Biermann Ch. Handbook of Pulping and Papermaking

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334 13. INTRODUCTORY CHEMISTRY REVIEW

Hydrogen bonding

In which of the following compounds is hy-

drogen bonding possible?

a) RNHjb) RCHO

d) H^S e) CO2

g) EtOHh) RCHjCl

j) RCH2OH

c) RCOOH

f) HF

i) HjSOj

Draw the partial charges involved with hy-

drogen bonding of HCl in water.

Which of the following three compounds is

expected to have the highest boiling point?

Why?

a) CH3CH2OCH2CH3

b) CH3CH2CH2OCH3

c) CH3CH2GH2CH2OH

The mole and mass percentage

Calculate the number of moles in each of the

following:

a) 98.3 g C5H12O6

b) 17.3gNa2SO4-10H2O

c) 210gNa2S

8. Calculate the mass percentage of sulfur in the

following chemicals:

a) NajS

d) SO2

b) Na2S04

e) NaHS03

c) H2SO3

9. 50 g of a mixture of NaCl and sucrose

(C12H22O11) was burned leaving 30.5 grams

of NaCl ash. How much

CO2

was produced?

Equivalency, molarity, and normality

10.

150 g of water contains 20 g of glucose,

C6HJ2O6, with MW = 180 g/mol. What is

the molality of the solution.

11.

How many grams of NaOH are required to

make 700 ml of 1.25 M NaOH solution?

What volume of 0.50 M H2SO4 is required to

neutralize this solution?

Acids, bases, and

the

pH scale

12.

What is the pH of the following solutions?

a) 0.047 M HCl

b) 0.0013

NaOH

c) O.OIMH2SO4

13.

What is the pH of a solution containing 0.5

M acetic acid and 0.2 M sodium acetate?

14.

WhatisthepHof0.1MNaHCO3?

Oxidation-reduction

reactions

15.

What is the oxidation number of chromium in

16.

In the reaction 2Na + 2 H2O ^2NaOH +

Hjt, what species is the reducing agent?

Electrochemistry and corrosion

17.

The voltage required to produce electrolysis

of NaCl is about 4 volts. How much electri-

cal energy is required to produce one pound

of NaOH?

18.

Describe how pitting leads to high localized

corrosion.

19.

Why is it important to periodically replace

the magnesium strip in hot water heaters

equipped with these?

20.

What is galvanized metal?

Properties of gases

21.

A pure liquid is analyzed as 85.6% C and

14.4%

H. Under STP conditions the density

of its vapor is 3.75 g/L. What is the molec-

ular formula of

the

substance? Show a possi-

ble structure.

ANALYTICAL AND COORDINATE CHEMISTRY

This chapter presents concepts of analytical

chemistry. Acid-base chemistry is important for

pulping and bleaching operations. Redox reactions

are important to understand bleaching, corrosion,

and ion-specific electrodes. Coordinate chemistry

is central to alum and wet end chemistry and

involves many principles of analytical chemistry.

For more detail on these concepts one should

consult an introductory text on quantitative analy-

sis or analytical chemistry.

14.1 STRONG ACID-STRONG BASE

TITRATIONS

Neutralization is the formation of water from

a strong acid and a strong base to form a neutral

compound, i.e., a compound that is neitiier an acid

nor a base, such as NaCl. Consider the reaction

of HCl + NaOH -^ H2O + NaCl. Since HCl,

NaOH, and NaCl are all completely ionized in

solution, this reaction can be written as: H"^ +

CI"

+ Na+ + OH- -* Na+ + CI" + HOH. Na+

and CI" are spectator ions, that is, they are not

involved in the reaction, so the overall reaction is

written simply as H"^ + OH- -* HOH; this is the

essence of neutralization reactions. The enthalpy

of neutralization of a strong acid and strong base

is about -57.2 kJ/mol. The term "neutralization"

is sometimes used for the reaction of any acid and

base,

such as the reaction of acetic acid and

sodium hydroxide to form sodium acetate and

water, but this is not strictly correct use of the

word, since sodium acetate is a weak base.

