16.1 Acids and Bases: A Brief Review - Yonseichem.yonsei.ac.kr/~mhmoon/pdf/GenChem_Brown/16_Acid-Base.pdf · 16.1 Acids and Bases: A Brief Review – Acids taste sour and cause certain

Chapter 16

Acid–Base Equilibria

Lecture Presentation

Yonsei University

© 2012 Pearson Education, Inc.

Acidsand

Bases

2

16.1 Acids and Bases: A Brief Review

– Acids taste sour and cause certain dyes to change color.

– Bases taste bitter and feel soapy.

• Arrhenius Model– An acid is a substance that, when dissolved in water,

increases the concentration of hydrogen ions(H+).

HCl(aq) H+(aq) + Cl(aq)

– A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions(OH-).

NaOH(aq) Na+(aq) + OH(aq)


Acidsand

Bases

3

The H+ Ion in Water

• The H+ ion is simply a proton.

• In water, clusters of hydrated H+(aq) ions form.

• The simplest cluster is H3O+(aq). – We call this a hydronium ion.

– Larger clusters are also possible (such as H5O2+ and H9O4

+)

– H+(aq) ≡ H3O+(aq)

A shell of 21 water molecules

Acidsand

Bases

4

16.2 Brønsted-Lowry Acids and Bases• Brønsted–Lowry

– An acid is a proton donor.

– A base is a proton acceptor.

:NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

A Brønsted–Lowry acid…

…must have a removable (acidic) proton.

A Brønsted–Lowry base…

…must have a pair of nonbonding electrons.


If it can be either, it is amphiprotic.

HCO3, HSO4

, H2O

Acidsand

Bases

5

What Happens When an Acid Dissolves in Water?

• Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid.

• As a result, the conjugate base of the acid and a hydronium ion are formed.


Acidsand

Bases

6

Conjugate Acids and Bases

• conjugate : “conjugare,” meaning “to join together.”

• Reactions between acids and bases alwaysyield their conjugate bases and acids.


Acidsand

Bases

7

Sample Exercise 16.1 Identifying Conjugate Acids and Bases

(a) What is the conjugate base of HClO4, H2S, PH4+, HCO3

?

(b) What is the conjugate acid of CN, SO42, H2O, HCO3

?

Solution(a) HClO4 less one proton H+ is ClO4

. The other conjugate bases are HS, PH3, and CO3

2.(b) CN plus one proton H+ is HCN. The other conjugate acids are HSO4

, H3O+, and H2CO3. Notice that the hydrogen carbonate ion (HCO3

) is amphiprotic. It can act as either an acid or a base.

Practice ExerciseWrite the formula for the conjugate acid of each of the following: HSO3

, F, PO43, CO.

Answers: H2SO3, HF, HPO42

, HCO+

Acidsand

Bases

8

Acid and Base Strength

• Strong acids are completelydissociated in water.– Their conjugate bases are quite

weak.

• Weak acids only dissociate partially in water.– Their conjugate bases are weak

bases.

• Substances with negligible acidity do not dissociate in water. CH4– Their conjugate bases are

exceedingly strong.


The stronger an acid is, the weaker its conjugate base will be.

Acidsand

Bases

9

Acid and Base Strength

• In any acid–base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

HCl(aq) + H2O(l) H3O+(aq) + Cl(aq)

• H2O is a much stronger base than Cl, so the equilibrium lies so far to the right that K is not measured (K >> 1).


• Acetate is a stronger base than H2O, so the equilibrium favors the left side (K < 1).

CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2(aq)

Acidsand

Bases

10

leveling effect of water

• In every acid-base reaction, the position of the equilibrium favors the transfer of a proton from the stronger acid to the stronger base. – H+ is the strongest acid that can exist in equilibrium in

aqueous solution.

– OH– is the strongest base that can exist in equilibrium in aqueous solution.

• Stronger acids react with water to produce hydronium ions and stronger bases react with water to form hydroxide ions.

: the leveling effect of water.

Acidsand

Bases

11

16.3 Autoionization of Water

• As we have seen, water is amphoteric.

• In pure water, a few molecules act as bases and a few act as acids : autoionization.


• ion product constant for water, Kw = [H3O+] [OH] =1.0 1014, at 25C,

Acidsand

Bases

12

Sample Exercise 16.5 Calculating [H+] from [OH]

Calculate the concentration of H+ (aq) in (a) a solution in which [OH] is 0.010 M, (b) a solution in which [OH] is 1.8 109 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 C .

