1 Ch16 and 17 - Acids and Bases Brady & Senese, 5th Ed.

1

Ch16 and 17 - Acids and Bases

Brady & Senese, 5th Ed

2

Index

15.1. Brønsted-Lowry acids and bases exchange protons15.2. Strengths of Brønsted acids and bases follow periodic trends15.3. Lewis acids and bases involve coordinate covalent bonds15.4. Elements and their oxides demonstrate acid-base properties15.5. pH is a measure of the acidity of a solution15.6. Strong acids and bases are fully dissociated in solution

15.1. Brønsted-Lowry acids and bases exchange protons3

Brønsted Acid/Base Reactions Transfer H+

• Products differ by one H+ from the reactants to form conjugate

• Conjugate acid-base pairs differ by one H+.

• HCN(aq) + OH-(aq)

↔H2O(l) + CN-(aq)

• Note that in the conjugate pairs, the acid has one more H+ than its conjugate base

Brønsted

acid

Brønsted

base

conjugate

acid

conjugate

base

15.1. Brønsted-Lowry acids and bases exchange protons 4

Learning Check

conjugate base conjugate acid

HCl

NH3

HC2H3O2

CN-

HF

Cl-

NH4+

C2H3O2-

HCN

F-

• Identify the Conjugate Partner for Each


Your Turn!

How many of the following pairs are conjugate pairs:

i. HCN/CN- ii. HCl/Cl- iii. H2S/S2-

A. 1

B. 2

C. 3

D. None of them are conjugate


Amphoteric Substances

• Amphoteric (aka amphiprotic) substances are able to act either as Brønsted acid or Brønsted base

• The following materials are amphoteric: Water Amino acids Anions of polyprotic acids


Acid/Base Strengths In Aqueous Solution

• Hydronium ion (H3O+) is the strongest acid in solution: stronger acids react completely with water to give H3O+

• Hydroxide ion (OH-) is the strongest possible base in solution: stronger bases react completely with water to give OH-

• This leveling effect causes very strong acids/bases to have nearly equal strength in aqueous solution


Conjugate Pairs Have Reciprocal Strengths

• The stronger the acid, the weaker its conjugate base

• The stronger the base, the weaker its conjugate acid

• Strong acids are ionized 100%: their anions are extraordinarily poor bases- most are essentially neutral

15.2. Strengths of Brønsted acids and bases follow periodic trends 9

Acid Strength

Binary Acids:

• Acidity increases from left to right in a period, increasing electronegativity makes the H-X bond more polar (weaker)

• Acidity increases from top to bottom in a group, the increasing length of the H-X bond leads to poor orbital overlap and a weaker bond

Oxoacids:

• Acidity increases as number of oxygens increases (presence of high EN oxygen draws electrons away from other bonds)

• Oxoacids with at least 2 more oxygens than hydrogens tend to be strong. Others tend to be weak.

15.2. Strengths of Brønsted acids and bases follow periodic trends 10

Learning Check

• Which is stronger?

• H2S or H2O

• CH4 or NH3

• HF or HI

δ-

δ+δ+

δ-δ-

δ+δ+

δ-

δ+

δ-

δ+

15.3. Lewis acids and bases involve coordinate covalent bonds 11

Lewis Acids and Bases

• Lewis acids (electron pair acceptors) molecules & ions with incomplete valence shells molecules or ions with central atoms that can

accommodate additional electrons

• Lewis bases (electron pair donors) molecules & ions that have complete valence shells with

unshared electrons generally donate lone pairs or pi bond electrons


Lewis Acid/Base Reactions

• Lewis acids accept an electron pair to form coordinate covalent bonds

• Lewis bases donate lone pairs of electron to form coordinate covalent bonds

• Neutralization is the formation of a coordinate covalent bond between the donor and acceptor


Learning Check

Identify the Lewis acid and base in the following

• NH3 + H+ ↔NH4+

• F- + BF3 ↔ BF4

-

+

:::

: --

15.4. Elements and their oxides demonstrate acid-base properties 14

Hydrated Metal Ions Can Act as Weak Acids

• Electron deficiency of metal cations causes them to inductively attract electrons from the hydrating water molecules

• M(H2O)m n+ + H2O↔M(H2O)m-1OH(n-1)+ + H3O+

• The higher the charge density (greater charge concentrated in smaller radius) the more acidic the metal.