During the start of a titration of a strong base

with a strong acid (Fig. 14-1), all of the acid will

be consumed by the base since there is more base

than acid. The concentration of the base will be

reduced slightly and the pH will decrease slightly.

As the base becomes depleted near the equivalence

point, its concentration decreases by several orders

of magnitude and the pH decreases rapidly. At

the equivalence point, all of the base is reacted

with all of the acid, and the pH is equal to 7. As

more acid is added, the pH decreases less and less

rapidly. Fig. 14-1 shows the pH during the titra-

tion of a strong base with a strong acid. For a

strong acid being titrated with a strong base, the

figure can be read with the pH scale inverted.

14-

10-

A-

""^^

0.2 0.4 0.6 0.8 1 1.2

Moles HQ per mole NaOH (or 100 ml HCl)

Fig. 14-1. Titration of 100 ml of 0.1 M NaOH with 0.1 M HCl.

1.4 1.6

335

336 14. ANALYTICAL AND COORDINATE CHEMISTRY

The endpoint is detected with a pH meter or

suitable indicator (usually phenolphthalein) as

discussed in the next section.

Since normality is equivalents per liter, if the

normality of a solution is multiplied by the volume

of solution, the number of equivalents is deter-

mined for that solution: normality x volume =

equivalents. Also, at the endpoint of a titration,

by definition, the number of equivalents in the

sample being analyzed (the analyte) is equal to the

number of equivalents in the titrant (the standard

solution used in the buret). This leads to the very

useful titration equation of

N,V,=N,V,

(14-1)

The normality of titrant multiplied by the volume

of titrant equals the normality of

analyte

multiplied

by the volume of analyte. During titrations the

volume is measured in ml; therefore it is useful to

remember that I N = I eq/L = 1 meq/ml.

EXAMPLE 1. 5.00 ml of HCl solution of un-

known strength required 15.35 ml of 0.500 N

NaOH to achieve the endpoint (neutraliza-

tion).

Calculate the following:

10.

Milliequivalents of NaOH consumed.

Milliequivalents of HCl consumed, or

neutralized by the NaOH.

Equivalents of HCl consumed.

The equivalent weight of HCl.

The milliequivalent weight of HCl.

The weight of HCl neutralized.

The normality of the original HCl solu-

tion.

The strength of the original HCl solu-

tion in g/L.

The strength of the original HCl solu-

tion in lb/gal.

The weight of the NaOH consumed.

SOLUTION:

15.35 ml NaOH x 0.5 meq/ml = 7.67

meq NaOH.

By definition of equivalency,

eq NaOH = eq HCl = 7.67 meq.

7.67 meq/(1000 meq/eq) = 0.00767 eq

HCl.

The equivalent weight of HCl is equal

to the formula weight; therefore, the

equivalent weight = 1 + 35.5 = 36.5.

(36.5 g/eq HC1)/(1000 meq/eq) =

0.0365 g/meq HCl.

0.00767 eq HCl x 36.5 g/eq = 0.280

g HCl.

Since

NiVi=N2V2,

with rearrangement

one obtains the following:

^1 =

N^xV^

^i^l5^

= L534NHCl

5.0

(0.280 g HCl)/(0.005 L) = 56 g/L HCl

Dimensional analysis of the concentra-

tion in g/L gives:

U lib

454 g 0.264 gal

=0.467-5!^

HCl

gal

10.

7.67 meq NaOH x 40 mg/meq X 1

g/(1000 mg) = 0.306 g NaOH

EXAMPLE 2. How many ml of concentrated

sulfuric acid (density = 1.86 g/ml) should

one add to make 1.00 L of 0.05 N H2SO4

solution? Note the notation of normality!

