CHAPTER 14CHAPTER 14

THE CHEMISTRY OF THE CHEMISTRY OF ACIDS AND BASESACIDS AND BASES

"ACID""ACID"

Latin word acidus, meaning Latin word acidus, meaning sour. sour.

(lemon)(lemon)

"ALKALI""ALKALI"

Arabic word for the ashes that Arabic word for the ashes that come come

from burning certain plants. from burning certain plants.

Water solutions feel slippery and Water solutions feel slippery and

taste bitter (soap).taste bitter (soap).

Acids and bases are extremely Acids and bases are extremely

important in many everyday important in many everyday

applications: our own bloodstream, applications: our own bloodstream,

our environment, cleaning materials, our environment, cleaning materials,

industry. industry.

(sulfuric acid is an economic (sulfuric acid is an economic indicator!)indicator!)

ACID-BASEACID-BASE

THEORIESTHEORIES

Arrhenius Definition

acidacid----donates a hydrogen ion donates a hydrogen ion (H(H++) in water) in water

basebase--donates a hydroxide ion --donates a hydroxide ion in water (OHin water (OH--))

This theory was limited to This theory was limited to

substances with those "parts"; substances with those "parts";

ammonia is a MAJOR exception!ammonia is a MAJOR exception!

Bronsted-Lowry Bronsted-Lowry DefinitionDefinitionacidacid----donates a proton in waterdonates a proton in water

basebase----accepts a proton in water accepts a proton in water

This theory is better; it explains This theory is better; it explains ammonia as a base! This is the main ammonia as a base! This is the main theory that we will use for our theory that we will use for our acid/base discussion.acid/base discussion.

Lewis DefinitionLewis Definition

acidacid----accepts an electron pairaccepts an electron pair

basebase--donates an electron pair--donates an electron pair

This theory explains This theory explains allall traditional traditional acids and bases + a host of acids and bases + a host of coordination compounds and is coordination compounds and is used used

widely in organic chemistry. Uses widely in organic chemistry. Uses coordinate covalent bondscoordinate covalent bonds

The Bronsted-Lowry The Bronsted-Lowry Concept of Acids and Bases Concept of Acids and Bases

Using this theory, you should be Using this theory, you should be

able to write weak acid/base able to write weak acid/base

dissociation equations and dissociation equations and identify identify

acid, base, conjugate acid and acid, base, conjugate acid and

conjugate base.conjugate base.

Conjugate Acid-Base PairConjugate Acid-Base Pair

A pair of compounds that differ A pair of compounds that differ by the presence of one Hby the presence of one H++ unit. unit.

This idea is critical when it comes This idea is critical when it comes to understanding buffer systems. to understanding buffer systems.

Pay close attention here! Pay close attention here!

AcidsAcids

donate a proton (Hdonate a proton (H++))

Neutral CompoundNeutral Compound

HNOHNO33 + H + H22O O H H33OO++ + + NONO33

--

acid base CA CBacid base CA CB

CationCation

NHNH44++ + H + H22O O H H33OO++ + NH + NH33

acid base CA CBacid base CA CB

AnionAnion

HH22POPO44-- + H + H220 0 H H33OO+ + + HPO + HPO44

2-2-

acid base CA CBacid base CA CB

In each of the acid examples---notice In each of the acid examples---notice

the formation of the formation of HH33OO++

This species is named the This species is named the hydronium hydronium

ionion. .

It lets you know that the solution is It lets you know that the solution is

acidicacidic!!

Hydronium, HHydronium, H33OO++

--H--H++ riding piggy-back on a water riding piggy-back on a water

molecule. molecule.

Water is polar and the + charge of Water is polar and the + charge of the the

naked proton is greatly attracted to naked proton is greatly attracted to

Mickey's chin!)Mickey's chin!)

BasesBases

accept a proton (Haccept a proton (H++))

Neutral CompoundNeutral Compound

NHNH33 + H + H22O O NH NH44++ + OH + OH--

base acid CA CBbase acid CA CB

AnionAnion

COCO332-2- + H + H22O O HCO HCO33

-- + + OHOH--

base acid CA base acid CA CBCB

AnionAnion

POPO443- 3- + H + H22O O HPO HPO44

2-2- + OH + OH--

base acid CA base acid CA CBCB

In each of the basic examples--In each of the basic examples--

notice the formation of notice the formation of OH-OH- -- this -- this

species is named the species is named the hydroxide hydroxide

ionion. It lets you know that the . It lets you know that the

solution is solution is basicbasic! !

You try!!You try!!

Exercise 1Exercise 1

In the following reaction, identify In the following reaction, identify

the acid on the left and its CB on the acid on the left and its CB on

the right. Similarly identify the the right. Similarly identify the base base

on the left and its CA on the right.on the left and its CA on the right.

