chapter 10 acids and bases acids and bases. arrhenius acids and bases in 1884, svante arrhenius...

Chapter 10

Acids and BasesAcids and Bases

Arrhenius Acids and Bases

In 1884, Svante Arrhenius proposed these definitions acid:acid: a substance that produces H3O+ ions

aqueous solution base:base: a substance that produces OH- ions in

aqueous solution


when HCl, for example, dissolves in water, its reacts with water to give hydronium ion and chloride ion

we use curved arrows to show the change in position of electron pairs during this reaction

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

H O

:

+ H Cl:

: : H O H

:

+H

+Cl -

::: ::

H


With bases, the situation is slightly different many bases are metal hydroxides such as KOH,

NaOH, Mg(OH)2, and Ca(OH)2

these compounds are ionic solids and when they dissolve in water, their ions merely separate

other bases are not hydroxides; these bases produce OH- by reacting with water molecules

NaOH(s) H2O Na+(aq) +OH-(aq)

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


we use curved arrows to show the transfer of a proton from water to ammonia

HO H

::+ H N H

H

H+ + O

:::

-H N

H

H: H

Acid and Base Strength Strong acid:Strong acid: one that reacts completely or almost

completely with water to form H3O+ ions Strong base:Strong base: one that reacts completely or almost

completely with water to form OH- ions here are the six most common strong acids and the four

most common strong bases

HClHBrHIHNO3

H2SO4

HClO4

LiOHNaOHKOH

Ba(OH)2

Hydrochloric acidHydrobromic acidHydroiodic acidNitric acidSulfuric acidPerchloric acid

Lithium hydroxideSodium hydroxidePotassium hydroxideBarium hydroxide

Formula Name Formula Name

Acid and Base Strength Weak acid:Weak acid: a substance that dissociates only

partially in water to produce H3O+ ions

acetic acid, for example, is a weak acid; in water, only 4 out every 1000 molecules are converted to acetate ions

Weak base:Weak base: a substance that dissociates only partially in water to produce OH- ions ammonia, for example, is a weak base

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Acetic acid Acetate ion

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

Brønsted-Lowry Acids & Bases Acid:Acid: a proton donor Base:Base: a proton acceptor Acid-base reaction:Acid-base reaction: a proton transfer reaction Conjugate acid-base pair:Conjugate acid-base pair: any pair of molecules or ions

that can be interconverted by transfer of a proton

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

WaterHydrogenchloride

Hydroniumion

Chlorideion

(base)(acid) (conjugateacid of water)

(conjugatebase of HCl)

conjugate acid-base pair


Brønsted-Lowry Acids & Bases

Brønsted-Lowry definitions do not require water as a reactant

NH4+CH3COOH CH3COO-

NH3

(base) (conjugate baseacetic acid)

(conjugate acidof ammonia)


+ +Acetic acid Ammonia

(acid)


Acetate ion

Ammoniumion


we can use curved arrows to show the transfer of a proton from acetic acid to ammonia

CH3-C-OO

H N HH

H CH3-C-O -

OH N H

H

H+ +

Acetic acid(proton donor)

Acetate ion

+:

:: ::

:: :: :

Ammonia(proton acceptor)

Ammoniumion

C2H5OH C2H5O-H2O OH-HPO4

2- PO43-

HCO3- CO3

2-

C6H5OH C6H5O-HCN CN-

NH3NH4+

H2PO4- HPO4

2-

H2S HS-H2CO3 HCO3

-CH3COOH CH3COO-H3PO4 H2PO4

-HSO4

- SO42-

H2OH3O+HNO3 NO3

-H2SO4 HSO4

-HCl Cl-HI I-Hydroiodic acid

Hydrochloric acidSulfuric acid

Dihydrogen phosphateAcetateBicarbonate

Hydrogen phosphateAmmonia

Phenoxide

Carbonate

PhosphateHydroxideEthoxide

Hydrogen sulfide

Nitric acidHydronium ion

Hydrogen sulfate ion

Name of acid Name of ion

Phosphoric acidAcetic acidCarbonic acid

Dihydrogen phosphateAmmonium ion

Phenol

Bicarbonate ion

Hydrogen phosphate ionWaterEthanol

Hydrogen sulfide

AcidConjugate Base

IodideChlorideHydrogen sulfateNitrateWater

Sulfate

StrongAcids

Weak Acids

Weak Bases

StrongBases

Hydrocyanic acid Cyanide


Note the following about the conjugate acid-base pairs in the table

1. an acid can be positively charged, neutral, or negatively charged; examples of each type are H3O

