acids and bases chapter 15. properties of acids sour to the taste. react with a variety of metals....

Acids andBases

Chapter 15

Properties of Acids • Sour to the taste.• React with a variety of metals.• Turn blue litmus pink.• Denature proteins.• Cancel the effect of a base (neutralization).

Properties of Bases • Bitter to the taste.• Slippery to the touch.• Turn pink litmus blue.• Denature proteins.• Cancel the effect of an acid (neutralization).

Acid-Base Properties

3

• Binary acids have acid hydrogens attached to a nonmetal atom: HCl, HF

• Carboxylic acids have COOH group

Structure of Acids

• Oxy acids have acid hydrogens attached to an oxygen atom: H2SO4, HNO3

4

Common AcidsChemical Name Formula Uses Strength

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

Carbonic Acid H2CO3 soda water Weak

Boric Acid H3BO3 eye wash Weak

5

Structure of Bases

• Most ionic bases contain OH ionsNaOH, Ca(OH)2

• Some contain CO32- ions

CaCO3 , NaHCO3

• Molecular bases contain structures that react with H+

mostly amine groups

http://wps.prenhall.com/wps/media/objects/477/489045/st1402.html

6

Common BasesChemical

Name Formula

Common Name

Uses Strength

sodium hydroxide

NaOH lye,

caustic soda soap, plastic,

petrol refining Strong

potassium hydroxide

KOH caustic potash

soap, cotton, electroplating

Strong

calcium hydroxide

Ca(OH)2 slaked lime cement Strong

sodium bicarbonate

NaHCO3 baking soda cooking, antacid Weak

magnesium hydroxide

Mg(OH)2 milk of

magnesia antacid Weak

ammonium hydroxide

NH4OH, {NH3(aq)}

ammonia water

detergent, fertilizer,

explosives, fibers Weak

Acid-Base Concepts

• Concepts of acid-base theory including:

The Arrhenius concept

The Bronsted Lowry concept

The Lewis concept

8

Arrhenius Theory

HCl ionizes in water,producing H+ and Cl– ions

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

Strong Acids and Bases

• In the Arrhenius concept, a strong acid is a substance that ionizes completely in aqueous solution to give H3O+(aq) and an anion.

)aq(ClO)aq(OH)l(OH)aq(HClO 4324

• In the Arrhenius concept, a strong base is a substance that ionizes completely in aqueous solution to give OH-(aq) and a cation.

)aq(OH)aq(Na)s(NaOH OH 2

The STRONG ACIDS (all dissociate completely in

water)• Hydrochloric acid: HCl Hydrobromic acid: HBr Hydroiodic acid: HI Sulfuric acid: H2SO4

Nitric acid: HNO3

Perchloric acid: HClO4

Some other acids that are sometimes considered strong are: chloric acid (HClO3), bromic acid (HBrO3), perbromic acid (HBrO4), iodic acid (HIO3), and per-iodic acid (HIO4).

The STRONG BASES (all dissociate completely in

water)• Lithium hydroxide: LiOH

Sodium hydroxide: NaOH Potassium hydroxide: KOH Rubidium hydroxide: RbOH Cesium hydroxide: CsOH Magnesium hydroxide: Mg(OH)2

Calcium hydroxide: Ca(OH)2

Strontium hydroxide: Sr(OH)2

Barium hydroxide: Ba(OH)2

• Note: Alkali and some alkaline earth metals

Weak Acids and Bases

• Most other acids and bases that you encounter are weak. They are not completely ionized and exist in reversible reaction with the corresponding ions.

Ammonia, NH3, is a weak base.

(aq)OHC(aq)OH 2323

An example is acetic acid, HC2H3O2.

)l(OH)aq(OHHC 2232

)aq(OH )aq(NH )l(OH )aq(NH 423

13

Arrhenius Acid-Base Reactions

• The H+ from the acid combines with the OH- from the base to make a molecule of H2Oit is often helpful to think of H2O as H-OH

• The cation from the base combines with the anion from the acid to make a salt

acid + base → salt + water

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

14

Brønsted-Lowry Concept of Acids and Bases

• A base is the species accepting the proton in a proton-transfer reaction.

o In any reversible acid-base reaction, both forward and reverse reactions involve proton transfer.

