Biermann Ch. Handbook of Pulping and Papermaking

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314 13. INTRODUCTORY CHEMISTRY REVffiW

chlorine to give NaCl. There is a complication

that 3d orbitals can be filled with bonding elec-

trons so that an element like sulfur commonly has

two valence {sharing or combining) electrons but

can also have four or six valence electrons; indeed

in the most stable compounds of sulfur, the sulfur

atoms share six electrons as in the case of H2SO4.

This,

of course, also happens in the higher rows.

Fig. 13-1 shows the valence electrons for some

atomic, ionic, and molecular structures; the next

section helps one predict the types of bonds

formed as shown in this figure. Depiction of

electrons in this fashion is credited to G.N. Lewis.

13.2 IONIC AND COVALENT BONDS

The elements of the lower left of

the

periodic

table donate electrons the most easily and are the

electropositive elements. The metallic elements

(element to the lower right and including the

elements of the diagonal formed by Al, Ge, Sb,

and Po) donate electrons easily. When electrons

are donated, the atom remains as a cation such as

Na"^, Ca^"^,

AP"^,

and so on. The elements in the

upper right, not including the inert gases, attract

electrons with high affinity and are the electroneg-

ative elements, including (in order of decreasing

electronegativity) F, O, N, CI, Br, Se, S, and I.

The periodic table of the elements gives values for

electronegativity of the elements (the values are

listed in the lower right of each element's box).

When electrons are accepted, the atom becomes an

anion, such as CI', I", S^', and so on. Chemical

bonds cover a spectrum of equally sharing elec-

trons to forming ion pairs or ionic bonds. Bonds

between the electronegative elements and the

electropositive ions are highly ionic in

nature;

such

compounds when dissolved in water will conduct

an electric current and are described as strong

electrolytes. Some examples of these are NaCl,

CaBrj,

and Na-OH. Bonds between elements of

intermediate electronegativity share the electrons,

forming covalent bonds, and do not disassociate

into individual ions and will not conduct electricity

in water

(nonelectrolytes)

or are weak conductors

(weak electrolytes). Some examples are CH4,

CH3OH, H2S, and CO2. Generally, if the differ-

ence in electronegativity of two elements sharing

electrons is 1.7 or greater, the bond formed

Valence electrons of atomic structures

H • • C •

0 : : CI :

Valence electrons of nnolecular structures

2 H' ~* H:H = H-H

2H'+:0:

-^ H:0:H = H-O-H

Valence electrons of ionic compounds

No-

4- :CI : —^ Na"^ + :CI :""

Fig. 13-1. Lewis structures showing the valence

electrons of various structures.

between the two elements behaves in a highly

ionic manner. If the difference is less than 1.0,

the bond behaves in a covalent manner. The

actual amount of ionization in a chemical contain-

ing intermediate bonds depends on many things,

such as the presence and type of solvent.

There are several particularly important

polyatomic ions encountered. For example, the

ammonium cation NH4^ behaves very similarly to

the Na^ ion, except the ammonium ion decom-

poses at pH above 9 and is subject to combustion.

Many polyatomic anions with several atoms of

oxygen have delocalized (that is, spread out over

two or more atoms) negative charges, which

increase the stability of the oxygen anion such as

in the cases of

RCOj-,

COa^", NO{, NO2", SO42-,

and

SOj^'.

Resonance structures are Lewis struc-

tures which differ only in the position of a double

bond and other electrons. The actual structure is

a resonance

hybrid,

which is an average of all of

the resonance structures. Fig. 13-2 shows how

two resonance structures contribute to the carbox-

ylate anion. Experiments indicate that the two C-

O bonds are identical and intermediate in prop-

erties between C-0 and C=0 bonds.

Fig. 13-2. Resonance structures for the carbox-

ylate anion.

IONIC AND COVALENT BONDS 315

Table 13-2. Approximate bond energies for

interatomic bonds.

109-5"

Fig. 13-3. Tetrahedron formed by CH4.

