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) or no. eq analyte present no. eq standard reagent added (A6-2) 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. A6-1
A6-2 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: eqw Ba(OH) 2 M 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: H 3 PO 4 OH : H 2 PO 4 H 2 O With certain other indicators, a color change occurs only after two hydrogen ions have reacted: H 3 PO 4 2OH : HPO 2 4 2H 2 O 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 PO 4.) 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: 5C 2 O 2 4 2MnO 4 16H : 10CO 2 2Mn 8H 2 O (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 Mn 2 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.
Web Appendix 6 A6-3 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 2 O 3 is to be determined by a titration based on the reaction given in Equation A6-3. That each manganese in the Mn 2 O 3 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 2 O 3 is then reduced from the 7 to the 2 state in the titration step. The equivalent weight is thus the molar mass of Mn 2 O 3 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 MnO 2 : MnO 4 3e 2H 2 O : MnO 2 (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 AgNO 3 : 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 AlCl 3 M AlCl 3 3 Because each mole of BiOCl reacts with only 1 Ag ion, eqw BiOCl M BiOCl 1 Note that Bi 3 (or Al 3 ) being trivalent has no bearing because the definition is based on the cation involved in the titration: Ag.
A6-4 Web Appendix 6 A6-2 THE DEFINITION OF NORMALITY The normality c N 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 c N(A) is given by the equations c N(A) c N(A) no. meq A no. ml solution no. eq A no. L solution (A6-4) (A6-5) A6-3 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 mass A (g) meqw A (g/meq) (A6-6) amount A no. eq A mass A (g) eqw A (g/eq) (A6-7) amount A no. meq A V (ml) c N(A) (meq/ml) amount A no. eq A V (L) c N(A) (eq/l) (A6-8) (A6-9) A6-4 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 2 CO 3 (105.99 g/mol) from the primary-standard solid, assuming that the solution is to be used for titrations in which the reaction is Applying Equation A6-9 gives CO 2 3 2H : H 2 O CO 2 amount Na 2 CO 3 V soln (L) c N(Na2 CO 3 )(eq/l) 5.000 L 0.1000 eq/l 0.5000 eq Na 2 CO 3 Rearranging Equation A6-7 gives mass Na 2 CO 3 no. eq Na 2 CO 3 eqw Na 2 CO 3
Web Appendix 6 A6-5 But 2 eq of Na 2 CO 3 are contained in each mole of the compound; therefore, mass Na 2 CO 3 0.5000 eq Na 2 CO 3 105.99 g Na 2CO 3 2 eq Na 2 CO 3 26.50 g 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, c Ba(OH)2 0.03926 meq ml 1 mmol 2 meq 0.01963 M 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 c N(HCl) (29.71 0.03926 1) meq HCl c N(HCl) (29.71 0.03926 1) meq HCl 50.00 ml HCl 0.02333 N
A6-6 Web Appendix 6 Example A6-4 A 0.2121-g sample of pure Na 2 C 2 O 4 (134.00 g/mol) was titrated with 43.31 ml of KMnO 4. What is the normality of the KMnO 4 solution? The chemical reaction is 2MnO 4 5C 2 O 2 4 16H : 2Mn 2 10CO 2 8H 2 O By definition, at the equivalence point in the titration, no. meq Na 2 C 2 O 4 no. meq KMnO 4 Substituting Equations A6-6 and A6-8 into this relationship gives V KMnO4 c N(KMnO4 ) mass Na 2 C 2 O 4 (g) meqw Na 2 C 2 O 4 (g/meq) 43.31 ml KMnO 4 c N(KMnO4 ) 0.2121 g Na 2 C 2 O 4 0.13400 g Na 2 C 2 O 4 /2 meq c N(KMnO4 ) 0.2121 g Na 2 C 2 O 4 43.31 ml KMnO 4 0.1340 g Na 2 C 2 O 4 /2 meq 0.073093 meq/ml KMnO 4 0.07309 N 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) KMnO 4 solution. Calculate the results of this analysis in terms of (a) percent Fe (55.847 g/mol) and (b) percent Fe 3 O 4 (231.54 g/mol). The reaction of the analyte with the reagent is described by the equation MnO 4 5Fe 2 8H : Mn 2 5Fe 3 4H 2 O (a) At the equivalence point, we know that no. meq KMnO 4 no. meq Fe 2 no. meq Fe 3 O 4 Substituting Equations A6-8 and A6-6 leads to V KMnO4 (ml) c N(KMnO4 )(meq/ml) mass Fe 2 (g) meqw Fe 2 (g/meq)
Web Appendix 6 A6-7 Substituting numerical data into the equation gives, after rearranging, mass Fe 2 47.22 ml KMnO 4 0.1121 meq ml KMnO 4 0.055847 g 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 percent Fe 2 Fe2 0.8040 g sample 36.77% 100% (b) Here, no. meq KMnO 4 no. meq Fe 3 O 4 and V KMnO4 (ml) c N(KMnO4 )(meq/ml) Substituting numerical data and rearranging give mass Fe 3 O 4 (g) meqw Fe 3 O 4 (g/meq) mass Fe 3 O 4 47.22 ml 0.1121 meq ml 0.23154 g Fe 3O 4 3 meq Note that the milliequivalent weight of Fe 3 O 4 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 3 O 4 is then percent Fe 3 O 4 (47.22 0.1121 0.23154/3) g Fe 3O 4 0.8040 g sample 50.81% 100% 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 ) 2 C 2 O 4 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 2 SO 4. The excess H 2 SO 4 was backtitrated with 11.13 ml of 0.1214 N NaOH. Calculate the percentage of N (14.007 g/mol) and of (NH 4 ) 2 C 2 O 4 (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 H 2 SO 4 no. meq NH 3 no. meq NaOH After rearranging, no. meq NH 3 no. meq N no. meq H 2 SO 4 no. meq NaOH
A6-8 Web Appendix 6 Substituting Equations A6-6 and A6-8 for the number of milliequivalents of N and H 2 SO 4, respectively, yields mass N (g) meqw N (g/meq) 50.00 ml H 2SO 4 0.1007 meq ml H 2 SO 4 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 percent N (50.00 0.1007 11.13 0.1214) 0.014007 g N 0.4755 g sample 10.85% 100% The number of milliequivalents of (NH 4 ) 2 C 2 O 4 is equal to the number of milliequivalents of NH 3 and N, but the milliequivalent weight of the (NH 4 ) 2 C 2 O 4 is equal to one-half its molar mass. Thus, mass (NH 4 ) 2 C 2 O 4 (50.00 0.1007 11.13 0.1214) meq 0.12410 g/2 meq percent (NH 4 ) 2 C 2 O 4 (50.00 0.1007 11.13 0.1214) 0.06205 g (NH 4) 2 C 2 O 4 0.4755 g sample 48.07% 100% Note that the results obtained here are identical to those obtained in Example 6-8.