HBr + NHHBr + NH33 NH NH44++ + Br + Br--

What is the conjugate base of What is the conjugate base of HH22S?S?

What is the conjugate acid of What is the conjugate acid of NONO33

--??

ACIDS ONLY DONATE ACIDS ONLY DONATE

ONE PROTON AT A ONE PROTON AT A

TIME!!!TIME!!!

– monoproticmonoprotic--acids donating one --acids donating one HH++ (ex. HC (ex. HC22HH33OO22))

– diproticdiprotic--acids donating two --acids donating two HH++'s (ex. H's (ex. H22CC22OO44))

– polyproticpolyprotic--acids donating --acids donating many Hmany H++'s (ex. H's (ex. H33POPO44))

Polyprotic BasesPolyprotic Bases

accept more than one Haccept more than one H++

anions with -2 and -3 charges anions with -2 and -3 charges

(example: PO(example: PO443-3- ; HPO ; HPO44

2-2-))

Amphiprotic or Amphiprotic or AmphotericAmphoteric

molecules or ions that can behave as molecules or ions that can behave as

EITHER acids or bases: EITHER acids or bases:

water, anions of weak acids water, anions of weak acids

(look at the examples above—sometimes (look at the examples above—sometimes

water was an acid, sometimes it acted as water was an acid, sometimes it acted as

a base)a base)

Exercise 2Exercise 2 Acid Acid Dissociation (Ionization) Dissociation (Ionization) ReactionsReactions

Write the simple dissociation Write the simple dissociation

(ionization) reaction (omitting (ionization) reaction (omitting

water) for each of the following water) for each of the following

acids.acids.

a. Hydrochloric acid (HCl)a. Hydrochloric acid (HCl)

b. Acetic acid (HCb. Acetic acid (HC22HH33OO22))

c. The ammonium ion (NHc. The ammonium ion (NH44++))

d. The anilinium ion (Cd. The anilinium ion (C66HH55NHNH33++))

e. The hydrated aluminum(III) ion e. The hydrated aluminum(III) ion

[Al(H[Al(H22O)O)66]]3+3+

SolutionSolution

A: HCl(aq) A: HCl(aq) H H++(aq) + Cl(aq) + Cl--(aq)(aq)

B: HCB: HC22HH33OO22(aq) (aq)

HH++(aq) + C(aq) + C22HH33OO22--(aq)(aq)

C: NHC: NH44++(aq) (aq) H H++(aq) + NH(aq) + NH33(aq)(aq)

Solution, cont.Solution, cont.

D: CD: C66HH55NHNH33++(aq) (aq) H H++(aq) + (aq) +

CC66HH55NHNH22(aq)(aq)

E: Al(HE: Al(H22O)O)663+3+(aq) (aq)

HH++(aq) + Al(H(aq) + Al(H22O)O)55OHOH22++(aq) (aq)

Relative Strengths of Relative Strengths of Acids and BasesAcids and Bases

Strength is determined by the Strength is determined by the

position of the "dissociation" position of the "dissociation"

equilibrium.equilibrium.

Strong acids/Strong basesStrong acids/Strong bases

dissociate completely in waterdissociate completely in water

have very large K valueshave very large K values

Weak acids/Weak basesWeak acids/Weak bases

dissociate only to a slight extent dissociate only to a slight extent in in

waterwater

dissociation constant is very dissociation constant is very small small

Strong

Weak

Do NotDo Not

confuse concentration confuse concentration

with strength! with strength!

Strong AcidsStrong Acids

Hydrohalic acids: Hydrohalic acids: HCl, HBr, HIHCl, HBr, HI

Nitric: HNONitric: HNO33

Sulfuric: HSulfuric: H22SOSO44

Perchloric: HClOPerchloric: HClO44

The more The more oxygen present oxygen present in the in the polyatomic ion, polyatomic ion,

the stronger its the stronger its acid WITHIN that acid WITHIN that group. group.

Strong BasesStrong Bases

Hydroxides OR oxides of IA Hydroxides OR oxides of IA and and

IIA metalsIIA metals

Solubility plays a role (those Solubility plays a role (those that that

are very soluble are strong!)are very soluble are strong!)

The stronger The stronger the acid, the the acid, the weaker its CB. weaker its CB.

The converse The converse is also true.is also true.

Weak Acids and Bases - Weak Acids and Bases - Equilibrium Equilibrium expressionsexpressions

The vast majority of acid/bases are The vast majority of acid/bases are

weak.weak.

Remember, this means they do Remember, this means they do not not

ionize much.ionize much.

The equilibrium expression for The equilibrium expression for acids acids

is known as the is known as the KKaa (the acid (the acid

dissociation constantdissociation constant). ).

It is set up the same way as in It is set up the same way as in

general equilibrium.general equilibrium.

Many common weak acids are Many common weak acids are

oxyacids, oxyacids,

like phosphoric acid and like phosphoric acid and

nitrous acid.nitrous acid.