It is useful to consider the strength of chemi-

cal bonds. The strength of chemical bonds is

measured as the enthalpy (energy) required to

break the bond between two elements. It is

reported in imits of energy per mole. For exam-

ple,

one mole of

CI2

gas (70.9 g) can be separated

(hypothetically) into two moles of chlorine radicals

CI-

by the addition of

242

kJ of energy as follows:

CI2 + energy -* 2 CI- . Other types of chemical

bonds occur that are weaker than these primary

bonds. These will be discussed in other sections.

Table 13-2 gives some average values for bond

energies. This table shows that the shorter the

bond length, the more stable it is (that is, the more

energy it takes to break it).

Electrostatic repulsion of electrons means the

valence electron pairs are more stable when

farther apart from each other. This causes the

bonds and lone electron pairs of an octet to,

approximately, form the comers of a tetrahedron.

Fig. 13-3 shows the tetrahedral shape of the four

carbon-hydrogen bonds of methane.

The tetrahedron form is characteristic of

elements that complete an octet with just eight

electrons. In the case of the oxygen atom of

water, two pairs of lone electrons occupy two

corners of the octet. Since lone electron pairs

have high repulsion, the H-O-H bond angle of

water is only 104.5°. While the carbon backbone

of materials like hexane is often drawn as C-C-C-

C-C-C, it is more closely approximated as:

13.3 HYDROGEN BONDING

Hydrogen bonding is one example of a sec-

ondary bond and is one of the stronger secondary

bonds. Secondary bonds are much weaker than

Bond

type

1 ci-ci

H-"

H-ci

C-H

C-C

^=^

C=C

1 benzene

C-0

|c=o

^-N

0-H

0-0

N-H

Bond energy

kJ/mol

242.

436.

431.

420

350

620

518

360

740

305

463

220

430

kcal/mol

57.8

104.

103.

100

150

124

175

110

105

Bond

length, nm

0.2

0.074

0.135

0.11

0.154

0.13

0.139

0.14

0.12

0.15

010 1

0.15

0.10

the primary bonds, being on the order of 1-10

kcal/mol compared to 50-200 kcal/mol for the

primary covalent chemical bonds. Hydrogen

bonds are on the order of 4-5 kcal/mol, although

hydrogen bonding of carboxylic acids to each

other is about 13 kcal/mol. Many hydrogen

bonds, however, can add up to a strong structure;

most papers are held together without adhesives,

relying only on hydrogen bonding to hold the

fibers together. Hydrogen bonding occurs when

hydrogen that is bonded to one of the four highly

electronegative elements (F, O, N, or CI) comes

near a second highly electronegative element.

Partial charges develop between the hydrogen and

electronegative atom since the bond is fairly ionic.

The partial positive charge on the side of the

hydrogen atom away from the shared electrons is

attracted to the partial negative charge of the

unshared electron pairs of the second electronega-

tive element, as shown in the following diagram

where 6 designates a partial charge:

316 13. INTRODUCTORY CHEMISTRY REVIEW

6+6-

R-0:H :OR

For water, the difference in electronegativities is

1.4 units, corresponding to a bond that is about

40%

ionic. While the length of the covalent bonds

of water is 0.1 nm, the length of a hydrogen bond

in water is almost 0.2 nm.

EXAMPLE 1. Generally the boiling points of

substances increase with increasing formula

weights for a related series. The boiling

points of the hydrides of the elements of row

6 of the periodic table are as follows (note

the temperatures are given in Kelvin):

Element

Hydride

b.p.,

Oxygen

Sulfur

Selenium

Tellurium

HjO

HjS

HjSe

HjTe

?????

212.5

231.6

271.0

From this, predict the boiling point of water,

H2O.

What factor should be considered to

explain its unusual boiling point?

SOLUTION. From the data, one might predict

a boiling point around 200 K for water. In

fact the boiling point is 373 K, due to rela-

tively strong hydrogen bonding. This means

that a water molecule has a high affinity for

the liquid, thereby decreasing its vapor pres-

sure and increasing the boiling point.

EXAMPLE 2. One mole of acetic acid at 273 K

occupying 22.4 liters should have a pressure

of LOO atmosphere according to ideal gas

theory. In fact, there is a huge deviation and

the pressure is about 0.60 atmospheres.

