18 Ionic Equilibria I 離子平衡 : Acids and Bases Acids and Bases 檸檬酸 Citric acid 維他命 C Ascorbic acid

1818Ionic EquilibriaIonic Equilibria I I 離子離子

平衡平衡 ::

Acids and BasesAcids and Bases

檸檬酸Citric acid

維他命 C Ascorbic acid

Chapter GoalsChapter Goals1.1.A Review of Strong Electrolytes (A Review of Strong Electrolytes ( 強電解質強電解質 ))2.2.The Autoionization of WaterThe Autoionization of Water (( 水的自身離子化 )3.3.The pH and pOH ScalesThe pH and pOH Scales (pH(pH 值和值和 pOHpOH 值值 ))4.4.Ionization Constants for Weak Monoprotic Acids and Ionization Constants for Weak Monoprotic Acids and BasesBases (( 弱單質子酸及鹼的解離常數弱單質子酸及鹼的解離常數 ))

5.5.Polyprotic AcidsPolyprotic Acids (( 多質子酸多質子酸 ))6.6.SolvolysisSolvolysis (( 溶劑分解 )7.7.Salts of Strong Bases and Strong AcidsSalts of Strong Bases and Strong Acids (( 強鹼及強酸形成的強鹼及強酸形成的鹽鹽 ))

8.8.Salts of Strong Bases and Weak AcidsSalts of Strong Bases and Weak Acids (( 強鹼及弱酸形成的鹽強鹼及弱酸形成的鹽 ) ) 9.9.Salts of Weak Bases and Strong AcidsSalts of Weak Bases and Strong Acids (( 弱鹼及強酸形成的鹽弱鹼及強酸形成的鹽 ) ) 10.10.Salts of Weak Bases and Weak Acids Salts of Weak Bases and Weak Acids (( 弱鹼及弱酸形成的鹽弱鹼及弱酸形成的鹽 ) ) 11.11.Salts That Contain Small, Highly Charged Cations Salts That Contain Small, Highly Charged Cations

2

A Review of Strong Electrolytes

•This chapter details the equilibria of weak acids and bases.–We must distinguish weak acids and bases from strong electrolytes.

•Weak acids and bases ionize or dissociate partially, much less than 100%.–In this chapter we will see that it is often less than 10%!

•Strong electrolytes ionize or dissociate completely. ( 強電解質完全解離 )–Strong electrolytes approach 100% dissociation in aqueous solutions. ( 強電解質在水溶液中幾近 100% 解離 )

3

Aqueous Solutions: An Aqueous Solutions: An IntroductionIntroduction

• The reason nonelectrolytes do not conduct electricity is because they do not form ions in solution.

• ions conduct electricity in solution

4

Acid: a substance that produces hydrogen ions, H+, in aqueous solutions

Base: a substance that produces hydroxide ions, OH-, in aqueous solutions

Salt: a compound that contains a cation other than H+, and an anion other than hydroxide ion, OH- , or oxide ion, O2-

4


•Classification of solutes–strong electrolytes 強電解質 - conduct electricity extremely well in dilute aqueous solutions

•Examples of strong electrolytes1.HCl, HNO3, etc.

– strong soluble acids

2.NaOH, KOH, etc.–strong soluble bases

3.NaCl, KBr, etc.–soluble ionic salts– ionize in water essentially 100%

5

檸檬酸

5


•Classification of solutes–weak electrolytes - conduct electricity poorly in dilute aqueous solutions

1.CH3COOH, (COOH)2

•weak acids

2.NH3, Fe(OH)3

•weak bases

3.some soluble covalent salts•ionize in water much less than 100%

66


Strong and Weak Acids• Acids are substances that generate H+ in

aqueous solutions.• Strong acids ionize 100% in water.

7

HCl(g) H+(aq) + Cl-

(aq)100%

HNO3 + H2O H3O+(aq) + NO-

3(aq)

HNO3 H+(aq) + NO-

3(aq)H2O

100%

7


8

氫氯酸氫溴酸

氫碘酸硝酸

氯酸

硫酸

過氯酸

strong acids--ionize almost 100%

8


9

Weak acids--Typically ionize 10% or less!

氫氟酸醋酸

亞硝酸氰化氫

碳酸

亞硫酸

磷酸

草酸9


Strong Bases, Insoluble Bases, and Weak Bases

• Strong Bases–Characteristic of common inorganic bases is that they produce OH- ions in solution.

–Similarly to strong acids, strong bases ionize 100% in water.

10

NaOH Na+(aq) + OH-

(aq)H2O

Ba(OH)2 Ba+2(aq) + 2OH-

(aq)

H2O

10

11

氫氧化鋰氫氧化鈉氫氧化鉀氫氧化銣

氫氧化銫

氫氧化鈣

氫氧化鍶

氫氧化鋇


Notice that they are all hydroxides of IA and IIA metals

11


Strong Bases, Insoluble Bases, and Weak Bases

• Insoluble bases– Ionic compounds that are insoluble in

water, consequently, not very basic.– Cu(OH)2, Zn(OH)2, Fe(OH)2 氫氧化亞鐵 , Fe(OH)3

• Weak bases– are covalent compounds that ionize slightly

in water.– Ammonia is most common weak base-- NH3

12

NH3(g) + H2O(l) NH4+

(aq) + OH-(aq)

12

13


Most Water Soluble SaltsThe solubility guidelines from Chapter 4

will help you remember these salts.