Solution

(a) Using Equation 16.16, we have

This solution is basic because [OH] > [H+]

(b) In this instance

This solution is acidic because [H+] > [OH]

Acidsand

Bases

13

16.4 The pH Scale

pH = log [H3O+]• In pure water,

Kw = [H3O+] [OH] = 1.0 1014

• Since in pure water, [H3O+] = [OH],

[H3O+] = 1.0 1014 = 1.0 107


Acidsand

Bases

14

pH

These are the pH values for several common substances Fig. 16.5.


Acidsand

Bases

15

Other “p” Scales• The “p” in pH tells us to take the negative

base-10 logarithm of the quantity (in this case, hydronium ions).

• Some similar examples are– pOH = log [OH] pKw = log Kw


Because [H3O+] [OH] = Kw = 1.0 1014

we know that

log [H3O+] + log [OH] = log Kw = 14.00

or, in other words,

pH + pOH = pKw = 14.00 (at 25°C)

Acidsand

Bases

16

How Do We Measure pH?• For less accurate measurements, one can use

– Litmus paper

• “Red” paper turns blue above ~pH = 8

• “Blue” paper turns red below ~pH = 5

– Or an indicator.


• For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Acidsand

Bases

17

16.5 Strong Acids and Bases

Strong Acids• You will recall that the seven strong acids

are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

• These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.

HNO3(aq) + H2O(l) H3O+(aq) + NO3–(aq).

HNO3(aq) H+(aq) + NO3–(aq).

• For the monoprotic strong acids,[H3O+] = [acid]initial.


Acidsand

Bases

18

Strong Bases

• Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).

NaOH(aq) Na+(aq) + OH–(aq)

• Again, these substances dissociate completely in aqueous solution.

cf) Ionic metal oxides: O2–(aq) + H2O(l) 2OH–(aq)


Acidsand

Bases

19

16.6 Weak Acids

• For a generalized acid dissociation,

[H3O+] [A][HA]

Ka =

HA(aq) + H2O(l) A(aq) + H3O+(aq)


acid-dissociation constant

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

Acidsand

Bases

20

Calculating Ka from the pH

Ex 16.10) The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature.

102.38=4.2 103= [H3O+] = [HCOO]

[H3O+] [HCOO][HCOOH]

Ka =


Acidsand

Bases

21

Calculating Ka from pH

Now we can set up a table…

[HCOOH], M [H3O+], M [HCOO], M

Initially 0.10 0 0

Change

At equilibrium


4.2 103 +4.2 103 +4.2 103

0.10 4.2 103

= 0.0958 = 0.104.2 103 4.2 103

[4.2 103] [4.2 103][0.10]

Ka =

= 1.8 104

Acidsand

Bases

22

Calculating Percent Ionization

• Percent ionization = 100

• In ex. 16.10,

[H3O+]eq = 4.2 103 M

[HCOOH]initial = 0.10 M

[H3O+]eq

[HA]initial

Percent ionization = 1004.2 103

0.10= 4.2%


Acidsand

Bases

23

Calculating pH from Ka

Ex) Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq)

Ka for acetic acid at 25C is 1.8 105.


[H3O+] [C2H3O2]

[HC2H3O2]Ka =

Acidsand

Bases

24


We next set up a table…

[C2H3O2], M [H3O+], M [C2H3O2], M

Initially 0.30 0 0

Change

At equilibrium

We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.


x +x +x

0.30 x 0.30 x x

Acidsand

Bases

25


Now, (x)2

(0.30)1.8 105 =

(1.8 105) (0.30) = x2

5.4 106 = x2

2.3 103 = x

pH = log (2.3 103)pH = 2.64


Acidsand

Bases

26

Sample Exercise 16.12 Using Ka to Calculate pH

Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of Ka.)

Solution

Acidsand

Bases

27

Sample Exercise 16.13 Using Ka to Calculate Percent IonizationCalculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution.

(a)Solution

Acidsand

Bases

28

(b) Proceeding similarly for the 0.010 M solution, we have

Solving the resultant quadratic expression, we obtain

The percentage of molecules ionized is

Sample Exercise 16.13 Using Ka to Calculate Percent IonizationContinued

Comment Notice that if we do not use the quadratic formula, we calculate 8.2% ionization for (a) and 26% ionization for (b). Notice also that in diluting the solution by a factor of 10, the percentage of molecules ionized increases by a factor of 3. This result is in accord with what we see in Figure 16.11. It is also what we expect from Le Châtelier’s principle. (Section 15.7) There are more “particles” or reaction components on the right side of the equation than on the left. Dilution causes the reaction to shift in the direction of the larger number of particles (toward the product side) because this counters the effect of the decreasing concentration of particles.