• Acidity increases left to right in a period.• Acidity decreases top to bottom in a group.

volumeioniccharge ionic

density charge

Al(H2O)5OH2+Al(H2O)63+

The acidic behavior of the hydrated Al3+ ion.

H2O H3O+

Electron density drawn toward Al3+

Nearby H2O acts as base

15.4. Elements and their oxides demonstrate acid-base properties 16

Oxides Can Be Acidic or Basic

• Most metal oxides are basic• Non-metal oxides are acidic anhydrides

SO2+H2O →H2SO3

15.5. pH is a measure of the acidity of a solution 17

• Water ionizes to a very small extent (Kw=1.0x10-

14 at room temperature) according to the following reaction: H2O(l) + H2O(l) ↔ H3O+

(aq) + OH-(aq)

• Since water is present in all aqueous solutions, the Kw equilibrium exists in all aqueous solutions. Kw=[H3O+ ][OH-]

Kw = 1.0 x 10-14 at 25°C

• When [H3O+]=[OH-], the solution is neutral.

Auto-ionization of Water (Kw)


pH and Kw

• p -log• pH is defined for aqueous solutions only, and is

temperature dependent• pH=-log[H3O+]• 10-pH = [H3O+]• It derives from the auto ionization of water.

Kw=[H3O][OH-] pKw= pH + pOH

• pH>7 is basic; pH=7 is neutral; pH<7 is acidic

Figure 18.5

The pH values of some familiar

aqueous solutions

pH = -log [H3O+]


Indicators Help Us Estimate pH


Learning Check

[OH-] [H3O+] pH

3.2 × 10-3 M

2.3 × 10-5 M

1.5 × 10-2 M

2.55 × 10-6 M

Complete the following with the missing data at 25 deg C

4.64

12.17

5.593

11.513.1×10-12

4.3×10-10

6.7×10-13

3.9×10-9

15.6. Strong acids and bases are fully dissociated in solution 22

Strong Acids Ionize 100% in Water

• As the substances are placed into water, they form H3O+ .

• Water’s contribution to [H3O+] is negligible• The pH can be calculated from the concentration

of H3O+ produced by the strong acid• The reaction of strong acids occurs irreversibly, so

we show the reaction with a → instead of using a double arrow


Learning Check

What is the pH of 0.1M HCl

•HCl(aq) + H2O(l) →H3O+(aq) + Cl-

(aq)

•0.1 - 0 0 I•-0.1 - 0.1 0.1 C•0 - 0.1 0.1 end•pH = -log(0.1) = 1.0


Strong Bases Dissociate 100% In Water

• They are strong electrolytes that form OH- when dissolved

• pOH can be calculated from the [OH-] from the solution

• Water’s contribution is negligible if the base is sufficiently concentrated [OH-]>10-7M


Learning Check

What is the pH of 0.5M Ca(OH)2?

• Ca(OH)2(aq) → Ca2+(aq) + 2OH-

(aq)

• 0.5 0 0 I•-0.5 +0.5 + 0.5×2 C•0 0.5 1 end•pOH = -log(1) =0.0•pH = 14.0

26

Index

16.1. Ionization constants can be defined for weak acids and bases 16.2. Calculations can involve finding or using Ka and Kb 16.3. Salt solutions are not neutral if the ions are weak acids

or bases 16.4. Simplifications fail for some equilibrium calculations16.5. Buffers enable the control of pH 16.6. Polyprotic acids ionize in two or more steps 16.7. Acid-base titrations have sharp changes in pH at the

equivalence point

16.1. Ionization constants can be defined for weak acids and bases 27

Weak Acids & (Ka)

• Weak acids only partially ionize

• The equilibrium constant Ka is used to indicate the degree of ionization

• Ka are tabulated in table 16.1, as are pKa

• Ka=10-pKa pKa=-log(Ka)

• Acid strength increases as Ka increases and pKa decreases

• For Conjugate Pairs Kw=Ka x Kb

(aq)(aq)3 (l)2(aq) AOHOHHA

16.1. Ionization constants can be defined for weak acids and bases 28

Bases & Kb

• Weak bases only partially ionize

• The equilibrium constant Kb is used to indicate the degree of ionization

• Kb and pKb are tabulated in table 16.2

• Kb=10-pKb pKb=-log(Kb)