SOLUTION:

ImolH^SO,

0.05 eq/5 x

x —- x 98g/mol H2SO4

eq H2SO4

H2SO4

L86gH2S04

1.32mC

cone. H^SO^

14.2 pH PROPERTIES OF WEAK ACID-

CONJUGATE BASE PAIRS

The pH of a weak acid can be expressed in

terms of the ionization constant and the concen-

trations of the weak acid, HA, and its conjugate

base.

A". The conjugate base is sometimes re-

ferred to as the salt of the weak acid. Rearrange-

ment of Eq. (13-2) gives the following equation

that is applicable when pKJ[C] > 0.1; i.e.,

greater than 10% ionization of A':

pH PROPERTIES OF WEAK ACID-CONJUGATE BASE PAIRS 337

[H^] = ^ [A]/[A-]

Taking the logarithm of each side gives:

pH = pi^ - log([A]/[A-])

Using the properties of logarithms, this equation

becomes:

pH = pi^ + log([A-]/[A]) (14-2)

Eq. 14-2 is called the

Henderson-Hasselbalch

equation. This equation should be used with some

care if the ratio of the salt to acid is very high or

very low. A

10"^

M solution of acetic acid without

sodium acetate might be expected to have a very

low pH as the ratio approaches 0 (and the log

approaches -00). However, as shown in Example

13-7,

the acid does ionize and appreciable amounts

of the conjugate base are present. Thus, Eq. 14-2

cannot be used near equivalence points (endpoints)

of titrations. Eq. 14-2 can be written in the form

of a weak base as follows:

pOH = vK, + log([BH^]/[B])

= pjR; + log([salt]/[base]) (14-3)

The Henderson-Hasselbalch equation is very

useful for describing the pH behaviors of mixtures

of weak acids and their conjugate bases in solu-

tion. If equal molar concentrations of an acid and

its conjugate base are in solution, the pH will be

equal to the pX^ of that acid, since -log(/z//i) = 0

and pH = pX^. This is a useful fact to remember

because it gives a rough estimate of pH of a

solution containing an acid and its conjugate base.

Furthermore, even if this solution is diluted, the

pH will not change. Also this solution can absorb

small amounts of acid or base without appreciably

changing the ratio of acid and salt form. There-

fore,

such a solution would be a good buffer, that

is,

a solution which resists changes in pH by

dilution or by addition of acid or base. Therefore,

any solution of a weak acid and its conjugate base

is a buffer. The amount of buffering capacity is

proportional to the concentration of these species.

The maximum buffering action is achieved for a

given concentration of weak acid when the ratio of

the concentration of salt to acid is unity (i.e., both

the acid and its conjugate base are of equal con-

centration). Later, it will be observed that this is

the flattest part of a titration curve for a weak acid

or base.

EXAMPLE 3. What is the pH of a solution of

0.1 M

H2SO3

and 0.2 M NaHSOa?

SOLUTION: Table 11-3 gives a p^, of 1.89 for

sulftirous acid at 25

^'C.

Therefore:

pH = 1.89 + log (0.2/0.1) = 2.19^

EXAMPLE 4. A solution that is 1 M acetic acid

and 1 M sodium acetate has a pH of 4.76,

which is the p^a fo^* acetic acid. What will

the pH be after 0.5 mole of HCl is added to

1 liter of this solution?

SOLUTION: The strong acid reacts completely

with the weak base to give 0.5 mol/L HAc

and leaving 0.5 mol/L Ac'. The total [HAc]

= 1.5 M. Therefore, pH = 4.76 +

log(0.5/1.5) = 4.28, a small change.

PROBLEM: What would be the pH of 1 L of

water with 0.5 moles HCl? Answer: The

pH would be 0.30, which shows the effect

of a buffer solution when compared to this

example.

