Bài giảng Chapter 15 Acids and Bases

Common Acids Nitric Acid HNO3 explosive, fertilizer, dye, glue Strong Sulfuric Acid H2SO4 explosive, fertilizer, dye, glue, batteries Strong Hydrochloric Acid HCl metal cleaning, food

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Stomach Acid & Heartburn

• the cells that line your stomach produce

hydrochloric acid

to kill unwanted bacteria

to help break down food

to activate enzymes that break down food

• if the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn

acid reflux

GERD = gastroesophageal reflux disease = chronic

leaking of stomach acid into the esophagus

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

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Curing Heartburn

neutralizing the acid in the esophagus

 swallowing saliva which contains bicarbonate ion

 taking antacids that contain hydroxide ions and/or carbonate ions

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Common Acids

Nitric Acid HNO3 explosive, fertilizer, dye, glue Strong Sulfuric Acid H2SO4 explosive, fertilizer, dye, glue, batteries Strong Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong Phosphoric Acid H3PO4 fertilizer, plastics & rubber, food preservation Moderate Acetic Acid HC2H3O2 plastics & rubber, food preservation, Vinegar Weak Hydrofluoric Acid HF metal cleaning, glass etching Weak

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Tro, Chemistry: A Molecular Approach 7

Structure of Acids

an oxygen atom

 H2SO4, HNO3

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• solutions feel slippery

• change color of vegetable dyes

different color than acid

red litmus turns blue

• react with acids to form ionic salts

neutralization

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ammonia water

detergent, fertilizer, explosives, fibers

Weak

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Arrhenius Theory

• bases dissociate in water to produce OH- ions and

cations

 ionic substances dissociate in water

NaOH(aq) → Na+(aq) + OH–(aq)

• acids ionize in water to produce H+ ions and anions

because molecular acids are not made of ions, they cannot dissociate

they must be pulled apart, or ionized, by the water

HCl(aq) → H+(aq) + Cl–(aq)

in formula, ionizable H written in front

HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)

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Arrhenius Theory

HCl ionizes in water,

producing H+ and Cl– ions

NaOH dissociates in water,producing Na+ and OH– ions

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Hydronium Ion

• the H+ ions produced by the acid are so reactive they cannot exist in water

H+ ions are protons!!

• instead, they react with a water molecule(s) to produce complex ions, mainly hydronium ion, H3O+

H+ + H2O  H3O+

there are also minor amounts of H+ with multiple water

molecules, H(H2O)n+

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Arrhenius Acid-Base Reactions

 it is often helpful to think of H2O as H-OH

anion from the acid to make a salt

acid + base → salt + water

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

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Problems with Arrhenius Theory

• does not explain why molecular substances, like NH3, dissolve in water to form basic solutions – even

though they do not contain OH– ions

• does not explain how some ionic compounds, like

Na2CO3 or Na2O, dissolve in water to form basic

solutions – even though they do not contain OH– ions

• does not explain why molecular substances, like CO2, dissolve in water to form acidic solutions – even

though they do not contain H+ ions

• does not explain acid-base reactions that take place outside aqueous solution

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Brønsted-Lowry Theory

H+ is transferred

 does not have to take place in aqueous solution

 broader definition than Arrhenius

 base structure must contain an atom with an

unshared pair of electrons

H–A + :B  :A– + H–B+

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Brønsted-Lowry Acids

 any material that has H can potentially be a

Brønsted-Lowry acid

 because of the molecular structure, often one H in

the molecule is easier to transfer than others

H2O, forming H3O+ ions

 water acts as base, accepting H+

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

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Brønsted-Lowry Bases

• Brønsted-Lowry bases are H+ acceptors

 any material that has atoms with lone pairs can

potentially be a Brønsted-Lowry base

 because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer

than others

• NH3(aq) is basic because NH3 accepts an H+

from H2O, forming OH–(aq)

 water acts as acid, donating H+

NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)

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Amphoteric Substances

• amphoteric substances can act as either an

acid or a base

 have both transferable H and atom with lone pair

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)

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Brønsted-Lowry Acid-Base Reactions

• one of the advantages of Brønsted-Lowry theory

is that it allows reactions to be reversible

:A– + H–B+  H–A + :B

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Conjugate Pairs

original base becomes an acid in the reverse

reaction, and the original acid becomes a base in the reverse process

and the original acid becomes the conjugate base

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Brønsted-Lowry Acid-Base Reactions