Acidsand

Bases

29

Polyprotic Acids• Polyprotic acids have more than one acidic proton.

• If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.


H2SO3(aq) H+(aq) + HSO3–(aq) Ka1 = 1.7 10–2

HSO3–(aq) H+(aq) + SO3

2–(aq) Ka2 = 6.4 10–8

Acidsand

Bases

30

Sample Exercise 16.14 Calculating the pH of a Polyprotic Acid Solution

The solubility of CO2 in water at 25 C and 0.1 atm is 0.0037 M. The common practice is to assume that all the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced in the reaction

What is the pH of a 0.0037 M solution of H2CO3?

Solution

pH = log[H+] = log(4.0 105) = 4.40

Acidsand

Bases

31

Comment If we were asked for [CO32], we

would need to use Ka2. Let’s illustrate that calculation. Using our calculated values of [HCO3

] and [H+] and setting [CO32] = y,

we have

Assuming that y is small relative to 4.0 105, we have

Sample Exercise 16.14 Calculating the pH of a Polyprotic Acid SolutionContinued

This value is indeed very small compared with 4.0 105, showing that our assumption was justified. It also shows that the ionization of HCO3

is negligible relative to that of H2CO3, as far as production of H+ is concerned. However, it is the only source of CO3

2, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO2 is in the form of CO2 or H2CO3, only a small fraction ionizes to form H+ and HCO3

, and an even smaller fraction ionizes to give Co3

2. Notice also that [CO32] is numerically equal to Ka2.

Acidsand

Bases

32

16.7 Weak Bases

Bases react with water to produce hydroxide ion.


The equilibrium constant expression for this reaction is [HB] [OH]

[B]Kb =

base-dissociation constant

Acidsand

Bases

33

Weak BasesNeutral substances (-N) & Anions


Acidsand

Bases

34

pH of Basic Solutions

Ex) What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH]

[NH3]Kb = = 1.8 105

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


Acidsand

Bases

35

pH of Basic Solutions

Tabulate the data.

[NH3], M [NH4+], M [OH], M

Initially 0.15 0 0

At equilibrium


0.15 x 0.15 x x

(x)2

(0.15)1.8 105 = 1.6 103 = x2

pOH = log (1.6 103) = 2.80 : pH = 14.00 2.80 = 11.20

Acidsand

Bases

36

Sample Exercise 16.16 Using pH to Determine the Concentration of a Salt

A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Using the information in Equation 16.37, calculate the number of moles of NaClO added to the water.

pOH = 14.00 pH = 14.00 10.50 = 3.50[OH] = 103.50 = 3.2 104 M

Solution

We say that the solution is 0.31 M in NaClO even though some of the ClO ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water.

Acidsand

Bases

37

16.8 Relationship Between Ka and Kb

Ka and Kb are related in this way:

Ka Kb = Kw

or pKa + pKb = pKw = 14.00 (at 25 °C)


Acidsand

Bases

38

Sample Exercise 16.17 Calculating Ka or Kb for a Conjugate Acid–Base Pair

Calculate (a) Kb for the fluoride ion, (b) Ka for the ammonium ion.

Solution(a)HF, Ka = 6.8 104. Kb for the conjugate base, F

(b)For NH3, Kb = 1.8 105, Ka for the conjugate acid, NH4+:

Practice Exercise(a) Which of these anions has the largest base-dissociation constant: NO2

, PO43, or N3

?(b) The base quinoline has the structure

Its conjugate acid is listed in handbooks as having a pKa of 4.90.What is the base-dissociation constant for quinoline?

Answer: (a) PO43 (Kb = 2.4 102), (b) Kb = 7.9 1010

Acidsand

Bases

39

16.9 Acid-Base Properties of Salt Solutions

Reactions of Anions with Water• Anions are bases.

• As such, they can react with water in a hydrolysis reaction to form OH and the conjugate acid:

X(aq) + H2O(l) HX(aq) + OH(aq)


Acidsand

Bases

40

Reactions of Cations with Water

• Cations with acidic protons (like NH4+) will

lower the pH of a solution.

• Most metal cations that are hydrated in solution also lower the pH of the solution.


Acidsand

Bases

41

• Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water.

• This makes the O–H bond more polar and the water more acidic.



Acidsand

Bases

42


• Greater charge and smaller size make a cation more acidic.


Acidsand

Bases

43

Effect of Cations and Anions

1. Anions – conjugate base of a strong acid : no effect– conjugate base of a weak acid : increase the pH.

2. Cations– conjugate acid of a strong Arrhenius base : no effect– the conjugate acid of a weak base : decrease the pH.– Other metal ions: decrease in pH.