14.3 pH INDICATORS

The pH of solutions can be monitored using

indicators. Indicators are themselves weak acids

or bases whose ionized forms differ in color from

the un-ionized forms. The color of these species

should be intense so only small amounts of indica-

tor are used; after all, the indicators are also

reacting with

the

titrant. Usually these compounds

involve conjugated aromatic systems. A common-

ly used indicator is phenolphthalein. Fig. 14-2

shows the un-ionized, colorless form and the

ionized, deep red-purple form. The p^a of pti^-

nolphthalein is about 9. Using Eq. 14-2, this

338 14. ANALYTICAL AND COORDINATE CHEMISTRY

"°-^0^?-o

^'-'

COO"

Na"^

Acid form - colorless Salt form - deep red-purple

Fig. 14-2. Phenolphthalein indicator forms.

means that at pH 9 about half the indicator will be

in the ionized form, while at pH 8 only 1/10 the

indicator will be ionized. This is the range in

which one observes the most color change. Most

pH indicators change color over a pH range of

about 2, going from 90% of one color to 90% of

the other color. If one of the forms is colorless,

the color change is over one pH unit.

Phenolphthalein is a suitable indicator for

strong acid-strong base reactions. It reacts with

CO2,

so the CO2 must be boiled off near the end-

point to give a sharp endpoint for titrations involv-

ing carbonate. This will be discussed in greater

detail in Chapter 16 on pulping liquor analysis.

Phenolphthalein is suitable for titration of borax if

glycerol is present to avoid color fading. The

color change goes from colorless up to pH 8.5 to

deep red-purple above pH 9. Phenolphthalein is

usually used as a few drops of a

1 %

ethanol-water

solution, since it is only slightly soluble in water

in its un-ionized form.

Methyl orange is another common pH indica-

tor and goes from red below pH 3.1 to yellow

above pH 4.4; it is used as a 0.1% aqueous solu-

tion. There are about 60 additional pH indicators

available.

14.4 TITRATION OF A WEAK ACID WITH

A STRONG BASE OR WEAK BASE

WITH A STRONG ACID

The tools are now available to describe the

titration curve of a weak acid by a strong base or,

more likely in kraft pulping liquor analysis, the

titration of a weak base by a strong acid. Consid-

er the titration of 100 ml of 0.1 iV acetic acid by

0.1 N NaOH (Fig. 14-3). With no addition of

4.76

L3n

Ll"

LU"

0.2 0.4 0.6 0.8 1 1.2

Moles NaOH per mole HAc (xlOOs ml NaOH)

1.4 1.6

Fig. 14-3. Titration of 100 ml of 0.1 M acetic acid with 0.1 M NaOH.

WEAK ACID OR BASE TITRATIONS 339

base Eq. 13-5 shows [H"*^] =

(K-C^^f'

or 1.32

X 10•^ a pH of 2.88; thus, the first part the

titration curve is concentration dependent. From

2 ml to 98 ml of NaOH the pH is determined firom

the Henderson-Hasselbalch equation, and the pH

is not concentration dependent. At the endpoint

the pH is determined from Eq. 13-6, which is

solved for this case in Example 13-8 and is pH of

8.89. Beyond the endpoint, excess alkali sup-

presses abstraction of protons by sodium acetate so

the hydroxide concentration is equal to the excess

equivalents of hydroxide divided by the volume of

solution. At 101 ml, [OH] = (101-100 ml)(0.1

mmol/ml)/(200 ml) = 5 X 10*^ mmol/ml (or

mol/L);

this corresponds to a pH of 10.70. Since

the endpoint occurs at pH 8.9, phenolphthalein is

an ideal pH indicator to use to detect this end-

point. Be sure to notice that when the acetic acid

is half titrated, the pH = 4.76 =

pK^

and, as this

is the maximum buffer region, the titration curve

is flattest here. This curve is similar to those of

other carboxylic acids, such as fortified rosin and

carboxylate salts in black liquor.

In a similar fashion, ammonia can be titrated

(Fig. 14-4). The format of the A:-axis is going to

be modified here, and the modified format will be

used throughout the remainder of the book, unless

stated otherwise. Instead of the ratio of acid to

base,

we will assume all of the acid reacts with all

of the base. The jc-axis will be reported in terms

of percent of acid and percent of base form. One

can use this to see where endpoints occur, but

more generally, one can quickly determine the pH

of a solution of mixed composition.