• Base strength increases as Kb increases and pKb decreases

• For Conjugate Pairs Kw=Ka x Kb

)()()(2)( aqaqlaq OHBHOHB

16.2. Calculations can involve finding or using Ka and Kb 29

Determining the pH Of Aqueous Weak Acid Solutions

• Dominant equilibrium is Ka reaction

write the net ionic equation look up the Ka value for the acid set up ICE table solve for x

• Calculate pH from the hydronium concentration at equilibrium


Simplifications: Dropping x Term

• Solving ICE tables sometimes requires the quadratic equation

• Other times simplifications may be made• If the K for the reaction is small, the degree of

ionization will also be small• The concentration of a reactant may sometimes be

simplified to the initial concentration• You must perform a proof to show that the

dropped x was justified (less than 5% error)

05.0binomial ofconstant

x term dropped ?


Learning Check

Determine the pH of 0.1M solutions of:

• HC2H3O2

• Ka=1.8×10-5

• HCN

• Ka=6.2×10-10

HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2

-(aq)

I

C

E

HCN + H2O↔ H3O+(aq) + CN-

(aq)

I

C

E

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

0.1M N/A 0 0

-x +x +x

(0.1-x)≈0.1 N/A x

-x

x

52

108.10.1

x

pH=2.87

102

102.60.1

x

pH=5.10

X=1.34(10-3)M

X=7.87(10-6)M


Determining The pH Of Base Solutions:

• Dominant equilibrium is Kb reaction write the net ionic equation set up ICE table for starting quantities solve for x

• Calculate pOH from the [OH-] at equilibrium, and convert to pH


Learning Check:

Determine the pH of 0.1M solutions of:• N2H4

• Kb=1.7×10-6

• NH3

• Kb=1.8×10-5

N2H4 + H2O ↔ OH-(aq) + N2H5

+(aq)

I

C

E

NH3 + H2O↔ OH-(aq) + NH4

+(aq)

I

C

E

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

62

107.10.1x

pOH=3.38

52

108.10.1

x

pH=11.13

pH=10.62

pOH=2.87

X=4.12(10-4)M

X=1.34(10-3)M


Percent Ionization

• Weak acids ionize some amount less than 100%.

• Generally calculated from Ka and ICE table

• Percentage ionized varies with starting concentration, even for the same acid

100[HA]

[A-]ionization %


Learning Check

Determine the % ionization of 0.2M solution of HC2H3O2

• Ka=1.8×10-5

HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2

-(aq)

I

C

E

0.2M N/A 0 0

-x +x +x

(0.2-x) ≈ 0.2 N/A x

-x

x

52

108.10.2

x

x=1.90×10-3M

%1000.2

x 100

literper moles availableliterper ionized moles

ionization %

0.95 % ionized


Learning Check

Determine the % ionization of 0.1M solution of HC2H3O2

• Ka=1.8×10-5HC2H3O2 + H2O↔ H3O+

(aq) + C2H3O2-(aq)

I

C

E

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

52

108.10.1x

x=1.34 x 10-3M

%1000.1x 100

literper moles availableliterper ionized moles

ionization %

1.3 % ionized

16.5. Buffers enable the control of pH 37

Buffers enable the control of pH

• Are a combination of a weak acid and its conjugate weak base or a weak base and its conjugate acid

• Are a special case of the common ion effect (the presence of the conjugate base inhibits the ionization of the Brønsted acid)

• Are resistant to changes in the pH Any acid added to the mixture is reacted by the base of the

buffer pair added bases are reacted by the conjugate acid


Simplification Of Buffer pH Calculations – Henderson-Hasslebach Equation

[HA]]][AO[H

Ka

AOHOHHA

3

(aq)(aq)3(l)2(aq)

)[HA]

][Alog()Olog(Hlog(Ka) 3

)log(log(Ka)])Olog([H [HA]][A

3

][

][log

HA

ApKapH


Learning Check

What is the pH of a buffer made from 0.5M HF (Ka=6.8×10-4) and 0.2M NaF?

][][

logHAA

pKapH

]5.0[]2.0[

log17.3pH

pKa=-log(6.8×10-4)

pH=2.8


Buffer Characteristics

• Buffer capacity is the ability to compensate for added acid or base, defined for each component acid buffer capacity=moles of conjugate base present base buffer capacity= moles of acid present.

• If acid is added, remove from base and add to the acid of the pair

• If base is added, remove from acid of the pair and add to the acid of the pair.

acid mol HAmolacid molA mol

logpKapH

base mol - HAmolbase molA mol

logpKapH


Your Turn!

