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1 Ch 17 — Acids and Bases

Jeffrey Mack

California State University,

Sacramento

Chapter 17 Principles of Chemical Reactivity: The Chemistry of Acids and Bases

• In Chapter 3, you were introduced to two

definitions of acids and bases: the Arrhenius and

the Brønsted–Lowry definition.

• Arrhenius acid: Any substance that when

dissolved in water increases the concentration of

hydrogen ions, H+.

• Arrhenius base: Any substance that increases

the concentration of hydroxide ions, OH, when

dissolved in water.

• A Brønsted–Lowry acid is a proton (H+) donor.

• A Brønsted–Lowry base is a proton acceptor.

Acids & Bases: A Review

• Generally divide acids and bases into STRONG or

WEAK ones.

STRONG ACID:

HNO3(aq) + H2O(liq) H3O+(aq) + NO3

-(aq)

HNO3 is about 100% dissociated in water.

Strong & Weak Acids/Bases

HNO3, HCl, H2SO4 and HClO4 are classified as

strong acids.


• Strong Base: 100% dissociated in water.

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

Other common strong

bases include KOH and

Ca(OH)2.

CaO (lime) + H2O

Ca(OH)2 (slaked lime) CaO


• Weak base: less than 100% ionized in water

An example of a weak base is ammonia

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



Weak acids are much less than 100% ionized

in water.

Example: acetic acid = CH3CO2H


• Proton donors may be molecular compounds,

cations or anions.

3

3 2 3 3

4 2 3 3

2

2 3 3

HNO (aq) H O(l) NO (aq) H O (aq)

NH (aq) H O(l) NH (aq) H O (aq)

HCO (aq) H O(l) CO (aq) H O (aq)

The Brønsted–Lowry Concept of Acids & Bases Extended

• Proton acceptors may be molecular

compounds, cations or anions.

3

3

3 2

2

2 25

3

2 6

2

3 2

NH (aq) H O(l) NH (aq) OH (aq)

Al H O OH (aq) H O(l)

Al H O (aq) OH (aq)

CO (aq) H O(l) HCO (aq) OH (aq)


Using the Brønsted definition, NH3 is a BASE in water and water is itself an ACID

Proton acceptor

Proton donor

33 2NH (aq) H O(l) NH (aq) OH (aq)+ -+ +


• Acids such as HF, HCl, HNO3, and CH3CO2H (acetic acid)

are all capable of donating one proton and so are called

monoprotic acids.

• Other acids, called polyprotic acids are capable of donating

two or more protons.


• A conjugate acid–base pair consists of two species that

differ from each other by the presence of one hydrogen ion.

• Every reaction between a Brønsted acid and a Brønsted

base involves two conjugate acid–base pairs

Conjugate Acid–Base Pairs


Conjugate Acid–Base Pairs

Water Autoionization and the Water Ionization Constant, Kw:

The water autoionization equilibrium lies far to the left side. In fact, in pure water at 25 °C, only about two out of a billion (109) water molecules are ionized at any instant.

Even in pure water, there is a small concentration of ions present at all times. [H3O

+] = [OH] = 1.00 107

2 2 3H O(l) H O(l) H O (aq) OH (aq)

14

w 3K H O OH 1.00 10+ - -é ù é ù= = ´ë û ë û

Water & the pH Scale

H2O can function as both an ACID and a BASE.

In pure water there can be AUTOIONIZATION.

Equilibrium constant for autoionization = Kw

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 °C


• In a neutral solution, [H3O+] = [OH]

• Both are equal to 1.00 10 7 M

• In an acidic solution, [H3O+] > [OH]

• [H3O+] > 1.00 10 7 M and [OH] < 1.00 10

7 M

• In a basic solution, [H3O+] < [OH]

• [H3O+] < 1.00 10 7 M and [OH] > 1.00 10

7 M


The pH Scale

• The pH of a solution is defined as the

negative of the base (10) logarithm (log) of

the hydronium ion concentration.

pH = log[H3O+]

• In a similar way, we can define the pOH of a

solution as the negative of the base - 10

logarithm of the hydroxide ion concentration.

pOH = log[OH]

pH + pOH = pKw = 14

The pH Scale


• The concentration of acid, [H3O+] is found by

taking the antilog of the solutions pH.

