.5 x 10−4

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

though not exact, the answer is reasonably close

[HF] [F-] [H3O+]




Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and

0.15 M HF



Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF

25

pH is unitless; the fact that the pH < 7 means the solution is acidic

[HCl] = 4.5 x 10−2 M, [HF] = 0.15 M

pH

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

[H3O+][HCl] pH

Because HCl is a strong acid and HF is a weak acid, [H3O+] = [HCl]



Strong Bases

26

NaOH Na+ + OH−• The stronger the base, the

more willing it is to accept Huse water as the standard acid

• For ionic bases, practically all units are dissociated into OH– or accept H’sstrong electrolytemulti-OH strong bases

completely dissociated

• [HO–] = [strong base] x (# OH)



Example 15.11: Calculate the pH at 25 °C of a 0.0015 M Sr(OH)2 solution and determine if the solution is

acidic, basic, or neutral

pH is unitless; the fact that the pH > 7 means the solution is basic

[Sr(OH)2] = 1.5 x 10−3 M

pH

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

[H3O+][OH] pH[Sr(OH)2]

[OH]=2[Sr(OH)2]

[OH] = 2(0.0015)= 0.0030 M



Practice – Calculate the pH of the following strong acid or base solutions

0.0020 M HCl

0.0015 M Ca(OH)2

[H3O+] = [HCl] = 2.0 x 10−3 MpH = −log(2.0 x 10−3) = 2.70

[OH−] = 2 x [Ca(OH)2] = 3.0 x 10−3 MpOH = −log(3.0 x 10−3) = 2.52pH = 14.00 − pOH = 14.00 − 2.52pH = 11.48



Weak Bases

30

NH3 + H2O NH4+ + OH−

• In weak bases, only a small fraction of molecules accept H’sweak electrolytemost of the weak base molecules do

not take H from watermuch 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



Base Ionization Constant, Kb

• Base strength measured by the size of the equilibrium constant when react with H2O

:Base + H2O OH− + H:Base+

• The equilibrium constant is called the base ionization constant, Kb

larger Kb = stronger base


Copyright 2011 Pearson Education, Inc.32Tro, Chemistry: A Molecular Approach, 2/e


Structure of Amines



[NH3] [NH4+] [OH]

initialchangeequilibrium

Example 15.12:Find the pH of 0.100 M NH3(aq)

34

write the reaction for the base with waterconstruct an ICE table for the reactionenter the initial concentrations – assuming the [OH] from water is ≈ 0

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

NH3 + H2O NH4+ + OH

[NH3] [NH4+] [OH]




[NH3] [NH4+] [OH]


Example 15.12: Find the pH of 0.100 M NH3(aq)

35


+x+xx0.100 x x x




36

determine the value of Kb from Table 15.8because Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x

Kb for NH3 = 1.76 x 10−5

[NH3] [NH4+] [OH]





37

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

Kb for NH3 = 1.76 x 10−5

[NH3] [NH4+] [OH]



x = 1.33 x 10−3




38


Kb for NH3 = 1.76 x 10−5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0change -x +x +xequilibrium 0.100 x x x

x = 1.33 x 10−3

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0change -x +x +xequilibrium 0.099 1.33E-3 1.33E-3




39

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


Kb for NH3 = 1.76 x 10−5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0change −x +x +xequilibrium 0.099 1.33E−3 1.33E−3




41

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

Kb for NH3 = 1.76 x 10−5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0change −x +x +xequilibrium 0.099 1.33E−3 1.33E−3




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



Practice – Find the pH of a 0.0015 M morphine solution

43

write the reaction for the base with waterconstruct an ICE table for the reactionenter the initial concentrations – assuming the [OH] from water is ≈ 0

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

B + H2O BH+ + OH

[B] [BH+] [OH]initial 0.0015 0 ≈ 0changeequilibrium



[B] [BH+] [OH]initial 0.0015 0 0changeequilibrium


44


+x+xx0.0015 x x x

B + H2O BH+ + OH




45

determine the value of Kb

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

Kb for morphine = 1.6 x 10−6

[B] [BH+] [OH]initial 0.0015 0 ≈ 0change −x +x +xequilibrium 0.0015 x x0.0015 x




46

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 ≈ 0change −x +x +xequilibrium 0.0015 x x