14

NaCl(s) Na+(aq) + Cl-

(aq)H2O

100%

Ca(NO3)2 Ca+2(aq) + 2NO3

-(aq)

H2O100%


•The calculation of ion concentrations in solutions of strong electrolytes is easy.

Example 18-1: Calculate the concentrations of ions in 0.050 M nitric acid, HNO3.

HNO3 + H2O H3O+(aq) + NO-

3(aq)100%

0.050M 0.050M 0.050M

15


Example 18-1 Calculation of Concentrations of IonsCalculation the molar concentration of Ba2+ and OH- ions in 0.03M barium hydroxide.

Example 18-1 Calculation of Concentrations of IonsCalculation the molar concentration of Ba2+ and OH- ions in 0.03M barium hydroxide.Ba(OH)2 Ba2+

(aq) + 2OH-(aq)

H2O

initial 0.03Mchange -0.03M +0.03M +2x(0.03)Mfinal 0.0M 0.03M +0.06M

[Ba2+]=0.03M [OH-]=0.06M

Exercise 4 and 6

16


Example 18-2: Calculate the concentrations of ions in 0.020 M strontium hydroxide, Sr(OH)2, solution.

Sr(OH)2 Sr2+(aq) + 2OH-

(aq)

H2O

initial 0.02Mchange -0.02M +0.02M +2x(0.02)Mfinal 0.0M 0.02M +0.04M

[Sr2+]=0.02M [OH-]=0.04M

17

The Autoionization of Water

水的自身離子化 [ 反應 ] •Pure water ionizes very slightly.

–The concentration of the ionized water is less than one-millionth molar at room temperature.

•We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter.

•Because the activity of pure water is 1, the equilibrium constant for this reaction is:

Kc=[H3O+][OH-]

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

(aq)

18


•Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC.–Note that this is at 25oC, not every temperature!

•We can determine the value of Kc from this information.

Kc=[H3O+][OH-]

= (1.0x10-7)x(1.0x10-7)

= 1.0x10-14

19


•This particular equilibrium constant is called the ion-product for water and given the symbol Kw.

–Kw is one of the recurring expressions for the remainder of this chapter and Chapters 19 and 20.

Kw=[H3O+][OH-]

= 1.0x10-14

20

Example 18-2 Calculation of Ions Concentrations Calculation the concentrations of H3O+ and OH- ions in 0.05M HNO3 solution.

Example 18-2 Calculation of Ions Concentrations Calculation the concentrations of H3O+ and OH- ions in 0.05M HNO3 solution.

initial 0.05Mchange -0.05M +0.05M +0.05MAt equil 0.0M 0.05M +0.05M

[H3O+]=[NO3-]=0.05M

Exercise 14

HNO3 + H2O H3O+(aq) + NO3

- (aq)Strong acid

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

(aq)

initial 0.05Mchange -2xM +xM +xMAt equil (0.05+x)M +xM

Kw=[H3O+][OH-]1.0x10-14 =(0.05+x)(x) (very small number)1.0x10-14 =(0.05)(x) x=2.0x10-13 M=[OH-] 21

Example 18-3: Calculate the concentrations of H3O+ and OH- in 0.050 M HCl.

HCl + H2O H3O+ + Cl-

initial 0.05Mchange -0.05M +0.05M+0.05MAt equil 0.0M 0.05M+0.05M

[H3O+]=[Cl-]=0.05M

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

(aq)

initial 0.05Mchange -2xM +xM +xMAt equil (0.05+x)M +xM

Kw=[H3O+][OH-]1.0x10-14 =(0.05+x)(x) (very small number)1.0x10-14 =(0.05)(x) x=2.0x10-13 M=[OH-] 22

The pH and pOH scales• A convenient way to express the acidity

and basicity of a solution is the pH and pOH scales.

• The pH of an aqueous solution is defined as: pH=-log[H3O+]

or [H3O+]=10-pH

pOH=-log[OH-]or [OH-]=10-pOH

23

The pH and pOH scales•If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated.Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M.

pH=-log[H3O+]

pH=-log(3.0x10-2)pH=1.52

24

The pH and pOH scalesExample 18-5: The pH of a solution is 4.597. What

is the concentration of H3O+?

pH=-log[H3O+]

4.597=-log[H3O+]

[H3O+]=10-4.597

[H3O+]=2.53x10-5M

25

Example 18-3 Calculation of pHCalculate the pH of a solution in which the H3O+ concentration is 0.050mol/L.

Example 18-3 Calculation of pHCalculate the pH of a solution in which the H3O+ concentration is 0.050mol/L.

pH=-log[H3O+]

pH=-log(5.0x10-2)pH=1.3

Example 18-4 Calculation of H3O+ concentration from pHThe pH of a solution is 3.301. What is the concentration of H3O+ in this solution?

