© 2008 brooks/cole 1 chapter 16:acids and bases chemistry: the molecular science moore, stanitski...

© 2008 Brooks/Cole 1

Chapter 16: Acids and BasesChapter 16: Acids and Bases

Chemistry: The Molecular ScienceChemistry: The Molecular ScienceMoore, Stanitski and Jurs


Arrhenius Definition

Arrhenius: any substance which ionizes in water to produce…

Better version of the Arrhenius definition:

Acid:Acid: hydronium ions (H3O+) in water are acidic.

BaseBase: hydroxide ions (OH-) in water are basic.

• protons (H+) is an acid.acid. • hydroxide ions (OH-) is a base.base.


Acid-Base Reactions

A strong acid is an acid that ionizes completely in water (product favored).

)aq(OH)aq(NO)aq(HNOOH

333

2

)aq(OH)aq(Cl )aq(HClO2H

3

There are six strong acids:

HCl - hydrochloric acid HBr – hydrobromic acid

HI – hydroiodic acid

H2SO4 - sulfuric acidHNO3 – nitric acid

HClO4 – perchloric acid


Acid-Base Reactions

Strong Acids and Bases

A strong base is a base that is present entirely as ions (product favored).

The hydroxides of Group IA, IIA (except Be and Mg hydroxides) are strong bases.

)aq(OH)aq(Na)s(NaOH O

2H

So, wSo, why is NH3 (aq) basic?


Brønsted-Lowry Concept

An alternative definition:

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

indirectly producedBase:H+ acceptor

Acid:H+ donor

acidacid = proton (H+) donor

basebase = proton (H+) acceptor

Works for non-aqueous solutions and explains why NH3 is basic:


WeakWeak acids and bases do not fully ionize (reactant favored).

Brønsted-Lowry ConceptStrongStrong acids and bases almost completely ionize (product favored).

HNO3(aq) + H2O(l) H3O+(aq) + NO3-

(aq)

Note: the products are a new acid and base pair.LeChatelier – a more dilute solution (more water), will ionize more.

HF(aq) + H2O(l) H3O+(aq) + F-(aq)

Base:H+ acceptor

Acid:H+ donor

Base:H+ acceptor

Acid:H+ donor

100%

3%


Water’s Role as Acid or Base

Water acts as a base when an acid dissolves in water:

HBr(aq) + H2O(l) H3O+(aq) + Br-(aq)acid base acid base

Water is amphiproticamphiprotic - it can donate or accept a proton (act as acid or base).

But water acts as an acid for some bases:

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

acid base acid base


Practice

• Complete the equations: (acid or base?, mass/charge balance?, single or double arrow?)

• HClO4 + H2O

• CN- + H2O

• H2S + H2O

• NO2- + H2O

• Ca(OH)2 + H2O

ClO4- + H3O+

HCN + HO-

HS- + H3O+

HNO2 + HO-

Ca+2 + 2OH-


Conjugate Acid-Base Pairs

Molecules or ions related by the loss/gain of oneone H+.

Conjugate Acid Conjugate Base

H3O+ H2O

CH3COOH CH3COO-

NH4+ NH3

H2SO4 HSO4-

HSO4- SO4

2-

HCl Cl-

donate H+

accept H+

NH4+ and NH2

- are notnot conjugate(conversion requires 2 H+)


Identify the base conjugate to HC2H3O2(aq) and the acid conjugate to HCO3

-(aq)

HCO3- + H2O OH-

+

HC2H3O2 + H2O H3O+ +

Conjugate Acid

Base:H+ acceptor

ConjugateBase

Acid:H+ donor

Conjugate Acid-Base Pairs

Show that HCO3- is amphoteric.

C2H3O2 -

H2CO3


Practice: Conjugate Acid-Base Pairs

Write the equation for the acid or base in water and identify the acid-base pairs.

HBr

HSO3- (as an acid)

HSO3- (as a base)

PH4+


Relative Strength of Acids & Bases

Strong acids are better H+ donors than weak acids.