Why? One mole of water under the same

conditions has a pressure of about 0.90

atmosphere. Why? What does this say

about the magnitude of hydrogen bonding in

hydroxyl groups relative to that of carboxylic

acids? Could this help explain why hemi-

celluloses increase the strength of papers?

SOLUTION. The strong hydrogen bonding of

carboxylate groups (about 13 kcal/mol)

means that most of the acetic acid under

these conditions will exist as dimers. The

relatively weak hydrogen bonding of water

(about 5 kcal/mol) means that a significant,

but smaller portion than acetic acid, of water

will exist as dimers. The carboxylic acid

groups of hemicellulose could form strong

hydrogen bonds (with other carboxylate

groups or with hydroxyl groups). Carboxyl-

ation of fibers or addition of carboxymethyl

cellulose to fibers is known to increase the

strength of paper made from these fibers.

PROBLEM: Ethanol (CH3CH2OH) has a boiling

point of 79°C; its isomer, dimethyl ether

(CH3OCH3) has a boiling point of

-25

°C.

Explain this phenomenon with the use of a

diagram showing partial charges.

13.4 THE MOLE AND MASS PERCENTAGE

Atoms almost always react with each other in

whole number ratios to form chemical compounds.

For instance carbon reacts with oxygen gas to give

carbon dioxide during combustion of charcoal.

C +

^ CO2

It is very useful, therefore, to have a measure of

atoms and molecules that reflects the number of

particles. (Of course, we usually deal with

weights when actually handling chemicals.) Since

the gram is the basis of the metric system, as well

as a convenient amount of material to handle in

the laboratory, it is used as the basis of the amu,

which is the scale used to identify the mass of

individual atoms and molecules. By using the

weight of an amu, we can calculate the number of

amu per g; this number, 6.0220-10^^ is called the

Avogadro constant and is denoted as

N^^,

lamu = L6606-10-2^g;

therefore, 1 g = 6.0220-102^ amu

This number of objects is known as the mole.

The concept of the mole is extremely important

because chemicals react in whole number relation-

ships when compared on a molar basis. Since

THE MOLE AND MASS PERCENTAGE 317

carbon is 12.01 amu, one mole of carbon is 12.01

g of carbon. This is often called the gram atomic

weight since it is the atomic weight of a substance

in grams. Similarly, one mole of carbon dioxide

(CO2) is 12.01 + (2 X 16.00) = 44.01 grams. In

the case of molecules with covalent bonds we

speak of gram molecular weights, and in the case

of compounds with ionic bonds we speak of gram

formula weights. The gram molecular weight of

CO2 is 44.01 grams, while the gram formula

weight of NaCl is 58.44 (from 22.99 + 35.45).

It is not so important to remember atomic versus

molecular versus formula weights as it is to under-

stand the concept of the mole, because one can

speak of 1 mole of carbon, carbon dioxide, and

sodium chloride.

The mass percentage of a substance is a

convenient method of expressing its composition.

Each component (such as "A") of the whole may

be expressed as a mass percentage as follows:

(13-1)

mass%A= '"^ of

in sample ^^^^^

total mass of sample

EXAMPLE 3. Calculate the formula weight of

Al2(S04)3 and the mass percentages of the

individual elements.

SOLUTION: This compound consists of two

aluminum cations and three sulfate anions.

The formula weight of this material is calcu-

lated as follows:

2 Al is 2 X 27 = 54

3 S is 3 X 32 = 96

12 O is 12 X 16 = 192

342 = formula weight

%A1 = 54/342 X 100% = 15.8%

%S = 96/342 X 100% = 28.1%

%0 = 192/342 X 100%

=56.1%

PROBLEM: Calculate the molecular weight for

glucose and the mass percentage of carbon in

glucose. The formula for glucose is

C6H12O6. Answers: 180 g/mol and 40.0%

carbon.

EXAMPLE 4. Given the (quantitative) chemi-

cal equation below, which is the conversion

of the kraft makeup chemical sodium sulfate

(Na2S04) into the active cooking species

sodium sulfide (NajS) in the recovery boiler,

how many grams of NajS will be produced

from 100 grams of Na2S04?

