D. Harvey - Modern Analytical Chemistry (794078), страница 101
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A 25-mL portion of the diluted sample was transferred by pipet into anErlenmeyer flask and treated with excess KI, oxidizing the OCl– to Cl–, andproducing I3–. The liberated I3– was determined by titrating with 0.09892 MNa2S2O3, requiring 8.96 mL to reach the starch indicator end point. Report the%w/v NaOCl in the sample of bleach.SOLUTIONReducing OCl– to Cl– requires two electrons, and oxidizing a single I– to I3–requires 2/3 of an electron; thus, conservation of electrons requires that2 × moles NaOCl = 2/3 × moles I–In the titration, two electrons are needed when I3– is reduced, but oxidizingS2O32– releases one electron; thus2 × moles I3– = moles S2O32–These two equations can be combined by recognizing that a conservation ofmass for iodine requires thatMoles I– = 3 × moles I3–Thus2 × moles NaOCl = moles S2O32–Substituting for moles of NaOCl and S2O32– leaves us with an equation2 × grams NaOCl= M S O 2 − × VS O 2 −2 32 3FW NaOCl1400-CH09 9/9/99 2:13 PM Page 349Chapter 9 Titrimetric Methods of Analysisthat is solved for the grams of NaOCl.MS2O 32−× VS2O 32−× FW NaOCl(0.09892 M)(0.00896 L)(74.44 g /mol)2= 0.03299 g NaOCl=2Thus, the %w/v NaOCl in the diluted sample isGrams NaOCl0.03299 g× 100 =× 100 = 0.132% w/v NaOClmL sample25.00 mLSince the bleach was diluted by a factor of 40 (25 mL to 1000 mL), theconcentration of NaOCl in the bleach is 5.28% (w/v).EXAMPLE 9.15The amount of ascorbic acid, C 6H8O6, in orange juice was determined byoxidizing the ascorbic acid to dehydroascorbic acid, C6H6O6, with a knownexcess of I3–, and back titrating the excess I3– with Na2S2O3.
A 5.00-mL sampleof filtered orange juice was treated with 50.00 mL of excess 0.01023 M I3–. Afterthe oxidation was complete, 13.82 mL of 0.07203 M Na2S2O3 was needed toreach the starch indicator end point. Report the concentration of ascorbic acidin milligrams per 100 mL.SOLUTIONOxidizing ascorbic acid requires two electrons, and reducing I3– to I– alsorequires two electrons. Thus(Moles I3–)ascorbic acid = moles C6H8O6For the back titration, the stoichiometric relationship between I3– and S2O32– is(see Example 9.14)(Moles I3–)back titration = 0.5 × moles S2O32–The total moles of I3– used in the analysis is the sum of that reacting withascorbic acid and S2O32–(Moles I3–)tot = (moles I3–)ascorbic acid + (moles I3–)back titrationorMoles I3– = moles C6H8O6 + 0.5 × moles S2O32–Making appropriate substitutions for the moles of I3–, C6H8O6, and S2O32–MI3−× VI3−=g C 6 H 8 O6+ 0.5 × M S O 2 − × VS O 2 −2 32 3FW C 6 H 8O6and solving for the grams of C6H8O6 gives(M I3−× VI3−– 0.5 × M S O232−× VS O232−) × FW C 6 H8O 6 =[(0.01023 M)(0.0500 L) – (0.5)(0.07203 M)(0.01382 L)](176.13 g/mol)= 0.00243 g C 6 H8O 63491400-CH09 9/9/99 2:13 PM Page 350350Modern Analytical ChemistryPotential (V)Thus, there is 2.43 mg of ascorbic acid in the 5.00-mL sample, or 48.6 mg/100mL of orange juice.9D.5 Evaluation of Redox Titrimetry1.8001.6001.4001.2001.0000.8000.6000.4000.2000.000080204060Volume of titrant (mL)Figure 9.40Titration curve for 50.00 mL of 0.0250 MSn2+ and 0.0250 M Fe2+ with 0.055 M Ce4+.precipitation titrationA titration in which the reaction betweenthe analyte and titrant involves aprecipitation.100The scale of operations, accuracy, precision, sensitivity, time, and cost ofmethods involving redox titrations are similar to those described earlier in thechapter for acid–base and complexometric titrimetric methods.
