General chemistry

Chemistry is a very wide field; however, only a very small portion, indeed, of this seemingly complex subject is used in this book. These include equivalents and equivalent mass, methods of expressing concentrations, activity and active concentration, equilibrium and solubility product constants, and acids and bases. This knowledge of chemistry will be used under the unit processes part of this book.

Equivalents and Equivalent Masses

The literature shows confused definitions of equivalents and equivalent masses and no universal definition exists. They are defined based on specific situations and are never unified. For example, in water chemistry, three methods of defining equivalent mass are used: equivalent mass based on ionic charge, equivalent mass based on acid-base reactions, and equivalent mass based on oxidation-reduction reactions (Snoeyink and Jenkins, 1980). This section will unify the definition of these terms by utilizing the concept of reference species; but, before the definition is unified, the aforementioned three methods will be discussed first. The result of the discussion, then, will form the basis of the unification.

Equivalent mass based on ionic charge. In this method, the equivalent mass is defined as (Snoeyink and Jenkins, 1980):

molecular weight

ionic charge

For example, consider the reaction,

Calculate the equivalent mass of Fe(HCO3)2.

When this species ionizes, the Fe will form a charge of plus 2 and the bicarbonate ion will form a charge of minus 1 but, because it has a subscript of 2, the total ionic charge is minus 2. Thus, from the previous formula, the equivalent mass is Fe(HCO3)2/2, where Fe(HCO3)2 must be evaluated from the respective atomic masses. The positive ionic charge for calcium hydroxide is 2. The negative ionic charge of OH- is 1; but, because the subscript is 2, the total negative ionic charge for calcium hydroxide is also 2. Thus, if the equivalent mass of Ca(OH)2 were to be found, it would be Ca(OH)2/2. It will be mentioned later that Ca(OH)2/2 is not compatible with Fe(HCO3)2/2 and, therefore, Equation (6) is not of universal application, because it ought to apply to all species participating in a chemical reaction. Instead, it only applies to Fe(HCO3)2 but not to Ca(OH)2, as will be shown later.

Equivalent mass based on acid-base reactions. In this method, the equivalent mass is defined as (Snoeyink and Jenkins, 1980):

where n is the number of hydrogen or hydroxyl ions that react in a molecule. For example, consider the reaction,

Now, calculate the equivalent mass of H3PO4. It can be observed that H3PO4 converts

to HPO4 . Thus, two hydrogen ions are in the molecule of H3PO4 that react and the equivalent mass of H3PO4 is H3PO4/2, using the previous equation. The number of hydroxyl ions that react in NaOH is one; thus, using the previous equation, the equivalent mass of NaOH is NaOH/1.

Equivalent mass based on oxidation-reduction reactions. For oxidation-reduction reactions, the equivalent mass is defined as the mass of substance per mole of electrons involved (Snoeyink and Jenkins, 1980).

For example, consider the reaction,

The ferrous is oxidized to the ferric form from an oxidation state of +2 to +3. The difference between 2 and 3 is 1, and because there are 4 atoms of Fe, the amount of electrons involved is 1 x 4 = 4. The equivalent mass of Fe(OH)2 is then 4Fe(OH)2/4. Note that the coefficient 4 has been included in the calculation. This is so, because in order to get the total number of electrons involved, the coefficient must be included. The electrons involved are not only for the electrons in a molecule but for the electrons in all the molecules of the balanced chemical reaction, and, therefore must account for the coefficient of the term. For oxygen, the number of moles electrons involved will also be found to be 4; thus, the equivalent mass of oxygen is O2/4.