Strong bases are better H+ acceptors than weak bases. Strong acids have weak conjugate bases. Weak acids have strong conjugate bases.

Strong acid + H2O H3O+ + weak conjugate baseFully ionized, reverse reaction essentially does not occur.

The conjugate base is weak.

Weakly ionized, reverse reaction readily occurs.

The conjugate base is strong.

Weak acid + H2O H3O+ + strong conjugate base


Conjugate acid Conjugate baseH2SO4 HSO4

-

HBr Br-

HCl Cl-

HNO3 NO3-

H3O+ H2O

H2SO3 HSO3-

HSO4- SO4

2-

H3PO4 H2PO4-

HF F-

CH3COOH CH3COO-

H2S HS-

H2PO4- HPO4

2-

NH4+ NH3

HCO3- CO3

2-

H2O OH-

OH- O2-

H2 H-

CH4 CH3-

Acid

streng

th in

crea

sing

Ba

se s

tre

ngth

incr

ea

sin

g

stong acids

strong baseextremelyweak acids

extremelyweak bases



HF is a stronger acid than NH4+.

(NH3 is a stronger base than F-)

HF has greater tendency to ionize than NH4

+. (NH3 more readily accepts H+ than F-)

Reactant FavoredReactant Favored

NH4+ + F- NH3 + HF

ProblemProblemIs the following aqueous reaction product or reactant favored?

Acid

streng

th in

crea

sing

Ba

se s

tre

ngth

incr

ea

sin

g

Conj acid. Conj. base

H2SO4 HSO4-

HBr Br-

HCl Cl-

HF F-

NH4+ NH3

OH- O2-

H2 H-

CH4 CH3-



Carboxylic Acids

H C

H

H

C

H

H

C

H

H

C

O

O H

nonacidic hydrogens

butanoic acid

acidic hydrogen

CH3 C

O

acetic acidOH acidic hydrogen CH3 C

O

O-acetate ion

C3H7COOH

CH3COOH


Amines (bases)

CH3NH2 (CH3)2NH (CH3)3N

proton acceptors


Autoionization of Water

Pure water conducts a very small electrical current.

AutoionizationAutoionization occurs:

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

Kw = ionization constant for water

Heavily reactant favored.

Kw = [ H3O+ ] [ OH- ]

Kw = 1.0 x 10-14 (at 25°C)

Base AcidAcid Base


Ionization Constant for WaterKw is a special equilibrium constant for the autoionization of water. It is T-

dependent.

T = 25°C (77°F) is usually used as the standard T.

T (°C) Kw

10 0.29 x 10-14

15 0.45 x 10-14

20 0.68 x 10-14

2525 1.01 x 101.01 x 10-14-14

30 1.47 x 10-14

50 5.48 x 10-14

Kw increases with temperature. Is it endo or exothermic?



Hydronium and hydroxide ions are produced in equal numbers in pure water,

Thus, the concentrations of H3O+ and OH- in pure water are both 1.0 x 10-7 M.

C 25at )x)(x(100.1 o14

714 100.1100.1x

Kw = [ H3O+ ] [ OH- ]

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


Autoionization of WaterH3O+ and OH- are present in allall aqueous solutions.

Neutral solution.Neutral solution.

Pure water (@ 25°C): [H3O+] = 10-7 M = [OH-]

Acidic solutionAcidic solution

If acid is added to water:• [ H3O+ ] is increased, disturbing the equilibrium:

2 H2O H3O+ + OH-

• LeChatelier: equilibrium shifts to the left. [OH-]

• Equilibrium is reestablished: [ H3O+ ] > 10-7 M > [OH-]

• Product of ion concentrations is the same

[ H3O+ ][OH-] = Kw = 1.0 x 10-14


Autoionization of WaterBasic solution.Basic solution.