As withacid–base titrations, redox titrations can be extended to the analysis of mixtures if there is a significant difference in the ease with which the analytes canbe oxidized or reduced. Figure 9.40 shows an example of the titration curvefor a mixture of Fe2+ and Sn2+, using Ce4+ as the titrant. The titration of amixture of analytes whose standard-state potentials or formal potentials differby at least 200 mV will result in a separate equivalence point for each analyte.9E Precipitation TitrationsThus far we have examined titrimetric methods based on acid–base, complexation,and redox reactions. A reaction in which the analyte and titrant form an insolubleprecipitate also can form the basis for a titration.
We call this type of titration a precipitation titration.One of the earliest precipitation titrations, developed at the end of the eighteenth century, was for the analysis of K2CO3 and K2SO4 in potash. Calcium nitrate,Ca(NO3)2, was used as a titrant, forming a precipitate of CaCO3 and CaSO4.
Theend point was signaled by noting when the addition of titrant ceased to generate additional precipitate. The importance of precipitation titrimetry as an analyticalmethod reached its zenith in the nineteenth century when several methods were developed for determining Ag+ and halide ions.9E.1 Titration CurvesThe titration curve for a precipitation titration follows the change in either the analyte’s or titrant’s concentration as a function of the volume of titrant.
For example,in an analysis for I– using Ag+ as a titrantAg+(aq) + I–(aq)t AgI(s)the titration curve may be a plot of pAg or pI as a function of the titrant’s volume.As we have done with previous titrations, we first show how to calculate the titration curve and then demonstrate how to quickly sketch the titration curve.Calculating the Titration Curve As an example, let’s calculate the titration curvefor the titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+. The reaction in thiscase isAg+(aq) + Cl–(aq)t AgCl(s)The equilibrium constant for the reaction isK = (Ksp)–1 = (1.8 × 10–10)–1 = 5.6 × 109Since the equilibrium constant is large, we may assume that Ag + and Cl– reactcompletely.1400-CH09 9/9/99 2:13 PM Page 351Chapter 9 Titrimetric Methods of AnalysisBy now you are familiar with our approach to calculating titration curves.
Thefirst task is to calculate the volume of Ag+ needed to reach the equivalence point.The stoichiometry of the reaction requires thatMoles Ag+ = moles Cl–orMAgVAg = MClVClSolving for the volume of Ag+M ClVCl(0.0500 M)(50.0 mL)== 25.0 mLM Ag(0.100 M)VAg =shows that we need 25.0 mL of Ag+ to reach the equivalence point.Before the equivalence point Cl– is in excess. The concentration of unreacted–Cl after adding 10.0 mL of Ag+, for example, is[Cl − ] ==M ClVCl − M AgVAgmoles excess Cl −=total volumeVCl + VAg(0.0500 M)(50.0 mL) − (0.100 M)(10.0 mL)50.0 mL + 10.0 mL= 2.50 × 10 −2 MIf the titration curve follows the change in concentration for Cl–, then we calculatepCl aspCl = –log[Cl–] = –log(2.50 × 10–2) = 1.60However, if we wish to follow the change in concentration for Ag+ then we mustfirst calculate its concentration. To do so we use the Ksp expression for AgClKsp = [Ag+][Cl–] = 1.8 × 10–10Solving for the concentration of Ag+[Ag + ] =Ksp[Cl − ]=1.8 × 10 −10= 7.2 × 10 −9 M2.50 × 10 −2gives a pAg of 8.14.At the equivalence point, we know that the concentrations of Ag+ and Cl– areequal.
Using the solubility product expressionKsp = [Ag+][Cl–] = [Ag+]2 = 1.8 × 10–10gives[Ag+] = [Cl–] = 1.3 × 10–5 MAt the equivalence point, therefore, pAg and pCl are both 4.89.After the equivalence point, the titration mixture contains excess Ag+. The concentration of Ag+ after adding 35.0 mL of titrant is[Ag + ] ==M AgVAg − M ClVClmoles excess Ag+=total volumeVCl + VAg(0.100 M)(35.0 mL) − (0.0500 M)(50.0 mL)50.0 mL + 35.0 mL= 1.18 × 10 −2 M3511400-CH09 9/9/99 2:13 PM Page 352pCl or pAg352Modern Analytical Chemistry9.008.007.006.005.004.003.002.001.000.00Table 9.21(a)(b)010402030Volume AgNO350Figure 9.41Precipitation titration curve for 50.0 mL of0.0500 M Cl– with 0.100 M Ag+.