• According to the Brønsted-Lowry concept, an acid is the species donating the proton in a proton-transfer reaction.

15


• Consider the reaction of NH3 and H2O.

o In the forward reaction, NH3 accepts a proton from H2O. Thus, NH3 is a base and H2O is an acid.

)aq(OH )aq(NH )l(OH )aq(NH 423

H+

base acid

16

Amphoteric Substances

• Amphoteric substances can act as either an acid or a basehave both transferable H and atom with lone

pair

• Water acts as base, accepting H+ from HCl

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

• Water acts as acid, donating H+ to NH3

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

17



In the Brønsted-Lowry concept:

2. Acids and bases can be ions as well as molecular substances.

3. Acid-base reactions are not restricted to aqueous solution.

4. Some species can act as either acids or bases depending on what the other reactant is.

1. A base is a species that accepts protons; OH- is only one example of a base.

18 18

Conjugated Acid-Base PairsConjugated Acid-Base Pairs

• The Brønsted-Lowry concept introduced the idea of conjugate acid-base pairs and proton-transfer reactions.

o If an acid loses its H+, the resulting anion is now in a position to reaccept a proton, making it a Brønsted-Lowry base.

o It is logical to assume that if an acid is considered strong, its conjugate base (that is, its anion) would be weak, since it is unlikely to accept a hydrogen ion.

conjugate acid-base pairs

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

acid acidbase base

Copyright © Houghton Mifflin Company. All rights reserved.

Label each species as an acid or base. Identify the conjugate acid-base pairs.

a. HCO3-(aq) + HF(aq) H2CO3(aq) + F-(aq)

b. HCO3-(aq) + OH-(aq) CO3

2-(aq) + H2O(l)

Base Acid Conjugate base

Conjugate acid

Acid Base Conjugate acid

Conjugate base

The conjugate acid of HSO4

- is

A. H2SO4

B. HSO3+

C. HSO4+

D. H+

E. SO42-

The conjugate base of HSO4

- is

A. H2SO4

B. HSO3+

C. HSO4+

D. H+

E. SO42-

24

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

H2O

NH3

CO32−

H2PO41−

25

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

H2O H3O+

NH3 NH4+

CO32− HCO3

−

H2PO41− H3PO4

26

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

H2O

NH3

CO32−

H2PO41−

27

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

H2O HO−

NH3 NH2−

CO32− since CO3

2− does not have an H, it cannot be an acid

H2PO41− HPO4

2−

28

Polyprotic Acids

• Acid molecules have more than one ionizable H

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

HCl , H2SO4, H3PO4

• 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

29

Strengths of Acids & Bases• Commonly, acid or base strength is measured by

determining the equilibrium constant of a substance’s reaction with water

HAcid + H2O Acid-1 + H3O+1

Base: + H2O HBase+1 + OH-1

• The farther the equilibrium position lies to the products, the stronger the acid or base

• The position of equilibrium depends on the strength of attraction between the base form and the H+

stronger attraction means stronger base or weaker acid

30

Acid Ionization Constant, Ka

• Acid strength measured by the size of the equilibrium constant when react with H2O

HAcid + H2O Acid-1 + H3O+1

• The equilibrium constant is called the acid ionization constant, Ka

larger Ka = stronger acid

[HAcid]

]O[H][Acid 13

1

a

K

32

Autoionization of WaterH2O H+ + OH–

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)• All aqueous solutions contain both H3O+ and OH–

the concentration of H3O+ and OH– are equal in water [H3O+] = [OH–] = 10-7M @ 25°C

• The ion product of water and has the symbol Kw

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

Self-ionization of WaterSelf-ionization of Water

Example– 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]

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

[H3O+] [OH]