If base is added to water.• LeChatelier: equilibrium shifts to the left, [H3O+]

• New equilibrium: [H3O+] < 10-7 M < [OH-]

• [ H3O+ ][OH-] = Kw = 1.0 x 10-14 at 25°C

ExampleExample

Calculate the hydronium and hydroxide ion concentrations at 25°C in a 6.0 M aqueous sodium hydroxide solution.

Kw = [H3O+][OH-] = 1.0 x 10-14

NaOH (aq) is strong (100% ionized) so [OH-] = 6.0 M.

[H3O+](6.0) = 1.0 x 10-14

[H3O+] = 1.7 x 10-15 M [ OH- ] = 6.0 M

2 H2O H3O+ + OH-



In a neutral solution, the concentrations of H3O+ (aq) and OH-(aq) are equal.

In an acidic solution, the concentration of H3O+ (aq) is greater than that of OH-(aq).

In a basic solution, the concentration of OH-(aq) is greater than that of H3O+ (aq).

The relative concentrations of H3O+ and OH- indicate acidic, neutral or basic character of a solution:


At 25 oC, we observe the following conditions.

Acid or Base

In an acidic solution, [H3O+ ] > 1.0 x 10-7 M.

In a neutral solution, [H3O+ ] = [OH-] = 1.0 x 10-7 M.

In a basic solution, [H3O+ ] < 1.0 x 10-7 M.


The pH scale

Can describe acidity in terms of [H3O+], but a logarithmic scale is more convenient:

pH = −logpH = −log1010[H[H33OO++]]

At 25°C a neutral aqueous solution has:

pH = −log10[1.0 x 10-7] = −(−7.00) = 7.00

acidic solutions: [H3O+]>1.0 x 10-7, pH < 7.00

basic solutions: [H3O+]<1.0 x 10-7, pH > 7.00


The pH of a Solution

For a solution in which the hydronium-ion concentration is 1.0 x 10-3, the pH is:

Note that the number of decimal places in the pH equals the number of significant

figures in the hydronium-ion concentration.

00.3)100.1log( 3 pH


Example – pH given [H3O+] ?

A sample of orange juice has a hydronium-ion concentration of 2.9 x 10-4 M. What is the pH?

]OHlog[pH 3

)109.2log(pH 4

54.3pH


Example - [H3O+] given pH?

]OHlog[pH 3

The pH of human arterial blood is 7.40 (acidic or basic?). What is the hydronium-ion concentration? (Guesss….> or < 1.0x10-7M?)

M100.410]OH[ 840.73

]OHlog[pH 3

]OHlog[pH 31010

pH3 10]OH[

10logx = x


The pOH scale

pOH = −logpOH = −log1010[OH[OH--]]

A neutral solution (25°C) has:

pOH = −log10[1.0 x 10-7] = −(−7.00) = 7.00

Since Kw = [ H3O+ ][ OH- ] = 1.0 x 10-14

−log(KW)= −log[H3O+] + (−log[OH-]) = −log(1.0 x 10-14)

pKw = pH + pOH = 14.00


Example

An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is the pH of the solution?

we first calculate the pOH:

72.2)109.1log(pOH 3

Then the pH is:

28.1172.200.14pH


pH Calculations

Solution A: pOH = −log[ OH-] = 3.37

pH + pOH = pKw = 14.00

pH = 10.63

Solution B: pH = −log[ H3O+] = 8.12

A has higher pH, B is more acidic

Given two aqueous solutions (25°C).

Solution A: [OH-] = 4.3 x 10-4 M,

Solution B: [H3O+] = 7.5 x 10-9 M.

Which has the higher pH? Which is more acidic?