Now, we are going to unify this equivalence using the concept of the reference species. The positive or negative charges, the hydrogen or hydroxyl ions, and the moles of electrons used in the above methods of calculation are actually references species. They are used as references in calculating the equivalent mass. Note that the hydrogen ion is actually a positive charge and the hydroxyl ion is actually a negative charge. From the results of the previous three methods of calculating equivalent mass, we can make the following generalizations:

1. The mass of any substance participating in a reaction per unit of the number of reference species is called the equivalent mass of the substance, and, it follows that

2. The mass of the substance divided by this equivalent mass is the number of equivalents of the substance.

The expression molecular weight/ionic charge is actually mass of the substance per unit of the reference species, where ionic charge is the reference species. The expression molecular weight n is also actually mass of the substance per unit of the reference species. In the case of the method based on the oxidation-reduction reaction, no equation was developed; however, the ratios used in the example are ratios of the masses of the respective substances to the reference species, where 4, the number of electrons, is the number of reference species.

From the discussion above, the reference species can only be one of two possibilities: the electrons involved in an oxidation-reduction reaction and the positive (or, alternatively, the negative) charges in all the other reactions. These species (electrons and the positive or negative charges) express the combining capacity or valence of the substance. The various examples that follow will embody the concept of the reference species.

Again, take the following reactions, which were used for the illustrations above:

4Fe(OH)2 + O2 + 2H2O ^ 4Fe(OH)3 Fe(HCO2)3 + 2Ca(OH)2 ^ Fe(OH)2 + 2CaCO3 + 2H2O

In the first reaction, the ferrous form is oxidized to the ferric form from an oxidation state of +2 to +3. Thus, in this reaction, electrons are involved, making them the reference species. The difference between 2 and 3 is 1, and since there are 4 atoms of Fe, the amount of electrons involved is 1 x 4 = 4. For the oxygen molecule, its atom has been reduced from 0 to -2 per atom. Since there are 2 oxygen atoms in the molecule, the total number of electrons involved is also equal to 4 (i.e., 2 x 2 = 4-. In both these cases, the number of electrons involved is 4. This number is called the number of reference species, combining capacity, or valence. (Number of reference species will be used in this book.- Thus, to obtain the equivalent masses of all the participating substances in the reaction, each term must be divided by 4: 4Fe(OH-2/4, O2/4, 2H2O/4, and 4Fe(OH-3/4. We had the same results for Fe(OH-2 and O2 obtained before.

If the total number of electrons involved in the case of the oxygen atom were different, a problem would have arisen. Thus, if this situation occurs, take the convention of using the smaller of the number of electrons involved as the number of reference species. For example if the number of electrons involved in the case of oxygen were 2, then all the participating substances in the chemical reaction would have been divided by 2 rather than 4. For any given chemical reaction or series of related chemical reactions, however, whatever value of the reference species is chosen, the answer will still be the same, provided this number is used consistently. This situation of two competing values to choose from is illustrated in the second reaction to be addressed below. Also, take note that the reference species is to be taken from the reactants only, not from the products. This is so, because the reactants are the ones responsible for the initiation of the interaction and, thus, the initiation of the equivalence of the species in the chemical reaction.

In the second reaction, no electrons are involved. In this case, take the convention that if no electrons are involved, either consider the positive or, alternatively, the negative oxidation states as the reference species. For this reaction, initially consider the positive oxidation state. Since the ferrous iron has a charge of +2, ferrous bicarbonate has 2 for its number of reference species. Alternatively, consider the negative charge of bicarbonate. The charge of the bicarbonate ion is -1 and because two bicarbonates are in ferrous bicarbonate, the number of reference species is, again, 1 x 2 = 2. From these analyses, we adopt 2 as the number of reference species for the reaction, subject to a possible modification as shown in the paragraph below. (Notice that this is the number of reference species for the whole reaction, not only for the individual term in the reaction. In other words, all terms and each term in a chemical reaction must use the same number for the reference species.-

In the case of the calcium hydroxide, since calcium has a charge of +2 and the coefficient of the term is 2, the number of reference species is 4. Thus, we have now two possible values for the same reaction. In this situation, there are two alternatives: the 2 or the 4 as the number of reference species. As mentioned previously, either can be used provided, when one is chosen, all subsequent calculations are based on the one particular choice; however, adopt the convention wherein the number of reference species to be chosen should be the smallest value. Thus, the number of reference species in the second reaction is 2, not 4—and all the equivalent masses of the participating substances are obtained by dividing each balanced term of the reaction by 2: Fe(HCO3-2/2, 2Ca(OH-2/2, Fe(OH-2/2, 2CaCO3/2 and 2H2O/2.