47


x = 4.9 x 10−5


[B] [BH+] [OH]initial 0.0015 0 ≈ 0change −x +x +xequilibrium 0.0015 x x x

[B] [BH+] [OH]initial 0.0015 0 ≈ 0change −x +x +xequilibrium 0.0015 4.9E−5 4.9E−5




48








50


the answer matches the given Kb





Acid–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 basicNaHCO3 solutions are basic

Na+ is the cation of the strong base NaOHHCO3

− is the conjugate base of the weak acid H2CO3

• Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidicNH4Cl solutions are acidic

NH4+ is the conjugate acid of the weak base NH3

Cl− is the anion of the strong acid HCl



Anions 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 is, the weaker the

conjugate base is• An anion that is the conjugate base of a strong

acid is pH neutralCl−(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)



Example 15.13: 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) NO2−

the conjugate base of a weak acid, therefore basic



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

when you add equations, you multiply the K’s



Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution

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

− is the anion of a weak acid – pH basicwrite the reaction for the anion with waterconstruct an ICE table for the reactionenter the initial concentrations – assuming the [OH] from water is ≈ 0

CHO2− + H2O HCHO2 + OH

[CHO2−] [HCHO2] [OH]

initial 0.100 0 ≈ 0

changeequilibrium



initial 0.100 0 ≈ 0

changeequilibrium 0.100 x

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution

56

represent the change in the concentrations in terms of xsum the columns to find the equilibrium concentrations in terms of x

Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5substitute into the equilibrium constant expression

+x+xxx x



Example 15.14: Find the pH of 0.100 M NaCHO2(aq)

57

because 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 +xequilibrium 0.100 x x0.100 x




58

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 +xequilibrium 0.100 x x




initial 0.100 0 ≈ 0

change −x +x +xequilibrium 0.100 2.4E-6 2.4E-6


59


x = 2.4 x 10−6

Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

change −x +x +xequilibrium 0.100 −x x x




60



Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

change −x +x +xequilibrium 0.100 2.4E-6 2.4E-6




62



Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

change −x +x +xequilibrium 0.100 2.4E−6 2.4E−6



Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the Ka of HA?



If a 0.0015 M NaA solution has a pOH of 5.45, what is the Ka of HA?

Na+ is the cation of a strong base – pOH neutral. Because pOH is < 7, the solutinon is basic. A− is basic.write the reaction for the anion with waterconstruct an ICE table for the reactionenter the initial concentrations – assuming the [OH] from water is ≈ 0

A− + H2O HA + OH

[A−] [HA] [OH]

initial 0.100 0 ≈ 0

changeequilibrium

64



use the pOH to find the [OH−]

use [OH−] to fill in other items

[A−] [HA] [OH]

initial 0.15 0 ≈ 0

change −3.6E−6 +3.6E−6 +3.6E−6

equilibrium 0.15 3.6E-6 3.6E-6

[A−] [HA] [OH]

initial 0.15 0 ≈ 0

changeequilibrium



calculate the value of Kb of A−

[A−] [HA] [OH]

initial 0.15 0 ≈ 0

change −3.6E−6 +3.6E−6 +3.6E−6


66




use Kb of A− to find Ka of HA

67


[A−] [HA] [OH]

initial 0.15 0 ≈ 0

change −3.6E−6 +3.6E−6 +3.6E−6


Kb 8.39 10 11



Polyatomic Cations as Weak Acids

• Some cations can be thought of as the conjugate acid of a weak baseothers are the counter-ions of a strong base

• Therefore, some cations can potentially be acidicMH+(aq) + H2O(l) MOH(aq) + H3O+(aq)

• The stronger the base is, the weaker the conjugate acid isa cation that is the counter-ion of a strong base is pH neutrala cation that is the conjugate acid of a weak base is acidic

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

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



Metal Cations as Weak Acids• Cations of small, highly charged metals are weakly

acidicalkali metal cations and alkali earth metal cations are pH

neutralcations are hydrated

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

(aq) + H3O+(aq)



Example 15.15: Determine if the given cation Is acidic or neutral

a) C5N5NH2+

the conjugate acid of the weak base pyridine, therefore acidic

b) Ca2+

the counter-ion of the strong base Ca(OH)2, therefore neutral

c) Cr3+

a highly charged metal ion, therefore acidic



Classifying Salt Solutions asAcidic, Basic, or Neutral

• If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solutionNaCl Ca(NO3)2 KBr

• If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solutionNaF 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 solutionNH4Cl

• 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 solutionAl(NO3)3




• 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 baseNH4F because HF is a stronger acid than NH4

+, Ka

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

solution will be acidic



Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral

a) SrCl2Sr2+ is the counter-ion 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 acidCl− is the conjugate base of a strong acid, pH neutralsolution will be acidic

c) 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



Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral

d) NaCHO2

Na+ is the counter-ion of a strong base, pH neutralCHO2

− is the conjugate base of a weak acid, basic

solution will be basice) NH4F

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

F− is the conjugate base of a weak acid, basicKa(NH4

+) > Kb(F−); solution will be acidic



the solution is pH neutralCo3+ is a highly charged cation, pH acidic

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

76

• KNO3

• CoCl3

• Ba(HCO3)2

• CH3NH3NO3

K+ is the counter-ion of a strong base, pH neutralNO3

− is the counter-ion of a strong acid, pH neutral

Cl− is the counter-ion of a strong acid, pH neutralthe solution is pH acidicBa2+ is the counter-ion of a strong base, pH neutral

HCO3− is the conjugate of a weak acid, pH basic

the solution is pH basic

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

NO3− is the counter-ion of a strong acid, pH neutral

the solution is pH acidic



Ionization in Polyprotic Acids• Because 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 pHmost pH problems just do first ionizationexcept H2SO4 use [H2SO4] as the [H3O+] for the second

ionization

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


Copyright 2011 Pearson Education, Inc.78Tro, Chemistry: A Molecular Approach, 2/e


Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?

(Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11)




80

write the reactions for the acid with water one H at a timeconstruct an ICE table for the reactionenter the initial concentrations – assuming the second ionization is negligible

H2CO3 + H2O HCO3 + H3O+

[HA] [A−] [H3O+]


HCO3− + H2O CO3

2− + H3O+



[HA] [A−] [H3O+]



81


+x+xx0.12 x x x





determine the value of Ka1

because Ka1 is very small, approximate the [HA]eq = [HA]init and solve for x

[HA] [A−] [H3O+]


Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11




83

check if the approximation is valid by seeing if x < 5% of [H2CO3]init

Ka1 for H2CO3 = 4.3 x 10−7


x = 2.27 x 10−4

[HA] [A−] [H3O+]





84


x = 2.3 x 10−4

[HA] [A−] [H3O+]

initial 0.12 0 ≈ 0change −x +x +xequilibrium 0.12−x x x




85


[HA] [A−] [H3O+]





86


the values matchwithin sig figs

[HA] [A−] [H3O+]


Ka1 for H2CO3 = 4.3 x 10−7



Practice – What is the [CO32−] in a 0.12 M

solution of carbonic acid, H2CO3? (Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11)




solution of carbonic acid, H2CO3?

88

write the reactions for the acid with water one H at a timeconstruct an ICE table for the reactionenter the initial concentrations for the second ionization using the equilibrium concentrations from first ionization


[HCO3−] [CO3

2−] [H3O+]

initial 0.00023 0 0.00023

changeequilibrium

HCO3− + H2O CO3

2− + H3O+



[HCO3−] [CO3

2−] [H3O+]

initial 0.00023 0 0.00023

changeequilibrium


solution of carbonic acid, H2CO3?

89


+x+xx2.3E−4 x x

HCO3− + H2O CO3

2− + H3O+

2.3E−4 +x



Practice – What is the [CO32−] in a 0.12 M solution of

carbonic acid, H2CO3?

determine the value of Ka2

because Ka2 is very small, approximate the [HA]eq = [HA]init, [H3O+]eq = [H3O+]init, and solve for xusing this approximation, it is seen that x = Ka2. Therefore [CO3

2−] = Ka2

Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11

[HCO3−] [CO3

2−] [H3O+]

initial 0.00023 0 0.00023

change −x +x +x

equilibrium 2.3E−4 − x x 2.3E−4

+ x

[HCO3−] [CO3

2−] [H3O+]

initial 0.00023 0 0.00023

change −x +x +x

equilibrium 2.3E−4 x 2.3E−4



Ionization in H2SO4

• The ionization constants for H2SO4 are

H2SO4 + H2O HSO4 + H3O+ strong

HSO4 + H2O SO4

2 + H3O+ Ka2 = 1.2 x 10−2

• For most sulfuric acid solutions, the second ionization is significant and must be accounted for

• Because the first ionization is complete, use the given [H2SO4] = [HSO4

−]initial = [H3O+]initial



Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C

92

write the reactions for the acid with waterconstruct an ICE table for the second ionization reactionenter the initial concentrations – assuming the [HSO4

−] and [H3O+] is ≈ [H2SO4]

[HSO4 ] [SO4

2 ] [H3O+]

initial 0.0100 0 0.0100

changeequilibrium

HSO4 + H2O SO4

2 + H3O+

H2SO4 + H2O HSO4 + H3O+





[HSO4 ] [SO4

2 ] [H3O+]

initial 0.0100 0 0.0100

change −x +x +xequilibrium 0.0100

−x x 0.0100 −x

Tro, Chemistry: A Molecular Approach, 2/e 93



expand and solve for x using the quadratic formula

Ka for HSO4− = 0.012





x = 0.0045

[HSO4 ] [SO4

2 ] [H3O+]

initial 0.0100 0 0.0100change −x +x +x

equilibrium 0.0100 −x x 0.0100 −x


[HSO4 ] [SO4

2 ] [H3O+]


equilibrium 0.0055 0.0045 0.0145






[HSO4 ] [SO4

2 ] [H3O+]


equilibrium 0.0055 0.0045 0.0145




97


the answer matches


[HSO4 ] [SO4

2 ] [H3O+]


equilibrium 0.0055 0.0045 0.0145



Strengths of Binary Acids

• The more + H─X − 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 periodacidity: H─C < H─N < H─O < H─F

• Binary acid strength increases down the columnacidity: H─F < H─Cl < H─Br < H─I



Relationship between Bond Strength and Acidity

AcidBond

EnergykJ/mol

Type of Acid

HF 565 weak

HCl 431 strong

HBr 364 strong



Strengths of Oxyacids, H–O–Y• The more electronegative the Y atom, the stronger the

oxyacid HClO > HIO acidity of oxyacids decreases down a group

same trend as binary acids helps weakens the H–O bond

• The larger the oxidation number of the central atom, the stronger the oxyacid H2CO3 > H3BO3

acidity of oxyacids increases to the right across a periodopposite trend of binary acids

• The more oxygens attached to Y, the stronger the oxyacid further weakens and polarizes the H–O bond HClO3 > HClO2



Relationship Between Electronegativity and Acidity

AcidH─O─Y

Electronegativity of Y

Ka

H─O─Cl 3.0 2.9 x 10−8

H─O─Br 2.8 2.0 x 10−9

H─O─I 2.5 2.3 x 10−11



Relationship Between Number of Oxygens on the Central Atom and Acidity



Practice – Order the Following

• By Acidity (Least to Most)H3PO4 HNO3H3PO3 H3AsO3

• By Acidity (Least to Most)HCl HBr H2S HS−

• By Basicity (Least to Most)CO3

2− NO3− HCO3

− BO33−



Practice – Order the Following• By Acidity (Least to Most)H3PO4 HNO3H3PO3 H3AsO3

H3AsO3 < H3PO3 < H3PO4 < HNO3

• By Acidity (Least to Most)HCl HBr H2S HS−

HS− < H2S < HCl < HBr• By Basicity (Least to Most)CO3

2− NO3− HCO3

− BO33−

NO3− < HCO3

− < CO32− < BO3

3−



Lewis Acid–Base Theory

• Lewis Acid–Base theory focuses on transferring an electron pairlone pair bondbond lone pair

