D. Harvey - Modern Analytical Chemistry (794078), страница 90
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As outlined in Section 2C, stoichiometric calculations may be simplifiedby focusing on appropriate conservation principles. In an acid–base reaction thenumber of protons transferred between the acid and base is conserved; thusmoles of H+ donatedmoles of H + accepted× moles acid =× moles basemole acidmole baseThe following example demonstrates the application of this approach in the directanalysis of a single analyte.EXAMPLE 9.2A 50.00-mL sample of a citrus drink requires 17.62 mL of 0.04166 M NaOH toreach the phenolphthalein end point (Figure 9.19a). Express the sample’sacidity in terms of grams of citric acid, C6H8O7, per 100 mL.SOLUTIONSince citric acid is a triprotic weak acid, we must first decide to whichequivalence point the titration has been carried.
The three acid dissociationconstants arepKa1 = 3.13pKa2 = 4.76pKa3 = 6.4014.012.010.0pH3048.06.04.02.00.00.005.0010.0015.0020.00Volume of NaOH25.0030.00(a)Cit3–pH = 6.40pHHCit2–pH = 4.76H2Cit–Figure 9.19(a) Titration curve for 50.00 mL of a0.00489 M solution of citric acid,using 0.04166 M NaOH as thetitrant; (b) ladder diagram for citricacid.pH = 3.13H3Cit(b)1400-CH09 9/9/99 2:12 PM Page 305Chapter 9 Titrimetric Methods of AnalysisThe phenolphthalein end point is basic, occurring at a pH of approximately 8.3and can be reached only if the titration proceeds to the third equivalence point(Figure 9.19b); thus, we write3 × moles citric acid = moles NaOHMaking appropriate substitutions for the moles of citric acid and moles ofNaOH gives the following equation3 × g citric acid= M b × VbFW citric acidwhich can be solved for the grams of citric acidM b × Vb × FW citric acid(0.04166 M)(0.01762 L)(192.13 g/mol)=33= 0.04701 g citric acidSince this is the grams of citric acid in a 50.00-mL sample, the concentration ofcitric acid in the citrus drink is 0.09402 g/100 mL.In an indirect analysis the analyte participates in one or more preliminary reactions that produce or consume acid or base.
Despite the additional complexity, thestoichiometry between the analyte and the amount of acid or base produced or consumed may be established by applying the conservation principles outlined in Section 2C. Example 9.3 illustrates the application of an indirect analysis in which anacid is produced.EXAMPLE 9.3The purity of a pharmaceutical preparation of sulfanilamide, C6H4N2O2S, canbe determined by oxidizing the sulfur to SO2 and bubbling the SO2 throughH2O2 to produce H2SO4. The acid is then titrated with a standard solution ofNaOH to the bromothymol blue end point, where both of sulfuric acid’s acidicprotons have been neutralized. Calculate the purity of the preparation, giventhat a 0.5136-g sample required 48.13 mL of 0.1251 M NaOH.SOLUTIONConservation of protons for the titration reaction requires that2 × moles H2SO4 = moles NaOHSince all the sulfur in H2SO4 comes from sulfanilamide, we use a conservationof mass on sulfur to establish the following stoichiometric relationship.Moles C6H4N2O2S = moles H2SO4Combining the two conservation equations gives a single equation relating themoles of analyte to the moles of titrant.2 × moles C6H4N2O2S = moles NaOHMaking appropriate substitutions for the moles of sulfanilamide and moles ofNaOH gives2 × g sulfanilamide= M b × VbFW sulfanilamide3051400-CH09 9/9/99 2:12 PM Page 306306Modern Analytical Chemistrywhich can be solved for the grams of sulfanilamideM b × Vb × FW sulfanilamide(0.1251 M)(0.04813 L)(168.18 g/mol)=22= 0.5063 g sulfanilamideThus, the purity of the preparation isg sulfanilamide0.5063 g× 100 =× 100 = 98.58% w/w sulfanilamideg sample0.5136 gThis approach is easily extended to back titrations, as shown in the followingexample.EXAMPLE 9.4The amount of protein in a sample of cheese is determined by a Kjeldahlanalysis for nitrogen.