Practice - WSs

[H+] Kw =1x10-14 = [H+] [OH-] [OH-]

pH 14 = pH + pOH pOH

pH= -log [H+][H+] = 10-pH

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


pH of Aqueous Solutions

Stomach Fluids

Lemon JuiceVinegar

WineTomatoesBlack Coffee

MilkPure waterBlood, seawaterSodium bicarbonateBorax solutionMilk of MagnesiaDetergentsAqueous ammonia

1 M NaOH

[H[H33OO++] [OH] [OH--]]

100 10-14

10-1 10-13

10-2 10-12

10-3 10-11

10-4 10-10

10-5 10-9

10-6 10-8

10-7 10-7

10-8 10-6

10-9 10-5

10-10 10-4

10-11 10-3

10-12 10-2

10-13 10-1

10-14 1

ExampleExample

Battery Acid

pHpH

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

Bleach


Measuring pH

H3O+ concentrations can be measured with an:

• Electronic pH meterpH meter: fast and accurate preferred method.

• Acid-base IndicatorIndicator: substance that changes color within a narrow pH range may have multiple color change (e.g. bromthymol blue) one “color” may be colorless (e.g. phenolphthalein) cheap and convenient.


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

The acid ionization constantacid ionization constant is used to report the degree of ionization:

[A-][H3O+][HA]

Ka = (Water omitted, as usual)

Strong acids have large Ka values

Weak acid have small Ka values

Ionization Constants of Acids and Bases

When an acid ionizes in water:


Base Ionization Constants

For a base in water:

The base ionization constantbase ionization constant, Kb, is:

Kb =[BH+][OH-]

[B]

A-(aq) + H2O(l) HA(aq) + OH-(aq)

Kb =

[HA][OH-][A-]

B(aq) + H2O(l) BH+(aq) + OH-(aq)

If the base is an anion:


Larger Ka = stronger acid Larger Kb = stronger base

Ionization Constants


Acid and Base Strength

]HA[

]A][OH[Ka

3

[B]

]][OH[HBK b

)aq(OH)aq(HB )l(OH)aq(B 2

)aq(A)aq(OH )l(OH)aq(HA 32

Small value of Ka indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak acid.

Small value of Kb indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak base.


Write the ionization equation and ionization constant expression for the following acids and bases:

a.) HF

b.) HBrO

c.)NH3

d.) CH3NH2

Example


Ka Values for Polyprotic Acids

Some acids can donate more than one H+

Formula Name Acidic H’s

H2S Hydrosulfuric Acid 2

H3PO4 Phosphoric Acid 3

H2CO3 Carbonic Acid 2

HOOC-COOH Oxalic acid 2

C3H5(COOH)3 Citric acid 3

Each H+ ionization has a different Ka.• The 1st proton is easiest to remove• The 2nd is harder, etc.


Ka Values for Polyprotic Acids

Phosphoric acid (H3PO4) is weak. It has three acidic protons:

H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4- (aq)

Ka = 7.5 x 10-3

H2PO4-(aq) + H2O(l) H3O+(aq) + HPO4

2- (aq)

Ka = 6.2 x 10-8

HPO42-(aq) + H2O(l) H3O+(aq) + PO4

3- (aq)

Ka = 3.6 x 10-13

Ka

decreasingH

+ is harder to rem

ove


Practice – Relative Strength (Ka and Kb)

The following reactions all have Keq>1.

NO2- + HF HNO2 + F-

CH3COO- + HF CH3COOH + F-

HNO2 + CH3COO HNO2- + CH3COOH

Arrange the substances based on their relative base strength.

F-

CH3COOHHFHNO2

-

CH3COO-

HNO2

1 – strongest base2 – intermediate base3 – weakest base4 – not a B-L base

344214


Molecular Structure and Acid Strength

What makes a strong acid?• A weak H–A bond, so H+ can be easily removed!

smaller bond energy

larger acid strength

HX Bond Energy (kJ) Ka

HF weak acid 566 7 x 10-4

HCl 431 1 x 107

HBr 366 1 x 108

HI 299 1 x 1010

stro

ng a

cids

Consider the binary acidsbinary acids HF, HCl, HBr, HI.

H-A, bond energy decreases down a group.Acid strength increases.



What makes a strong acid?• A weak H–A bond, so H+ can be easily removed!

Consider the binary acidsbinary acids SiH4, PH3, H2S, HCl.