Example 18-4 Calculation of H3O+ concentration from pHThe pH of a solution is 3.301. What is the concentration of H3O+ in this solution?pH=-log[H3O+]

3.301=-log[H3O+][H3O+]=10-3.301

[H3O+]=5.0x10-4M

Exercise 22

Exercise 24 26

The pH and pOH scales•A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C.

•Taking the logarithm of both sides of this equation gives:

[H3O+][OH-]= 1.0x10-14

log[H3O+]+log[OH-]= -14.0

•Multiplying both sides of this equation by -1 gives:

•Which can be rearranged to this form:

-log[H3O+]+(-log[OH-])= 14.0

pH + pOH = 14.0

27

The pH and pOH scales

•Remember these two expressions!! –They are key to the next three chapters!

[H3O+][OH-]= 1.0x10-14

pH + pOH = 14.0

28

The pH and pOH scales•The usual range for the pH scale is 0 to 14.

•And for pOH the scale is also 0 to 14 but inverted from pH.–pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0.

[H3O+]=1.0 M to [H3O+]= 1.0x10-14M

pH=0 to pH=14.0

[OH-]= 1.0x10-14 up to [OH-]=1.0M

pOH=14.0 to pOH=0

29

The pH and pOH scalesExample 18-6: Calculate the [H3O+], pH, [OH-], and pOH

for a 0.020 M HNO3 solution.– Is HNO3 a weak or strong acid?– What is the [H3O+] ?

initial 0.02Mchange -0.02M +0.02M +0.02M

At equil 0.0M 0.02M +0.02M[H3O+]= 2x10-2 M

HNO3 + H2O H3O+(aq) + NO3

- (aq)

Strong acid

pH=-log(2x10-2 M)pH=1.70

Kw=[H3O+][OH-]=1.0x10-14

[OH-]=1.0x10-14/[H3O+] =1.0x10-14/2.0x10-2

=5.0x10-13 M pOH=-log[OH-] =-log(5.0x10-13) =12.30

30

The pH and pOH scales•To help develop familiarity with the pH and pOH scale we can look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.

[H3O+] [OH-] pH pOH

1.0 M 1.0 x 10-14 M

0.00 14.00

1.0 x 10-3 M

1.0 x 10-11 M

3.00 11.00

1.0 x 10-7 M 1.0 x 10-7 M

7.00 7.00

2.0 x 10-12

M5.0 x 10-3

M11.70 2.30

1.0 x 10-14

M1.0 M 14.00 0.00 31

Example 18-6 Calculations Involving pH and pOHCalculate the [H3O+], pH, [OH-], and pOH for a 0.015 M Ca(OH)2 solution.

Example 18-6 Calculations Involving pH and pOHCalculate the [H3O+], pH, [OH-], and pOH for a 0.015 M Ca(OH)2 solution.

initial 0.015Mchange -0.015M +0.015M+2x(0.015)M

At equil 0.0M 0.015M+0.03M[OH-]= 3x10-2 M

Ca(OH)2 Ca2+ + 2OH-

(aq)

pOH=-log(3x10-2 M)pOH=1.52pH=14-pOH

=14-1.52 =12.48

Kw=[H3O+][OH-]=1.0x10-14

[H3O+]=1.0x10-14/[OH-] =1.0x10-14/3.0x10-2

=3.3x10-13 M Exercise 26 and 37 32

33

胃酸

醋檸檬

炭酸飲料

蕃茄啤酒

尿液牛奶唾液

血漿蛋白

鎂乳 ( 瀉藥 ) 氨水

34

pH meter

Universal indicator

Universal pH paper35

Ionization Constants for Weak Monoprotic Acids 單質子酸 and

BasesStrong acids ionize completely in dilute aqueous, whereas weak acids ionize only slightly

•Let’s look at the dissolution of acetic acid, a weak acid, in water as an example.

•The equation for the ionization of acetic acid is:

•The equilibrium constant for this ionization is expressed as:

Kc=[H3O+] [CH3COO-][CH3COOH] [H2O]

CH3COOH + H2O H3O+ + CH3COO-

36

Ionization Constants for Weak Monoprotic Acids and

Bases•The water concentration in dilute aqueous solutions is very high.

•1 L of water contains 55.5 moles of water.•Thus in dilute aqueous solutions:

[H2O] 55.5M

37


Bases•The water concentration is many orders of magnitude greater than the ion concentrations.

•Thus the water concentration is essentially that of pure water.–Recall that the activity of pure water is 1.

Kc=[H3O+] [CH3COO-][CH3COOH] [H2O]

Kc [H2O] =[H3O+] [CH3COO-]

[CH3COOH]

K [H3O+] [CH3COO-]

[CH3COOH]38


Bases•We can define a new equilibrium constant for weak acid equilibria that uses the previous definition.–This equilibrium constant is called the acid ionization constant 酸解離常數 .

–The symbol for the ionization constant is Ka.