]-][OHOH[ 3wK

]OH[]-OH[

]-OH][OH[

3

w

3w

K

K

M 107.6105.1

100.1]-[OH 6

9

14

34

pH• pH = -log[H3O+], [H3O+] = 10-pH

exponent on 10 with a positive signpHwater = -log[10-7] = 7

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

• Another way of expressing the acidity/basicity of a solution is pOH

• pOH = -log[OH], [OH] = 10-pOH

• pH + pOH = 14.00

Example– 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 M

pH

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

]-][OHOH[ 3wK

2

14

3

3w

103.1

100.1]OH[

]-OH][OH[

K M 107.7]O[H 133

[H3O+][OH] pH

]OH[log pH 3-

12.11 pH

107.7log- pH 13

36

pK

• A way of expressing the strength of an acid or base is pK

• pKa = -log(Ka), Ka = 10-pKa

• pKb = -log(Kb), Kb = 10-pKb

• The stronger the acid, the smaller the pKa

larger Ka = smaller pKa

37

The pH of a Strong Acid

• There are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water

• The contribution of the water to the total [H3O+] is negligibleshifts the Kw equilibrium to the left so far that

[H3O+]water is too small to be significant

• For a monoprotic strong acid [H3O+] = [HAcid]for polyprotic acids, the other ionizations can

generally be ignored

• 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00

3838

• Calculate the concentration of H+ ion in 0.010 M NaOH.

Because you started with 0.010 M NaOH (a strong base) the reaction will produce 0.010 M OH-(aq).

– Substituting [OH-]=0.010 into the ion-product expression, we get:

)aq(OH)aq(Na)s(NaOH OH 2

MH 12-14-

101.0 010.0101.0

][

The pH of a Strong BaseThe pH of a Strong Base

39

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

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

]OH][HA[]A][OH[

K2

3c

40

Acid-Ionization Equilibria• Since the concentration of water remains

relatively constant, we rearrange the equation to get:

]HA[]A][OH[

K]OH[K 3c2a

• Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.

]HA[]A][OH[

K 3a

41

[HNO2] [NO2-] [H3O+]

initial

change

equilibrium

Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HNO2 + H2O NO2 + H3O+

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change

equilibrium

Ka for HNO2 = 4.6 x 10-4

42

[HNO2] [NO2-] [H3O+]

initial 0.200 0 0

change

equilibrium

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+xx

0.200 x x x

x

xxK

12

3-2

a 1000.2HNO

]OH][[NO


Ka for HNO2 = 4.6 x 10-4

43

x

xxK

12

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

1

2

3-2

a 1000.2HNO

OHNO

xxK

1

24

1000.2106.4

x

3

14

106.9

1000.2106.4

x

x

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200 x x0.200 x


44

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200 x x

check if the approximation is valid by seeing if x < 5% of [HNO2]init

%5%8.4%1001000.2

106.91

3

the approximation is valid

x = 9.6 x 10-3


45

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200-x x x

x = 9.6 x 10-3

substitute x into the equilibrium concentration definitions and solve

M 190.0106.9200.0200.0HNO 32 x

M 106.9OHNO 33

-2

x

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096


46

Ka for HNO2 = 4.6 x 10-4

substitute [H3O+] into the formula for pH and solve

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096

02.2106.9log

OH-logpH3

3


A solution has a hydroxide ion concentration of 0.001 M. What is the pOH of the solution?

A. 3

B. 7

C. 11

D. 13

47

The pH of a solution is 5.50. What is its hydronium concentration?

A. 5.50 M

B. 3.2 x 10-5 M

C. 3.2 x 10-6 M

D. 3.2 x 10-7 M

48

49

Base-Ionization Equilibria• Equilibria involving weak bases are treated

similarly to those for weak acids.