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Ex 15.1a – Identify the Brønsted-Lowry Acids and

Bases and Their Conjugates in the Reaction

H2SO4 + H2O  HSO4– + H3O+

acid base conjugate conjugate

H2SO4 + H2O  HSO4– + H3O+

When the H2SO4 becomes HSO4, it lost an H+  so

H2SO4 must be the acid and HSO4 its conjugate base

When the H2O becomes H3O+, it accepted an H+  so

H2O must be the base and H3O+ its conjugate acid

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Ex 15.1b – Identify the Brønsted-Lowry Acids and

Bases and Their Conjugates in the Reaction

When the H2O becomes OH, it donated an H+  so

H2O must be the acid and OH its conjugate base

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Practice – Write the formula for the

conjugate acid of the following

H2O

NH3

CO32−

H2PO41−

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Practice – Write the formula for the

conjugate acid of the following

H2O H3O+

CO32− HCO3−

H2PO41− H3PO4

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Practice – Write the formula for the conjugate base of the following

H2O

NH3

CO32−

H2PO41−

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Practice – Write the formula for the conjugate base of the following

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Arrow Conventions

• chemists commonly use two kinds of

arrows in reactions to indicate the

degree of completion of the reactions

• a single arrow indicates all the reactant

molecules are converted to product

molecules at the end

• a double arrow indicates the reaction

stops when only some of the reactant

molecules have been converted into

products

  in these notes

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Strong or Weak

• a strong acid is a strong electrolyte

 practically all the acid molecules ionize, →

• a strong base is a strong electrolyte

 practically all the base molecules form OH– ions, either through dissociation or reaction with water, →

• a weak acid is a weak electrolyte

 only a small percentage of the molecules ionize, 

• a weak base is a weak electrolyte

 only a small percentage of the base molecules form OH–

ions, either through dissociation or reaction with water,



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Strong Acids

• The stronger the acid, the

more willing it is to donate H

use water as the standard base

• strong acids donate

practically all their H’s

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-Tro, Chemistry: A Molecular Approach 35

Weak Acids

fraction of their H’s

 most of the weak acid

molecules do not donate H

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-Tro, Chemistry: A Molecular Approach 36

Polyprotic Acids

• often acid molecules have more than one ionizable H – these are called polyprotic acids

the ionizable H’s may have different acid strengths or be equal

1 H = monoprotic, 2 H = diprotic, 3 H = triprotic

 HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic

• polyprotic acids ionize in steps

each ionizable H removed sequentially

• removing of the first H automatically makes removal of the second H harder

H2SO4 is a stronger acid than HSO4

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Acids Conjugate Bases

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General Trends in Acidity

weaker the conjugate base is at accepting H

• higher oxidation number = stronger oxyacid

 H2SO4 > H2SO3; HNO3 > HNO2

• cation stronger acid than neutral molecule;

neutral stronger acid than anion

 H3O+1 > H2O > OH-1; NH4+1 > NH3 > NH2-1

 base trend opposite

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Acid Ionization Constant, K a

[H ]

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41

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Arsenic H 3 AsO 4 6.0 x 10-3 1.05 x 10-7 3.0 x 10-12Arsenious H3AsO3 6.0 x 10-10 3.0 x 10-14 very small Perchloric HClO4 > 108

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Autoionization of Water

 therefore there must be a few ions present

form ions through a process called

autoionization

H2O  H+ + OH–

H2O + H2O  H3O+ + OH–

 the concentration of H3O+ and OH– are equal in water

 [H3O+] = [OH–] = 10-7M @ 25°C

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Ion Product of Water

concentrations is always the same number

• [H3O+] x [OH–] = Kw = 1 x 10-14 @ 25°C

 if you measure one of the concentrations, you

can calculate the other

• as [H3O+] increases the [OH–] must decrease

so the product stays constant

 inversely proportional

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Acidic and Basic Solutions

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Example 15.2b – Calculate the [OH] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is

acidic, basic, or neutral

The units are correct The fact that the [H3O+] < [OH] means the solution is basic

[H3O+] = 1.5 x 10-9 M[OH]

-H [ 3

OH[

-]

OH][

-OH[

3 w

7

610

5.1

100

1]

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Complete the Table

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Sig Figs & Logs

• when you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number

log(2.0 x 106) = log(106) + log(2.0)