3. When a solution contains botha weak acid and a weak base, the affect on pH depends on the Ka and Kb values.


Acidsand

Bases

44

Sample Exercise 16.18 Determining Whether Salt Solutions areAcidic, Basic, or Neutral

Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba(CH3COO)2, (b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al(ClO4)3.

Solution(a) basic (combination 2).(b) acidic (combination 3).(c) acidic (combination 3).(d) neutral (combination 1).(e) acidic (combination 3).

Practice ExerciseIndicate which salt in each of the following pairs forms the more acidic (or less basic) 0.010 M solution:

(a) NaNO3 or Fe(NO3)3, (b) KBr or KBrO, (c) CH3NH3Cl or BaCl2, (d) NH4NO2 or NH4NO3.

Answers: (a) Fe(NO3)3, (b) KBr, (c) CH3NH3Cl, (d) NH4NO3

Acidsand

Bases

45

Sample Exercise 16.19 Predicting Whether the Solution of anAmphiprotic Anion Is Acidic or Basic

Predict whether the salt Na2HPO4 forms an acidic solution or a basic solution when dissolved in water.

Solution

This Kb >> Ka for HPO42; thus, the solution is basic.

Acidsand

Bases

46

16.10 Acid-Base Behavior and Chemical Structure

• Polarity & Bond Strength– The more polar the H–X bond

and/or the weaker the H–X bond, the more acidic the compound.

– So acidity increases from left to right across a row and from top to bottom down a group.

– Bond Strength

• HF < HCl < HBr < HI» larger X (lower electron

density)


Factors Affecting Acid Strength : binary acids : HX

Acidsand

Bases

47

Factors Affecting Acid Strength: binary acids

HI HBr HCl HF

160.9 > 141.4 > 127.4 > 91.7 pm

297 < 368 < 431 < 569 kJ/mol

Bond length

Bond energy

109 > 108 > 1.3x106 >> 6.6x10-4Acid strength

HF + H2O → [F-·····H3O+] F- + H3O+

ion pairH-bonding

free ions

Acidsand

Bases

48

Factors Affecting Acid Strengthoxyacids : -Y-O-H

In oxyacids, in which an –OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.


Acidsand

Bases

49

Factors Affecting Acid Strengthoxyacids

For a series of oxyacids, acidity increases with the number of oxygens (or oxidation number).


Acidsand

Bases

50

S OO

O

O

H H····

····

-

2+

··

···· ···· ··

-

S OO

O

H H····

····

-

+

··

···· ··

S OO

O

O

H H····

····

·· ···· ··

S OO

O

H H····

···· ··

·· ··

Ka 103 Ka =1.3x10-2

Factors Affecting Acid Strengthoxyacids

Acidsand

Bases

51

Factors Affecting Acid StrengthResonance

Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the proton more labile.


Acidsand

Bases

52

C

O

C

O

H

-H

H

Ka = 1.8x10-5

Ka = 1.4x10-3

C

O

C

O

H

-H

Cl

Factors Affecting Acid Strengthorganic acids

Acidsand

Bases

53

C

H

H

C

O

C

O

H

-H

H

CH

H

H

C

O

O

-

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

Ka = 1.8x10-5

Ka = 1.3x10-5

Acidsand

Bases

54

Factors Affecting Base StrengthStrengths of Amines

CH

H

H

C

H

H

CH

H

H

C

H

H

CH

H

H

C

H

H

pKb = 4.74 pKb = 3.38 pKb = 3.37

methylamine ethylamine propylamine

NH2 NH2 NH2

Acidsand

Bases

55

16.11 Lewis Acids and Bases

• Lewis acids are defined as electron-pair acceptors.

• Atoms with an empty valence orbital can be Lewis acids.


• Lewis bases are defined as electron-pair donors.

• Anything that could be a Brønsted–Lowry base is a Lewis base.

• Lewis bases can interact with things other than protons, however.

Acidsand

Bases

56

Anhydrides in Aqueous Solution

• Basic Anhydride

– A binary compound formed between metals with very low electronegativity and oxygen.

Na2O(s) + H2O(l) 2NaOH(aq)

• Acid Anhydride

– A binary compound formed between nonmetals and oxygen.

CO2(g) + H2O(l) H2CO3(aq)

Acidsand

Bases

57

Problems

• 20, 32, 58, 76, 84, 92, 94, 98, 124

Documents

16.1 Acids and Bases: A Brief Review - Yonseichem.yonsei.ac.kr/~mhmoon/pdf/GenChem_Brown/16_Acid-Base.pdf · 16.1 Acids and Bases: A Brief Review – Acids taste sour and cause certain