• In a similar way, [OH] can be found from:

pH

3H O 10 + -é ù =ë û

pOHOH 10 - -é ù =ë û

The pH Scale

Once [H3O+] is known, [OH] can be found

from:

And vice versa.

w

3

K[OH ]

[H O ]

-

+=

w3

K[H O ]

[OH ]

+

-=

The pH Scale

• In Chapter 3, it was stated that acids and bases

can be divided roughly into those that are strong

electrolytes (such as HCl, HNO3, and NaOH)

and those that are weak electrolytes (such as

CH3CO2H and NH3)

• In this chapter we will discuss the quantitative

aspects of dissociation of weak acids and bases.

• The relative strengths of weak acids and bases

can be ranked based on the magnitude of

individual equilibrium constants.

Equilibrium Constants for Acids & Bases

• Strong acids and bases almost completely

ionize in water (~100%):

Kstrong >> 1

(product favored)

• Weak acids and bases almost completely

ionize in water (<<100%):

Kweak << 1

(Reactant favored)


• The relative strength of an acid or base can also be

expressed quantitatively with an equilibrium

constant, often called an ionization constant. For

the general acid HA, we can write:

[ ]

2 3

3

a

HA(aq) H O(l) H O (aq) A (aq)

H O AK

HA

+ -

+ -

+ +

é ù é ùë û ë û=

Conjugate

acid

Conjugate

base


• The relative strength of an acid or base can

also be expressed quantitatively with an

equilibrium constant, often called an

ionization constant. For the general base B,

we can write:

[ ]

2

b

B(aq) H O(l) BH (aq) OH (aq)

BH OHK

B

+ -

+ -

+ +

é ù é ùë û ë û=

Conjugate

base

Conjugate

Acid



Acids Conjugate

Bases Increase

strength

Increase

strength

Ionization Constants for Acids/Bases

• The strongest acids are at the upper left.

They have the largest Ka values.

• Ka values become smaller on descending the

chart as the acid strength declines.

• The strongest bases are at the lower right.

They have the largest Kb values.

• Kb values become larger on descending the

chart as base strength increases.


• The weaker the acid, the stronger its

conjugate base: The smaller the value of Ka,

the larger the value of Kb.

• Aqueous acids that are stronger than H3O+

are completely ionized.

• Their conjugate bases (such as NO3) do not

produce meaningful concentrations of OH

ions, their Kb values are “very small.”

• Similar arguments follow for strong bases

and their conjugate acids.


Increasing Acid Strength ¾¾¾¾¾¾¾¾¾¾¾®

Acid HCO3 HClO HF

Ka 4.8 1011 3.5 108 7.2 104

Increasing Base Strength ¬¾¾¾¾¾¾¾¾¾¾¾

Base CO32 ClO F

Kb 2.1 104 2.9 107 1.4 1011



Ka Values for Polyprotic Acids

In general, each successive dissociation

produces a weaker acid.

2 2 3

7

a(1)

2

2 3

19

a(2)

H S(aq) H O(l) H O (aq) HS (aq)

K 1 10

HS (aq) H O(l) H O (aq) S (aq)

K 1 10



Logarithmic Scale of Relative Acid

Strength, pKa

• Many chemists use a logarithmic scale to

report and compare relative acid strengths.

pKa = log(Ka)

The lower the pKa, the stronger the acid.