Other common Other common weak acids are weak acids are organic acids,organic acids,those that contain a those that contain a carboxyl groupcarboxyl group

COOH group COOH group

like acetic acid and like acetic acid and benzoic acid. benzoic acid.

For Weak Acid Reactions:For Weak Acid Reactions:

HA + HHA + H22O O H H33OO+ + + A + A--

KKaa = = [H [H33OO++][A][A--]] < 1 < 1

[HA][HA]

Write the KWrite the Kaa expression for acetic expression for acetic

acid using Bronsted-Lowry. acid using Bronsted-Lowry.

(Note: Water is a pure liquid and (Note: Water is a pure liquid and

is thus, left out of the equilibrium is thus, left out of the equilibrium expression.)expression.)

Weak bases (bases without Weak bases (bases without OHOH--) )

react with water to produce a react with water to produce a

hydroxide ion.hydroxide ion.

Common examples of weak bases are Common examples of weak bases are

ammonia (NHammonia (NH33), methylamine ), methylamine

(CH(CH33NHNH22), and ethylamine (C), and ethylamine (C22HH55NHNH22). ).

The lone pair on N forms a bond with The lone pair on N forms a bond with

an Han H++. Most weak bases involve N.. Most weak bases involve N.

The equilibrium expression The equilibrium expression for for

basesbases is known as the is known as the KKbb..

For Weak Base For Weak Base Reactions:Reactions:

B + HB + H22O O HB HB+ + + OH + OH--

KKbb = = [H [H33OO++][OH][OH--] ] <1 <1

[B][B]

Set up the KSet up the Kbb expression for expression for

ammonia using Bronsted-ammonia using Bronsted-Lowry.Lowry.

Notice that KNotice that Kaa and K and Kbb expressions expressions

look very similar. look very similar.

The difference is that a base The difference is that a base

produces the hydroxide ion in produces the hydroxide ion in

solution, while the acid produces the solution, while the acid produces the

hydronium ion in solution. hydronium ion in solution.

Another note on this Another note on this point:point:

HH++ and H and H33OO++ are both equivalent are both equivalent terms here. Often water is left terms here. Often water is left completely out of the equation completely out of the equation since since

it does not appear in the it does not appear in the equilibrium. equilibrium.

This has become an accepted This has become an accepted practice. practice.

(*However, water is very important (*However, water is very important in causing the acid to dissociate.)in causing the acid to dissociate.)

Exercise 3Exercise 3 Relative Base StrengthRelative Base Strength

Using table 14.2, arrange the Using table 14.2, arrange the

following species according to following species according to their their

strength as bases: strength as bases:

HH22O, FO, F--, Cl, Cl--, NO, NO22--, and CN, and CN--

SolutionSolution

ClCl-- < H < H22O < FO < F-- < NO < NO22-- < CN < CN--

WATER WATER

THE HYDRONIUM ION THE HYDRONIUM ION

AUTO-IONIZATION AUTO-IONIZATION

THE pH SCALETHE pH SCALE

Fredrich Kohlrausch, around Fredrich Kohlrausch, around

1900, found that no matter how 1900, found that no matter how

pure water is, it still conducts a pure water is, it still conducts a

minute amount of electric minute amount of electric

current. This proves that water current. This proves that water

self-ionizes. self-ionizes.

Since the water molecule is Since the water molecule is

amphoteric, it may dissociate amphoteric, it may dissociate

with itself to a slight extent. with itself to a slight extent.

Only about 2 out of a billion Only about 2 out of a billion

water molecules are ionized at water molecules are ionized at

any instant!any instant!

HH22O(l) + HO(l) + H22O(l) <=> HO(l) <=> H33OO++(aq) + OH(aq) + OH--

(aq)(aq)

The equilibrium expression used The equilibrium expression used

here is referred to as the here is referred to as the KKww

(ionization constant for (ionization constant for water)water). .

In pure water or dilute aqueous In pure water or dilute aqueous

solutions, the concentration of water solutions, the concentration of water

can be considered to be a constant can be considered to be a constant

(55.4 M), so we include that with the (55.4 M), so we include that with the

equilibrium constant and write the equilibrium constant and write the

expression as:expression as:

KKeqeq[H[H22O]O]22 = K = Kww = [H = [H33OO++][OH][OH--]]

KKww = 1.0 x 10 = 1.0 x 10-14-14

(K(Kww = 1.008 x 10 = 1.008 x 10-14-14 @ 25° Celsius) @ 25° Celsius)

Knowing this value allows us to Knowing this value allows us to

calculate the OHcalculate the OH-- and H and H++

concentration for concentration for various various situationsituations.s.