(a) pClversus volume of titrant; (b) pAg versusvolume of titrant.Data for Titration of 50.0 mL of0.0500 M Cl– with 0.100 M Ag+Volume AgNO3(mL)pClpAg0.005.0010.0015.0020.0025.0030.0035.0040.0045.0050.001.301.441.601.812.154.897.547.827.978.078.14—8.318.147.937.604.892.201.931.781.681.60or a pAg of 1.93. The concentration of Cl– is[Cl − ] =Ksp[Ag + ]=1.8 × 10 −10= 1.5 × 10 −8 M1.18 × 10 −2or a pCl of 7.82.
Additional results for the titration curve are shown in Table 9.21and Figure 9.41.Sketching the Titration Curve As we have done for acid–base, complexometrictitrations, and redox titrations, we now show how to quickly sketch a precipitationtitration curve using a minimum number of calculations.EXAMPLE 9.16Sketch a titration curve for the titration of 50.0 mL of 0.0500 M Cl– with 0.100M Ag + . This is the same titration for which we previously calculated thetitration curve (Table 9.21 and Figure 9.41).SOLUTIONWe begin by drawing axes for the titration curve (Figure 9.42a). Having shownthat the equivalence point volume is 25.0 mL, we draw a vertical lineintersecting the x-axis at this volume (Figure 9.42b).Before the equivalence point, pCl and pAg are determined by theconcentration of excess Cl–. Using values from Table 9.21, we plot either pAgor pCl for 10.0 mL and 20.0 mL of titrant (Figure 9.42c).After the equivalence point, pCl and pAg are determined by theconcentration of excess Ag+.
Using values from Table 9.21, we plot points for30.0 mL and 40.0 mL of titrant (Figure 9.42d).To complete an approximate sketch of the titration curve, we drawseparate straight lines through the two points before and after the equivalencepoint (Figure 9.42e). Finally, a smooth curve is drawn to connect the threestraight-line segments (Figure 9.42f).1400-CH09 9/9/99 2:13 PM Page 353Chapter 9 Titrimetric Methods of AnalysisPercent titrated50100Percent titrated15020008.08.06.06.0pCl or pAgpCl or pAg04.0502002.00.00.0020.0010.0030.0040.00Volume of titrant (mL)50.00(a)20.0010.0030.0040.00Volume of titrant (mL)050.00(b)Percent titrated050100Percent titrated15020008.08.06.06.0pCl or pAgpCl or pAg1504.02.04.02.0501001502004.02.00.00.0010.0030.0040.0020.00Volume of titrant (mL)050.00(c)10.0030.0040.0020.00Volume of titrant (mL)50.00(d)Percent titrated050100Percent titrated15020008.08.06.06.0pCl or pAgpCl or pAg1004.02.0501001502004.02.00.00.0010.0030.0040.0020.00Volume of titrant (mL)50.00( )(e)Figure 9.42How to sketch a precipitation titration curve; see text for explanation.0(f)10.0030.0040.0020.00Volume of titrant (mL)50.003531400-CH09 9/9/99 2:13 PM Page 354354Modern Analytical Chemistry9E.2 Selecting and Evaluating the End PointInitial attempts at developing precipitation titration methods were limited by apoor end point signal.
Finding the end point by looking for the first addition oftitrant that does not yield additional precipitate is cumbersome at best. The feasibility of precipitation titrimetry improved with the development of visual indicatorsand potentiometric ion-selective electrodes.Finding the End Point with a Visual Indicator The first important visual indicatorto be developed was the Mohr method for Cl– using Ag+ as a titrant.
By adding asmall amount of K2CrO4 to the solution containing the analyte, the formation of aprecipitate of reddish-brown Ag2CrO4 signals the end point. Because K2CrO4 imparts a yellow color to the solution, obscuring the end point, the amount of CrO42–added is small enough that the end point is always later than the equivalence point.To compensate for this positive determinate error an analyte-free reagent blank isanalyzed to determine the volume of titrant needed to effect a change in the indicator’s color.
The volume for the reagent blank is subsequently subtracted from theexperimental end point to give the true end point. Because CrO42– is a weak base,the solution usually is maintained at a slightly alkaline pH. If the pH is too acidic,chromate is present as HCrO4–, and the Ag2CrO4 end point will be in significanterror. The pH also must be kept below a level of 10 to avoid precipitating silverhydroxide.A second end point is the Volhard method in which Ag+ is titrated with SCN–in the presence of Fe3+.
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