Mixtures of acids or bases can be titrated

sequentially if they differ by a factor of about 10"^

in their

K^,

values. Sometimes a species may

be removed from solution to prevent interference.

Other times a species may react with an added

compound to change its acid or base characteris-

tics.

Examples of such titrations are given in the

section on analysis of pulping liquors.

14.5 REDUCTION-OXIDATION TITRATIONS

Reduction-oxidation (redox) titrations are

performed in a manner similar to those of acid-

Iz-

11-

10-

100

SS-

Ammonia, %

80 60 40 20 0

' ]vn

20 30

90 100

40 50 60 70

Ammonium ion, %

Fig. 14-4. The pH as a function of ammonia-ammoniiun (NH3-NH4*) composition (0.1 M total

concentration) at 25°C.

340 14. ANALYTICAL AND COORDINATE CHEMISTRY

base titrations. The titration is monitored by the

potential (voltage) with respect to a standard elec-

trode or by the use of indicators that are oxidized

or reduced after the analyte. In redox titrations,

equivalents are based on moles of electrons trans-

ferred in the reaction. A compound with a certain

equivalent weight for an acid-base reaction may

have a different equivalent weight in a redox

reaction; it may even have an equivalent weight

that varies depending on the particular redox

reaction.

In order for an analyte to be titrated with a

titrant there should be at least 0.2 to 0.3 V differ-

ence between the standard two electrode poten-

tials.

This does not say the reaction will occur

quickly, only that thermodynamically the reaction

is predicted to occur.

When the two materials have reacted, i.e.,

they are in equilibrium, both of their actual

half-

reaction potentials are equal. The general form of

the titration reaction with species (1) and (2) with

stoichiometric reaction coefficients of a, b, c, and

^is:

aOxi +

bRed2

^ cRed^ +

dOx2

Activity coefficients are ignored here since

the concentrations are low and the change in

potential at the endpoint will be sharp.

Consider the titration of I2, which usually

occurs as I3', with Na2S203 (sodium thiosulfate)

that is used in lignin determinations with

permanganate or determination of bleaching

chemicals with iodine. The reaction becomes:

I3-

+ 2S2O32 ^ 31 + SA^-

At any time during the titration the E of the

iodine/iodide cell is equal to the E of the thiosul-

fate/tetrathionate cell. The Nernst

equation

for the

half reaction of the compound that is reduced

(iodine) is :

n [OxV

at25°C

The iodine is generated from excess iodide

that is oxidized by permanganate or bleaching

chemicals in many pulp and paper laboratory

determinations. Therefore, let us consider the

titration of 100 ml of 0.1

M13"

in the presence of

0.2 M iodide with 0.1 M thiosulfate (Fig. 14-5).

£° for the iodine/iodide cell is 0.534 V. The

E""

for the thiosulfate cell is 0.08 so the £° (cell) is

0.454 V, enough of a difference to give a sharp

endpoint. As soon as the first 0.5 ml or so is

added, the [I] will be known (if none had been

present initially) and the cell potential can be

calculated.

EXAMPLE 5. What is E of the iodine/iodide

cell just described after 0.5 ml of thiosulfate

is added in the reaction above?

SOLUTION: For each I3- reacted 3 l form.

Ignoring the small dilution:

E = £^-0.05915/2 X log(0.2015)V0.0995

= 0.534 + 0.032 = 0.566 V.

EXAMPLE 6. Knowing that the £eeii of the thio-

sulfate cell is 0.566 V, what is the concentra-

tion of thiosulfate?

SOLUTION: Assume all of the thiosulfate

reacted except for a minuscule amount in

equilibrium.

0.566 = 0.08 -0.05915/2 X log(x/0.0005)

3.69 X 10-^^ = (x/0.0005)

X = 1.83 X

10'^^

M (correct assumption)

PROBLEM: Redo this example with 2 ml of

titrant added. The answers are 0.565 V and

2 X 10-2« M

EXAMPLE 7. Shortly before the endpoint. Fig.