2C + Na2S04 -* Na2S + 2CO2

SOLUTION: The formula weight of sodium

sulfate is 142 g/mole and that of sodium

sulfide is 78 g/mole.

1 mol Na^SO. 1 mol Na,S

lOOgNa^SO.

-^—^

-?—

^ ^ * 142 g Na^SO^ 1 mol Na2S04

78 g Na2S

1 mol Na^S

54.9gNa2S

PROBLEM: In this example, how many grams

of carbon were reacted? Answer: 16.9 g C.

13.5 EQUIVALENCY, MOLARITY AND

NORMALITY

Molarity

(A/)

is a measure of concentration in

terms of the mole and is moles of a substance per

liter (mol/L) of solution. One M solution of NaCl

would be 58.44 g/L of NaCl. An equivalent is

one mole of reactive species, whether it be an H^

of acid-base reactions, an electron of reduction-

oxidation reactions, or some other unit. In acid-

base reactions the basis of the equivalent is one

mole of hydrogen ions, H"^. For a monoprotic

acid,

which has one ionizable proton, such as

HCl, one equivalent is the same as one mole. For

a diprotic

acid,

which has two ionizable protons,

such as H2SO4 ?^ 2 H+ + S04^-, the equivalent

weight is one-half the molecular weight. One

mole of

H2SO4

is 98 g; one equivalent of

H2SO4

49 g. In electron-transfer (oxidation/reduction)

reactions the electron is the basis of the mole.

Potassium permanganate (KMn04) transfers five

electrons in most reactions where it is used. The

molecular weight of KMn04 is 158.0, and the

equivalent weight is 31.6 g/eq. Normality is

analogous to molarity and is concentration ex-

pressed as equivalents per liter. For example,

318

13.

INTRODUCTORY CHEMISTRY REVIEW

H2SO4 is 49 g/L of H2SO4 in an acid-base reac-

tion. In titrations, the volume is measured in ml.

Note that 1 eq/L = 1 meq/ml and 1 mol/L = 1

mmol/ml; these are useful relationships for

titrations.

EXAMPLE 5. If 17 grams of sodium chloride is

dissolved in water to make 600 ml of solu-

tion, what is the molarity of this solution?

SOLUTION: The formula weight of NaCl is

58.44.

Start with 17 grams NaCl per 0.6 L

and convert this to mol/L:

17 g NaCl 1 mole NaCl

0 X

0.6

58.44 g NaCl

0.485 mol/^ NaCl

0.485 M NaCl

PROBLEM: 2.0 g HCl is dissolved in water to

make 1 L of solution. What is the normality

of this solution? Same question with sulfuric

acid? Answers: 0.0549 iVHCl and 0.0408 iV

H2SO4.

13.6 ACIDS, BASES, AND THE pH SCALE

For the purposes of pulp and paper science it

is useful to use the Bronsted-Lowry theory of acids

and bases. According to this theory, an acid is a

compound that is a proton

(H"*^)

donor and a base

is a compound that is a proton acceptor. There

must be a proton acceptor to accept a donated

proton since protons are not stable in an uncom-

bined form. Thus, an acidic solution has a rela-

tively high concentration of protons (H"^ ions) in

hydrated form, since water is the proton acceptor.

The hydrated proton is known as the hydronium

ion,

HjO^. H^ and HsO"*" are often used inter-

changeably, although H"^ does not exist in water.

Alkaline or basic solutions have very low

concentrations of protons which, in water, neces-

sarily means a high concentration of

OH"

ion (to

be shown in Eq. 13-4). The empirical formula of

acids are often written with leading protons. For

example, CH3CH2OH, C2H4, and C3H8 are not

acids,

whereas HjS, H2SO3, and HC2H3O2 are

acids.

There are many exceptions to this rule.

For example, HC2H3O2 is acetic acid and is often

written as CH3COOH to show that it is a carbox-

ylic acid.

Another rule of

thumb

is that the elements H,

Al,

Ga, Sn, and Pb form oxides that are ampho-

teric, that is, compounds that are both acidic and

basic.