– In general, a weak base B with the base ionization

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

has a base ionization constant equal to

]B[

]OH][HB[Kb

50

[NH3] [NH4+] [OH]

initial

change

equilibrium

Example - Find the pH of 0.100 M NH3(aq) solution

Write the reaction for the base with water


Enter the initial concentrations – assuming the [OH] from water is ≈ 0

NH3 + H2O NH4+ + OH

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change

equilibrium

Kb for NH3 = 1.76 x 10-5

51

[NH3] [NH4+] [OH]

initial 0.100 0 0

change

equilibrium





+x+xx

0.100 x x x

x

xxK

13

4b 1000.1NH

]OH][[NH

Kb for NH3 = 1.76 x 10-5

52

x

xxK

13

4b 1000.1NH

OHNH

determine the value of Kb from Table 15.8

since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x

1

3

4b 1000.1NH

OHNH

xxK

1

25

1000.11076.1

x

3

15

1033.1

1000.11076.1

x

x

Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x



53

Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 x x

check if the approximation is valid by seeing if x < 5% of [NH3]init

%5%33.1%1001000.1

1033.11

3

the approximation is valid

x = 1.33 x 10-3


54


M 099.01033.1100.0100.0NH 33 x

M 1033.1][OH]NH[ 3-4

x

Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 x x x

x = 1.33 x 10-3

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3


55

use the [OH-] to find the [H3O+] using Kw


124.111052.7log

OH-logpH12

3

Kb for NH3 = 1.76 x 10-5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.099 1.33E-3 1.33E-3

12-3

3-

14-

3

-3w

107.52]O[H

101.33

101.00]O[H

]][OHO[H

K


57

• To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.1. A salt of a strong base and a strong acid.

The salt gives a neutral aqueous solution. NaCl.

2. A salt of a strong base and a weak acid. The salt gives a basic solution. NaCN

3. A salt of a weak base and a strong acid. The salt gives an acidic solution. NH4Cl

4. A salt of a weak base and a weak acid.Ka > Kb the solution is acidic. Kb > Ka the solution is basic.

Classifying Salt Solutions asAcidic, Basic, or Neutral

58

Example- Determine whether a solution of the following salts is acidic, basic, or neutral

a) SrCl2

Sr2+ is the counterion of a strong base, pH neutralCl− is the conjugate base of a strong acid, pH neutralsolution will be pH neutral

b) CH3NH3NO3

CH3NH3+ is the conjugate acid of a weak base, acidic

NO3− is the conjugate base of a strong acid, pH neutral

solution will be acidic

59

Example- Determine whether a solution of the following salts is acidic, basic, or neutral

c) NaCHO2

Na+ is the counterion of a strong base, pH neutral

CHO2− is the conjugate base of a weak acid, basic

solution will be basic

d) NH4F

NH4+ is the conjugate acid of a weak base, acidic

F− is the conjugate base of a weak acid, basic

Ka(NH4+) > Kb(F−); solution will be acidic

60

•If Ka > Kb, the solution is acidic. If Ka < Kb, the solution is basic. If Ka = Kb, the solution is neutral.

105

14

b

wa4 10 5.6

10 1.8

10 1.0 :NH For

K

KK

114

14

a

wb 10 1.5

10 6.8

10 1.0 :F For

K

KK

F for KNH for K b4a

acidic. is FNH4

An aqueous solution of __________ will produce a basic solution.

A. LiCl

B. NH4ClO4

C. CaSO4

D. KBr

E. Na2CO3

61

62

Ka and Kb Relationship

• Reference tables only provide Ka values because Kb values can be found from them

][A

]OHA][H[ )(OH )HA( )O(H )(A

[HA]

]O][HA[ )(OH )(A )O(H )HA(

3b2

3a32

Kaqaqlaq

Kaqaqlaqwhen you add equations, you multiply the K’s )(OH )(OH )O(H 2 32 aqaql

w3ba

3ba

]OH][OH[

]A[

]OH][HA[

]HA[

]OH][A[

KKK

KK

63

Example- Find the pH of 0.100 M NaCHO2(aq) solution

Na+ is the cation of a strong base – pH neutral. The CHO2

− is the anion of a weak acid – pH basic

Write the reaction for the anion with water


Enter the initial concentrations – assuming the [OH] from water is ≈ 0

CHO2− + H2O HCHO2 + OH

[CHO2−] [HCHO2] [OH]

initial 0.100 0 ≈ 0

change

equilibrium

64

0.100 x




Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5


+x+xxx x

x

xxK

12

2b 1000.1CHO

]OH][[HCHO


initial 0.100 0 ≈ 0

change

equilibrium

114

14

b

wba

106.5108.1

100.1

K

KKK

65


since Kb is very small, approximate the [CHO2

−]eq = [CHO2−]init

and solve for x

1

211

1000.1106.5

x

6

111

104.2

1000.1106.5

x

x

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x


x

xxK

12

2b 1000.1CHO

]OH][[HCHO

12

2b 1000.1CHO

]OH][[HCHO

xxK

66


M 100.0104.2100.0100.0CHO 62 x

M 104.2][OH]HCHO[ 6-2

x

x = 2.4 x 10-6

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 −x x x


initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100 2.4E-6 2.4E-6


67

use the [OH-] to find the [H3O+] using Kw


38.8102.4log

OH-logpH9

3

9-3

6-

14-

3

-3w

102.4]O[H

104.2

101.00]O[H

]][OHO[H

K

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x



69

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb

11

26

2

2b

108.5100.0

104.2

CHO

OHHCHO

Kthough not exact, the answer is reasonably close

Kb for CHO2− = 5.6 x 10-11


initial 0.100 0 ≈ 0

change -x +x +x



70

Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

– The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO4

-.

– Sulfuric acid, for example, can lose two protons in aqueous solution.

)aq(HSO)aq(OH)l(OH)aq(SOH 43242

)aq(SO)aq(OH )l(OH)aq(HSO 24324

71

Polyprotic Acids• For a weak diprotic acid like carbonic acid,

H2CO3, two simultaneous equilibria must be considered.

)aq(HCO)aq(OH )l(OH)aq(COH 33232

)aq(CO)aq(OH )l(OH)aq(HCO 23323

7

32

331a 103.4

]COH[]HCO][OH[

K

11

3

233

2a 108.4]HCO[

]CO][OH[K

73

Strengths of Binary Acids

• The more + H-X - polarized the bond, the more acidic the bond

• The stronger the H-X bond, the weaker the acid

• Binary acid strength increases to the right across a periodH-C < H-N < H-O < H-F

• Binary acid strength increases down the columnH-F < H-Cl < H-Br < H-I

XH + -

74

Strengths of Oxyacids, H-O-Y

• The more electronegative the Y atom, the stronger the acidhelps weakens the H-O bond

• The more oxygens attached to Y, the stronger the acidfurther weakens and polarizes the H-O bond

HOIHOBrHOCl

432 HClOHClOHClOHClO

75

Acid StrengthsAcid Strengths

76

Acid StrengthsAcid Strengths

77

Lewis Acid - Base Theory

• Electron donor = Lewis Base = nucleophilemust have a lone pair of electrons

• Electron acceptor = Lewis Acid = electrophileelectron deficient

• Covalent bond forms between the moleculesNucleophile: + Electrophile Nucleophile:Electrophile

• Product called an adduct• Other acid-base reactions also Lewis

78

Example - Complete the Following Lewis Acid-Base ReactionsLabel the Nucleophile and Electrophile

OH

H C H

+ OH-1

OH

H C H

OH

OH

H C H

+ OH-1

ElectrophileNucleophile

••

••••

Which of the following is a Lewis base?

A. SiF4

B. AlF3

C. H2O

D. C5H12

79

80

Practice - Complete the Following Lewis Acid-Base Reactions

Label the Nucleophile and Electrophile

• CaO + SO3

• KI + I2

81

Practice - Complete the Following Lewis Acid-Base Reactions

Label the Nucleophile and Electrophile

• CaO + SO3 Ca+2SO4-2

• KI + I2 KI3

O

S O

O

••

••Ca+2 O -2

•• •• +

O

S O

O

O

-2

Ca+2

ElecNuc

I I K+1 I -1••

••

•• •• +ElecNuc K+1 I I I -1

acids and bases chapter 15. properties of acids sour to the taste. react with a variety of metals....

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