+, H2CO3, and H2PO4-

2. a base can be negatively charged or neutral; examples are OH-, Cl-, and NH3

3. acids are classified a monoprotic, diprotic, or triprotic depending on the number of protons each may give up; examples are HCl, H2CO3, and H3PO4

Brønsted-Lowry Acids & Bases carbonic acid, for example can give up one proton to become

bicarbonate ion, and then the second proton to become carbonate ion

4. several molecules and ions appear in both the acid and conjugate base columns; that is, each can function as either an acid or a base

Carbonic acid

Bicarbonateion

Bicarbonateion

Carbonateion

H2CO3 H2O

HCO3- H2O

HCO3-

CO32-

H3O+

H3O+

+ +

+ +

Brønsted-Lowry Acids & Bases the HCO3

- ion, for example, can give up a proton to become CO3

2-, or it can accept a proton to become H2CO3

a substance that can act as either an acid or a base is said to be amphiproticamphiprotic

the most important amphiprotic substance in Table 8.2 is H2O; it can accept a proton to become H3O+, or lose a proton to become OH-

5. a substance cannot be a Brønsted-Lowry acid unless it contains a hydrogen atom, but not all hydrogen atoms in most compounds can be given up acetic acid, for example, gives up only one proton


6. there is an inverse relationship between the strength of an acid and the strength of its conjugate base the stronger the acid, the weaker its conjugate base HI, for example, is the strongest acid in Table 8.2, and its

conjugate base, I-, is the weakest base in the table CH3COOH (acetic acid) is a stronger acid that H2CO3

(carbonic acid); conversely, CH3COO- (acetate ion) is a weaker base that HCO3

- (bicarbonate ion)

Acid-Base Equilibria we know that HCl is a strong acid, which means that the

position of this equilibrium lies very far to the right

in contrast, acetic acid is a weak acid, and the position of its equilibrium lies very far to the left

but what if the base is not water? How can we determine which are the major species present?

HCl + H2O H3O++Cl-

H3O+CH3COO-H2OCH3COOH + +


CH3COOH NH3 CH3COO-NH4

++ +


?

Ammonia Ammonium ion(conjugate baseof CH3COOH

(conjugate acidof NH3

(acid) (base)

Acid-Base Equilibria

To predict the position of an acid-base equilibrium such as this, we do the following identify the two acids in the equilibrium; one on the left and one

on the right using the information in Table 10.1, determine which is the

stronger acid and which is the weaker acid also determine which is the stronger base and which is the

weaker base; remember that the stronger acid gives the weaker conjugate base, and the weaker acid gives the stronger conjugate base

the stronger acid reacts with the stronger base to give the weaker acid and weaker base; equilibrium lies on the side of the weaker acid and weaker base

Acid-Base Equilibria identify the two acids and bases, and their relative

strengths

the position of this equilibrium lies to the right

CH3COOH NH3 CH3COO-NH4

++ +

Acetic acid(stronger acid)

Acetate ion(weaker base)

Ammonia(stronger base)

Ammonium ion(weaker acid)

?

CH3COOH NH3 CH3COO- NH4++ +

Acetic acid(stronger acid)

Acetate ion(weaker base)




Example:Example: predict the position of equilibrium in this acid-base reaction

H2CO3 OH- HCO3- H2O+ +

?


Example:Example: predict the position of equilibrium in this acid-base reaction

Solution:Solution: the position of this equilibrium lies to the right


?