14-5 shows the potential as 0.40 V. What is

the concentration of thiosulfate if

tetrathionate is 0.1 Af?

0.40 = 0.08 -0.05915/2 X log(x/0.1000)

1.51 X 10" = (jc/0.1000)

x= 1.51 X

10-^0

REDUCTION-OXIDATION TITRATIONS 341

u.o-

U.4D-

"•'

.2 U.^D-

n 9-

u.z

1^-

U.ID"

0.1-

~~^^

0 20 40 60 80 100 120

ml of

0.1

M Thiosulfate

Fig. 14-5. Titration of 100 ml of 0.1 M !{in the presence of 0.2 M

with 0.1 M

SiOj^.

During the titration up to the equivalence

point the potential is calculated from the cell

reaction of the analyte. Beyond the endpoint the

potential is calculated from the cell reaction of the

titrant.

lodometry

The technique just demonstrated of using the

conversion of iodide to iodine by moderately

strong oxidizing agents followed by the titration of

the iodine is termed iodometry. This indirect

titration is a common technique in pulp and paper

measurements as well. One might ask why iodide

is not used directly as the titrant to measure these

oxidizing agents. The reason is that starch cannot

be used for the endpoint determination and the

reduction of iodide is fairly slow unless excess

iodide is present.

The level of a wide variety of oxidants such

as hydrogen peroxide, ozone, oxygen, Clj,

permanganate, and HCIO can be determined by

this method.

Indicators

The potential of a solution during a redox

reaction can be followed with electrodes. The

potential is usually measured across a platinum

electrode and an electrode that supplies a standard

potential. Chemical indicators can be used and are

compounds that are oxidized or reduced and

change colors between the two forms. The end-

point of iodine titrations is conveniently followed

with starch indicator, since small amounts of

iodine form a dark blue complex with it. (This is

also the basis of a qualitative test for the presence

of starch in paper.) Titrants or analytes that are

highly colored can be used as self indicators.

Miscellaneous aspects

This section has only been a cursory intro-

duction to redox titrations. Many standard poten-

tials vary with pH, the presence of complexing

species, and other factors. These considerations

may cause deviations from expected results in

experiments.

342 14. ANALYTICAL AND COORDINATE CHEMISTRY

14.6 COLORIMETRIC ANALYSIS

Colorimetric analysis is important in the area

of effluent color of mill discharges and some

determinations of concentrations of chemical spe-

cies.

Glucose liberated by enzymatic hydrolysis of

starch is measured colorimetrically in the assay of

starch in corrugated board (TAPPI Standard T

532).

In liquid solutions, the amount of mono-

chromatic light absorbed by a solution is described

Beer^s

law. Consider Fig. 14-6 where incident

light Po travels through a sample of liquid of path

length b and concentration c. Transmittance, T, is

defined as the fraction of light transmitted through

the sample and is equal to PIPQ. The transmit-

tance of light decreases exponentially. For exam-

ple,

90%

of the light is transmitted through cell

length b, then 90% of 90% (or 81%) will be

transmitted through length lb. Concentration

behaves similarly.

A constant is introduced called the molar

absorptivity, e, and the term absorbance, A, is

defined to give Eq. 14-4 (below) where c is

usually expressed in moles per liter and b is

Fig. 14-6. Absorbance of light in a solution.

usually 1 cm, the pathlength commonly used in

spectrophotometers. Fig. 14-7 shows the relation-

ship between absorbance and transmittance.

Sometimes the

absorptivity,

a, is expressed in g/L

if the molecular weight is unknown; in this case c

is expressed in g/L. The percent transmittance,

%T, is equal to 100% X T, and percent absor-

bance = 100% - %T.

Optical density is an obsolete term of

colorimetric analysis and should not be used.