The first two of these elements are particu-

larly relevant to pulp and paper. Elements (not

including the noble gases) to the upper left of

these on the periodic table of the elements form

acidic oxides (for example, SO2 forms the acid

H2SO3

in water), while those elements to the lower

right of these form basic oxides [for example,

CaO forms the base Ca(0H)2 in water].

Examples of proton donors are HCl, which

becomes CI"; HNO3, which becomes NO3"; and

RCOOH, which becomes RCOO". Examples of

proton acceptors other than water are OH" from

bases such as NaOH, which becomes water since

H+ + OH ^ H2O; carboxylate anion RCOO",

which becomes RCOOH; and NH3, which be-

comes

NH4'*'.

A proton donor and the correspond-

ing base formed after the donation of a proton are

called

conjugate

pairs. The compound left after

donating a proton is called a conjugate base since,

in principle, it should be able to accept a proton.

A compound like HCl is highly ionic and com-

pletely dissociates in water to form H"^ (as Yi^O^)

and CI" ions; consequently, it is a very good

hydrogen donor

{strong

acid). On the other hand,

the conjugate base, CI", would not be expected to

have a high affinity for protons. Therefore, the

conjugate

base of a strong acid is a weak base. In

the same manner the conjugate acid of a strong

base is a weak

acid.

For this reason Na"*^ is a

very weak acid because its conjugate base (NaOH)

is a very strong base. The group lA ions of Li to

Fr are very poor bases and act as spectator

ions

acid-base reactions, that is, they are ions not

involved in a reaction.

In the case of acetic acid, CH3COOH, there

is partial ionization to form the acetate and proton

ions in solution. This is termed a weak

acid.

Table 13-3 lists all of the commonly encountered

strong acids, which are completely ionized in

water, and some weak acids with their ionization

constants. The acid

ionization

constant,

K^, is the

equilibrium constant for the degree of ionization.

ACIDS BASES AND THE pH SCALE 319

Table 13-3. Strong and weak electrolytes in water at 25^C (77^F).

Substance

Strong

acids

HCl

H2SO4

HNO3

HCIO4

Weak

acids of general interest

Abietic acid

HC2H3O2 (acetic acid)

NH/

C^HjOH

(phenol)

H3PO4 (phosphoric acid)

CF3COOH (trifluoroacetic

K,=

acid)

1000+

10*

1000

2.40-10-«

1.75-10'

5.75-lO"

l.MO-'"

0.011

4.8-10"

1.698

Weak

acids of

significance

pulping

H2CO3

H2SO3

(at 140°C)

HjS

H2S4

polysnlfide

Weak

acids of

significance

HOC!

HCIO2

H,0,

4.3-10-'

1.3-10-^

8.010^

9.1-10-'

1.610^

bleaching

l.MO-«

•

0.011

2.2-10''

P^i

Below -3

-6

-3

-1.32

7.62

4.76

9.24

9.96

1.96

p^3 = 12.32

0.23

6.37

1.89

3.10

7.04

3.8

7.96'

1.95

11.65

0.012

7.5-10-«

4.8-10"

2-10'

1-10" '

5.0-10-'

VKi

1.92

7.12

10.32

6.7

17'

6.3

Strong

bases

LiOH

NaOH

Strong

electrolytes

All salts of group lA and HA elements

Sizeable variation was found for these

values.

Older sources report p^a of 12 for NajS, another around

15,

but the most recent work indicates it is at least 17. CRC's Handbook of Chemistry and Physics

reports a value of 4.53 for HOCl, although Cotton and Wilkinson give 7.50.

320 13. INTRODUCTORY CHEMISTRY REVIEW

In the case of polyprotic acids, they are listed as

ATj,

K2,

K^,,..,

There may be considerable varia-

tion in the reported values of some acid dissocia-

tion

constants!