Strongeracid

Strongerbase

Weakerbase

Weakeracid

Acid Ionization Constants

when a weak acid, HA, dissolves in water

the equilibrium constant, Keq, for this ionization is

because water is the solvent and its concentration changes very little when we add HA to it, we treat [H2O] as a constant equal to 1000 g/L or 55.5 mol/L

we combine the two constants to give a new constant, which we call an acid ionization constant, Ka

HA H2O A- H3O++ +

[HA][H2O]

[A-][H3O+]Keq =

[HA]

[A-][H3O+]Ka = Keq[H2O] =

Acid Ionization Constants Ka for acetic acid, for example is 1.8 x 10-5

because the acid ionization constants for weak acids are numbers with negative exponents, we commonly express acid strengths as pKa where

the value of pKa for acetic acid is 4.75

values of Ka and pKa for some weak acids are given in Table 10.2

as you study the entries in this table, note the inverse relationship between values of Ka and pKa

the weaker the acid, the smaller its Ka, but the larger its pKa

pKa = -log Ka

H3PO4

HCOOH

CH3CH(OH)COOH

CH3COOH

H2CO3

H2PO4-

H3BO3

NH4+

C6H5OH

HPO42-

HCO3-

HCN

Phosphoric acid

Formic acid

Lactic acid

Acetic acid

Carbonic acid

Dihydrogen phosphate ion

Name

7.21

pKa

9.14

9.25

9.89

12.66

10.25

Boric acid

Ammonium ion

Phenol

Hydrogen phosphate ion

Bicarbonate ion

Acid

7.5 x 10-3

1.8 x 10-4

8.4 x 10-4

1.8 x 10-5

4.3 x 10-7

6.2 x 10-8

Ka

7.3 x 10-10

5.6 x 10-10

1.3 x 10-10

2.2 x 10-13

5.6 x 10-11

2.12

3.75

3.08

4.75

6.37

Hydrocyanic acid 4.9 x 10-10 9.31

Properties of Acids & Bases

Neutralization acids and bases react with each other in a process

called neutralization. Reaction of acids with metals

strong acids react with certain metals (called active metals) to produce a salt and hydrogen gas, H2

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)Magnesium Hydrochloric

acidMagnesium

chlorideHydrogen


Reaction with metal hydroxides reaction of an acid with a metal hydroxide gives a salt plus water

the reaction is more accurately written as

omitting spectator ions gives this net ionic equation

HCl(aq)Hydrochloric

acid

+ KOH(aq)

Water

+KCl(aq)Potassiumchloride

Potassiumhydroxide

H2O

H3O+ Cl- K+ OH- 2H2O Cl- K++ + + + +

H3O+ OH- 2H2O+


Reaction with metal oxides strong acids react with metal oxides to give water

plus a salt

2H3O+(aq) + CaO(s) 3H2O(l) + Ca2+(aq)Calciumoxide

Properties of Acids & Bases Reaction with carbonates and bicarbonates

strong acids react with carbonates to give carbonic acid, which rapidly decomposes to CO2 and H2O

strong acids react similarly with bicarbonates

2H3O+(aq) + CO32-(aq) H2CO3(aq) + 2H2O(l)

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

2H3O+(aq) + CO32-(aq) CO2(g) + 3H2O(l)

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)

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

H3O+(aq) + HCO3-(aq) CO2(g) + 2H2O(l)

Properties of Acids & Bases Reaction with ammonia and amines

any acid stronger than NH4+ is strong enough to react

with NH3 to give a salt

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

Self-Ionization of Water

pure water contains a very small number of H3O+ ions and OH- ions formed by proton transfer from one water molecule to another

the equilibrium expression for this reaction is

we can treat [H2O] as a constant = 55.5 mol/L

H2O+H2O H3O++OH-

BaseAcid Conjugateacid of H2O

Conjugatebase of H2O

[H2O]2

[H3O+][HO-]Keq =

Self-Ionization of Water combining these constants gives a new constant called the ion ion

product of water, Kproduct of water, Kww

in pure water, the value of Kw is 1.0 x 10-14

this means that in pure water

[H3O+][OH-]Kw = Keq[H2O]2 =

Kw = 1.0 x 10-14

[H3O+]

[OH-]

= 1.0 x 10-7 mol/L

= 1.0 x 10-7 mol/Lin pure water

Self-Ionization of Water

the product of [H3O+] and [OH-] in any aqueous solution is equal to 1.0 x 10-14 for solutions as well.