Acid HCO3 HClO HF

pKa 10.32 7.46 3.14

Increasing Acid Strength ¾¾¾¾¾¾¾¾¾¾¾®


Relating the Ionization Constants for an

Acid and Its Conjugate Base

a

b

K

2 2 3

K

2 2


HS (aq) H O(l) H S(aq) OH (aq)

+ -

- -

¾¾®+ +¬¾¾

¾¾®+ +¬¾¾




a

b

K

2 2 3

K

2 2



+ -

- -

¾¾®+ +¬¾¾

¾¾®+ +¬¾¾




a

b

K

2 2 3

K

2 2

2 3



2H O(l) H O (aq) OH (aq)

+ -

- -

+ -

¾¾®+ +¬¾¾

¾¾®+ +¬¾¾

+




a

b

K

2 2 3

K

2 2

2 3




+ -

- -

+ -

¾¾®+ +¬¾¾

¾¾®+ +¬¾¾

+

3 2

a b 3 w

2

H O HS H S OHK K H O OH K

H S HS




[ ]

[ ]3 2

a b 3 w

2

a b w

H O HS H S OHK K H O OH K

H S HS

K K K

+ - -

+ -

-

é ù é ù é ùë û ë û ë û é ù é ù´ = ´ = =ë û ë ûé ùë û

´ =

a

b

K

2 2 3

K

2 2

2 3




+ -

- -

+ -

¾¾®+ +¬¾¾

¾¾®+ +¬¾¾

+

When adding equilibria,

multiply the K values.



Acid–Base Properties of Salts

Anions that are conjugate bases of strong

acids (for examples, Cl or NO3.

These species are such weak bases that they

have no effect on solution pH.

3 2NO (aq) H O(l) No Reaction


Anions such as CO3 that are the conjugate

bases of weak acids will raise the pH of a

solution.

Hydroxide ions are produced via “Hydrolysis”.

2

3 2 3CO (aq) H O(l) HCO (aq) OH (aq)


Anions such as CO3 that are the conjugate

bases of weak acids will raise the pH of a

solution.

Hydroxide ions are produced via “Hydrolysis”.

A partially deprotonated anion (such as HCO3)

is amphiprotic. Its behavior will depend on the

other species in the reaction.

2

3 2 3CO (aq) H O(l) HCO (aq) OH (aq)


Alkali metal and alkaline earth cations have no

measurable effect on solution pH.

Since these cations are conjugate acids of

strong bases, hydrolysis does not occur.

2Na (aq) H O(l) No Reaction


Basic cations are conjugate bases of acidic cations

such as [Al(H2O)6]3+.

Acidic cations fall into two categories: (a) metal

cations with 2+ and 3+ charges and (b) ammonium

ions (and their organic derivatives).

All metal cations are hydrated in water, forming ions

such as [M(H2O)6]n+.

3

2 26

2

2 35

4

Al H O (aq) H O(l)

Al H O (OH ) (aq) H O (aq)

Ka 7.9 10



Salt pH of (aq) solution

CaCl2

NH4Br

NH4F

KNO3

KHCO3

Acid–Base Properties of Salts: Practice

Acid–Base Properties of Salts: Practice

Salt pH of (aq) solution

CaCl2 Neutral

NH4Br Acidic

NH4F Basic

KNO3 Neutral

KHCO3 Basic

• According to the Brønsted–Lowry theory, all acid–

base reactions can be written as equilibria involving

the acid and base and their conjugates.

• All proton transfer reactions proceed from the

stronger acid and base to the weaker acid and

base.

Acid Base Conjugate base of the acid + Conjugate acid of the base

Predicting the Direction of Acid–Base Reactions

• When a weak acid is in solution, the products are a

stronger conjugate acid and base. Therefore

equilibrium lies to the left.

• All proton transfer reactions proceed from the

stronger acid and base to the weaker acid and

base.


• Will the following acid/base reaction occur spontaneously?

3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)


• Will the following acid/base reaction occur spontaneously?

Ka = 7.5 105 Ka = 1.8 105

Kb = 5.6 1010 Kb = 1.3 1012




• Will the following acid/base reaction occur

spontaneously?

• Equilibrium lies to the right since all proton

transfer reactions proceed from the stronger

acid and base to the weaker acid and base.

Ka = 7.5 105 Ka = 1.8 105

Kb = 5.6 1010 Kb = 1.3 1012

Stronger Acid + Stronger Base Weaker Base + Weaker Acid



Strong acid (HCl) + Strong base (NaOH)

Net ionic equation

Mixing equal molar quantities of a strong acid

and strong base produces a neutral solution.