Since the concentration of

water

not included in the definition of the equilibrium

constant (since it is approximately constant in

dilute solutions), using A' as a general term for the

conjugate base (such as Ac for CH3COO", from

acetic acid, HAc) gives the following relationship:

(13-2)

H^O^-HjO^+A"

K^ =

[H301[A-]

[HA]

Analogously, the base ionization

constant,

K^, is

the equilibrium constant for the production of OH",

such as from

NH3

(abbreviated as B) to give

NH4'^

(abbreviated as B"*") as shown in the following

equation:

(13-3)

B+HLjO^-BH^+OH-

^ _ [OH-][BH1

' [B]

A base can begin as a negative ion such as HSOj",

in which case BH"^ is H2SO3. These ionization

constants are examples of the law of mass action,

which is explained in more detail in the following

section. Strictly speaking, the

effective

concentra-

tion of ionic species is lower than the actual

concentration. The ratio of effective concentration

to the actual concentration is known as the activity

coefficient of the ion. Generally, as the ionic

strength of

solution increases, the activity coeffi-

cient of a given species decreases from unity.

Most pH calculations assume activity coefficients

unity.

The activity is the

effective concentration

ionic species. All equilibrium constants are

correctly stated only in terms of activity. Also,

pH meters and indicators actually measure hydro-

gen ion activity. Fortunately, in dilute solutions

the activity coefficients are approximately equal to

unity, and one can use concentration directly in-

stead of activities to simplify the calculations.

Even concentrated solutions are analyzed (such as

by titration) under relatively dilute conditions, al-

though in pulp and paper mills the pH of concen-

trated solutions is sometimes measured.

Water is both an acid and a base since it can

accept a proton to form Rf>^ or donate a proton

to become OH". The ionization constant for

water, K^, is 1 x 10"^^ at 25°C; p^ is 14.00.

Substituting water as the acid into Eq. (13-2), and

remembering that water as the solvent is not

included in the definition of an equilibriimi con-

stant, one obtains:

[H3O+] [OH] = 10-^^ = K^ (25°C) (13-4)

Lange's Handbook gives the p^ for water as

14.94 at 0\ 13.26 at 50°, 12.26 at 100°, and

11.91 at 130°C, while CRCs Handbook gives

11.64 at 150°C and 11.29 at 200°C.

This means that for pure water at 25°C, the

[H3O+] = [OH] = 10-^. The concentration of

H30"^

in water can vary from above 10^ to less

than

10"^^

molar. Because the acidity can vary by

over 16 orders of magnitude, it becomes conve-

nient to use a logarithmic scale to describe how

acidic or basic a solution is! The pH scale was

devised for that purpose. The pH is the negative

value

of the

base-10 logarithm

of the

concentration

hydrogen

ions. The pH of pure water is 7, the

pH of

0.1

(10-^

M) HCl is 1, and the pH of

10"

M NaOH is 13. By definition the pH scale goes

from 0 to 14, but it is sometimes useful to go

beyond a pH of 0 or 14. For example, one can

(secretly) imagine

M HCl as having a pH of

-1.

In general, the p function is quite useful as

will be seen in the next chapter. The p function is

the negative logarithm of a value. For example:

p(x) = - log (x)

We will use the p(OH), the p(^,), and the p(^b)

functions. In water p^a + P^b (of the conjugate

base) = 14; for example, the p^^ of ammonium

ion (4.76) plus the pXj, of ammonia (9.24) equal

14.

Table 13-4 shows the relationship between

pH, pOH,

[H"^],

and [OH] in dilute, aqueous

solutions.

By taking the logarithm of both sides of Eq.

(13-4) and using the definition of the p function,

we obtain the relationship pH 4- pOH = 14 for

water at 25°C.

ACIDS BASES AND THE pH SCALE 321

Table 13-4A. From strong, concentrated acid decreasing to neutral pH.

|pH

pOH

^"^^

[OH]

-1

10-''

10-'*

0.1

10-'^

10-^

10-'^

10-^

10"

10-"

10-10

10-'

10-*

io-«

10-'

io-«

Table 13-4B. From neutral pH increasing to strong, concentrated base.

IfpH

P^H

^"^^

1 [OH]

10'

10-'

10-*

10-'

10-10

10-*

10"

10-'

10-'^

10-^

10'^

0.1

10-'*

-1 1

10"

EXAMPLE 6. What is the pH of

0.025

M NaOH?

SOLUTION: NaOH is a strong electrolyte, so

[OH] = 0.025M. From Eq. 13-4 it follows

that [H+] = 4.16•10-^^ so the pH = -log

[H"*^] = 12.38. This is a reasonable answer

if one looks at the pH versus [OH] table

above.