[OH[OH--] = [H] = [H++] : solution is neutral (in ] : solution is neutral (in

pure water, each of these is 1.0 x 10pure water, each of these is 1.0 x 10--

77))

[OH[OH--] > [H] > [H++] : solution is basic ] : solution is basic

[OH[OH--] < [H] < [H++] : solution is acidic ] : solution is acidic

KKww = K = Kaa x K x Kbb

another very beneficial another very beneficial equationequation

Exercise 5Exercise 5 Autoionization of WaterAutoionization of Water

At 60°C, the value of KAt 60°C, the value of Kww is 1 X 10 is 1 X 10-13-13..

a. Using Le Chatelier’s principle, a. Using Le Chatelier’s principle,

predict whether the reaction predict whether the reaction

2H2H22O(l) O(l) H H33OO++(aq) + OH(aq) + OH--(aq)(aq)

is exothermic or endothermic.is exothermic or endothermic.

Exercise 5, cont.Exercise 5, cont.

b. Calculate [Hb. Calculate [H++] and [OH] and [OH--] in a ] in a

neutral solution at 60°C.neutral solution at 60°C.

SolutionSolution

A: endothermicA: endothermic

B: [HB: [H++] = [OH] = [OH--] = 3 X 10] = 3 X 10-7-7 MM

The pH ScaleThe pH Scale

Used to Used to designate the designate the [H[H++] in most ] in most aqueous aqueous solutions solutions where Hwhere H++ is is small.small.

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

pOH = - log [OHpOH = - log [OH--] ]

pH + pOH = 14 pH + pOH = 14

pH = 6.9 and lower (acidic) pH = 6.9 and lower (acidic)

= 7.0 (neutral) = 7.0 (neutral)

= 7.1 and greater = 7.1 and greater (basic)(basic)

Use as many decimal places as Use as many decimal places as

there are sig.figs. in the problem!there are sig.figs. in the problem!

The negative base 10 logarithm of The negative base 10 logarithm of

the hydronium ion concentration the hydronium ion concentration

becomes the whole number; becomes the whole number;

therefore, only the decimals to the therefore, only the decimals to the

right are significant. right are significant.

Exercise 6Exercise 6 Calculating [HCalculating [H++] and ] and [OH[OH--]]Calculate [HCalculate [H++] or [OH] or [OH--] as required ] as required

for for

each of the following solutions at each of the following solutions at

25°C, and state whether the solution 25°C, and state whether the solution

is neutral, acidic, or basic.is neutral, acidic, or basic.

a. 1.0 X 10a. 1.0 X 10-5-5 MM OH OH--

b. 1.0 X 10b. 1.0 X 10-7-7 MM OH OH--

c. 10.0c. 10.0 M M H H++

SolutionSolution

A: [HA: [H++] = 1.0 X 10] = 1.0 X 10-9-9 M, M, basic basic

B: [HB: [H++] = 1.0 X 10] = 1.0 X 10-7-7 M,M, neutral neutral

C: [OHC: [OH--] = 1.0 X 10] = 1.0 X 10-15-15 M, M, acidicacidic

Exercise 7Exercise 7 Calculating pH and Calculating pH and pOHpOHCalculate pH and pOH for each of Calculate pH and pOH for each of

the following solutions at 25°C.the following solutions at 25°C.

a. 1.0 X 10a. 1.0 X 10-3-3 MM OH OH--

b. 1.0b. 1.0 M M H H++

SolutionSolution

A: pH = 11.00A: pH = 11.00

pOH = 3.00pOH = 3.00

B: pH = 0.00B: pH = 0.00

pOH = 14.00pOH = 14.00

Exercise 8Exercise 8 Calculating Calculating pHpH

The pH of a sample of human The pH of a sample of human blood blood

was measured to be 7.41 at 25°C. was measured to be 7.41 at 25°C.

Calculate pOH, [HCalculate pOH, [H++], and [OH], and [OH--] for ] for

the sample.the sample.

SolutionSolution

pOH = 6.59pOH = 6.59

[H[H++] = 3.9 X 10] = 3.9 X 10-8-8

[OH[OH--] = 2.6 X 10] = 2.6 X 10-7-7 MM

Exercise 9Exercise 9 pH of Strong Acids pH of Strong Acids

Calculate the pH of:Calculate the pH of:

a. 0.10 a. 0.10 MM HNO HNO33

b. 1.0 X 10b. 1.0 X 10-10-10 MM HCl HCl

SolutionSolution

A: pH = 1.00A: pH = 1.00

B: pH = 7.00B: pH = 7.00

Exercise 10Exercise 10 The pH of Strong The pH of Strong BasesBases

Calculate the pH of a 5.0 X 10Calculate the pH of a 5.0 X 10-2-2 MM

NaOH solution.NaOH solution.

SolutionSolution

pH = 12.70pH = 12.70

Calculating pH of Weak Calculating pH of Weak Acid SolutionsAcid Solutions

Calculating pH of weak acids Calculating pH of weak acids

involves setting up an involves setting up an equilibrium. equilibrium.

Always start by…Always start by…

1) writing the equation… 1) writing the equation…

2) setting up the acid equilibrium 2) setting up the acid equilibrium

expression (Kexpression (Kaa)…)…

3) defining initial concentrations, 3) defining initial concentrations, changes, and final changes, and final concentrations in terms of X …concentrations in terms of X …

4) substituting values and 4) substituting values and variables variables

into the Kinto the Kaa expression… expression…

5) solving for X 5) solving for X

(use the (use the RICERICE diagram learned in diagram learned in

general equilibrium!)general equilibrium!)

Example:Example:

Calculate the pH of a 1.00 x 10Calculate the pH of a 1.00 x 10--

44 M M

solution of acetic acid.solution of acetic acid.

The KThe Kaa of acetic acid is 1.8 x 10 of acetic acid is 1.8 x 10-5-5

HCHC22HH33OO22 H H++ + C + C22HH33OO22--

KKaa = = [H[H++][C][C22HH33OO22--]] = 1.8 x 10 = 1.8 x 10-5-5

[HC[HC22HH33OO22]]

RReaction HCeaction HC22HH33OO22 H+ + C H+ + C22HH33OO22--

IInitial 1.00 x 10nitial 1.00 x 10-4-4 0 0 0 0

CChange -x +x +xhange -x +x +x

EEquilibrium 1.00 x 10quilibrium 1.00 x 10-4 -4 - x x x- x x x

1.8 x 101.8 x 10-5-5 = = (x)(x) _ (x)(x) _

1.00x101.00x10-4-4 - - xx

1.8 x 10-5 1.8 x 10-5 (x)(x) _ (x)(x) _

1.00 x 101.00 x 10-4-4

x = 4.2 x 10x = 4.2 x 10-5-5

Often, the -x in a KOften, the -x in a Kaa expression can be treated as expression can be treated as negligible.negligible.

When you assume that x is When you assume that x is

negligible, you must check the negligible, you must check the

validity of this assumption. validity of this assumption.

To be valid, x must be less than To be valid, x must be less than

5% of the number that it was to be 5% of the number that it was to be

subtracted from.subtracted from.

% dissociation = % dissociation = "x" "x" x 100 x 100

[original][original]

In this example, 4.2 x 10In this example, 4.2 x 10-5-5 is greater is greater

than 5% of 1.00 x 10than 5% of 1.00 x 10-4-4. .

This means that the assumption that This means that the assumption that

x was negligible was invalid and x x was negligible was invalid and x

must be solved for using the must be solved for using the

quadratic equation or the method of quadratic equation or the method of

successive approximation.successive approximation.

Use of the Use of the Quadratic Equation Quadratic Equation

a

acbbx

2

42

ax2 + bx + c = 0

0108.1108.1 952 xx

5105.3 x 5102.5 x

)1(2

)108.1)(1(4)108.1(108.1 9255 x

and

Using the values: a = 1, b = 1.8x10-5, c= -1.8x10-9

Since a concentration Since a concentration can not be negative…can not be negative…

x = 3.5 x 10x = 3.5 x 10-5-5 M M

x = [Hx = [H++] = 3.5 x 10] = 3.5 x 10-5-5

pH = -log 3.5 x 10pH = -log 3.5 x 10-5-5 = 4.46 = 4.46

Another method which some people Another method which some people

prefer is the method of successive prefer is the method of successive

approximations. In this method, you approximations. In this method, you

start out assuming that x is start out assuming that x is

negligible, solve for x, and repeatedly negligible, solve for x, and repeatedly

plug your value of x into the plug your value of x into the

equation again until you get the equation again until you get the

same value of x two successive times.same value of x two successive times.

Exercise 11Exercise 11 The pH of Weak Acids The pH of Weak Acids

The hypochlorite ion (OClThe hypochlorite ion (OCl--) is a ) is a

strong oxidizing agent often found strong oxidizing agent often found

in household bleaches and in household bleaches and

disinfectants. It is also the active disinfectants. It is also the active

ingredient that forms when ingredient that forms when

swimming pool water is treated swimming pool water is treated with with

chlorine. chlorine.

In addition to its oxidizing abilities, In addition to its oxidizing abilities,

the hypochlorite ion has a relatively the hypochlorite ion has a relatively

high affinity for protons (it is a high affinity for protons (it is a

much stronger base than Clmuch stronger base than Cl--, for , for

example) and forms the weakly example) and forms the weakly

acidic hypochlorous acid (HOCl, acidic hypochlorous acid (HOCl,

KKaa = 3.5 X 10 = 3.5 X 10-8-8).).

Calculate the pH of a 0.100Calculate the pH of a 0.100 M M

aqueous solution of aqueous solution of hypochlorous hypochlorous

acid.acid.

SolutionSolution

pH = 4.23pH = 4.23

Determination of the pH Determination of the pH of a Mixture of Weak of a Mixture of Weak AcidsAcidsOnly the acid with the largest KOnly the acid with the largest Kaa

value will contribute an value will contribute an appreciable appreciable

[H[H++]. ].

Determine the pH based on Determine the pH based on

this acid and ignore any others.this acid and ignore any others.

Exercise 12Exercise 12 The pH The pH of Weak Acid Mixturesof Weak Acid Mixtures

Calculate the pH of a solution that Calculate the pH of a solution that

contains: contains:

1.00 1.00 MM HCN (K HCN (Kaa = 6.2 X 10 = 6.2 X 10-10-10) ) and and

5.00 5.00 MM HNO HNO2 2 (K(Kaa = 4.0 X 10 = 4.0 X 10-4-4). ).

Exercise 12, cont.Exercise 12, cont.

Also, calculate the Also, calculate the concentration of concentration of

cyanide ion (CNcyanide ion (CN--) in this ) in this solution at solution at

equilibrium.equilibrium.

SolutionSolution

pH = 1.35pH = 1.35

[CN[CN--] = 1.4 X 10] = 1.4 X 10-8-8 MM

Exercise 13Exercise 13 Calculating Calculating Percent DissociationPercent Dissociation

Calculate the percent dissociation Calculate the percent dissociation of of

acetic acid (Kacetic acid (Kaa = 1.8 X 10 = 1.8 X 10-5-5) in ) in each of the following solutions.each of the following solutions.

a. 1.00 a. 1.00 MM HC HC22HH33OO22

b. 0.100 b. 0.100 MM HC HC22HH33OO22

SolutionSolution

A: = 0.42 %A: = 0.42 %

B: = 1.3 %B: = 1.3 %

Exercise 14Exercise 14 Calculating Calculating KKaa from Percent from Percent DissociationDissociation

Lactic acid (HCLactic acid (HC33HH55OO33) is a waste ) is a waste

product that accumulates in product that accumulates in muscle muscle

tissue during exertion, leading tissue during exertion, leading to to

pain and a feeling of fatigue. pain and a feeling of fatigue.

Exercise 14, cont.Exercise 14, cont.

In a 0.100 In a 0.100 M M aqueous solution, aqueous solution,

lactic acid is 3.7% dissociated. lactic acid is 3.7% dissociated.

Calculate the value of KCalculate the value of Kaa for for this this

acid.acid.

SolutionSolution

KKaa= 1.4 X 10= 1.4 X 10-4-4

Determination of the pH of a Determination of the pH of a weak weak

basebase is very similar to the is very similar to the

determination of the pH of a weak determination of the pH of a weak

acid. acid.

Follow the same steps. Follow the same steps.

Remember, however, that Remember, however, that xx is is the the

[OH[OH--]] and taking the negative and taking the negative log log

of of xx will give you the will give you the pOH pOH and and not not

the pH!the pH!

Exercise 15Exercise 15 The pH of Weak Bases I The pH of Weak Bases I

Calculate the pH for a 15.0 Calculate the pH for a 15.0 MM

solution of NHsolution of NH33 (K (Kbb = 1.8 X 10 = 1.8 X 10-5-5).).

SolutionSolution

pH = 12.20pH = 12.20

Exercise 16Exercise 16 The pH of Weak Bases The pH of Weak Bases IIIICalculate the pH of a 1.0 Calculate the pH of a 1.0 MM

solution solution

of methylamine (Kof methylamine (Kbb = 4.38 X = 4.38 X 1010-4-4).).

SolutionSolution

pH = 12.32pH = 12.32

Calculating pH of Calculating pH of polyprotic acidspolyprotic acids

Acids with more than one Acids with more than one ionizable ionizable

hydrogen will ionize in steps. hydrogen will ionize in steps.

Each dissociation has its own KEach dissociation has its own Kaa

value. value.

The The firstfirst dissociation will be dissociation will be the the

greatestgreatest and subsequent and subsequent

dissociations will have much dissociations will have much smaller equilibrium smaller equilibrium constants.constants.

As each H is removed, the As each H is removed, the

remaining acid gets weaker remaining acid gets weaker and and

therefore has a smaller Ktherefore has a smaller Kaa..

As the negative charge on the As the negative charge on the acid acid

increases, it becomes more increases, it becomes more difficult difficult

to remove the positively to remove the positively charged charged

proton.proton.

Example:Example:

Consider the dissociation of Consider the dissociation of phosphoric acid. phosphoric acid.

HH33POPO4(aq)4(aq) + H + H22OO(l)(l) <=> <=>

HH33OO++(aq)(aq) + H + H22POPO44

-- (aq)(aq)

KKa1a1 = 7.5 x 10 = 7.5 x 10-3-3

HH22POPO44--(aq)(aq) + H + H22OO(l) (l) <=> <=>

HH33OO++(aq)(aq) + HPO + HPO44

2-2-

(aq)(aq)

KKa2a2 = 6.2 x 10 = 6.2 x 10-8-8

HPOHPO442-2-

(aq)(aq) + H + H22OO(l) (l) <=> <=>

HH33OO++(aq)(aq) + PO + PO44

3-3-(aq)(aq)

KKaa33 = 4.8 x 10 = 4.8 x 10-13-13

Looking at the KLooking at the Kaa values, it is values, it is

obvious that only the first obvious that only the first

dissociation will be important in dissociation will be important in

determining the pH of the solution.determining the pH of the solution.

Except for HExcept for H22SOSO44, polyprotic acids , polyprotic acids

have Khave Ka2a2 and K and Ka3a3 values so much values so much

weaker than their Kweaker than their Ka1a1 value that value that

the 2nd and 3rd (if applicable) the 2nd and 3rd (if applicable)

dissociation can be ignored. dissociation can be ignored.

The [HThe [H++] obtained from this 2nd ] obtained from this 2nd

and 3rd dissociation is and 3rd dissociation is negligible negligible

compared to the [Hcompared to the [H++] from the ] from the 1st 1st

dissociation.dissociation.

Because HBecause H22SOSO44 is a strong acid in its is a strong acid in its

first dissociation and a weak acid in first dissociation and a weak acid in

its second, we need to consider both its second, we need to consider both

if the concentration is more dilute if the concentration is more dilute

than 1.0 M. than 1.0 M.

The quadratic equation is needed to The quadratic equation is needed to

work this type of problem.work this type of problem.

Exercise 17Exercise 17 The pH of a Polyprotic The pH of a Polyprotic AcidAcidCalculate the pH of a 5.0 Calculate the pH of a 5.0 MM H H33POPO44

solution and the equilibrium solution and the equilibrium

concentrations of the species: concentrations of the species:

HH33POPO44, H, H22POPO44--, HPO, HPO44

2-2-, and PO, and PO443-3-

SolutionSolution

pH = 0.72pH = 0.72

[H[H33POPO44] = 4.8 ] = 4.8 MM

[H[H22POPO44--] = 0.19] = 0.19 M M

[HPO[HPO442-2-] = 6.2 X 10] = 6.2 X 10-8 -8 MM

[PO[PO443-3-] = 1.6 X 10] = 1.6 X 10-19-19 MM

Exercise 18Exercise 18 The pH of a Sulfuric The pH of a Sulfuric AcidAcidCalculate the pH of a 1.0 Calculate the pH of a 1.0 MM

HH22SOSO44

solution.solution.

SolutionSolution

pH = 0.00pH = 0.00

Exercise 19Exercise 19 The pH of a Sulfuric The pH of a Sulfuric AcidAcidCalculate the pH of a 1.0 X 10Calculate the pH of a 1.0 X 10-2-2

MM

HH22SOSO44 solution. solution.

SolutionSolution

pH = 1.84pH = 1.84

ACID-BASE ACID-BASE PROPERTIES OF PROPERTIES OF

SALTS:SALTS:

HYDROLYSISHYDROLYSIS

Salts are produced from the reaction Salts are produced from the reaction

of an acid and a base. of an acid and a base. (neutralization) (neutralization)

Salts are Salts are not alwaysnot always neutral. Some neutral. Some

hydrolyze with water to produce hydrolyze with water to produce

acidic and basic solutions.acidic and basic solutions.

Neutral SaltsNeutral Salts

Salts that are formed from the Salts that are formed from the cation of a strong base and the cation of a strong base and the anion of a strong acid form anion of a strong acid form neutral solutions when dissolved neutral solutions when dissolved in water. in water.

A salt such as NaNOA salt such as NaNO33 gives a gives a neutral solution.neutral solution.

Basic SaltsBasic Salts

Salts that are formed from the Salts that are formed from the

cation of a strong base and the cation of a strong base and the

anion of a weak acid form basic anion of a weak acid form basic

solutions when dissolved in solutions when dissolved in water.water.

The anion hydrolyzes the The anion hydrolyzes the water water

molecule to produce molecule to produce hydroxide hydroxide

ions and thus a basic ions and thus a basic solution. solution.

KK22S should be basic since SS should be basic since S-2-2 is is

the CB of the very weak acid the CB of the very weak acid HSHS--, ,

while Kwhile K++ does not hydrolyze does not hydrolyze

appreciably.appreciably.

SS2-2- + H + H22O O OH OH-- + HS + HS--

strong base weak strong base weak acidacid

Acid SaltsAcid Salts

Salts that are formed from the Salts that are formed from the

cation of a weak base and the cation of a weak base and the

anion of a strong acid form anion of a strong acid form

acidic solutions when acidic solutions when dissolved in dissolved in

water.water.

The cation hydrolyzes the The cation hydrolyzes the water water

molecule to produce molecule to produce hydronium hydronium

ions and thus an acidic ions and thus an acidic solution. solution.

NHNH44Cl should be weakly acidic, Cl should be weakly acidic,

since NHsince NH44++ hydrolyzes to give an hydrolyzes to give an

acidic solution, while Clacidic solution, while Cl-- does not does not

hydrolyze.hydrolyze.

NHNH44++ + H + H22O O H H33OO++ + NH + NH33

strong acid weak strong acid weak basebase

If both the cation If both the cation

and the anion and the anion

contribute to the contribute to the

pH situation, pH situation,

compare Kcompare Kaa to K to Kbb..

If KIf Kbb is larger, basic! is larger, basic!

The converse is also true.The converse is also true.

The following will help predict The following will help predict

acidic, basic, or neutral. acidic, basic, or neutral.

However, you must explain However, you must explain using using

appropriate equations as appropriate equations as proof!!!proof!!!

1. Strong acid + Strong base 1. Strong acid + Strong base

= =

Neutral saltNeutral salt

2. Strong acid + Weak base 2. Strong acid + Weak base

= =

Acidic salt Acidic salt

3. Weak acid + Strong base 3. Weak acid + Strong base

= =

Basic salt Basic salt

4. Weak acid + Weak base 4. Weak acid + Weak base

= =

??? ???

(must look at K values to decide)(must look at K values to decide)

Exercise 20Exercise 20 The Acid-The Acid-Base Properties of SaltsBase Properties of Salts

Predict whether an aqueous Predict whether an aqueous solution solution

of each of the following salts will be of each of the following salts will be

acidic, basic, or neutral. Prove with acidic, basic, or neutral. Prove with

appropriate equations. appropriate equations.

a. NHa. NH44CC22HH33OO22

b. NHb. NH44CNCN

c. Alc. Al22(SO(SO44))33

SolutionSolution

A: neutralA: neutral

B: basicB: basic

C: acidicC: acidic

Exercise 21Exercise 21 Salts as Weak Bases Salts as Weak Bases

Calculate the pH of a 0.30 Calculate the pH of a 0.30 MM NaF NaF

solution. solution.

The KThe Kaa value for HF is 7.2 X 10 value for HF is 7.2 X 10-4-4..

SolutionSolution

pH = 8.31pH = 8.31

Exercise 22Exercise 22 Salts as Weak Acids I Salts as Weak Acids I

Calculate the pH of a 0.10 Calculate the pH of a 0.10 MM NHNH44Cl Cl

solution. solution.

The KThe Kbb value for NH value for NH33 is 1.8 X 10 is 1.8 X 10--

55..

SolutionSolution

pH = 5.13 pH = 5.13

Exercise 23Exercise 23 Salts as Weak Acids II Salts as Weak Acids II

Calculate the pH of a 0.010 Calculate the pH of a 0.010 MM AlClAlCl33

solution. solution.

The KThe Kaa value for Al(H value for Al(H22O)O)663+3+ is is

1.4 X 101.4 X 10-5-5..

SolutionSolution

pH = 3.43pH = 3.43

The Lewis Concept of The Lewis Concept of Acids and BasesAcids and Bases

acidacid--can --can acceptaccept a pair of a pair of

electrons to form a coordinate electrons to form a coordinate

covalent bondcovalent bond

basebase--can --can donatedonate a pair of a pair of

electrons to form a coordinate electrons to form a coordinate

covalent bondcovalent bond

Yes, this is the dot guy and the Yes, this is the dot guy and the

structures guy.structures guy.

He was extremely busy He was extremely busy making making

your life difficult!your life difficult!

BFBF3 3 — the most famous of — the most famous of all!!all!!

Exercise 24Exercise 24

Tell whether each of the following is Tell whether each of the following is

a Lewis acid or base. a Lewis acid or base.

Draw structures as proof. Draw structures as proof.

a) PHa) PH33 c) H c) H22SS

b) BClb) BCl33 d) SF d) SF44

Exercise 25Exercise 25Lewis Acids and BasisLewis Acids and Basis

For each reaction, identify the For each reaction, identify the Lewis Lewis

acid and base.acid and base.

a. Nia. Ni22++(aq) + 6NH(aq) + 6NH33(aq) (aq) Ni(NHNi(NH33))66

2+2+(aq)(aq)

b. Hb. H++(aq) + H(aq) + H22O(aq) O(aq) H H33OO++(aq) (aq)

SolutionSolution

A: Lewis acid = nickel(II) ionA: Lewis acid = nickel(II) ion

Lewis base = ammoniaLewis base = ammonia

B: Lewis acid = protonB: Lewis acid = proton

Lewis base = water Lewis base = water moleculemolecule