Ka =[H3O+] [CH3COO-]

[CH3COOH]=1.8x10-5

for acetic acid

39


Bases•In simplified formsimplified form the dissociation equation and acid ionization expression are written as:

Ka =[H+] [CH3COO-]

[CH3COOH]=1.8x10-5

CH3COOH H+ + CH3COO-

40

Ionization Constants for Weak Monoprotic Acids 單質子酸 and

Bases

Carbonic acid H2CO3

Citric acid C3H5O(COOH)3

41


Bases•From the above table we see that the order of increasing acid strength for these weak acids is:

•The order of increasing base strength of the anions (conjugate bases) of these acids is:

HF > HNO2 > CH3COOH > HClO > HCN

F- < NO2- < CH3COO- < ClO- < CN-

42


BasesExample 18-8: Write the equation for the

ionization of the weak acid HCN and the expression for its ionization constant.

Ka =[H+] [CN-]

[HCN]=4.0x10-10

HCN H+ + CN-

43

Example 18-9: In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid.

Ka =[H+] [Y-]

[HY]

HY H+ + Y-

•Since the weak acid is 5.0% ionized, it is also 95% unionized.

•Calculate the concentration of all species in solution.[H+]=[Y-]=0.05 x (0.12M) =0.006M =6x10-3M

[HY]=0.95 x (0.12M) =0.11M

Ka =[H+] [Y-]

[HY] =(6.0x10-3)(6.0x10-3)

(0.11)

Ka =3.3x10-4

44

Example 18-10: The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant?

pH = 2.97 so [H+]= 10-pH

[H+]= 10-2.79 [H+]=1.1x 10-3

•Use the [H3O+] and the ionization reaction to determine concentrations of all species.

Ka =[H+] [A-]

[HA]

HA H+ + A-

At equil (0.1-1.1x10-3)M 1.1x10-3M 1.1x10-3M 0.1M

•Calculate the ionization constant from this information.

=(1.1x10-3)(1.1x10-3)

0.1Ka =1.2x10-5

45

Example 18-7 Calculations of Ka and pKa from equilibrium ConcentrationsNicotinic acid is a weak monoprotic oganic acid that we can represent as HA.

A dilute solution of nicotinic acid was found to contain the following concentrations at equilibrium at 25oC. What are the Ka and pKa values? [HA]=0.049M; [H3O+]=[A-]=8.4x10-4M.

Example 18-7 Calculations of Ka and pKa from equilibrium ConcentrationsNicotinic acid is a weak monoprotic oganic acid that we can represent as HA.

A dilute solution of nicotinic acid was found to contain the following concentrations at equilibrium at 25oC. What are the Ka and pKa values? [HA]=0.049M; [H3O+]=[A-]=8.4x10-4M.

HA + H2O H3O+ + A-

Ka =[H3O+] [A-]

[HA](8.4x10-4)(8.4x10-4)

0.049

Ka =1.4x10-5

=

pKa =-log(1.4x10-5) =4.85

Exercise 30

46

Example 18-8 Calculations of Ka from Percent IonizationIn 0.01M solution, acetic acid is 4.2% ionized. Calculate its ionization constant.

Example 18-8 Calculations of Ka from Percent IonizationIn 0.01M solution, acetic acid is 4.2% ionized. Calculate its ionization constant.

•Since the weak acid is 4.2% ionized, it is also 95.8% unionized.

•Calculate the concentration of all species in solution.[H+]=[CH3COO-]=0.042 x (0.01M) =4.2x10-4M

[CH3COOH]=0.958 x (0.01M) =9.58x10-3M

=(4.2x10-4)(4.2x10-4)

(9.58x10-3)

Ka=[H3O+] [CH3COO-]

[CH3COOH]


= 1.8x10-5

Exercise 38

47

Example 18-9 Calculations of Ka from pHThe pH of a 0.115M solution of chloroacetic acid, ClCH2COOH, is measure to be 1.92. Calculate Ka from this weak momoprotic acid.

Example 18-9 Calculations of Ka from pHThe pH of a 0.115M solution of chloroacetic acid, ClCH2COOH, is measure to be 1.92. Calculate Ka from this weak momoprotic acid.

=(0.012)(0.012)(0.103)

Ka=[H3O+] [A-]

[HA]

= 1.4x10-3

Exercise 40

HA + H2O H3O+ + A-

pH=-log[H3O+][H3O+]=10-pH=10-1.92=0.012M

initial 0.115Mchange -0.012M +0.012M+0.012M

At equil 0.103M 0.012M 0.012M

48

Example 18-11: Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution.

Ka=[H3O+] [CH3COO-]

[CH3COOH]=1.8x10-5


initial 0.15Mchange -xM +xM +xMAt equil(0.15-x)M xM xM

Ka=(x)(x)

(0.15-x)=1.8x10-5 x is small

enough to ignore compare to 0.15M

x2(0.15)x(1.8x10-5)x22.7x10-6

x=1.6x10-3 M=[H3O+]=[CH3COO-][CH3COOH]=0.15-1.6x10-3 0.15M

49


Bases•Let us now calculate the percent ionization for the 0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is

% ionization=[CH3COOH]ionized

[CH3COOH]original

x100%

% ionization=(1.6x10-3M)

0.15Mx100%

=1.1%

50

Example 18-12: Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution.