= 6 + 0.30303… = 6.30303

• since the part of the scientific notation number that

determines the significant figures is the decimal part, the

sig figs are the digits after the decimal point in the log

log(2.0 x 106) = 6.30

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human blood 7.3 to 7.4 egg whites 7.6 to 8.0 milk of magnesia (sat’d Mg(OH)2) 10.5

household ammonia 10.5 to 11.5

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Example 15.3b – Calculate the pH at 25°C when the [OH]

= 1.3 x 10-2 M, and determine if the solution is acidic,

basic, or neutral

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

[OH] = 1.3 x 10-2 MpH

-H [ 3

3 w

103

.1

100

1]

OH[

]

OH][

-OH[



12.11 pH

10 7

7 log -





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Relationship between pH and pOH

pH

00 14 ]

[OH log

-] O H [ log

10 0

1 log ]

][OH O

-H [ log

10 0

1 ]

][OH O

-H [

3

14 3

14 w

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Finding the pH of a Strong Acid

HA  H+ + A

-2H2O  H3O+ + OH

to the total [H3O+] is negligible

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Finding the pH of a Weak Acid

HA  H+ + A

-2H2O  H3O+ + OH

the fact that the acid only undergoes partial

ionization

• calculating the [H3O+] requires solving an

equilibrium problem for the reaction that

defines the acidity of the acid

HAcid + H2O  Acid + H3O+

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[HNO2] [NO2-] [H3O+] initial

change equilibrium

Ex 15.6 Find the pH of 0.200 M HNO2(aq)

solution @ 25°CWrite the reaction

for the acid with

water

Construct an ICE

table for the reaction

Enter the initial

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[HNO2] [NO2-] [H3O+]

change equilibrium

solution @ 25°C

represent the change

in the concentrations

in terms of x

sum the columns to

find the equilibrium

3

2 a

-10 00

2 HNO

] O H ][

[NO

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3

2 a

-1000

.2HNO

OHNO

determine the value of

Ka from Table 15.5

since Ka is very small,

approximate the

[HNO2]eq = [HNO2]init

and solve for x

-1000

.2HNO

OH

10 00

2

10 6

9

10 00

2 10

6 4

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O H

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Practice - What is the pH of a 0.012 M solution of

nicotinic acid, HC6H4NO2?

(Ka = 1.4 x 10-5 @ 25°C)

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nicotinic acid, HC6H4NO2?Write the reaction

for the acid with

water

Construct an ICE

Enter the initial

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[HA] [A-] [H3O+]

change equilibrium

nicotinic acid, HC6H4NO2?

in terms of x

sum the columns to

4 6

3

2 4

-6 a

102

.1NO

HHC

]OH][

NOH

[C

HC6H4NO2 + H2O  C6H4NO2 + H3O+

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a

-102

.1HA

OHA

determine the value of

-102

.1HA

OH

10 2

1

10 4

10 1

4

10 2

1 10

4 1

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% 4 3

%

100 10

2 1

10 1

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O H

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Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C

check by substituting

the equilibrium

concentrations back into

the equilibrium constant

2 4

6

3

2 4

-6 a

10 4

.

1 10

2 1

10 1

4

NO H

HC

] O H ][

NO H

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[HClO2] [ClO2-] [H3O+]

change equilibrium

Ex 15.7 Find the pH of 0.100 M HClO2(aq)

solution @ 25°CWrite the reaction

for the acid with

water

Construct an ICE

Enter the initial

concentrations –

assuming the [H3O+]

from water is ≈ 0

HClO2 + H2O  ClO2 + H3O+

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solution @ 25°C

in terms of x

sum the columns to

3

2 a

-10 00

1 HClO

] O H ][

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[HClO2]eq = [HClO2]init

and solve for x

-1000

.1HClO

OH

10 00

1

10 1

10 3

3

10 00

1 10

1 1

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% 33

%

100 10

00 1

10 3

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3

2 a

-1000

.1HClO

OHClO

10 00

1

10 1

1

0.039 -

or 028

0

) 1 ( 2

) 0011

0 )(

1 ( 4 011

0 011

0

0011

0 011

0 0

2 2

x x

Ka for HClO2 = 1.1 x 10-2

if the approximation

is invalid, solve for x

using the quadratic

formula

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Tiêu đề	Chapter 15 Acids and Bases
Tác giả	Julia Burdge
Người hướng dẫn	Mr. Truong Minh Chien
Trường học	NKMB Co., Ltd.
Chuyên ngành	Chemistry
Thể loại	Giáo trình
Năm xuất bản	2011
Thành phố	N/A

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Số trang	136
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