• Does NOT require H atoms• The electron donor is called the Lewis Base

electron rich, therefore nucleophile• The electron acceptor is called the Lewis Acid

electron deficient, therefore electrophile



Lewis Bases

• Lewis Base has electrons it is willing to give away to or share with another atom

• Lewis Base must have lone pair of electrons on it that it can donate

• Anions are better Lewis Bases than neutral atoms or moleculesN: < N:−

• Generally, the more electronegative an atom, the less willing it is to be a Lewis BaseO: < S:



Lewis Acids• Electron deficient, either from

being attached to electronegative atom(s)not having an octet

• Must have empty orbital willing to accept the electron pair

• H+ has empty 1s orbital• B in BF3 has empty 2p orbital and an incomplete octet• Many small, highly charged metal cations have empty

orbitals they can use to accept electrons• Atoms that are attached to highly electronegative

atoms and have multiple bonds can be Lewis Acids



Lewis Acid–Base Reactions

• The base donates a pair of electrons to the acid

• Generally results in a covalent bond forming H3N: + BF3 H3N─BF3

• The product that forms is called an adduct• Arrhenius and Brønsted-Lowry acid–base

reactions are also Lewis



Examples of Lewis Acid–Base Reactions



Examples of Lewis Acid–Base Reactions

Ag+(aq) + 2 :NH3(aq) Ag(NH3)2

+(aq)

LewisAcid

LewisBase

Adduct



Practice – Identify the Lewis Acid and Lewis Base in Each Reaction

111

Lewis Base

Lewis Base

Lewis Acid

Lewis Acid



U.S. Fuel Consumption• Over 85% of the energy use in the United States comes

from the combustion of fossil fuelsoil, natural gas, coal

• Combustion of fossil fuels produces CO2

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

• Natural fossil fuels also contain small amounts of S that burn to produce SO2(g)

S(s) + O2(g) → SO2(g)

• The high temperatures of combustion allow N2(g) in the air to combine with O2(g) to form oxides of nitrogen

N2(g) + 2 O2(g) → 2 NO2(g)



What Is Acid Rain?

• Natural rain water has a pH of 5.6naturally 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



What Causes Acid Rain?• Many natural and pollutant gases dissolved in the air

are nonmetal oxidesCO2, SO2, NO2

• Nonmetal oxides are acidicCO2(g) + H2O(l) H2CO3(aq)

2 SO2(g) + O2(g) + 2 H2O(l) 2 H2SO4(aq)

4 NO2(g) + O2(g) + 2 H2O(l) 4 HNO3(aq)

• Processes that produce nonmetal oxide gases as waste increase the acidity of the rainnatural – volcanoes and some bacterial actionman-made – combustion of fuel



pH of Rain in Different Regions



Weather Patterns

• The prevailing winds in the United States travel west to east

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

• Much of the northeast United States has rain of very low pH, even though it has very low sulfur emissions, due in part to the general weather patterns



Sources of SO2 from Utilities



Damage 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 affects aquatic life • Soil acidity causes more dissolving of minerals

and leaching more minerals from soil making it difficult for trees



Damage from Acid Rain



Acid Rain Legislation

• 1990 Clean Air Act attacks acid rainforces utilities to reduce SO2

• Result is acid rain in the Northeast stabilized and beginning to be reduced


Documents

Copyright 2011 Pearson Education, Inc. Chapter 15 Acids and Bases: Part B. Roy Kennedy Massachusetts…