Ka= 4.0 x 10-10 for HCN

Ka=[H3O+] [CN-]

[HCN]=4.0x10-10

HCN + H2O H3O+ + CN-


=(x)(x)

(0.15-x)x is small enough to ignore compare to 0.15M

x2(0.15)x(4.0x10-10)x=7.7x10-6 M=[H3O+]=[CN-][HCN]=(0.15-7.7x10-6) 0.15M % ionization=

[HCN]ionized

[HCN]original

x100%

=(7.7x10-6/0.15)x100%=0.0051%

51


Bases•Let’s look at the percent ionization of two weak acids as a function of their ionization constants. Examples 18-11 and 18-12 will suffice.

• Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

Solution Ka [H+] pH%

ionization

0.15 M acetic acid

1.8 x 10-5 1.6 x 10-3 2.80 1.1

0.15 M HCN 4.0 x 10-

10

7.7 x 10-6 5.11 0.0051

52

•All of the calculations and understanding we have at present can be applied to weak acids and weak bases!

Example 18-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia.

Kb=[NH4

+] [OH-][NH3]

=1.8x10-5

NH3 + H2O NH4+ + OH-


=(x)(x)


x2(0.15)x(1.8x10-5)x=1.6x10-3 M=[NH4

+]=[OH-][NH3]=(0.15-1.6x10-3) 0.15M % ionization=

[NH3]ionized

[NH3]original

x100%

=(1.6x10-3/0.15)x100%=1.1%

53

Example 18-14: The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution.pH=11.37

pOH+pH=14 pOH=14-11.37=2.63[OH-]=10-pOH=10-2.63=2.3x10-3M

Kb=[NH4

+] [OH-][NH3]

=1.8x10-5

NH3 + H2O NH4+ + OH-

initial xMchange2.3x10-3M

At equil(x-2.3x10-3)M

=(2.3x10-3)x(2.3x10-3)

(x-2.3x10-3)

x(2.3x10-3)(2.3x10-3)/(1.8x10-5)x=0.3 M=[NH3]

2.3x10-3M2.3x10-3M2.3x10-3M2.3x10-3M

54

55

Example 18-10 Calculation of Concentrations from Ka(a) Calculate the concentration of the various species in 0.10M hypochlorous acid, HOCl.For HOCl, Ka=3.5x10-8. (b) What is the pH of this solution?

Example 18-10 Calculation of Concentrations from Ka(a) Calculate the concentration of the various species in 0.10M hypochlorous acid, HOCl.For HOCl, Ka=3.5x10-8. (b) What is the pH of this solution?

=(x)(x)

(0.10-x)Ka=

[H3O+] [OCl-][HOCl] = 3.5x10-8

Exercise 42

HOCl + H2O H3O+ + OCl-


x2(0.1)(3.5x10-8)x5.9x10-5=[H3O+]=[OCl-]

[HOCl]=(0.10-5.9x10-5)=0.10MpH=-log(5.9x10-5) = 4.23

56

Example 18-11 Percent IonizationCalculate the percent ionization of a 0.10 M solution of acetic acid.

Example 18-11 Percent IonizationCalculate the percent ionization of a 0.10 M solution of acetic acid.

Exercise 48

Ka=[H3O+] [CH3COO-]

[CH3COOH]=1.8x10-5

CH3COOH + H2O H+ + CH3COO-


=(x)(x)


x2(0.1M)x(1.8x10-5)x=1.3x10-3 M=[H3O+]=[CH3COO-][CH3COOH]=0.10-1.3x10-3 0.10M % ionization=

[CH3COOH]ionized

[CH3COOH]original

x100%

=(1.3x10-3M)

0.10Mx100%=1.3%

57

Example 18-12 pKa ValueThe Ka values for acetic acid and hydrofluoric acid are 1.8x10-5 and 7.2x10-4, respectively. What are their pKa value?

For CH3COOH pKa=-logKa =-log(1.8x10-5) =4.74

Example 18-12 pKa ValueThe Ka values for acetic acid and hydrofluoric acid are 1.8x10-5 and 7.2x10-4, respectively. What are their pKa value?

For CH3COOH pKa=-logKa =-log(1.8x10-5) =4.74 Exercise 50

For HF pKa=-logKa =-log(7.2x10-4) =3.14

Example 18-13 Acid Strength and Ka ValueGiven the following lost of weal acids and their Ka values, arrange the acids in order of (a) increasing acid strength and (b) increasing pKa values.Acid KaHOCl 3.5x10-8

HCN 4.0x10-10

HNO2 4.5x10-4

Example 18-13 Acid Strength and Ka ValueGiven the following lost of weal acids and their Ka values, arrange the acids in order of (a) increasing acid strength and (b) increasing pKa values.Acid KaHOCl 3.5x10-8

HCN 4.0x10-10

HNO2 4.5x10-4

(a) Increasing acid strength HCN<HOCl<HNO2

(b) Increasing pKa values HNO2<HOCl<HCN

Exercise 109

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Polyprotic Acids•Many weak acids contain two or more acidic hydrogens.