After digesting a 0.9814-g sample of cheese, the nitrogenis oxidized to NH 4 + , converted to NH 3 with NaOH, and distilled into acollection flask containing 50.00 mL of 0.1047 M HCl. The excess HCl is thenback titrated with 0.1183 M NaOH, requiring 22.84 mL to reach thebromothymol blue end point. Report the %w/w protein in the cheese giventhat there is 6.38 g of protein for every gram of nitrogen in most dairyproducts.SOLUTIONIn this procedure, the HCl reacts with two different bases; thusMoles HCl = moles HCl reacting with NH3 + moles HCl reacting with NaOHConservation of protons requires thatMoles HCl reacting with NH3 = moles NH3Moles HCl reacting with NaOH = moles NaOHA conservation of mass on nitrogen gives the following equation.Moles NH3 = moles NCombining all four equations gives a final stoichiometric equation ofMoles HCl = moles N + moles NaOHMaking appropriate substitutions for the moles of HCl, N, and NaOH givesM a × Va =gN+ M b × VbAW Nwhich we solve for the grams of nitrogen.g N = (Ma × Va – Mb × Vb) × AW N(0.1047 M × 0.05000 L – 0.1183 M × 0.02284 L) × 14.01 g/mol = 0.03549 g N1400-CH09 9/9/99 2:12 PM Page 307Chapter 9 Titrimetric Methods of AnalysisThe mass of protein, therefore, is6.38 g protein× 0.03549 g N = 0.2264 g proteingNand the %w/w protein isg protein0.2264 g× 100 =× 100 = 23.1% w/w proteing sample0.9814 gEarlier we noted that an acid–base titration may be used to analyze a mixture ofacids or bases by titrating to more than one equivalence point.
The concentration ofeach analyte is determined by accounting for its contribution to the volume oftitrant needed to reach the equivalence points.EXAMPLE 9.5The alkalinity of natural waters is usually controlled by OH–, CO32–, andHCO 3 – , which may be present singularly or in combination. Titrating a100.0-mL sample to a pH of 8.3 requires 18.67 mL of a 0.02812 M solution ofHCl. A second 100.0-mL aliquot requires 48.12 mL of the same titrant toreach a pH of 4.5. Identify the sources of alkalinity and their concentrationsin parts per million.SOLUTIONSince the volume of titrant needed to reach a pH of 4.5 is more than twice thatneeded to reach a pH of 8.3, we know, from Table 9.8, that the alkalinity of thesample is controlled by CO32– and HCO32–.Titrating to a pH of 8.3 neutralizes CO32– to HCO3–, but does not lead to areaction of the titrant with HCO3– (see Figure 9.14b).
ThusMoles HCl to pH 8.3 = moles CO32–orM a × Va,pH 8.3 =g CO32 −FW CO32 −Solving for the grams of carbonate givesg CO32– = Ma × Va, pH 8.3 × FW CO32–0.02812 M × 0.01867 L × 60.01 g/mol = 0.03151 g CO32–The concentration of CO32–, therefore, is31.51 mgmg CO32 −== 315.1 ppm CO32 −0.1000 LliterTitrating to the second end point at pH 4.5 neutralizes CO32– to H2CO3, andHCO3– to H2CO3 (see Figures 9.18b,c).
The conservation of protons, therefore,requires thatMoles HCl to pH 4.5 = 2 × moles CO32– + moles HCO3–or3071400-CH09 9/9/99 2:12 PM Page 308308Modern Analytical ChemistryM a × Va, pH 4.5 =2 × g CO32 −g HCO3−+FW CO32 −FW HCO3−Solving for the grams of bicarbonate gives2 × g CO32 − g HCO3− = M a × Va, pH 4.5 −× FW HCO3−2− FWCO32 × 0.03151 g 2− 0.02812 M × 0.04812 L − × 61.02 g/mol = 0.01849 g HCO360.01 g/mol The concentration of HCO3–, therefore, is18.49 mgmg HCO3−== 184.9 ppm HCO3−liter0.1000 L9B.6 Qualitative ApplicationsWe have already come across one example of the qualitative application ofacid–base titrimetry in assigning the forms of alkalinity in waters (see Example 9.5).This approach is easily extended to other systems.