PROBLEM: What is the pH of 0.33 M HCl?

Answer: pH = 0.48

Eq. 13-5 shows that the pH of a weak acid is

dependent upon its concentration. The [OH] may

be determined for a "pure" conjugate base of a

weak acid with K^ and is shown in Eq. 13-6.

Remember that

= KJK^, where b is the conju-

gate base of the acid a.

For a solution of weak base alone

[0H-]= 5^(C.--[0H-])«

(13-6)

N^a

The pH of

"pure" weak acid can be approx-

imated by Eq. 13-5 as long as less than 10% to

15 %

of the acid is ionized. Example 7 shows why

this works.

For a solution of weak acid alone

An interesting rearrangement of

the

approximation

is:

pH«

pJ^^-logC^

for a 1 Af solution of weak

acid:

pH« —-

EXAMPLE 7. The pH of a weak acid by

itself.

Calculate the pH of 0.01 M acetic acid.

SOLUTION: If x equals the concentration of

acetic acid that ionizes, then x = [H"^] =

[Ac];

[HAc] = (0.01 M-x). Using Eq. 13-

5 leads to the following equation.

;c2/(0.01 -x) = 1.75 X 10-^

By rearrangement and use of the quadratic

equation (see Example 9), this could be

easily solved precisely. Since this is a weak

acid, however, the amount of ionization will

be small and [HAc] « 0.01 Af; therefore,

322 13. INTRODUCTORY CHEMISTRY REVIEW

jc^

= 1.75 X 10-^ X 0.01;

X = [H+] = 4.18 X 10-"; pH = 3.38

Notice that the assumption that x will have a

negligible effect on [HAc] is approximately

true; otherwise

would have to be recalculat-

ed.

Use of the quadratic equation would

have given a pH of 3.39, which is not appre-

ciably different.

EXAMPLE 8. What is the pH of 0.1 M sodium

acetate?

SOLUTION: Using the approximation in Eq.

13-6 gives:

[OH] = ((IO-^VI.75 X 10-5)(0.1))«'^

[OH] = 7.56 X

10-^;

pH = 8.89

Since the [OH] is small relative to sodium

acetate, the approximation is warranted.

EXAMPLE 9. ^3 for phosphoric acid, H3PO4, is

4.8 X 10-^\ What is the pH of 0.1 M

trisodium phosphate (TSP)?

SOLUTION. Use of Eq. 13-5 with the approxi-

mation that

[H]"^

is small gives:

[OH] = ((10-^V4.8 X

10-^3)(0.1)f^

[OH] = 0.046; Note that the approximation

is not valid since this acid is nearly 50%

ionized!

Therefore, let x = [TSP] that reacts with

H+, so that

= [OH] = [CJ. At equilib-

rium, [TSP] = (0.1 Af -

jc).

Using Eq. 13-5

leads to the following equation.

A^ = (0.1 -

Jc)

KJK^ = (0.0208X0.1 -

Jc)

+ 0.0208

- 0.00208 = 0

By use of the quadratic equation this is

solved as

= 0.0364 (the negative solution

or root, -0.057, is discarded since concentra-

tions must be positive). This corresponds to

a pH of 12.56, which is very alkaline.

For polyprotic acids the determination of

the

intermediate salt, such as NaHSOj, is more

complex since the bisulfite ion can act as a base or

acid. In this case, the [H"^] for a solution of the

salt alone can be shown to be equal to (^1^2)°'^ ^

long as Ki> >K2.

{K^

should be 10* or 10,000

times as large as

K2.)

By taking the log of both

sides this is rewritten as pH = (p^i + pK^/2.

Notice that the concentration of the salt does not

affect the pH; however, this is not a buffer solu-

tion as it represents an equivalence point of a

titration. See the section on pulp liquor analysis

for more details on this point. Therefore, for a

solution of

conjugate base which itself is an acid

and

J^i

> > K2:

EXAMPLE 10.

KHSO3?

