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Web Appendix 6
Volumetric Calculations Using
Normality and Equivalent Weight
The normality of a solution expresses the number of equivalents of solute contained
in 1 L of solution or the number of milliequivalents in 1 mL. The equivalent and
milliequivalent, like the mole and millimole, are units for describing the amount of a
chemical species. The former two, however, are defined so that we may state that, at
the equivalence point in any titration,
no. meq analyte present ⫽ no. meq standard reagent added
(A6-1)
no. eq analyte present ⫽ no. eq standard reagent added
(A6-2)
or
As a consequence, stoichiometric ratios such as those described in Section 2C-2
(page 31) need not be derived every time a volumetric calculation is performed.
Instead, the stoichiometry is taken into account by how the equivalent or milliequivalent weight is defined.
A6-1
THE DEFINITIONS OF EQUIVALENT
AND MILLIEQUIVALENT
In contrast to the mole, the amount of a substance contained in one equivalent can vary
from reaction to reaction. Consequently, the weight of one equivalent of a compound
can never be computed without reference to a chemical reaction in which that compound is, directly or indirectly, a participant. Similarly, the normality of a solution can
never be specified without knowledge about how the solution will be used.
Equivalent Weights in Neutralization Reactions
One equivalent weight of a substance participating in a neutralization reaction is
that amount of substance (molecule, ion, or paired ion such as NaOH) that either
reacts with or supplies 1 mol of hydrogen ions in that reaction.1 A milliequivalent
is simply 1/1000 of an equivalent.
1
An alternative definition, proposed by the International Union of Pure and Applied Chemistry, is as
follows: An equivalent is “that amount of substance, which, in a specified reaction, releases or
replaces that amount of hydrogen that is combined with 3 g of carbon-12 in methane 12 CH 4” (see
Information Bulletin No. 36, International Union of Pure and Applied Chemistry, August 1974). This
definition applies to acids. For other types of reactions and reagents, the amount of hydrogen referred
to may be replaced by the equivalent amount of hydroxide ions, electrons, or cations. The reaction to
which the definition is applied must be specified.
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Web Appendix 6
Once again, we find ourselves using the term
weight when we really mean mass. The term
“equivalent weight” is so firmly engrained in
the literature and vocabulary of chemistry
that we retain it in this discussion.
The relationship between equivalent weight (eqw) and the molar mass (M) is
straightforward for strong acids or bases and for other acids or bases that contain
a single reactive hydrogen or hydroxide ion. For example, the equivalent weights
of potassium hydroxide, hydrochloric acid, and acetic acid are equal to their
molar masses because each has but a single reactive hydrogen ion or hydroxide
ion. Barium hydroxide, which contains two identical hydroxide ions, reacts with
two hydrogen ions in any acid/base reaction, and so its equivalent weight is onehalf its molar mass:
MBa(OH)2
eqw Ba(OH)2 ⫽
2
The situation becomes more complex for acids or bases that contain two or more
reactive hydrogen or hydroxide ions with different tendencies to dissociate. With
certain indicators, for example, only the first of the three protons in phosphoric acid
is titrated:
H3PO4 ⫹ OH⫺ : H2PO⫺
4 ⫹ H2O
With certain other indicators, a color change occurs only after two hydrogen ions
have reacted:
H3PO4 ⫹ 2OH⫺ : HPO2⫺
4 ⫹ 2H2O
For a titration involving the first reaction, the equivalent weight of phosphoric
acid is equal to the molar mass; for the second, the equivalent weight is onehalf the molar mass. (Because it is not practical to titrate the third proton, an
equivalent weight that is one-third the molar mass is not generally encountered for H 3 PO4 .) If it is not known which of these reactions is involved, an unambiguous definition of the equivalent weight for phosphoric acid cannot
be made.
Equivalent Weights in Oxidation/Reduction Reactions
The equivalent weight of a participant in an oxidation/reduction reaction is that
amount that directly or indirectly produces or consumes 1 mol of electrons. The
numerical value for the equivalent weight is conveniently established by dividing
the molar mass of the substance of interest by the change in oxidation number
associated with its reaction. As an example, consider the oxidation of oxalate ion
by permanganate ion:
⫺
⫹
⫹
5C2O2⫺
4 ⫹ 2MnO4 ⫹ 16H : 10CO2 ⫹ 2Mn ⫹ 8H2O
(A6-3)
In this reaction, the change in oxidation number of manganese is 5 because the
element passes from the ⫹ 7 to the ⫹ 2 state; the equivalent weights for MnO⫺
4
and Mn2⫹ are therefore one-fifth their molar masses. Each carbon atom in the
oxalate ion is oxidized from the ⫹ 3 to the ⫹ 4 state, leading to the production of two electrons by that species. Therefore, the equivalent weight of sodium
oxalate is one-half its molar mass. It is also possible to assign an equivalent
weight to the carbon dioxide produced by the reaction. Since this molecule
contains but a single carbon atom and since that carbon undergoes a change in
oxidation number of 1, the molar mass and equivalent weight of the two are
identical.