For example, the composition ofsolutions containing one or two of the following speciesH3PO4H2PO4–HPO42–PO43–NaOHHClcan be determined by titrating with either a strong acid or a strong base to themethyl orange and phenolphthalein end points. As outlined in Table 9.11, eachspecies or mixture of species has a unique relationship between the volumes oftitrant needed to reach these two end points.Table 9.11Relationship Between End Point Volumes for Solutions of Phosphate Specieswith HCl and NaOHSolutionCompositionRelationship Between End PointVolumes with Strong Base TitrantaRelationship Between End PointVolumes with Strong Acid TitrantaH3PO4H2PO4–HPO42–PO43–HClNaOHHCl and H3PO4H3PO4 and H2PO4–H2PO4– and HPO42–HPO42– and PO43–PO43– and NaOHVPH = 2 × VMOVPH > 0; VMO = 0——VPH = VMO—VPH < 2 × VMOVPH > 2 × VMOVPH > 0; VMO = 0———b—VMO > 0; VPH = 0VMO = 2 × VPH—VMO = VPH——VMO > 0; VPH = 0VMO > 2 × VPHVMO < 2 × VPHaVMO and VPH are, respectively, the volume of titrant needed to reach the methyl orange and phenolphthaleinbWhen no information is given, the volume of titrant needed to reach either end point is zero.end points.1400-CH09 9/9/99 2:13 PM Page 309Chapter 9 Titrimetric Methods of Analysis9B.7 Characterization ApplicationsTwo useful characterization applications involving acid–base titrimetry are the determination of equivalent weight, and the determination of acid–base dissociationconstants.Equivalent Weights Acid–base titrations can be used to characterize the chemicaland physical properties of matter.
One simple example is the determination of theequivalent weight* of acids and bases. In this method, an accurately weighed sampleof a pure acid or base is titrated to a well-defined equivalence point using a monoprotic strong acid or strong base. If we assume that the titration involves the transfer of n protons, then the moles of titrant needed to reach the equivalence point isgiven asMoles titrant = n × moles analyteand the formula weight isFW =g analyteg analyte=n×moles analytemoles titrantSince the actual number of protons transferred between the analyte and titrant isuncertain, we define the analyte’s equivalent weight (EW) as the apparent formulaweight when n = 1.
The true formula weight, therefore, is an integer multiple of thecalculated equivalent weight.FW = n × EWThus, if we titrate a monoprotic weak acid with a strong base, the EW and FW areidentical. If the weak acid is diprotic, however, and we titrate to its second equivalence point, the FW will be twice as large as the EW.EXAMPLE 9.6A 0.2521-g sample of an unknown weak acid is titrated with a 0.1005 Msolution of NaOH, requiring 42.68 mL to reach the phenolphthalein end point.Determine the compound’s equivalent weight. Which of the followingcompounds is most likely to be the unknown weak acid?ascorbic acidmalonic acidsuccinic acidcitric acidC6H8O6C3H4O4C4H6O4C6H8O7FW = 176.1FW = 104.1FW = 118.1FW = 192.1monoproticdiproticdiprotictriproticSOLUTIONThe moles of NaOH needed to reach the end point isMb × Vb = 0.1005 M × 0.04268 L = 4.289 × 10–3 mol NaOHgiving an equivalent weight ofEW =g analyte0.2521 g== 58.78 g/molmoles titrant4.289 × 10 −3 mol*See Section 2B.2 for a review of chemical equivalents and equivalent weights.3091400-CH09 9/9/99 2:13 PM Page 310310Modern Analytical ChemistryThe possible formula weights for the unknown weak acid arefor n = 1:for n = 2:for n = 3:FW = EW = 58.78 g/molFW = 2 × EW = 117.6 g/molFW = 3 × EW = 176.3 g/molIf the weak acid is monoprotic, then the FW must be 58.78 g/mol, eliminatingascorbic acid as a possibility.
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