3HCl (aq) NaOH (aq) H O (aq) NaCl (aq)

3 2H O (aq) OH (aq) 2H O(l)

Types Acids–Base Reactions

Weak acid (HCN) + Strong base (NaOH)

Mixing equal amounts (moles) of a strong base

and a weak acid produces a salt whose anion

is the conjugate base of the weak acid. The

solution is basic, with the pH depending on Kb

for the anion.

2HCN (aq) + OH (aq) CN (aq) + H O (l)

2CN (aq) + H O (l) HCN (aq) + OH (aq)


Strong acid (HCl) + Weak base (NH3)

Mixing equal amounts (moles) of a weak base

and a strong acid produces a conjugate acid of

the weak base. The solution is basic, with the

pH depending on Ka for the acid.

3 3 2 4H O (aq) + NH (aq) H O (l) + NH (aq)

4 2 3 3NH (aq) + H O (l) H O (aq) + NH (aq)


Weak acid (CH3CO2H) + Weak base (NH3)

Mixing equal amounts (moles) of a weak acid

and a weak base produces a salt whose cation

is the conjugate acid of the weak base and

whose anion is the conjugate base of the weak

acid. The solution pH depends on the relative

Ka and Kb values.

3 2 3 3 2 4CH CO H (aq) + NH (aq) CH CO + NH (aq)


Weak acid + Weak base

• Product cation = conjugate acid of weak base.

• Product anion = conjugate base of weak acid.

• pH of solution depends on relative strengths of

cation and anion.



Types Acids–Base Reactions Summary

Determining K from Initial Concentrations and

pH

2 2 3 2

2

HNO (aq) H O(l) H O (aq) NO (aq)

0.10 M HNO (aq) pH = 2.17

+ -+ +

[H2S] [H3O+] [HS]

Initial 0.10

Change

Equilibrium

Calculations with Equilibrium Constants


pH

2 2 3 2

2


0.10 M HNO (aq) pH = 2.17

+ -+ +

[H2S] [H3O+] [HS]

Initial 0.10 0 0

Change 0.10 - x + x + x

Equilibrium 0.10 - x x x



pH

2 2 3 2

2


0.10 M HNO (aq) pH = 2.17

+ -+ +

[H2S] [H3O+] [HS]

Initial 0.10 0 0

Change 0.10 - x + x + x


23 2

a

2

H O NO x x xK

HNO 0.10 x 0.10 x



pH 2 2 3 2

2


0.10 M HNO (aq) pH = 2.17

+ -+ +

[H3O+] = [NO2

] = 6.76 103

23

4

a 3

6.76 10K 4.9 10

0.10 6.76 10


23 2

a

2

H O NO x x xK

HNO 0.10 x 0.10 x


pH

2 2 3

7

2 a


1.00 M H S(aq) K 1.0 10

+ -

-

+ +

= ´

[H2S] [H3O+] [HS]

Initial 1.00

Change

Equilibrium



[H2S] [H3O+] [HS]

Initial 1.00 0 0

Change - x + x + x


2 2 3

7

2 a


1.00 M H S(aq) K 1.0 10

+ -

-

+ +

= ´

23 7

a

2

H O HS x x xK 1.0 10

H S 1.00 x 1.00 x



pH


pH

27

Since x << 1.00, the equation reduces to:

x1.0 10

1.00

x = 3.2 104 pH = 3.50

23 7

a

2

H O HS x x xK 1.0 10

H S 1.00 x 1.00 x



pH

In general, the approximation that

[HA]equilibrium = [HA]initial x [HA]initial

is valid whenever [HA]initial is greater than or

equal to 100 Ka.

If this is not the case, the quadratic equation

must by used.


Determining pH after an acid/base reaction:

Calculate the hydronium ion concentration and

pH of the solution that results when 22.0 mL of

0.15 M acetic acid, CH3CO2H, is mixed with 22.0

mL of 0.15 M NaOH.



Determining pH after an acid/base reaction:

Calculate the hydronium ion concentration and

pH of the solution that results when 22.0 mL of

0.15 M acetic acid, CH3CO2H, is mixed with 22.0

mL of 0.15 M NaOH.