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It is important to note that in evaluating the equivalent weight of a substance, only its change in oxidation number during the titration is considered.
For example, suppose the manganese content of a sample containing Mn 2O3 is
to be determined by a titration based on the reaction given in Equation A6-3.
That each manganese in the Mn 2O3 has an oxidation number of ⫹ 3 plays no
part in determining equivalent weight. Thus, we must assume that, by suitable
treatment, all the manganese is oxidized to the ⫹ 7 state before the titration is
begun. Each manganese from the Mn 2O3 is then reduced from the ⫹ 7 to the
⫹ 2 state in the titration step. The equivalent weight is thus the molar mass of
Mn 2O3 divided by 2 ⫻ 5 ⫽ 10.
As in neutralization reactions, the equivalent weight for a given oxidizing or
reducing agent is not invariant. Potassium permanganate, for example, reacts
under some conditions to give MnO2 :
⫺
⫺
MnO⫺
4 ⫹ 3e ⫹ 2H2O : MnO2(s) ⫹ 4OH
The change in the oxidation state of manganese in this reaction is from ⫹ 7 to
⫹ 4, and the equivalent weight of potassium permanganate is now equal to its
molar mass divided by 3 (instead of 5 as in the earlier example).
Equivalent Weights in Precipitation
and Complex-Formation Reactions
The equivalent weight of a participant in a precipitation or a complex-formation
reaction is the weight that reacts with or provides one mole of the reacting cation
if it is univalent, one-half mole if it is divalent, one-third mole if it is trivalent, and
so on. It is important to note that the cation referred to in this definition is always
the cation directly involved in the analytical reaction and not necessarily the
cation contained in the compound whose equivalent weight is being defined.
Example A6-1
Define equivalent weights for AlCl 3 and BiOCl if the two compounds are determined by a precipitation titration with AgNO3 :
Ag⫹ ⫹ Cl⫺ : AgCl(s)
In this instance, the equivalent weight is based on the number of moles of
silver ions involved in the titration of each compound. Since 1 mol of Ag ⫹
reacts with 1 mol of Cl⫺ provided by one-third mole of AlCl 3 , we can write
eqw AlCl3 ⫽
MAlCl3
3
Because each mole of BiOCl reacts with only 1 Ag ⫹ ion,
eqw BiOCl ⫽
MBiOCl
1
Note that Bi3⫹ (or Al3⫹) being trivalent has no bearing because the definition is
based on the cation involved in the titration: Ag ⫹.
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A6-2
THE DEFINITION OF NORMALITY
The normality cN of a solution expresses the number of milliequivalents of solute
contained in 1 mL of solution or the number of equivalents contained in 1 L.
Thus, a 0.20 N hydrochloric acid solution contains 0.20 meq of HCl in each milliliter of solution or 0.20 eq in each liter.
The normal concentration of a solution is defined by equations analogous to
Equations 2-1 and 6-3. Thus, for a solution of the species A, the normality cN(A)
is given by the equations
A6-3
cN(A) ⫽
no. meq A
no. mL solution
(A6-4)
cN(A) ⫽
no. eq A
no. L solution
(A6-5)
SOME USEFUL ALGEBRAIC RELATIONSHIPS
Two pairs of algebraic equations, analogous to Equations 6-1 and 6-2 as well as
6-3 and 6-4 in Chapter 6, apply when normal concentrations are used:
amount A ⫽ no. meq A ⫽
amount A ⫽ no. eq A ⫽
A6-4
mass A (g)
meqw A (g/meq)
mass A (g)
eqw A (g/eq)
(A6-6)
(A6-7)
amount A ⫽ no. meq A ⫽ V (mL) ⫻ cN(A)(meq/mL)
(A6-8)
amount A ⫽ no. eq A ⫽ V (L) ⫻ cN(A)(eq/L)
(A6-9)
CALCULATION OF THE NORMALITY
OF STANDARD SOLUTIONS
Example A6-2 shows how the normality of a standard solution is computed from
preparatory data. Note the similarity between this example and Example 6-1.