Note that Ca(OH)2 has now an equivalent mass of 2Ca(OH)2/2 which is different from Ca(OH)2/2 obtained before. This means that the definition of equivalent mass in Equation (6) is not accurate, because it does not apply to Ca(OH)2. The equivalent mass of Ca(OH)2/2 is not compatible with Fe(HCO3)2/2, Fe(OH)2/2, 2CaCO3/2, or 2H2O/2. Compatibility means that the species in the chemical reaction can all be related to each other in a calculation; but, because the equivalent mass of Ca(OH)2 is now made incompatible, it could no longer be related to the other species in the reaction in any chemical calculation. In contrast, the method of reference species that is developed here applies in a unified fashion to all species in the chemical reaction: Fe(HCO3)2, Ca(OH)2, Fe(OH)2, CaCO3, and H2O and the resulting equivalent masses are therefore compatible to each other. This is so, because all the species are using the same number of reference species.

Take the two reactions of phosphoric acid with sodium hydroxide that follow. These reactions will illustrate that the equivalent mass of a given substance depends upon the chemical reaction in which the substance is involved.

Consider the positive electric charge and the first reaction. Because Na+ of NaOH (remember that only the reactants are to be considered in choosing the reference species) has a charge of +1 and the coefficient of the term is 2, two positive charges (2 x 1 = 2) are involved. In the case of H3PO4, the equation shows that the acid breaks up into HPO^ and other substances with one H still "clinging" to the PO4 on the right-hand side of the equation. This indicates that two H+'s are involved in the breakup. Because the charge of H+ is +1, two positive charges are accordingly involved. In both the cases of Na+ and H+, the reference species are the two positive charges and the number of reference species is 2. Therefore, in the first reaction, the equivalent mass of a participating substance is obtained by dividing the term (including the coefficient) by 2. Thus, for the acid, the equivalent mass is H3PO4/2; for the base, the equivalent mass is 2NaOH/2, etc.

In the second reaction, again, basing on the positive electric charges and performing similar analysis, the number of reference species would be found to be +3. Thus, in this reaction, for the acid, the equivalent mass is H3PO4/3; for the base, the equivalent mass is 3NaOH/3, etc., indicating differences in equivalent masses with the first reaction for the same substances of H3PO4 and NaOH. Thus, the equivalent mass of any substance depends upon the chemical reaction in which it participates.

In the previous discussions, the unit of the number of reference species was not established. A convenient unit would be the mole (i.e., mole of electrons or mole of positive or negative charges). The mole can be a milligram-mole, gram-mole, etc. The mass unit of measurement of the equivalent mass would then correspond to the type of mole used for the reference species. For example, if the mole used is the gram-mole, the mass of the equivalent mass would be expressed in grams of the substance per gram-mole of the reference species; and, if the mole used is the milligram-mole, the equivalent mass would be expressed in milligrams of the substance per milligrammole of the reference species and so on.

Because the reference species is used as the standard of reference, its unit, the mole, can be said to have a unit of one equivalent. From this, the equivalent mass of a participating substance may be expressed as the mass of the substance per equivalent of the reference species; but, because the substance is equivalent to the reference species, the expression "per equivalent of the reference species" is the same as the expression "per equivalent of the substance." Thus, the equivalent mass of a substance may also be expressed as the mass of the substance per equivalent of the substance.