Solution: From the volume and concentration of

each solution, the moles of acid and base can be

calculated. Knowing the moles after the reaction

and the equilibrium constants, the concentration

of H3O+ and pH can be calculated.

Calculate the hydronium ion concentration and pH of the

solution that results when 22.0 mL of 0.15 M acetic acid,

CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.

All of the acetic acid is converted to acetate ion.

3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)

3 23

1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH

10 mL 1 L





All of the acetic acid is converted to acetate ion.

3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)

3 23

1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH

10 mL 1 L

3

3 2

0.0033 mols 10 mLCH CO 0.075 M

44.0 mL 1 L




3

3 2

0.0033 mols 10 mLCH CO 0.075 M

44.0 mL 1 L

-é ù = ´ =ë û

[CH3CO2-] [CH3CO2H] [OH-]

Initial 0.075 0 0

Change

Equilibrium

3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)




3

3 2

0.0033 mols 10 mLCH CO 0.075 M

44.0 mL 1 L

-é ù = ´ =ë û

3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)


Initial 0.075 0 0

Change - x + x + x






Initial 0.075 0 0

Change - x + x + x


[ ] 2 143 2 10w

b 5

a3 2

CH CO H OH Kx 1.00 10K 5.6 10

0.075 x K 1.8 10CH CO

- --

--

é ù ´ë û = = = = = ´- ´é ùë û




Since Kb100 > [CH3CO2-]initial, the quadratic equation

is not needed.


Initial 0.075 0 0

Change - x + x + x


[ ] 2 143 2 10w

b 5

a3 2

CH CO H OH Kx 1.00 10K 5.6 10

0.075 x K 1.8 10CH CO

- --

--

é ù ´ë û = = = = = ´- ´é ùë û




210

6

6

pH 9

3

x5.6 10

0.075

x OH 6.4 10

pOH log(6.4 10 ) 5.19

pH 14 pOH 8.81 H O 10 1.5 10 M

-

- -

-

+ - -

= ´

é ù= = ´ë û

= - ´ =

é ù= - = = = ´ë û


• Because polyprotic acids are capable of donating

more than one proton they present us with

additional challenges when predicting the pH of

their solutions.

• For many inorganic polyprotic acids, the ionization

constant for each successive loss of a proton is

about 104 to 106 smaller than the previous step.

• This implies that the pH of many inorganic

polyprotic acids depends primarily on the hydronium

ion generated in the first ionization step.

• The hydronium ion produced in the second step can

be neglected.

Polyprotic Acuids & Bases

Sulfurous acid, H2SO3, is a weak acid capable of providing two

H+ ions.

(a) What is the pH of a 0.45 M solution of H2SO3?

(b) What is the equilibrium concentration of the sulfite ion,

SO32- in the 0.45 M solution of H2SO3?

Polyprotic Acids & Bases


H+ ions.




2–2 3 3

2 3

[HSO ][H O ] x1.2 10

[H SO ] 0.45 – x



H+ ions.




Since 100 Ka is not << 0.45M, the quadratic equation must

be used

2–2 3 3

2 3

[HSO ][H O ] x1.2 10

[H SO ] 0.45 – x




H+ ions.




(a)

x = [H3O+] = 0.0677 M

pH = 1.17


H+ ions.




(a) in part a we found that x = [H3O+] = 0.0677 M

(b)

[ ]

[ ]

2

3 2 3 3

2 2

3 3 38

a2

3

2 8

3 a2

HSO (aq) H O(l) SO (aq) H O (aq)

SO H O SO 0.0677K 6.2 10

0.0677HSO

SO K 6.2 10 M

- - +

- + -

-

-

- -

+ +

é ù é ù é ùë û ë û ë û= ´ = =é ùë û

é ù = = ´ë û



Halide Acid Strengths

• Experiments show that the acid strength increases in the

order: HF << HCl < HBr < HI.

• Stronger acids result when the HX bond is readily broken

(as signaled by a smaller, positive value of H for bond

dissociation) and a more negative value for the electron

attachment enthalpy of X.