Example A6-2
Describe the preparation of 5.000 L of 0.1000 N Na 2CO3 (105.99 g/mol) from
the primary-standard solid, assuming that the solution is to be used for titrations
in which the reaction is
⫹
CO2⫺
3 ⫹ 2H : H2O ⫹ CO2
Applying Equation A6-9 gives
amount Na2CO3 ⫽ V soln (L) ⫻ cN(Na2CO3)(eq/L)
⫽ 5.000 L ⫻ 0.1000 eq/L ⫽ 0.5000 eq Na2CO3
Rearranging Equation A6-7 gives
mass Na2CO3 ⫽ no. eq Na2CO3 ⫻ eqw Na2CO3
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But 2 eq of Na 2CO3 are contained in each mole of the compound; therefore,
mass Na2CO3 ⫽ 0.5000 eq Na2CO3 ⫻
105.99 g Na2CO3
⫽ 26.50 g
2 eq Na2CO3
Therefore, dissolve 26.50 g in water and dilute to 5.000 L.
It is worth noting that when the carbonate ion reacts with two protons, the
weight of sodium carbonate required to prepare a 0.10 N solution is just one-half
that required to prepare a 0.10 M solution.
A6-5
THE USE OF NORMALITIES TO
TREAT TITRATION DATA
Calculating Normalities from Titration Data
Examples A6-3 and A6-4 illustrate how normality is obtained from standardization data. Note that these examples are similar to Examples 6-4 and 6-5 in
Chapter 6.
Example A6-3
A 50.00 mL volume of an HCl solution required 29.71 mL of 0.03926 N
Ba(OH)2 to give an end point with bromocresol green indicator. Calculate the
normality of the HCl.
Note that the molarity of Ba(OH)2 is one-half its normality. That is,
cBa(OH)2 ⫽ 0.03926
meq
1 mmol
⫻
⫽ 0.01963 M
mL
2 meq
Because we are basing our calculations on the milliequivalent, we write
no. meq HCl ⫽ no. meq Ba(OH)2
The number of milliequivalents of standard is obtained by substituting into
Equation A6-8:
amount Ba(OH)2 ⫽ 29.71 mL Ba(OH)2 ⫻ 0.03926
meq Ba(OH)2
mL Ba(OH)2
To obtain the number of milliequivalents of HCl, we write
amount HCl ⫽ (29.71 ⫻ 0.03926) meq Ba(OH)2 ⫻
1 meq HCl
1 meq Ba(OH)2
Equating this result to Equation A6-8 yields
amount HCl ⫽ 50.00 mL ⫻ cN(HCl)
⫽ (29.71 ⫻ 0.03926 ⫻ 1) meq HCl
cN(HCl) ⫽
(29.71 ⫻ 0.03926 ⫻ 1) meq HCl
⫽ 0.02333 N
50.00 mL HCl
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Web Appendix 6
Example A6-4
A 0.2121-g sample of pure Na 2C 2O4 (134.00 g/mol) was titrated with 43.31
mL of KMnO4 . What is the normality of the KMnO4 solution? The chemical
reaction is
2⫺
⫹
2⫹
2MnO⫺
⫹ 10CO2 ⫹ 8H2O
4 ⫹ 5C2O4 ⫹ 16H : 2Mn
By definition, at the equivalence point in the titration,
no. meq Na2C2O4 ⫽ no. meq KMnO4
Substituting Equations A6-6 and A6-8 into this relationship gives
VKMnO4 ⫻ cN(KMnO4) ⫽
43.31 mL KMnO4 ⫻ cN(KMnO4) ⫽
mass Na2C2O4(g)
meqw Na2C2O4 (g/meq)
0.2121 g Na2C2O4
0.13400 g Na2C2O4/2 meq
0.2121 g Na2C2O4
43.31 mL KMnO4 ⫻ 0.1340 g Na2C2O4/2 meq
⫽ 0.073093 meq/mL KMnO4 ⫽ 0.07309 N
cN(KMnO4) ⫽
Note that the normality found here is five times the molarity computed in
Example 6-5.
CALCULATING THE QUANTITY
OF ANALYTE FROM TITRATION DATA
The examples that follow illustrate how analyte concentrations are calculated when
normalities are involved. Note that these examples are similar to Examples 6-6
and 6-8.