Which of the following could be used as a buffering pair?

A. NH4Cl/NH3

B. HCl/NaCl

C. NaOH/H2O

D. All of these


Buffer CharacteristicsConsider 500. mL of a sodium acetate/acetic acid buffer in which

the concentration of sodium acetate is 0.500M and that of acetic acid is 0.100M. What is the pH after 10.0 mL of 0.100M NaOH is added? (Ka acetic acid= 1.8×10-5)


Selecting A Buffer

• To choose buffer system, try to find a system whose pKa is close to desired system pH (pH = pKa is ideal)

• Then, calculate necessary ratio of components to obtain desired pH.

• Consider toxicity of the materials!


Learning Check

Choose a buffer system to maintain a pH of 5.5. Describe its composition completely. select acid with pKa≈5.5.

thus, look for Ka ≈3.2×10-

6

H2C8H4O4, phthalic acid has Ka2=3.9×10-6

. We could use NaHC8H4O4 and Na2C8H4O4.

448

2448

OH HCmolOHC mol

log41.55.5

HAmolA mol

logpK5.5 a

23.1OH HCmol

OHC mol448

2448

16.6. Polyprotic acids ionize in two or more steps 45

Polyprotic Acids

• Have more than one ionizable hydrogen• Each successive ionization has a specific

ionization constant (Ka)

• Each successive proton is significantly less acidic than the last

• For H3PO4: H3PO4(aq) + H2O(l) ↔H3O+

(aq) + H2PO4-(aq) Ka1

H2PO4-(aq) + H2O(l) ↔H3O+

(aq) + HPO42-

(aq) Ka2

HPO42-

(aq) + H2O(l) ↔H3O+(aq) + PO4

3-(aq) Ka3

Ka1 >> Ka2 >> Ka3

16.6. Polyprotic acids ionize in two or more steps 46

Learning Check:

What is the pH of a 0.1M solution of H3PO4? Ka1=7.1×10-3; Ka2=6.3×10-8; Ka3=4.5×10-13

16.7. Acid-base titrations have sharp changes in pH at the equivalence point 47

Titration

• The controlled addition of one substance (titrant) to a known quantity of another substance (analyte) until the stoichiometric requirements are met

• Equivalence point - the volume of titrant needed to achieve an equimolar concentration of titrant and of titrate

• Endpoint - the volume of titrant necessary to achieve the stoichiometric ratio of reactants

• The endpoint may be the last of several equivalence points in the case of polyprotic acids

Curve for a strong acid-strong base titration

Curve for a weak acid-

strong base titration

Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH

[HPr] = [Pr-]

pH = 8.80 at equivalence point

pKa of HPr = 4.89

methyl red

Calculating the pH During a Weak Acid-Strong Base Titration

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000M NaOH:

(a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mLPLAN: The amounts of HPr and Pr- will be changing during the titration.

Remember to adjust the total volume of solution after each addition.

SOLUTION: (a) Find the starting pH using the methods of Chapter 18.

Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] [Pr-] = x = [H3O+]

x (1.3x105)(0.10) x = 1.1x10-3 ; pH = 2.96

(b)

Before addition

Addition

After addition

0.004000

0.003000

0.0030000.001000

0 -

-

-0

-

- -

HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)Amount (mol)

Calculating the pH During a Weak Acid-Strong Base Titration

continued

[H3O+] = 1.3x10-50.001000 mol

0.003000 mol= 4.3x10-6M pH = 5.37

(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be (0.004000 mol)

0.04000L + 0.04000L= 0.05000M

Ka x Kb = Kw Kb = Kw / Ka = 1.0x10-14 / 1.3x10-5 = 7.7x10-10

[H3O+] = Kw / = 1.6x10-9M

Kbx[Pr ] pH = 8.80

(d) 50.00mL of NaOH will produce an excess of OH-.

mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100molM = (0.001000mol)

(0.09000L)

M = 0.01111[H3O+] = 1.0x10-14/0.01111 = 9.0x10-11M

pH = 12.05

Curve for a weak base-strong acid

titration

Titration of 40.00mL of 0.1000M NH3 with 0.1000M HCl

pH = 5.27 at equivalence point

pKa of NH4+ =

9.25

pKa = 7.19

pKa = 1.85

Curve for the titration of a weak polyprotic acid.

Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH

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1 Ch16 and 17 - Acids and Bases Brady & Senese, 5th Ed.

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