–Examples include H3PO4 and H3AsO4.•The calculation of equilibria for polyprotic acids is done in a stepwise fashion.–There is an ionization constant for each step.

•Consider arsenic acid, H3AsO4, which has three ionization constants.

1. Ka1 = 2.5 x 10-4

2. Ka2 = 5.6 x 10-8

3. Ka3 = 3.0 x 10-13

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Polyprotic Acids•The first ionization step for arsenic acid is:

Ka1=[H+] [H2AsO4

-][H3AsO4]

H3AsO4 H+ + H2AsO4-

=2.5x10-4

Ka2=[H+] [HAsO4

2-][H2AsO4

-]

H2AsO4- H+ + HAsO4

2-

=5.6x10-8

•The second ionization step for arsenic acid is:

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Polyprotic Acids•The third ionization step for arsenic acid is:

Ka3=[H+] [AsO4

3-][H2AsO4

2-]

HAsO42- H+ + AsO4

3-

=3.0x10-13

•Notice that the ionization constants vary in the following fashion:

•This is a general relationship.

–For weak polyprotic acids the Ka1 is always > Ka2, etc.

Ka1>Ka2>Ka3

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Polyprotic AcidsExample 18-15: Calculate the concentration of all

species in 0.100 M arsenic acid, H3AsO4, solution.

H3AsO4 H+ + H2AsO4-

At equil (0.10-x)M xM xM

Ka1=[H+] [H2AsO4

-][H3AsO4]

=2.5x10-4=(x)(x)

(0.10-x)X2+2.5x10-4X-2.5x10-5=0

x=-5.1x10-3 and x=4.9x10-3

[H+]=[H2AsO4-]=4.9x10-3M

[H3AsO4]=(0.10-4.9x10-3) M =0.095M

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x= -2.5x10-4 (2.5x10-4)2-4(1)(-2.5x10-5)

2x1

Polyprotic AcidsNext, write the equation for the second step

ionization and represent the concentrations.

H2AsO4- H+ + HAsO4

-2

[] from 1st step(4.9x10-3)M At equil (4.9x10-3-y)M yM yM

Ka2=[H+] [HAsO4

-2][H2AsO4

-]=5.6x10-8

(4.9x10-3)M

= (4.9x10-3-y) (4.9x10-3+y)(y)

y<<4.9x10-3 Thus (4.9x10-3 –y) 4.9x10-3

Ka2= (4.9x10-3) (4.9x10-3)(y)

=5.6x10-8

y=5.6x10-8=[H+]2nd=[HAsO42-]

Note the [H+]1st >>[H+]2nd

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Polyprotic AcidsFinally, repeat the entire procedure for the third

ionization step.HAsO4

2- H+ + AsO43-

[] from 1st and 2nd step(5.6x10-8)M At equil(5.6x10-8-z)M zM zM

(4.9x10-3+5.6x10-8)M

Ka3=[H+] [AsO4

-3][HAsO4

-]=3.0x10-13

= (5.6x10-8-z) (4.9x10-3+5.6x10-8+z)(z)

z<<5.6x10-8Thus (5.6x10-8 –z) 5.6x10-8

Ka3= (5.6x10-8) (4.9x10-3)(z)

=3.0x10-13

z=3.4x10-18=[H+]3rd=[AsO43-]

•Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.[H+][OH-] =1.0x10-14

(4.9x10-3)x[OH-]=1.0x10-14

[OH-]=2.0x10-1264

Polyprotic Acids• A comparison of the various species in 0.100 M

H3AsO4 solution follows.

Species Concentration

H3AsO4 0.095 M

H+ 0.0049 M

H2AsO4- 0.0049 M

HAsO42- 5.6 x 10-8 M

AsO43- 3.4 x 10-18 M

OH- 2.0 x 10-12 M

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Example 18-18 Solutions Strong Polyprotic AcidCalculate concentration of all species present in 0.10M H2SO4. Ka2=1.2x10-2.

Example 18-18 Solutions Strong Polyprotic AcidCalculate concentration of all species present in 0.10M H2SO4. Ka2=1.2x10-2.

[H3O+]2nd=[SO42-]=0.01M

=(0.1+x)(x)(0.1-x)

Ka2=[H3O+] [SO4

-2][HSO4

-]

H2SO4 + H2O H3O+ + HSO4-

= 1.2x10-2

Exercise 38

0.10M 0.10M 0.10M100% ionization

HSO4- + H2O H3O+ + SO4

-2

0.10M 0.10M-xM +xM +xM

(0.10-x)M (0.10+x)MxM

x2+0.1x= 1.2x10-3 – (1.2x10-2)xx=0.01 or x=-0.12

[H2SO42-] 0M[HSO4

-]=0.1-0.01M=0.09 M[H3O+]=0.10+0.01M=0.11 M[OH-]=1.0x10-14/0.11=9.1x10-14 M

66

Solvolysis 溶劑分解溶劑分解 [[ 作用作用 ]]•This reaction process is the most difficult concept in this chapter.