Determine the pH of 0.1 M

Since the approximation cannot be used, the SOLUTION: Using Eq. 13-7 gives: pH =

quadratic equation is used to get a more exact

answer. A quadratic

equation

is in the form

ax^

+ &c + c = 0 where a, b and c are

coefficients

and

x is the variable. The gener-

al solution to the quadratic equation is:

JC=

-'b±yjb^-4ac

(pK, + pK2)/2 = (1.89 +

5.3)/2

= 3.59.

EXAMPLE 11. Determine the pH of 0.10 M

HK2PO3?

SOLUTION: Use of Eq. 13-7, but with

and

^3,

gives a pH of 9.72.

THE LAW OF MASS ACTION 323

13.7 THE LAW OF MASS ACTION

All chemical reactions are reversible (al-

though many to an infmitesimally small degree),

and an equilibrium condition is eventually estab-

lished where the rate of the forward reaction is

equal to the rate of the reverse reaction. A gener-

alized form of a chemical reaction is given as ah

+ fcB 4± cC + dD, where the lowercase letters in

italics are stoichiometric coefficients and the

capital letters are reacting species or products.

Strictly speaking, the effective concentration

of ionic species is lower than the actual concentra-

tion. The ratio of effective concentration to the

actual concentration is known as the activity

coefficient of the ion. Generally, as the total ionic

strength of a solution increases, the activity coeffi-

cient of a given species decreases from unity. The

activity is the effective concentration of ionic spe-

cies.

All equilibrium constants are correctly stated

only in terms of activity. Fortunately, in dilute

solutions the activity coefficients are approximately

equal to unity, and one can use concentration

directly instead of activities to simplify the calcula-

tions.

Even concentrated solutions are analyzed

(such as by titration) under relatively dilute condi-

tions.

Assuming activity coefficients of unity, the

molar concentrations, [A], [B], etc. are related at

equilibrium as follows:

[AfEBf

(13-8)

K^ is called the equilibrium constant and

varies with temperature. Solubility products and

acid ionization constants are specific examples of

the law of mass action. This law applies to gas-

eous or solution reactions. A key concept is that

the law of mass action does not predict the reac-

tion rate, only the final equilibrium concentration.

For example, a gaseous mixture of hydrogen and

oxygen at 25 °C is predicted to be almost com-

pletely reacted to form water; however, at 25 °C

the reaction will take thousands of years to reach

equilibrium. Catalysts are used to increase reac-

tion rates, but do not affect equilibrium constants.

Sometimes it is difficult to know whether there is

no change observed in the concentration of reac-

tants because equilibrium is achieved or because

the reactions are very slow.

With gases, it is often easier to express con-

centrations as partial pressures (the pressure

exerted by an individual species). When this is

done, the term

is used and often the units are

atmospheres. K^ is usually highly dependent on

the temperature. For example, the equilibrium

constant for the oxidation of sulftir dioxide to

sulfiir trioxide is 4.0-10'" atm'^ at 298 K, 2.5-10'°

atm-' at 500 K, and 3.4 atm'^ at 1000 K. (Of

course, K^ can be determined from K^ using the

ideal gas law.) When the reaction species are

expressed as their pressures, P, Eq. 13-8 becomes:

(13-9)

EXAMPLE 12. Nitrogen gas reacts with hy-

drogen gas at elevated temperature in the

presence of a catalyst to produce ammonia in

the Haber synthesis. At 500°C and equilibri-

um the concentration of ammonia is 0.43 M,

the concentration of hydrogen gas is 1.09 M,

and the concentration of nitrogen gas is 1.3

M. Calculate the equilibrium constant for the

formation of ammonia.

SOLUTION: The reaction is Nj + 3H2 ^ 2NH3.

Using Eq. 13-8 with a=l,Z? = 3andc =

and substituting in the concentrations in

molarity, it is found that K^ = 0.11 M.

PROBLEM: During the production of SO2 for

use in sulfite pulping, if excess oxygen is

present, SO3 may be formed and interferes

with the pulping process. At 1000 K, K^ =

3.4 atm-^ for the reaction 2SO2 + 02^

2SO3.

If the partial pressures of

SO2

and O2

are 0.25 atm and 0.001 atm, respectively,

what is the partial pressure of

SO3

at equilib-

rium? Answer: 0.0146 atm.