Each term of a balanced chemical reaction, represents the mass of a participating substance. Thus, the general formula for finding the equivalent mass of a substance is

number of equivalents of substance = term in balanced reaction number of moles of reference species

Example 5 Water containing 2.5 moles of calcium bicarbonate and 1.5 moles of calcium sulfate is softened using lime and soda ash. How many grams of calcium carbonate solids are produced (a) using the method of equivalent masses and (b) using the balanced chemical reaction? Pertinent reactions are as follows:

Ca(HCO3)2 + Ca(OH)2 ^ 2CaCO31 + 2H2O CaSO4 + Na2CO3 ^ CaCO3 + Na2SO4

Solution:

(a) Ca(HCO3)2 + Ca(OH)2 ^ 2CaCO31 + 2H2O number of reference species = 2

eq. mass of Ca (HCO3 )2 = ^a("CCO3)2 = 4400 + 2[1 + j2 + 3(16)-j16-2 = 81

g of CaCO3 solids = 5( 100) = 500 g Ans CaSO4 + Na2CO3 ^ CaCO3 + Na2SO4

number of reference species = 2

CaSO4 40 + 32 + 4( 16) 136 ,fi eq. mass of CaSO4 = —^— = -^—-—- = — = 68

g of CaCO3 solids = 3 (50) = 150 g Ans Total grams of CaCO3 = 500 + 150 = 650 Ans

g of CaCO3 solids = ----A (2.5)(162) = 500 Ca (HCO3 )2

CaSO4 + Na2CO3 ^ CaCO3 + Na2SO4

CaSO4

Methods of Expressing Concentrations

Several methods are used to express concentrations in water and wastewater and it is appropriate to present some of them here. They are molarity, molality, mole fraction, mass concentration, equivalents concentration, and normality.

Molarity. Molarity is the number of gram moles of solute per liter of solution, where from the general knowledge of chemistry, gram moles is the mass in grams divided by the molecular mass of the substance in question. Take note that the statement says "per liter of solution." This means the solute and the solvent are taken together as a mixture in the liter of solution. The symbol used for molarity is M.

For example, consider calcium carbonate. This substance has a mass density Pcaco3 equals 2.6 g/cc. Suppose, 35 mg is dissolved in a liter of water, find the corresponding molarity.

The mass of 35 mg is 0.035 g. Calcium carbonate has a molecular weight of 100 g/mol; thus, 0.035 g is 0.035/100 = 0.00035 gmol. The volume corresponding to 0.35 g is 0.035/2.6 = 0.0135 cc = 0.0000135 L and the total volume of the mixture then is 1.0000135 L. Therefore, by the definition of molarity, the corresponding molarity of 35 mg dissolved in one liter of water is 0.00035/1.0000135 = 0.00035 M.

Molality. Molarity is the number of gram moles of solute per 1000 g of solvent. Take note of the drastic difference between this definition of molality and the difinition of molarity. The solvent is now "separate" from the solute. The symbol used for molality is m.

Now, consider the calcium carbonate example above, again, and find the corresponding molality. The only other calculation we need to do is to find the number of grams of the liter of water. To do this, an assumption of the water temperature must be made. Assuming it is 5°C, its mass density is 1.0 g/cc and the mass of one liter is then 1000 g. Thus, the 0.00035 gmol of calcium carbonate is dissolved in 1000 gm of water. This is the very definition of molality and the corresponding molality is therefore 0.00035m. Note that there is really no practical difference between molarity and molality in this instance. Note that the mass density of water does not vary much from the 5°C to 100°C.

Mole fraction. Mole fraction is a method of expressing the molar fractional part of a certain species relative to the total number of moles of species in the mixture. Letting ni be the number of moles of a particular species i, the mole fraction of this species xt is

N is the total number of species in the mixture.

Example 6 The results of an analysis in a sample of water are shown in the table below. Calculate the mole fractions of the respective species.

Ions

Ca(HCO3)2

Mg(HCO3)2

Na2SO4

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