•Solvolysis is the reaction of a substance with the solvent in which it is dissolved.

•Hydrolysis refers to the reaction of a substance with water or its ions.

•Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.

67

Solvolysis•Hydrolysis refers to the reaction of a substance with water or its ions.–Hydrolysis is solvolysis in aqueous solutions.

•The combination of a weak acid’s anion with H3O+ ions, from water, to form nonionized weak acid molecules is a form of hydrolysis.A- + H3O+ HA +

H2O Recall H2O + H2O H3O+ + OH-

•The reaction of the anion of a weak monoprotic acid with water is commonly represented as:

A- + H2O HA + OH- The removal of H3O+ upsets the water equilibrium

68

Solvolysis•Recall that at 25oC•in neutral solutions: [H3O+] = 1.0 x 10-7 M = [OH-]

•in basic solutions:[H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M

•in acidic solutions:

[OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M

69

Solvolysis•Remember from BrØnsted-Lowry acid-base theory:–The conjugate base of a strong acid is a very weak base.

–The conjugate base of a weak acid is a stronger base.

70

Solvolysis

• The conjugate base of HCl, the Cl- ion, is a very weak base.– The chloride ion is such a weak base that it will not react

with the hydronium ion.

• This fact is true for all strong acids and their anions.

Hydrochloric acid, a typical strong acid, is essentially completely ionized in dilute aqueous solutions.

HCl +H2O H3O++Cl-~100%

Cl- + H3O+ → No rxn. In dilute aqueous solutions

71

Solvolysis•HF, a weak acid, is only slightly ionized in dilute aqueous solutions.

•Its conjugate base, the F- ion, is a much stronger base than the Cl- ion.

•The F- ions combine with H3O+ ions to form nonionized HF.–Two competing equilibria are established.

HF +H2O H3O++F-

Only slightly

H3O++F- HF +H2ONearly completely

72

Solvolysis•Dilute aqueous solutions of salts that contain no free acid or base come in four types:1. Salts of Strong Bases and Strong Acids

2. Salts of Strong Bases and Weak Acids 3. Salts of Weak Bases and Strong Acids 4. Salts of Weak Bases and Weak Acids

73

Salts of Strong Bases and Weak Acids

• Salts made from strong acids and strong soluble bases form neutral aqueous solutions.

• An example is potassium nitrate, KNO3, made from nitric acid and potassium hydroxide.

KNO3(s) K++NO3-~100% in H2O

H2O +H2O H3O++OH-

The ions that are in solutionThe KOH and HNO3 are present in equal amountsThere is no reaction to upset [H3O+][OH-]Thus the solution is neutral

HNO3 KOH

74


•Salts made from strong soluble bases and weak acids hydrolyze to form basic solutions.–Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions.

•An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid.

NaClO(s) Na++ClO-~100% in H2O

H2O +H2O H3O++OH-

The ions that are in solutionWhich is the stronger acid or base?

HClO NaOH

75


•We can combine these last two equations into one single equation that represents the total reaction.

NaClO(s) Na++ClO-~100% in H2O

H2O +H2O H3O++OH-

ClO- + H3O+ HClO + H2O

ClO- + H2O HClO + OH-

•The equilibrium constant for this reaction, called the hydrolysis constant, is written as:

Kb=[HClO] [OH-]

[ClO-]76


• Algebraic manipulation of the previous expression give us a very useful form of the expression.

• Multiply the expression by one written as [H+]/ [H+].

H+/H+ = 1Kb=

[HClO][OH-][ClO-]

x[H+][H+]

Kb=[HClO]

[ClO-][H+] x[H+][OH-]

1

Kb=1

Ka for HClO

x Kw= Kw

Ka for HClO= 1x10-14

3.5x10-8

Kb=[HClO] [OH-]

[ClO-]= 2.9x10-7

77


• This same method can be applied to the anion of any weak monoprotic acid.

A- + H2O HA + OH-

= Kw

Ka for HAKb=

[HA] [OH-][A-]

Kw=KaKb

78


Example 18-16: Calculate the hydrolysis constants for the following anions of weak acids.

The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4.

F- + H2O HF + OH-

= Kw

Ka for HFKb=

[HF] [OH-][F-]

Kb=1.0x10-14

7.2x10-4=1.4x10-11

79


• The cyanide ion, CN-, the anion of hydrocyanic acid, HCN. For HCN, Ka = 4.0 x 10-10.

CN- + H2O HCN + OH-

= Kw

Ka for HCNKb=

[HCN] [OH-][CN-]

Kb=1.0x10-14

4.0x10-10=2.5x10-5

80

Example 18-17: Calculate [OH-], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

NaClO(s) Na++ ClO-~100% in H2O

0.10M 0.10M 0.10MClO- + H2O HClO + OH-


Kb=[HClO] [OH-]

[ClO-]=2.9x10-7=

(x)(x)(0.1-x)

x<<0.1 so (0.1-x)0.1x=1.7x10-4M=[ClO-]=[OH-]pOH=-log(1.7x10-4)=3.77pH=14-3.77=10.23

% hydrolysis=[ClO-]hydrolyzed

[ClO-]originalx100%

% hydrolysis= 1.7x10-4

0.10M x100%=0.17% 81


• If a similar calculation is performed for 0.10 M NaF solution and the results from 0.10 M sodium fluoride and 0.10 M sodium hypochlorite compared, the following table can be constructed.