Molecular Structure, Bonding, & Acid–Base Behavior

Comparing Oxoacids: HNO2 and HNO3

• In all the series of related oxoacid compounds, the

acid strength increases as the number of oxygen

atoms bonded to the central element increases.

• Thus, nitric acid (HNO3) is a stronger acid than

nitrous acid (HNO2).


Why Are Carboxylic Acids Brønsted Acids?

• There is a large class of organic acids, all like acetic

acid (CH3CO2H) have the carboxylic acid group,

CO2H

• They are collectively called carboxylic acids.



• The carboxylate anion is stabilized by

resonance.



• The acidity of carboxylic acids is enhanced if

electronegative substituents replace the hydrogen

atoms in the alkyl (–CH3 or –C2H5) groups.

• Compare, for example, the pKa values of a series of

acetic acids in which hydrogen is replaced sequentially

by the more electronegative element chlorine.


• Trichloroacetic acid is a much stronger acid

owing to the high electronegativity of Cl.

• Cl withdraws electrons from the rest of the

molecule.

• This makes the O—H bond highly polar. The H

of O—H is very positive.

Acetic acid Trichloroacetic acid

Ka = 1.8 x 10-5 Ka = 0.3


• The concept of acid–base behavior advanced by

Brønsted and Lowry in the 1920’s works well for

reactions involving proton transfer.

• However, a more general acid– base concept,

was developed by Gilbert N. Lewis in the

1930’s. • A Lewis acid is a substance that can accept a

pair of electrons from another atom to form a

new bond.

• A Lewis base is a substance that can donate a

pair of electrons to another atom to form a new

bond.

The Lewis Concept of Acids & Bases

A + B: BA

Acid Base Adduct

• The product is often called an acid–base adduct.

In Section 8.3, this type of chemical bond was

called a coordinate covalent bond.

• Lewis acid-base reactions are very common. In

general, they involve Lewis acids that are cations or

neutral molecules with an available, empty valence

orbital and bases that are anions or neutral

molecules with a lone electron pair.


Lewis acid

a substance that accepts an electron pair

Lewis base

a substance that donates

an electron pair


• New bond formed

using electron pair

from the Lewis base.

• Coordinate covalent

bond

• Notice geometry

change on reaction.

Reaction of a Lewis Acid & Lewis Base

The formation of a hydronium ion is an example of a Lewis acid / base reaction

H H

H

B A S E

• • • • • •

O — H O — H

H +

A C I D

The H+ is an electron pair acceptor.

Water with it’s lone pairs is a Lewis acid

donor.


Lewis AcidBase Reactions


Metal cations often act as Lewis acids because of open d-orbitals.

Lewis Acids & Bases

The combination of metal ions (Lewis acids) with

Lewis bases such as H2O and NH3 leads to

Coordinate Complex ions.

Lewis Acids & Bases

Aqueous solutions of Fe3+, Al3+, Cu2+, Pb2+,

etc. are acidic through hydrolysis.

This interaction weakens this bond

Another H2O pulls

this H away as H+

[Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O

+(aq)

Lewis Acids & Bases

• Because oxygen is more electronegative than C,

the CO bonding electrons in CO2 are polarized

away from carbon and toward oxygen.

• This causes the carbon atom to be slightly positive,

and it is this atom that the negatively charged Lewis

base OH can attack to give, ultimately, the

bicarbonate ion.

Molecular Lewis Acids

• Ammonia is the parent compound of an enormous

number of compounds that behave as Lewis and

Brønsted bases. These molecules all have an

electronegative N atom with a partial negative

charge surrounded by three bonds and a lone pair

of electrons.

• This partially negative N atom can extract a proton

from water.

Molecular Lewis Acids

Many complex ions containing water undergo

HYDROLYSIS to give acidic solutions.

2

2 4 2 2 3 3[Cu(H O) ] (aq) H O(l) [Cu(H O) (OH)] (aq) H O (aq)+ + ++ +

2+ + +

Lewis Acids & Bases


Reaction of NH3 with Cu2+(aq)

• The heme group in

hemoglobin can interact

with O2 and CO.

• The Fe ion in

hemoglobin is a Lewis

acid

• O2 and CO can act as

Lewis bases

Heme group

Lewis Acid–Base Interactions in Biology

Slides from Chang Book

99

pH – A Measure of Acidity

pH = -log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7

pH < 7

pH > 7

At 250C

pH [H+]

100

percent ionization =

Ionized acid concentration at equilibrium

Initial concentration of acid x 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0 x 100% [HA]0 = initial concentration

% Ionization =

101

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

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

- (aq)

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M 0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M 0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6 102

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka = [H+][F-]

[HF] = 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka = x2

0.50 - x = 7.1 x 10-4

Ka x2

0.50 = 7.1 x 10-4

0.50 – x 0.50 Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M


103

When can I use the approximation?

0.50 – x 0.50 Ka << 1

When x is less than 5% of the value from which it is

subtracted. x = 0.019

0.019 M 0.50 M

x 100% = 3.8% Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka x2

0.05 = 7.1 x 10-4 x = 0.006 M

0.006 M 0.05 M

x 100% = 12% More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or

method of successive approximations.

Ka *100 << [HA]0

104

What is the pH of a 0.122 M monoprotic acid whose

Ka is 5.7 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka = x2

0.122 - x = 5.7 x 10-4

Ka x2

0.122 = 5.7 x 10-4

0.122 – x 0.122 Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M 0.122 M

x 100% = 6.8% More than 5%

Approximation not ok.

105

Ka = x2

0.122 - x = 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0 -b ± b2 – 4ac

2a x =

x = 0.0081 x = - 0.0081

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

106

Acid-Base Properties of Salts

Neutral Solutions:

Salts containing an alkali metal or alkaline earth

metal ion (except Be2+) and the conjugate base

of a strong acid (e.g. Cl-, Br-, and NO3-).

NaCl (s) Na+ (aq) + Cl- (aq) H2O

Basic Solutions:

Salts derived from a strong base and a weak

acid. NaCH3COOH (s) Na+ (aq) + CH3COO- (aq)

H2O

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

107


Acid Solutions:

Salts derived from a strong acid and a weak

base.

NH4Cl (s) NH4+ (aq) + Cl- (aq)

H2O

NH4+ (aq) NH3 (aq) + H+ (aq)

Salts with small, highly charged metal cations

(e.g. Al3+, Cr3+, and Be2+) and the conjugate

base of a strong acid. Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)

3+ 2+

108


Solutions in which both the cation and the anion hydrolyze:

• Kb for the anion > Ka for the cation, solution will be basic

• Kb for the anion < Ka for the cation, solution will be acidic

• Kb for the anion Ka for the cation, solution will be neutral


109

What is the molarity of an NH4NO3(aq) solution that has

a pH = 4.80?

Strategy

Ammonium nitrate is the salt of a strong acid (HNO3) and a weak base

(NH3). In NH4NO3(aq), NH4+ hydrolyzes and NO3

– does not. The ICE format

must be based on the hydrolysis equilibrium for NH4+(aq). In that format

[H3O+], derived from the pH, will be a known quantity, and the initial

concentration of NH4+ will be the unknown.

Solution

We begin by writing the equation for the hydrolysis equilibrium and the

equation for Ka in terms of Kw and Kb.

As usual, we can calculate [H3O+] from the pH of the solution.

log[H3O+] = –pH = –4.80

[H3O+] = 10–4.80 = 1.6 x 10–5 M

110

Example 15.15 continued

Solution continued

If we assume that all the hydronium ion comes from the hydrolysis reaction,

we can set up an ICE format in which x represents the unknown initial

concentration of NH4+.

We now substitute equilibrium concentrations into the ionization constant

expression for the hydrolysis reaction.

We can assume that the ammonium ion is mostly nonhydrolyzed and that

the change in [NH4+] is much smaller than the initial [NH4

+], so that 1.6 x 10–

5 << x and we can replace (x – 1.6 x 10–5) by x. Then we can solve for x.

The solution is 0.46 M NH4NO3.

The Chemistry of Acids and Bases - sunysuffolk.edu · The Chemistry of Acids and Bases ... aspects...

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