for example, if we add 0.010 mole of HCl to 1 liter of pure water, it reacts completely with water to give 0.010 mole of H3O+

in this solution, [H3O+] is 0.010 or 1.0 x 10-2

this means that the concentration of hydroxide ion is

[OH-] = 1.0 x 10-14

1.0 x 10-2= 1.0 x 10-12

pH and pOH

we commonly express these concentrations as pH, where

pH = -log [H3O+] we can now state the definitions of acidic and basic

solutions in terms of pH acidic solution:acidic solution: one whose pH is less than 7.0 basic solution:basic solution: one whose pH is greater than 7.0 neutral solution:neutral solution: one whose pH is equal to 7.0

pH and pOH

just as pH is a convenient way to designate the concentration of H3O+, pOH is a convenient way to designate the concentration of OH-

pOH = -log[OH-] the ion product of water, Kw, is 1.0 x 10-14

taking the logarithm of this equation gives

pH + pOH = 14 thus, if we know the pH of an aqueous solution, we can

easily calculate its pOH

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

pH and pOH pH of some common materials

pH

Battery acidGastric juiceLemon juiceVinegarTomato juiceCarbonated beveragesBlack coffee

UrineRain (unpolluted)

Milk

SalivaPure waterBloodBilePancreatic fluidSeawaterSoap

Milk of magnesiaHousehold ammonia

Lye (1.0 M NaOH)

0.51.0-3.02.2-2.42.4-3.44.0-4.44.0-5.05.0-5.1

5.5-7.56.2

6.3-6.6

6.5-7.57.0

7.35-7.456.8-7.07.8-8.08.0-9.08.0-10.0

10.511.7

14.0

Material pHMaterial

pH of Salt Solutions

When some salts dissolve in pure water, there is no change in pH from that of pure water

Many salts, however, are acidic or basic and cause a change the pH when they dissolve

We are concerned in this section with basic salts and acidic salts


Basic salt: Basic salt: raises the pH as an example of a basic salt is sodium acetate when this salt dissolves in water, it ionizes; Na+ ions

do not react with water, but CH3COO- ions do

the position of equilibrium lies to the left nevertheless, there are enough OH- ions present in

0.10 M sodium acetate to raise the pH to 8.88

OH-CH3COOHH2OCH3COO- + +Acetic acid

(stronger acid)Acetate ion

(weaker baseHydroxide ion(stronger base)

Water(weaker base)


Acidic salt:Acidic salt: lowers the pH an example of an acidic salt is ammonium chloride chloride ion does not react with water, but the

ammonium ion does

although the position of this equilibrium lies to the left, there are enough H3O+ ions present to make the solution acidic

NH4+ + H2O NH3 + H3O+



Hydronium ion(stronger acid

Water(weaker base)

Acid-Base Titrations

Titration:Titration: an analytical procedure in which a solute in a solution of known concentration reacts with a known stoichiometry with a substance whose concentration is to be determined


An acid-base titration must meet these requirement1. we must know the equation for the reaction so that we

can determine the stoichiometric ratio of reactants to use in our calculations

2. the reaction must be rapid and complete

3. there must be a clear-cut change in a measurable property at the end pointend point (when the reagents have combined exactly)

4. we must have precise measurements of the amount of each reactant


As an example, let us use 0.108 M H2SO4 to determine the concentration of a NaOH solution requirement 1:requirement 1: we know the balanced equation

requirement 2:requirement 2: the reaction between H3O+ and OH- is rapid and complete

requirement 3:requirement 3: we can use either an acid-base indicator or a pH meter to observe the sudden change in pH that occurs at the end point of the titration

requirement 4:requirement 4: we use volumetric glassware

2NaOH(aq)+H2SO4(aq) Na2SO4(aq) + 2H2O(l)(concentration

known)(concentrationnot known)