Solution Ka Kb

[OH-] (M) pH

% hydrolysis

NaF 7.2 x 10-4 1.4 x 10-

11

1.2 x 10-6 8.08 0.0012

NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23

0.17

82

Salts of Weak Bases and Strong Acids

• Salts made from weak bases and strong acids form acidic aqueous solutions.

• An example is ammonium bromide, NH4Br, made from ammonia and hydrobromic acid.

NH4Br (s) NH4++ Br-~100% in H2O

H2O +H2O OH- + H3O+

The ions that are in solutionWhich is the stronger acid or base?

HBrNH3

83


• The reaction may be more simply represented as:

NH4+ + OH- NH3 + H2O

The relatively strong acid, NH4+, reacts with

the OH- ion removing it from solution leaving excess H3O+

NH4+ + H2O NH3 + H3O+

generates excess H3O+

NH4+ NH3 + H+

• The hydrolysis constant expression for this process is:

•more simply as:

Ka=[NH3][H3O+]

[NH4+]

or Ka=[NH3][H+]

[NH4+]

84


• Multiplication of the hydrolysis constant expression by [OH-]/ [OH-] gives:

Ka=[NH3][H3O+]

[NH4-]

x[OH-][OH-]

Ka=[NH3]

[NH4+][OH-] x

[H3O+][OH-]1

Ka=1

Kb for NH3

x Kw= Kw

Kb for NH3

= 1x10-14

1.8x10-5

Ka=[NH3] [H3O+]

[NH4-]

= 5.6x10-10

85


• In its simplest form for this hydrolysis:

Ka=[NH3] [H+]

[NH4-]

= 5.6x10-10

NH4+ NH3 + H+

86

Example 18-18: Calculate [H+], pH, and percent hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH4Br, solution.NH4

+ NH3 + H+


Ka=[NH3] [H+]

[NH4+]

= 5.6x10-10=(x)(x)

(0.1-x)x<<0.1 so (0.1-x)0.1x2=5.6x10-11M

X=7.5x10-6M=[NH3]=[H+]pH=-log(7.5x10-6)=5.12

% hydrolysis=[NH4

+]originalx100%

% hydrolysis= 7.5x10-6

0.10M x100%=0.0075%

[NH4+]hydrolyzed

87

Salts of Weak Bases and Weak Acids

• Salts made from weak acids and weak bases can form neutral, acidic or basic aqueous solutions.– The pH of the solution depends on the

relative values of the ionization constant of the weak acids and bases.

1.Salts of weak bases and weak acids for which parent Kbase =Kacid make neutral solutions.– An example is ammonium acetate,

NH4CH3COO, made from aqueous ammonia, NH3,and acetic acid, CH3COOH.

Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.

88


• The ammonium ion hydrolyzes to produce H+ ions. Its hydrolysis constant is:

Ka=[NH3] [H+]

[NH4-]

= 5.6x10-10

NH4+ NH3 + H+

• The acetate ion hydrolyzes to produce OH- ions. Its hydrolysis constant is:

Ka=[OH-] [CH3COOH]

[CH3COO-]=5.6x10-10

CH3COO- + H2O CH3COOH + OH-

89


•Because the hydrolysis constants for both ions are equal, their aqueous solutions are neutral.

•Equal numbers of H+ and OH- ions are produced.

NH4+CH3COO- NH4

++CH3COO-~100% in H2O

H2O +H2O OH- + H3O+

The ions that are in solutionA weak acid and base are formed in solution

NH4OH CH3COOH

90


2.Salts of weak bases and weak acids for which parent Kbase > Kacid make basic solutions.

• An example is ammonium hypochlorite, NH4ClO, made from aqueous ammonia, NH3,and hypochlorous acid, HClO.

Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8

91


3.Salts of weak bases and weak acids for which parent Kbase < Kacid make acidic solutions.

• An example is trimethylammonium fluoride,(CH3)3NHF, made from trimethylamine, (CH3)3N,and hydrofluoric acid acid, HF.–Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4

92


• Summary of the major points of hydrolysis up to now.

1.The reactions of anions of weak monoprotic acids (from a salt) with water to form free molecular acids and OH-.

A- + H2O HA + OH-

= Kw

Ka for HAKb=

[HA] [OH-][A-]

93


2.The reactions of anions of weak monoprotic acids (from a salt) with water to form free molecular acids and OH-.

BH+ + H2O B + H3O+

Ka=Kw

Kb (waek base)

94


• Aqueous solutions of salts of strong acids and strong bases are neutral.

• Aqueous solutions of salts of strong bases and weak acids are basic.

• Aqueous solutions of salts of weak bases and strong acids are acidic.

• Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic.

The values of Ka and Kb determine the pH.

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