D. Harvey - Modern Analytical Chemistry (794078), страница 9
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The molar concentration of a solution of glucose, for example, is the same as its formality.For substances that ionize in solution, such as NaCl, molarity and formality aredifferent. For example, dissolving 0.1 mol of NaCl in 1 L of water gives a solutioncontaining 0.1 mol of Na+ and 0.1 mol of Cl–. The molarity of NaCl, therefore,is zero since there is essentially no undissociated NaCl in solution. The solution,molarityThe number of moles of solute per literof solution (M).formalityThe number of moles of solute,regardless of chemical form, per liter ofsolution (F).151400-CH02 9/8/99 3:48 PM Page 1616Modern Analytical ChemistryTable 2.4Common Units for ReportingConcentrationNameUnitsaSymbolmolaritymoles soluteliters solutionMformalitynumber FWs soluteliters solutionFnormalitynumber EWs soluteliters solutionNmolalitymoles solutekg solventmweight %g solute100 g solution% w/wvolume %mL solute100 mL solution% v/vweight-to-volume %g solute100 mL solution% w/vparts per milliong solute10 6 g solutionppmparts per billiong solute10 9 g solutionppbaFW= formula weight; EW = equivalent weight.instead, is 0.1 M in Na+ and 0.1 M in Cl–.
The formality of NaCl, however, is 0.1 Fbecause it represents the total amount of NaCl in solution. The rigorous definitionof molarity, for better or worse, is largely ignored in the current literature, as it is inthis text. When we state that a solution is 0.1 M NaCl we understand it to consist ofNa+ 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 oftenused to simplify its expression in equations and writing. The use of square bracketsaround 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 NormalitynormalityThe number of equivalents of solute perliter of solution (N).Normality is an older unit of concentration that, although once commonly used, isfrequently ignored in today’s laboratories. Normality is still used in some handbooks of analytical methods, and, for this reason, it is helpful to understand itsmeaning.
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 onechemical species reacting stoichiometrically with another chemical species. Notethat this definition makes an equivalent, and thus normality, a function of thechemical reaction in which the species participates. Although a solution of H2SO4has a fixed molarity, its normality depends on how it reacts.1400-CH02 9/8/99 3:48 PM Page 17Chapter 2 Basic Tools of Analytical ChemistryThe number of equivalents, n, is based on a reaction unit, which is that part ofa 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; thusfor the reactionPb2+(aq) + 2I–(aq)17equivalentThe moles of a species that can donateone reaction unit.t PbI2(s)n = 2 for Pb2+ and n = 1 for I–. In an acid–base reaction, the reaction unit is thenumber of H+ ions donated by an acid or accepted by a base.
For the reaction between sulfuric acid and ammoniaH2SO4(aq) + 2NH3(aq)t 2NH4+(aq) + SO42–(aq)we find that n = 2 for H2SO4 and n = 1 for NH3. For a complexation reaction, thereaction unit is the number of electron pairs that can be accepted by the metal ordonated by the ligand. In the reaction between Ag+ and NH3Ag+(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–reductionreaction the reaction unit is the number of electrons released by the reducing agentor accepted by the oxidizing agent; thus, for the reaction2Fe3+(aq) + Sn2+(aq)t Sn4+(aq) + 2Fe2+(aq)n = 1 for Fe3+ and n = 2 for Sn2+.
Clearly, determining the number of equivalentsfor 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 theratio of a chemical species’ formula weight (FW) to the number of its equivalentsEW =FWnConsequently, the following simple relationship exists between normality andmolarity.N=n×MExample 2.1 illustrates the relationship among chemical reactivity, equivalentweight, and normality.EXAMPLE2.1Calculate the equivalent weight and normality for a solution of 6.0 M H3PO4given the following reactions:t PO43–(aq) + 3H2O(l)H3PO4(aq) + 2NH3(aq) t HPO42–(aq) + 2NH4+(aq)H3PO4(aq) + F–(aq) t H2PO4–(aq) + HF(aq)(a) H3PO4(aq) + 3OH–(aq)(b)(c)SOLUTIONFor phosphoric acid, the number of equivalents is the number of H+ ionsdonated to the base.
For the reactions in (a), (b), and (c) the number ofequivalents are 3, 2, and 1, respectively. Thus, the calculated equivalent weightsand normalities areequivalent weightThe mass of a compound containing oneequivalent (EW).formula weightThe mass of a compound containing onemole (FW).1400-CH02 9/8/99 3:48 PM Page 1818Modern Analytical ChemistryFW97.994== 32.665n3FW97.994(b) EW === 48.997n2FW97.994(c) EW === 97.994n1(a) EW =N = n × M = 3 × 6.0 = 18 NN = n × M = 2 × 6.0 = 12 NN = n × M = 1 × 6.0 = 6.0 N2B.3 MolalitymolalityThe number of moles of solute perkilogram of solvent (m).Molality is used in thermodynamic calculations where a temperature independentunit of concentration is needed.
