Modern Analytical Chemistry Table 2. 4 Common Units for Reporting Concentration molarity moles solute formality umber Fws solute iters solutio normality number ews solute liters solution moles solute volume mL solute v/ weight-to-volume g solute parts per milli ppm parts per billion g solute 10 g solution aFW= formula EW=equivalent weight. instead, is 0.1 M in Na* and 0. 1 M in CI. The formality of NaCl, however, is 0.1 F because it represents the total amount of NaCl in solution. The rigorous definition of molarity, for better or worse, is largely ignored in the current literature, as it is in this text. when we state that a solution is o1 M Nacl we understand it to consist of Na+ and CH ions. The unit of formality is used only when it provides a clearer de- scription of solution chemistry Molar concentrations are used so frequently that a symbolic notation is often used to simplify its expression in equations and writing. The use of square brackets around a species indicates that we are referring to that species molar concentration Thus, [Na*] is read as the "molar concentration of sodium ions. 2B.2 Normality Normality is an older unit of concentration that, although once commonly used,is frequently ignored in todays laboratories. Normality is still used in some hand- books of analytical methods, and, for this reason, it is helpful to understand its neaning. For example, normality is the concentration unit used in Standard Meth- ods for the Examination of water and Wa lytical methods for environmental laboratories. Normality makes use of the chemical equivalent, which is the amount of one The number of equivalents of solute per chemical species reacting stoichiometrically with another chemical species. Note ter of solution(N) that this definition makes an equivalent, and thus normality, a function of the chemical reaction in which the species participates. Although a solution of H2So4 has a fixed molarity, its normality depends on how it reacts
instead, is 0.1 M in Na+ and 0.1 M in Cl–. The formality of NaCl, however, is 0.1 F because it represents the total amount of NaCl in solution. The rigorous definition of molarity, for better or worse, is largely ignored in the current literature, as it is in this text. When we state that a solution is 0.1 M NaCl we understand it to consist of Na+ and Cl– ions. The unit of formality is used only when it provides a clearer description of solution chemistry. Molar concentrations are used so frequently that a symbolic notation is often used to simplify its expression in equations and writing. The use of square brackets around a species indicates that we are referring to that species’ molar concentration. Thus, [Na+] is read as the “molar concentration of sodium ions.” 2B.2 Normality Normality is an older unit of concentration that, although once commonly used, is frequently ignored in today’s laboratories. Normality is still used in some handbooks of analytical methods, and, for this reason, it is helpful to understand its meaning. For example, normality is the concentration unit used in Standard Methods for the Examination of Water and Wastewater,1 a commonly used source of analytical methods for environmental laboratories. Normality makes use of the chemical equivalent, which is the amount of one chemical species reacting stoichiometrically with another chemical species. Note that this definition makes an equivalent, and thus normality, a function of the chemical reaction in which the species participates. Although a solution of H2SO4 has a fixed molarity, its normality depends on how it reacts. 16 Modern Analytical Chemistry Table 2.4 Common Units for Reporting Concentration Name Unitsa Symbol molarity M formality F normality N molality m weight % % w/w volume % % v/v weight-to-volume % % w/v parts per million ppm parts per billion ppb aFW = formula weight; EW = equivalent weight. moles solute liters solution number F solute liters solution Ws number E solute liters solution Ws m solute k solvent oles g g solute 100 g solution m solute solution L 100 mL g solute 100 mL solution g solute 10 g solution 6 g solute 10 g solution 9 normality The number of equivalents of solute per liter of solution (N). 1400-CH02 9/8/99 3:48 PM Page 16