Example A6-5
A 0.8040-g sample of an iron ore was dissolved in acid. The iron was then
reduced to Fe 2⫹ and titrated with 47.22 mL of 0.l121 N (0.02242 M) KMnO4
solution. Calculate the results of this analysis in terms of (a) percent Fe (55.847
g/mol) and (b) percent Fe 3O4 (231.54 g/mol). The reaction of the analyte with
the reagent is described by the equation
2⫹
MnO⫺
⫹ 8H⫹ : Mn2⫹ ⫹ 5Fe3⫹ ⫹ 4H2O
4 ⫹ 5Fe
(a) At the equivalence point, we know that
no. meq KMnO4 ⫽ no. meq Fe2⫹ ⫽ no. meq Fe3O4
Substituting Equations A6-8 and A6-6 leads to
VKMnO4(mL) ⫻ cN(KMnO4)(meq/mL) ⫽
mass Fe2⫹ (g)
meqw Fe2⫹ (g/meq)
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Substituting numerical data into the equation gives, after rearranging,
mass Fe2⫹ ⫽ 47.22 mL KMnO4 ⫻ 0.1121
meq
0.055847 g
⫻
mL KMnO4
1 meq
Note that the milliequivalent weight of the Fe 2⫹ is equal to its millimolar
mass. The percentage of iron is
(47.22 ⫻ 0.1121 ⫻ 0.055847) g Fe2⫹
⫻ 100%
0.8040 g sample
⫽ 36.77%
percent Fe2⫹ ⫽
(b) Here,
no. meq KMnO4 ⫽ no. meq Fe3O4
and
VKMnO4(mL) ⫻ cN(KMnO4)(meq/mL) ⫽
mass Fe3O4(g)
meqw Fe3O4(g/meq)
Substituting numerical data and rearranging give
mass Fe3O4 ⫽ 47.22 mL ⫻ 0.1121
meq
g Fe3O4
⫻ 0.23154
mL
3 meq
Note that the milliequivalent weight of Fe 3O4 is one-third its millimolar
mass because each Fe 2⫹ undergoes a one-electron change and the compound is converted to 3Fe 2⫹ before titration. The percentage of Fe 3O4 is
then
percent Fe3O4 ⫽
(47.22 ⫻ 0.1121 ⫻ 0.23154/3) g Fe3O4
⫻ 100%
0.8040 g sample
⫽ 50.81%
Note that the answers to this example are identical to those in Example 6-6.
Example A6-6
A 0.4755-g sample containing (NH 4)2C 2O4 and inert compounds was dissolved
in water and made alkaline with KOH. The liberated NH 3 was distilled into
50.00 mL of 0.1007 N (0.05035 M) H 2SO4 . The excess H 2SO4 was backtitrated with 11.13 mL of 0.1214 N NaOH. Calculate the percentage of N
(14.007 g/mol) and of (NH 4)2C 2O4 (124.10 g/mol) in the sample.
At the equivalence point, the number of milliequivalents of acid and base
are equal. In this titration, however, two bases are involved: NaOH and NH 3 .
Thus,
no. meq H2SO4 ⫽ no. meq NH3 ⫹ no. meq NaOH
After rearranging,
no. meq NH3 ⫽ no. meq N ⫽ no. meq H2SO4 ⫺ no. meq NaOH
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Substituting Equations A6-6 and A6-8 for the number of milliequivalents of
N and H 2SO4 , respectively, yields
mass N (g)
meq
⫽ 50.00 mL H2SO4 ⫻ 0.1007
meqw N (g/meq)
mL H2SO4
⫺11.13 mL NaOH ⫻ 0.1214
meq
mL NaOH
mass N ⫽ (50.00 ⫻ 0.1007 ⫺ 11.13 ⫻ 0.1214) meq ⫻ 0.014007 g N/meq
(50.00 ⫻ 0.1007 ⫺ 11.13 ⫻ 0.1214) ⫻ 0.014007 g N
⫻ 100%
0.4755 g sample
⫽ 10.85%
percent N ⫽
The number of milliequivalents of (NH 4)2C 2O4 is equal to the number of milliequivalents of NH 3 and N, but the milliequivalent weight of the (NH 4)2C 2O4
is equal to one-half its molar mass. Thus,
mass (NH4)2C2O4 ⫽ (50.00 ⫻ 0.1007 ⫺ 11.13 ⫻ 0.1214) meq
⫻ 0.12410 g/2 meq
percent (NH4)2C2O4
(50.00 ⫻ 0.1007 ⫺ 11.13 ⫻ 0.1214) ⫻ 0.06205 g (NH4)2C2O4
⫻ 100%
0.4755 g sample
⫽ 48.07%
⫽
Note that the results obtained here are identical to those obtained in Example 6-8.