Molarity, formality and normality are based on thevolume 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 inplace of its volume, the resulting concentration becomes independent of temperature.2B.4 Weight, Volume, and Weight-to-Volume Ratiosweight percentGrams of solute per 100 g of solution.(% w/w).volume percentMilliliters of solute per 100 mL ofsolution (% v/v).weight-to-volume percentGrams of solute per 100 mL of solution(% w/v).parts per millionMicrograms of solute per gram ofsolution; for aqueous solutions the unitsare often expressed as milligrams ofsolute per liter of solution (ppm).parts per billionNanograms of solute per gram ofsolution; for aqueous solutions the unitsare often expressed as micrograms ofsolute per liter of solution (ppb).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 whicha 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 ofsolute to one million or one billion grams of sample, respectively. For example, a steelthat 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 canbe expressed in parts per million or parts per billion using the following relationships.mgµg=litermLµgngppb ==litermLppm =For gases a part per million usually is a volume ratio. Thus, a helium concentrationof 6.3 ppm means that one liter of air contains 6.3 µL of He.2B.5 Converting Between Concentration UnitsThe units of concentration most frequently encountered in analytical chemistry aremolarity, weight percent, volume percent, weight-to-volume percent, parts per million, and parts per billion.
By recognizing the general definition of concentrationgiven in equation 2.1, it is easy to convert between concentration units.EXAMPLE 2.2A concentrated solution of aqueous ammonia is 28.0% w/w NH3 and has adensity of 0.899 g/mL. What is the molar concentration of NH3 in this solution?SOLUTION28.0 g NH 30.899 g solution 1 mole NH 3 1000 mL×××= 14.8 M100 g solutionmL solution17.04 g NH 3liter1400-CH02 9/8/99 3:48 PM Page 19Chapter 2 Basic Tools of Analytical ChemistryEXAMPLE 2.3The maximum allowed concentration of chloride in a municipal drinkingwater supply is 2.50 × 102 ppm Cl–. When the supply of water exceeds thislimit, it often has a distinctive salty taste. What is this concentration in molesCl–/liter?SOLUTION2.50 × 102 mg Cl –1g1 mole Cl –××= 7.05 × 10 –3 ML1000 mg 35.453 g Cl –2B.6 p-FunctionsSometimes 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, wemight wish to plot the reactant’s concentration as a function of time or as a functionof 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 afunction of the volume of NaOH added to a solution of HCl. The initial [H+] is 0.10M, and its concentration after adding 75 mL of NaOH is 5.0 × 10–13 M.
We can easilyfollow 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 isoften more convenient to express the concentration as a p-function. The p-function of a number X is written as pX and is defined aspX = –log(X)Thus, the pH of a solution that is 0.10 M H+ ispH = –log[H+] = –log(0.10) = 1.00and the pH of 5.0 × 10–13 M H+ ispH = –log[H+] = –log(5.0 × 10–13) = 12.30Figure 2.1 shows how plotting pH in place of [H+] provides more detail about howthe concentration of H+ changes following the addition of NaOH.EXAMPLE 2.4What is pNa for a solution of 1.76 × 10–3 M Na3PO4?SOLUTIONSince each mole of Na3PO4 contains three moles of Na+, the concentration ofNa+ is[Na + ] =3 mol Na +× 1.76 × 10 –3 M = 5.28 × 10 –3 Mmol Na3 PO4and pNa ispNa = –log[Na+] = –log(5.28 × 10–3) = 2.277p-functionA function of the form pX, wherepX = -log(X).191400-CH02 9/8/99 3:48 PM Page 2020Modern Analytical Chemistry0.1214[H+]pH0.1012108pH[H+] (M)0.080.0660.0440.02Figure 2.1Graph of [H+] versus volume of NaOH andpH versus volume of NaOH for the reactionof 0.10 M HCl with 0.10 M NaOH.20.0000204060Volume NaOH (mL)80EXAMPLE 2.5What is the [H+] in a solution that has a pH of 5.16?SOLUTIONThe concentration of H+ ispH = –log[H+] = 5.16log[H+] = –5.16[H+] = antilog(–5.16) = 10–5.16 = 6.9 × 10–6 M2C Stoichiometric CalculationsA balanced chemical reaction indicates the quantitative relationships between themoles of reactants and products.
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