Chapter 2 Basic Tools of Analytical Chemistry The number of equivalents, n, is based on a reaction unit, which is that part of a chemical species involved in a reaction. In a precipitation reaction, for example, quivalent The moles of a species that can donat the reaction unit is the charge of the cation or anion involved in the reaction; thus one reaction unit. for the reaction Pb2+(aq)+21-(ag) Pbl2(s) n=2 for Pb2+ and n= l for I-. In an acid-base reaction the reaction unit is the number of H+ ions donated by an acid or accepted by a base. For the reaction be- tween sulfuric acid and ammonia H2so,(ag)+ 2NH3(ag)*(ag)+so2(aq) we find that n=2 for H2SO4 and n=l for NH3. For a complexation reaction, the reaction unit is the number of electron pairs that can be accepted by the metal or donated by the ligand In the reaction between Ag* and Nh Ag*(aq)+ 2NH3(ag)Ag(NH3)2(ag) the value of n for Agt is 2 and that for NH3 is 1. Finally, in an oxidation-reduction reaction the reaction unit is the number of electrons released by the reducing agent or accepted by the oxidizing agent; thus, for the reaction ag)e Sn**(ag)+2Fe2+(ag) n=I for Fe+ and n= 2 for Sn2t. Clearly, determining the number of equivalents for a chemical species requires an understanding of how it reacts Normality is the number of equivalent weights(EW) per unit volume and quivalent weight like formality, is independent of speciation. An equivalent weight is defined as the The I ratio of a chemical species formula weight(FW) to the number of its equivalents quivalent(EW) EW= FW formula weight Consequently, the following simple relationship exists between normality and molarity Example 2. 1 illustrates the relationship among chemical reactivity, equivalent weight, and normality. EXAMPLE 2. Calculate the equivalent weight and normality for a solution of 6.0 M H3 PO4 a given the following reactions C(a)H3 PO,(aq)+30H-(aq)ePO()+3H20(e) (b)H3PO,(ag)+2NH3(ag)HPO4(aq)+ 2NH*(aq) (c)H3PO4(ag)+ F(ag)e H2PO4(ag)+ Hf(ag) SOLUTION For phosphoric acid, the number of equivalents is the number of H* ions donated to the base. For the reactions in(a),(b), and (c) the number of equivalents are 3, 2, and 1, respectively. Thus, the calculated equivalent weights and normalities are
The number of equivalents, n, is based on a reaction unit, which is that part of a chemical species involved in a reaction. In a precipitation reaction, for example, the reaction unit is the charge of the cation or anion involved in the reaction; thus for the reaction Pb2+(aq) + 2I–(aq) t PbI2(s) n = 2 for Pb2+ and n = 1 for I–. In an acid–base reaction, the reaction unit is the number of H+ ions donated by an acid or accepted by a base. For the reaction between sulfuric acid and ammonia H2SO4(aq) + 2NH3(aq) t 2NH4 +(aq) + SO4 2–(aq) we find that n = 2 for H2SO4 and n = 1 for NH3. For a complexation reaction, the reaction unit is the number of electron pairs that can be accepted by the metal or donated by the ligand. In the reaction between Ag+ and NH3 Ag+(aq) + 2NH3(aq) t Ag(NH3)2 +(aq) the value of n for Ag+ is 2 and that for NH3 is 1. Finally, in an oxidation–reduction reaction the reaction unit is the number of electrons released by the reducing agent or accepted by the oxidizing agent; thus, for the reaction 2Fe3+(aq) + Sn2+(aq) t Sn4+(aq) + 2Fe2+(aq) n = 1 for Fe3+ and n = 2 for Sn2+. Clearly, determining the number of equivalents for a chemical species requires an understanding of how it reacts. Normality is the number of equivalent weights (EW) per unit volume and, like formality, is independent of speciation. An equivalent weight is defined as the ratio of a chemical species’ formula weight (FW) to the number of its equivalents Consequently, the following simple relationship exists between normality and molarity. N = n × M Example 2.1 illustrates the relationship among chemical reactivity, equivalent weight, and normality. EXAMPLE 2.1 Calculate the equivalent weight and normality for a solution of 6.0 M H3PO4 given the following reactions: (a) H3PO4(aq) + 3OH–(aq) t PO4 3–(aq) + 3H2O(l) (b) H3PO4(aq) + 2NH3(aq) t HPO4 2–(aq) + 2NH4 +(aq) (c) H3PO4(aq) + F–(aq) t H2PO4 –(aq) + HF(aq) SOLUTION For phosphoric acid, the number of equivalents is the number of H+ ions donated to the base. For the reactions in (a), (b), and (c) the number of equivalents are 3, 2, and 1, respectively. Thus, the calculated equivalent weights and normalities are EW = FW n Chapter 2 Basic Tools of Analytical Chemistry 17 equivalent The moles of a species that can donate one reaction unit. equivalent weight The mass of a compound containing one equivalent (EW). formula weight The mass of a compound containing one mole (FW). 1400-CH02 9/8/99 3:48 PM Page 17
Modern Analytical Chemistry (a)Ew=FW_97.99 =32.665N=n×M=3×6.0=18N (b) EW= FW97.994 48.997N=nxM=2×6.0=12N 2 FW9799 (c) EW =97994N=nXM=1×6.0=6.0N 2B.3 Molality Molality is used in thermodynamic calculations where a temperature independent The number of moles of solute per unit of concentration is needed. Molarity, formality and normality are based on the kilogram of solvent (m) volume of solution in which the solute is dissolved. Since density is a temperature de- pendent property a solution s volume, and thus its molar, formal and normal concen trations,will change as a function of its temperature. By using the solvent's mass in place of its volume, the resulting concentration becomes independent of temperature 2B. 4 Weight, Volume and Weight-to-Volume Ratios eight percent Weight percent(%w/w), volume percent(% v/v)and weight-to-volume percent Grams of solute per 100 g of (%w/v)express concentration as units of solute per 100 units of sample. A solution in which (%w/w) a solute has a concentration of 23%w/v contains 23 g of solute per 100 mL of solution. Parts per million(ppm)and parts per billion(ppb)are mass ratios of grams of solute per 100 m solute to one million or one billion grams of sample, respectively. For example, a steel solution(% v/v) mate the density of an aqueous solution as. 00 g/ml, then gram of steel. If we approxi- olution concentrations can ght-to-volume percent be expressed in parts per million or parts per billion using the following relationships ns of solute per 100 mL of solution PPI 吗gng liter mL as milligrams of For gases a part per million usually is a volume ratio. Thus, a helium concentration solute per liter of solution(ppm). of 6.3 ppm means that one liter of air contains 6.3 uL of He. Nanograms of solute per gram of B5 Converting Between Concentration Units are often expressed as microgramorits lution; for aqueous solution The units of concentration most frequently encountered in analytical chemistry are solute per liter of solution(ppb). molarity, weight percent, volume percent, weight-to-volume percent, parts per mil lion, and parts per billion. By recognizing the general definition of concentration given in equation 2.1, it is easy to convert between concentration units EXAMPLE 2.2 A concentrated solution of aqueous ammonia is 28.0% w/w NH3 and has a density of 0.899 g/mL. What is the molar concentration of NH, in this solution? SOLUTION 28.0gNHs-x0899 g solution I mole NH,* liter+. 8M 100 g solution mL solution 17.04 g NH
18 Modern Analytical Chemistry 2B.3 Molality Molality is used in thermodynamic calculations where a temperature independent unit of concentration is needed. Molarity, formality and normality are based on the volume of solution in which the solute is dissolved. Since density is a temperature dependent property a solution’s volume, and thus its molar, formal and normal concentrations, will change as a function of its temperature. By using the solvent’s mass in place of its volume, the resulting concentration becomes independent of temperature. 2B.4 Weight, Volume, and Weight-to-Volume Ratios Weight percent (% w/w), volume percent (% v/v) and weight-to-volume percent (% w/v) express concentration as units of solute per 100 units of sample. A solution in which a solute has a concentration of 23% w/v contains 23 g of solute per 100 mL of solution. Parts per million (ppm) and parts per billion (ppb) are mass ratios of grams of solute to one million or one billion grams of sample, respectively. For example, a steel that is 450 ppm in Mn contains 450 µg of Mn for every gram of steel. If we approximate the density of an aqueous solution as 1.00 g/mL, then solution concentrations can be expressed in parts per million or parts per billion using the following relationships. For gases a part per million usually is a volume ratio. Thus, a helium concentration of 6.3 ppm means that one liter of air contains 6.3 µL of He. 2B.5 Converting Between Concentration Units The units of concentration most frequently encountered in analytical chemistry are molarity, weight percent, volume percent, weight-to-volume percent, parts per million, and parts per billion. By recognizing the general definition of concentration given in equation 2.1, it is easy to convert between concentration units. EXAMPLE 2.2 A concentrated solution of aqueous ammonia is 28.0% w/w NH3 and has a density of 0.899 g/mL. What is the molar concentration of NH3 in this solution? SOLUTION 28 0 100 0 899 1 17 04 1000 14 8 3 3 3 . . . . L g NH g solution g solution m solution mole NH g NH mL liter × × ×= M ppm = mg liter ppb = g liter = = µ µ g mL ng mL (a) EW = FW = 97.994 3 = 32.665 N = M = 3 6.0 = 18 N (b) EW = FW = 97.994 2 = 48.997 N = M = 2 6.0 = 12 N (c) EW = = 97.994 1 = 97.994 N = M = 1 6.0 = 6.0 N n n n n n n × × × × × × FW molality The number of moles of solute per kilogram of solvent (m). weight percent Grams of solute per 100 g of solution. (% w/w). volume percent Milliliters of solute per 100 mL of solution (% v/v). weight-to-volume percent Grams of solute per 100 mL of solution (% w/v). parts per million Micrograms of solute per gram of solution; for aqueous solutions the units are often expressed as milligrams of solute per liter of solution (ppm). parts per billion Nanograms of solute per gram of solution; for aqueous solutions the units are often expressed as micrograms of solute per liter of solution (ppb). 1400-CH02 9/8/99 3:48 PM Page 18
Chapter 2 Basic Tools of Analytical Chemistry EXAMPLE 2.3 The maximum allowed concentration of chloride in a municipal drinking E water supply is 2.50 x 102 ppm CI. When the supply of water exceeds this limit, it often has a distinctive salty taste. What is this concentration in mole Cl/liter? SOLUTION 250×102mgdl 1 g I mole cl =7.05×10-3M 1000mg 35.453gCl 2B.6 p-Functions ometimes it is inconvenient to use the concentration units in table 2. 4. for exam- ple, during a reaction a reactant,s concentration may change by many orders of mag- nitude. If we are interested in viewing the progress of the reaction graphically, we might wish to plot the reactant's concentration as a function of time or as a function of the volume of a reagent being added to the reaction. Such is the case in Figure 2. where the molar concentration of H* is plotted (y-axis on left side of figure)as a function of the volume of Naoh added to a solution of HCl. The initial [H*) is 0.10 M, and its concentration after adding 75 mL of NaOH is 5.0X 10-M. We can easily follow changes in the [H+] over the first 14 additions of NaOH. For the last ten addi tions of NaoH, however, changes in the [H*] are too small to be seen ders of often more convenient to express the concentration as a p-function. The p-func- p-function tion of a number X is written as p X and is defined as A function of the form pX, wl pX=-log(X) PX=-log(X) Thus, the pH of a solution that is 0. 10 M H* is g[H+=-og(0.10)=100 nd the pH of 5. X 10-3M H* is pH=-logH=-log(5.0×10-13)=1230 Figure 2. 1 shows how plotting pH in place of [H*] provides more detail about how the concentration of H* changes following the addition of NaOH. EXAMPLE 2.4 a What is pNa for a solution of 1. x 10-3MNasPO2 SOLUTION Since each mole of Na PO4 contains three moles of Na*, the concentration of 3 mol Na+ ×1.76×10-3M=5.28×10 mol Na3 PO4 nd pNa is pNa=- -log[ Na*=-log(5.28×103)=2277
Chapter 2 Basic Tools of Analytical Chemistry 19 EXAMPLE 2.3 The maximum allowed concentration of chloride in a municipal drinking water supply is 2.50 × 102 ppm Cl–. When the supply of water exceeds this limit, it often has a distinctive salty taste. What is this concentration in moles Cl–/liter? SOLUTION 2B.6 p-Functions Sometimes it is inconvenient to use the concentration units in Table 2.4. For example, during a reaction a reactant’s concentration may change by many orders of magnitude. If we are interested in viewing the progress of the reaction graphically, we might wish to plot the reactant’s concentration as a function of time or as a function of the volume of a reagent being added to the reaction. Such is the case in Figure 2.1, where the molar concentration of H+ is plotted (y-axis on left side of figure) as a function of the volume of NaOH added to a solution of HCl. The initial [H+] is 0.10 M, and its concentration after adding 75 mL of NaOH is 5.0 × 10–13 M. We can easily follow changes in the [H+] over the first 14 additions of NaOH. For the last ten additions of NaOH, however, changes in the [H+] are too small to be seen. When working with concentrations that span many orders of magnitude, it is often more convenient to express the concentration as a p-function. The p-function of a number X is written as pX and is defined as pX = –log(X) Thus, the pH of a solution that is 0.10 M H+ is pH = –log[H+] = –log(0.10) = 1.00 and the pH of 5.0 × 10–13 M H+ is pH = –log[H+] = –log(5.0 × 10–13) = 12.30 Figure 2.1 shows how plotting pH in place of [H+] provides more detail about how the concentration of H+ changes following the addition of NaOH. EXAMPLE 2.4 What is pNa for a solution of 1.76 × 10–3 M Na3PO4? SOLUTION Since each mole of Na3PO4 contains three moles of Na+, the concentration of Na+ is and pNa is pNa = –log[Na+] = –log(5.28 × 10–3) = 2.277 [] . . Na mol Na – – mol Na PO + M M + = ×× = × 3 1 76 10 5 28 10 3 4 3 3 2 50 10 1 1 1 35 453 7 05 10 2 3 . . . – – – – × × × =× 000 g mg Cl L g m mole Cl g Cl M p-function A function of the form pX, where pX = -log(X). 1400-CH02 9/8/99 3:48 PM Page 19
20 Modern Analytical Chemistry 0.08 0.02 Figure 2.1 0.00 Graph of [H*] versus volume of NaOH and volume of naoh for the reaction of 0.10 M HCI with 0.10 M NaOH olume NaoH(mL) EXAMPLE 2.5 What is the [H*] in a solution that has a pH of 5.16? SOLUTION The PH=-log[H*]=5.16 log[H+]=-5.16 H+]= antilog(-5.16)=10516=6.9×106M 2CStoichiometric Calculations a balanced chemical reaction indicates the quantitative relationships between the moles of reactants and products. These stoichiometric relationships provide the basis for many analytical calculations. Consider, for example, the problem of dete lining the amount of oxalic acid, H2C2O4, in rhubarb. One method for this analy sis uses the following reaction in which we oxidize oxalic acid to CO2 2Fe+(ag)+H2C2O (ag)+ 2H2O(e)-2Fe2(ag)+ 2CO2(g)+ 2H3O*(aq) 2.2 The balanced chemical reaction provides the stoichiometric relationship between the moles of Fe3+ used and the moles of oxalic acid in the sample being analyzed- pecifically, one mole of oxalic acid reacts with two moles of Fe+. As shown in Ex ample 2.6, the balanced chemical reaction can be used to determine the amount of oxalic acid in a sample, provided that information about the number of moles of
20 Modern Analytical Chemistry 0.12 0.10 0.08 0.06 0.04 0.02 0.00 [H+] (M) Volume NaOH (mL) 0 20 40 60 [H+] pH 80 pH 14 12 10 8 6 4 2 0 Figure 2.1 Graph of [H+] versus volume of NaOH and pH versus volume of NaOH for the reaction of 0.10 M HCl with 0.10 M NaOH. EXAMPLE 2.5 What is the [H+] in a solution that has a pH of 5.16? SOLUTION The concentration of H+ is pH = –log[H+] = 5.16 log[H+] = –5.16 [H+] = antilog(–5.16) = 10–5.16 = 6.9 × 10–6 M 2C Stoichiometric Calculations A balanced chemical reaction indicates the quantitative relationships between the moles of reactants and products. These stoichiometric relationships provide the basis for many analytical calculations. Consider, for example, the problem of determining the amount of oxalic acid, H2C2O4, in rhubarb. One method for this analysis uses the following reaction in which we oxidize oxalic acid to CO2. 2Fe3+(aq) + H2C2O4(aq) + 2H2O(l) → 2Fe2+(aq) + 2CO2(g) + 2H3O+(aq) 2.2 The balanced chemical reaction provides the stoichiometric relationship between the moles of Fe3+ used and the moles of oxalic acid in the sample being analyzed— specifically, one mole of oxalic acid reacts with two moles of Fe3+. As shown in Example 2.6, the balanced chemical reaction can be used to determine the amount of oxalic acid in a sample, provided that information about the number of moles of Fe3+ is known. 1400-CH02 9/8/99 3:48 PM Page 20