The higher electronegativity, across a period, makes the bond more polar and hence more ionic – a stronger acid.


Oxoacids hydrogen, oxygen and one other element

H O-E higher the electronegativity on E, stronger the acid as

this weakens the bond between the O and H


HOCl HOBr HOI

Electronegativity: Cl = 3.0, Br = 2.8, I = 2.5

Ka: 3.5 × 10-8 2.5 × 10-9 2.3 × 10-11


The acid strength increases as the number of oxygen atoms bonded to E increases. Strong oxoacid has at least two oxygen atoms per acidic H atoms.


HClO HClO2 HClO3 *HClO4

Ka: 3.5 × 10-8 1.1 × 10-2 ≈ 103 ≈ 108


Problem Solving Using Ka and Kb

Example: Example: KKaa from pH from pH

Lactic acid is monoprotic. The pH of a 0.100 M solution was 2.43 at 25°C. Determine Ka for this acid.


Using the pH information:- log [H3O+] = 2.43 [H3O+] = 10-2.43 = 0.0037 M

These hydronium ions are produced by:

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

and from the autoionization of water:


Ka = = 1.4 x 10-4 (0.0037)(0.0037)(0.100 – 0.0037)


[ ]initial 0.100 1.0 x 10-7 0 0 (water autoionization)

change -0.0037 +0.0037 +0.0037

[ ]equil 0.100 – 0.0037 0.0037 0.0037

Ka =[H3O+][A-]

[HA]




Example: pH from Example: pH from KKaa

Determine the pH of a 0.100 M propanoic acid solution at 25°C. Ka = 1.4 x 10-5. What % of acid is ionized?

C2H5COOH + H2O(l) H3O+(aq) + C2H5COO-(aq)

Ka = = 1.4 x 10-5[H3O+][C2H5COO-] [C2H5COOH]



[ ]initial 0.100 0* 0

change -x +x +x

[ ]equil 0.100 – x x* x

*Ignore the water contribution

1.4 x 10-5=x2

(0.100 – x)

[H3O+][A-][HA]

Ka = 1.4 x 10-5 =


Determine the pH of 0.001 M propanoic acid soln. at 25°C. Ka = 1.4 x10-5. What % of acid is ionized?



Because Ka is small, assume x < [HA]

Ka = 1.4 x 10-5 = ≈

x2

(0.100 – x)x2

0.100

x = (1.4 x 10-5 )(0.100) = 0.00118 M

(0.00118 << 0.100 M; the approximation is good)

pH = -log(0.00118) = 2.93

%-ionized = x100 = 1.18 %0.001180.100

<5% rulecan assume[HA]eq~[HA]0


What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.

Example 3 – calculating pH from known Kb

)aq(OH)aq(NHHC )l(OH)aq(NHC 55255

Initial 0.20 0 0Change -x +x +x

Equilibrium 0.20-x x x




)x20.0(

x104.1

29

x+x0 00.20Initial

x0.20-xEquilibrium+x-xChange

x+x0 00.20Initial

x0.20-xEquilibrium+x-xChange

b55

55 K]NHC[

]OH][NHHC[

)aq(OH)aq(NHHC )l(OH)aq(NHC 55255




)20.0(104.1

29 x

Solving for x we get

)104.1()20.0(x 92

][107.1)104.1()20.0( 59 OHx




Solving for pOH

77.4)107.1log(]OHlog[pOH 5

23.977.400.14pH


Practice

Ka from pHPara-aminobenzoic acid (PABA), HC7H6NO2, is used in some sunscreen agents. A solution is made by dissolving 0.263 mol of PABA in enough water to make 750.0 mL of solution. The solution’s pH = 2.59. What is the Ka for PABA? (1.9x10-5)

pH from Ka Barbituric acid (Ka=1.1x10-4) is used in the manufacture of some sedatives. What is the pH for a 0.673 M solution of barbituric acid? (2.07)


Relationship between Ka and Kb values

For an acid-base conjugate pair: HA and A-

[HA][OH-][A-]Ka x Kb =

[H3O+][A-][HA]

= [H3O+][OH-] = Kw

)aq(CN)aq(OH )l(OH)aq(HCN 32

)aq(OH)aq(HCN )l(OH)aq(CN 2

Ka

Kb

)aq(OH)aq(OH )l(OH2 32

When two reactions are added, their equilibrium constants

are multiplied (chapter 14).wba KKK


Relationship between Ka and Kb values

Phenol, C6H5OH, is a weak acid, Ka = 1.3 x 10-10 at 25°C. Calculate Kb for the phenolate ion C6H5O-

Ka x Kb = 1.0 x 10-14

Kb = = 7.7 x 10-51.0 x 10-14

1.3 x 10-10


Acid Base Neutralization Reactions

SaltSalt = ionic compound formed in acid + base reaction

Strong acid + strong baseStrong acid + strong base form neutral salts.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)

Na+(aq) + Cl-(aq) + H2O(l)

HX(aq) + MOH(aq) MX(aq) + H2O(l)acid base salt

Net ionic equation: H+(aq) + OH-(aq) H2O(l)

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


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

Keq = 1/Kw = 1x1014

strongbase

strongacid

weakbase

weakacid

• Na+ and Cl- are spectator ions.• Na+ is the v. weak conjugate acid of a v. strong base. • Cl- is the v. weak conjugate base of a v. strong acid.• Final solution has pH = 7.

Salts of Strong Bases and Strong Acids

H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)

Na+(aq) + Cl-(aq) + 2 H2O(l)

100%


Salts of Strong Bases and Weak Acids

Weak acid + strong baseWeak acid + strong base form basic salts.

CH3COOH(aq) + NaOH(aq)

CH3COONa(aq) + H2O(l)

Net ionic: CH3COOH(aq) + OH-(aq)

CH3COO- + H2O(l)

CH3COOH(aq) + Na+(aq) + OH-(aq)

Na+(aq) + CH3COO-(aq) + H2O

100%

final solution


• acetate is a weak base (much stronger than water)• solution is basic (pH > 7.0).

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-

(aq)

Calculate pH of solution from Kb of CH3COO-.

A hydrolysishydrolysis reaction – water is broken apart.

CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(l)

baseacid base acid

Salts of Strong Bases and Weak Acids100%


pH of a Salt Solution

For a weak acid + strong base, pH depends on Kb

Larger Kb = stronger base

ExampleExample

What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb (CO3

2-) = 2.1 x 10-4.

(or equal volumes of 1.5M H2CO3 and 3.0 M NaOH are mixed, what is the pH of the mixture?)


Na+ is the conjugate acid of NaOH (strong base). It remains 100% ionized Na+ has no effect on pH.

CO32- is the conjugate base of HCO3

- (weak acid)

Most CO32- ions undergo hydrolysis

Generates HCO3- and OH-

Changes the pHCO3

2-(aq) + H2O(l) HCO3-(aq) + OH-(aq)

What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4



CO32-(aq) + H2O(l) HCO3

-(aq) + OH-(aq)

[ ]initial 1.50 0 0change - x +x +x[ ]equil 1.50 - x x x

Kb = 2.1 x 10-4 = = ≈[HCO3-][OH-]

[CO32-]

x2

(1.50 – x)x2

1.50

x = 1.77 x 10-2 pOH = −log(1.77 x 10-2) = 1.75

pH = 14.00 - 1.75 = 12.25

What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4



Salts of Weak Bases and Strong Acids

NH3(aq) + HCl(aq) NH4+(aq) + Cl-(aq)

The products of the neutralization reaction form an acidic solution:

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

NH3(aq) + H3O+(aq) + Cl-(aq)

NH4+(aq) + Cl-(aq) +

H2O(l)

Net ionic: NH3(aq) + H3O+(aq) NH4+(aq) + H2O(l)

The final pH depends on the Ka , larger Ka = more acidic



Find the pH of a 0.132 M NH4Br. Is this solution acidic, basic or neutral? Ka(NH4

+) = 5.6 x 10-10.

• The solution contains NH4+ ions and Br- ions.

• NH4+ is the conjugate acid of NH3 (a weak base).

It will partially ionize in solution giving NH3 and H3O+

Forms an acidic solution.

• Br- is the conjugate base of HBr (a strong acid) Br- ions stay fully ionized Does not change the pH.



[ ]initial 0.132 0 0change - x +x +x[ ]equil 0.132 - x x x

x = 8.57 x 10-6 pH = −log(8.57 x 10-6) = 5.07

Ka = 5.6 x 10-10 = = ≈[NH3][H3O+][NH4

+]

x2

(0.132 – x)x2

0.132

pH of a Salt SolutionpH of a 0.132 M aq. solution of NH4Br? Is this acidic, basic or neutral? Ka(NH4

+) = 5.6 x 10-10.

Acidic!


Salts of Weak Bases and Weak Acids

The most difficult. Consider qualitative results. e.g.

F-(aq) + H2O(l) HF(aq) + OH-(aq)

NH3 (aq) + HF(aq) NH4+ (aq) + F- (aq)


Ka(NH4+) = 5.6 x 10-10 ; Kb(F-) = 1.4 x 10-11

Ka (weak acid) > Kb (weak base)

The ammonium reaction is more favorable.

The acid wins! The solution will be acidic!


Acids, Bases and Salts

Strong acidStrong acid + strong basestrong base → salt solution, pH = 7pH = 7

Strong acidStrong acid + weak baseweak base → salt solution, pH < 7pH < 7

Weak acidWeak acid + strong basestrong base → salt solution, pH > 7pH > 7

Weak acidWeak acid + weak baseweak base → salt solution, pH = ?pH = ?

Need K’s “strongest wins”


Neutral Ions in Water (Table 16.5)

Anions Cl- NO3-

Br- ClO4-

I-

Cations Li+ Ca+2

Na+ Sr+2

K+ Ba+2


What is the pH of a 0.10 M NaCN solution at

25 oC? The Kb for CN- is 2.5 x 10-5.

Example

Sodium cyanide gives Na+ ions and CN- ions in solution.

Only the CN- ion hydrolyzes.

)aq(OH)aq(HCN )l(OH)aq(CN 2

pH =11.2


What is the pH of a 0.15 M NH4NO3 solution at

25 oC? The Ka for NH4+ is 5.6 x 10-10.

Example

NH4NO3 gives NH4+ ions and NO3

- ions in solution.

Only the NH4+ ion hydrolyzes.

)aq(OH)aq(NH )l(OH)aq(NH 3324


The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

The Common Ion Effect

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232

Consider a solution of acetic acid (HC2H3O2),

in which you have the following equilibrium.


The Common Ion Effect

The equilibrium composition would shift to the left

(LeChatelier Principle) and the degree of ionization of the acetic acid would decrease.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232

If we were to add NaC2H3O2 to this solution, it would provide C2H3O2

- ions which are present on the right side of the equilibrium.

This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.


What is the pH of a solution that is 0.025M in formic acid?

The Ka for formic acid is 1.7 x 10-4.

What is the pH of the same solution after 0.018 M in sodium formate, NaCH2O is added?

Example

(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22

Initial

Change

Equilibrium

0.025 0

-x +x +x

0.025-x x 0.018+x

From NaCH2O

0.018


Initial 0.025 0 0.018Change -x +x +x

Equilibrium 0.025-x x 0.018+x

4107.1)025.0()018.0( x

Example

4

2

23a 101.7

x)(0.025x)x(0.018

O][HCH]O][CHO[H

K

(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22

x [H3O+]= 2.4 × 10-4

pH = -log(H3O+) = -log(2.4 x 10-4) = 3.62

© 2008 brooks/cole 1 chapter 16:acids and bases chemistry: the molecular science moore, stanitski...

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