Acid-Base Titrations experimental measurements

doing the calculations

Trial I

Volume of 0.108 M H2SO4

Volumeof NaOH

25.0 mL 33.48 mLTrial II 25.0 mL 33.46 mL

Trial III 25.0 mL 33.50 mL

average = 33.48 mL

2 mol NaOH1 mol H2SO4

= 0.161 mol NaOHL NaOH

= 0.161 M

mol NaOHL NaOH = 0.108 mol H2SO4

1 L H2SO4x x0.0250 L H2SO4

0.03348 L NaOH

pH Buffers

pH buffer:pH buffer: a solution that resists change in pH when limited amounts of acid or base are added to it a pH buffer as an acid or base “shock absorber” a pH buffer is common called simply a buffer the most common buffers consist of approximately equal

molar amounts of a weak acid and a salt of the conjugate base of the weak acid

for example, if we dissolve 1.0 mole of acetic acid and 1.0 mole of its conjugate base (in the form of sodium acetate) in water, we have an acetate buffer

pH Buffers

How an acetate buffer resists changes in pH if we add a strong acid, such as HCl, added H3O+ ions

react with acetate ions and are removed from solution

if we add a strong base, such as NaOH, added OH- ions react with acetic acid and are removed from solution

CH3COO- H3O+ CH3COOH H2O+ +

CH3COOH OH- CH3COO- H2O+ +

CH3COOH H2O CH3COO- H3O++ +

Added asCH3COOH

Added asCH3COO-Na+

pH Buffers

The effect of a buffer can be quite dramatic consider a phosphate buffer prepared by

dissolving 0.10 mole of NaH2PO4 (a weak acid) and 0.10 mole of Na2HPO4 (the salt of its conjugate base) in enough water to make 1 liter of solution

waterpH

0.10 M phosphate buffer7.07.21

2.0 12.07.12 7.30

pH afteraddition of

0.010 mole HCl

pH afteraddition of

0.010 mole NaOH

pH Buffers

Buffer pHBuffer pH if we mix equal molar amounts of a weak acid and

a salt of its conjugate base, the pH of the solution will be equal to the pKa of the weak acid

if we want a buffer of pH 9.14, for example, we can mix equal molar amounts of boric acid (H3BO3), pKa 9.14, and sodium dihydrogen borate (NaH2BO3), the salt of its conjugate base

pH Buffers

Buffer capacity depends both its pH and its concentration

pH The closer the pH of the buffer is to the pKaof the weak acid, the greater the buffer capacity

Concentration The greater the concentration of the weak acid and its conjugate base, the greater the buffer capacity

Blood Buffers

The average pH of human blood is 7.4 any change larger than 0.10 pH unit in either direction

can cause illness To maintain this pH, the body uses three buffer

systems carbonate buffer:carbonate buffer: H2CO3 and its conjugate base,

HCO3-

phosphate buffer:phosphate buffer: H2PO4- and its conjugate base,

HPO42-

proteins:proteins: discussed in Chapter 21

Henderson-Hasselbalch Eg. Henderson-Hasselbalch equation:Henderson-Hasselbalch equation: a mathematical

relationship between pH, pKa of the weak acid, HA concentrations HA, and its conjugate base, A-

It is derived in the following way

taking the logarithm of this equation gives

HA H2O A- H3O++ +

[HA]

[A-][H3O+]Ka =

[HA]log [H3O+] + log

[A-]log Ka =

Henderson-Hasselbalch Eg. multiplying through by -1 gives

-log Ka is by definition pKa, and -log [H3O+] is by definition pH; making these substitutions gives

rearranging terms gives

[HA]-log [H3O+] - log

[A-]-log Ka =

[HA]

[A-]+ logpH = pKa Henderson-Hasselbalch Equation

[HA][A-]

pKa = pH - log

Henderson-Hasselbalch Eg.

Example:Example: what is the pH of a phosphate buffer solution containing 1.0 mole of NaH2PO4 and 0.50 mole of Na2HPO4 dissolved in enough water to make 1.0 liter of solution

Henderson-Hasselbalch Eg. Example:Example: what is the pH of a phosphate buffer

solution containing 1.0 mole of NaH2PO4 and 0.50 mole of Na2HPO4 in enough water to make one liter of solution

SolutionSolution the equilibrium we are dealing with and its pKa are

substituting these values in the H-H equation gives

H2PO4- H2O HPO4

2- H3O++ + pKa = 7.21

1.0 mol/L 0.50 mol/L

= 7.21 - 0.30 = 6.91

+ logpH = 7.21 0.501.0

chapter 10 acids and bases acids and bases. arrhenius acids and bases in 1884, svante arrhenius...

Documents