1 S-

l.o

1 #^-

l.O

1 A-

1.4

1 9-

8 ^-^

•S 1-

U.o

U.4

U.Z"'

10 20 30 40 50 60

% Transmittance

90 100

Fig. 14-7. Absorbance versus

transmittance.

COLORIMETRIC ANALYSIS 343

Absorbance, and not transmittance nor

percent absorbance, is proportional to the concen-

tration and path length. Absorbance much above

2.5 cannot be measured accurately since this

corresponds to only 0.31% transmittance.

.4 = - log r = log 1/r

= log PJP = ebc

log 1/r = log PJP = abc

(14-4)

EXAMPLE 8. A sample in a 1.0 cm cell absorbs

10%

of the light at a certain wavelength. If

the absorptivity of the material is 3.0

L(g-cm) at this wavelength, what is the

concentration of the material?

SOLUTION: The transmittance is 0.90,

-log 0.90 = 3.0 L(gxm) x 1.0 cm x c;

c = 0.0153 g/L.

EXAMPLE 9. 10% of the light of a certain

wavelength of a pulp mill effluent is trans-

mitted through a distance of 1 feet. What

will the percent transmittance be at the same

wavelength through the same distance be if

the effluent is diluted

1:100?

SOLUTION: A^ = abc, = 1; ^2 = cibc^ =

abcJ\OQ\ A2 = 0.01; foT^ = 97.7%

PROBLEM: For this same problem what would

the apparent transmittance be for a river 2.5

feet deep? Remember the effective distance

will be 5 feet since the light will reflect off

the bottom and back through the water before

reaching the eye of the observer. Answer:

89.1%.

Beers law ahnost always applies. However,

apparent deviations may occur. For example, the

color of pulp mill effluents are pH dependent. If,

in the problem above, the pH were different

before and after dilution, then Beer's law may not

appear to hold. This is why the pH*s of mill

effluents are adjusted to that at which they will

ultimately be discharged before the absorbance

(using color units. Section 11.3) is measured.

14.7 COORDINATE CHEMISTRY

Introduction

Coordinate chemistry involves the for-

mation of complexes when two atoms share an

electron

pair.

One species makes the donation and

the second species accepts the electron pair.

Coordinate covalent bonds are formed when one

atom donates both electrons of the electron pair;

with other covalent bonds each atom donates one

electron to the pair. The distinction is arbitrary,

but has historical precedence. More information

and detail on this important subject may be found

in advanced inorganic chemistry textbooks.

Coordinate chemistry

explains

the behavior of

alum (which is central to many aspects of wet end

chemistry) in aqueous solutions. Aluminum ions

will be used to demonstrate some aspects of

coordinate chemistry. Also, many size press

formulations contain zirconium compounds or

synthetic polymers containing carboxylated poly-

mers that are "set" with metal ions. Applications

of coordinate chemistry will undoubtedly receive

much wider recognition in the future by the pulp

and paper industry.

The first explanation of the actual nature of

complexes is credited to the classic 1906 work of

Werner, for which Werner received the Nobel

Prize in 1913. Werner showed that neutral com-

pounds were bound directly to metal atoms in

many complexes. Thus, CuCl2*6NH3 is correctly

written as Cu(NH3)6Cl2. In general, coordination

complexes are formed upon the reaction of Lewis

acids, compounds which accept electron pairs,

with Lewis bases, compounds which donate elec-

tron pairs, to form the complex. Acid-base reac-

tions involving proton transfers are examples of

coordinate chemistry reactions. The reaction of

H^ and OH" to form the product H2O proceeds as

H^ + :0H' -• H:OH, where the electron pair is

shared in the product. Lewis greatly expanded the

class of acid-base reactions over that of the

Bronsted-Lowry theory by making electron pairs

of central importance rather than the proton.

Many other reactions besides acid-base

reactions are included as well. One classic exam-

ple is the reaction of the two gases BF3 and :NH3

to form the complex H3N:BF3, which is a white

solid: