Modern Analytical Cheymistry - Chapter 9 pdf

CChhaapptteerr 9 273 Titrimetric Methods of Analysis Titrimetry, in which we measure the volume of a reagent reacting stoichiometrically with the analyte, first appeared as an analytical method in the early eighteenth century. Unlike gravimetry, titrimetry initially did not receive wide acceptance as an analytical technique. Many prominent late-nineteenth century analytical chemists preferred gravimetry over titrimetry and few of the standard texts from that era include titrimetric methods. By the early twentieth century, however, titrimetry began to replace gravimetry as the most commonly used analytical method. Interestingly, precipitation gravimetry developed in the absence of a theory of precipitation. The relationship between the precipitate’s mass and the mass of analyte, called a gravimetric factor, was determined experimentally by taking known masses of analyte (an external standardization). Gravimetric factors could not be calculated using the precipitation reaction’s stoichiometry because chemical formulas and atomic weights were not yet available! Unlike gravimetry, the growth and acceptance of titrimetry required a deeper understanding of stoichiometry, thermodynamics, and chemical equilibria. By the early twentieth century the accuracy and precision of titrimetric methods were comparable to that of gravimetry, establishing titrimetry as an accepted analytical technique. 1400-CH09 9/9/99 2:12 PM Page 273 274 Modern Analytical Chemistry 9 A Overview of Titrimetry Titrimetric methods are classified into four groups based on the type of reaction in- volved. These groups are acid–base titrations, in which an acidic or basic titrant re- acts with an analyte that is a base or an acid; complexometric titrations involving a metal–ligand complexation reaction; redox titrations, where the titrant is an oxidiz- ing or reducing agent; and precipitation titrations, in which the analyte and titrant react to form a precipitate. Despite the difference in chemistry, all titrations share several common features, providing the focus for this section. 9 A.1 Equivalence Points and End Points For a titration to be accurate we must add a stoichiometrically equivalent amount of titrant to a solution containing the analyte. We call this stoichiometric mixture the equivalence point. Unlike precipitation gravimetry, where the precipitant is added in excess, determining the exact volume of titrant needed to reach the equiv- alence point is essential. The product of the equivalence point volume, V eq , and the titrant’s concentration, C T , gives the moles of titrant reacting with the analyte. Moles titrant = V eq × C T Knowing the stoichiometry of the titration reaction(s), we can calculate the moles of analyte. Unfortunately, in most titrations we usually have no obvious indication that the equivalence point has been reached. Instead, we stop adding titrant when we reach an end point of our choosing. Often this end point is indicated by a change in the color of a substance added to the solution containing the analyte. Such sub- stances are known as indicators. The difference between the end point volume and the equivalence point volume is a determinate method error, often called the titra- tion error. If the end point and equivalence point volumes coincide closely, then the titration error is insignificant and can be safely ignored. Clearly, selecting an ap- propriate end point is critical if a titrimetric method is to give accurate results. 9 A.2 Volume as a Signal* Almost any chemical reaction can serve as a titrimetric method provided that three conditions are met. The first condition is that all reactions involving the titrant and analyte must be of known stoichiometry. If this is not the case, then the moles of titrant used in reaching the end point cannot tell us how much ana- lyte is in our sample. Second, the titration reaction must occur rapidly. If we add titrant at a rate that is faster than the reaction’s rate, then the end point will ex- ceed the equivalence point by a significant amount. Finally, a suitable method must be available for determining the end point with an acceptable level of accu- racy. These are significant limitations and, for this reason, several titration strate- gies are commonly used. A simple example of a titration is an analysis for Ag + using thiocyanate, SCN – , as a titrant. Ag + (aq) + SCN – (aq) t AgSCN(s) equivalence point The point in a titration where stoichiometrically equivalent amounts of analyte and titrant react. end point The point in a titration where we stop adding titrant. indicator A colored compound whose change in color signals the end point of a titration. titration error The determinate error in a titration due to the difference between the end point and the equivalence point. titrant The reagent added to a solution containing the analyte and whose volume is the signal. *Instead of measuring the titrant’s volume we also can measure its mass. Since the titrant’s density is a measure of its mass per unit volume, the mass of titrant and volume of titrant are proportional. titrimetry Any method in which volume is the signal. 1400-CH09 9/9/99 2:12 PM Page 274 This reaction occurs quickly and is of known stoichiometry. A titrant of SCN – is easily prepared using KSCN. To indicate the titration’s end point we add a small amount of Fe 3+ to the solution containing the analyte. The formation of the red- colored Fe(SCN) 2+ complex signals the end point. This is an example of a direct titration since the titrant reacts with the analyte. If the titration reaction is too slow, a suitable indicator is not available, or there is no useful direct titration reaction, then an indirect analysis may be possible. Sup- pose you wish to determine the concentration of formaldehyde, H 2 CO, in an aque- ous solution. The oxidation of H 2 CO by I 3 – H 2 CO(aq) + 3OH – (aq)+I 3 – (aq) t HCO 2 – (aq)+3I – (aq)+2H 2 O(l) is a useful reaction, except that it is too slow for a direct titration. If we add a known amount of I 3 – , such that it is in excess, we can allow the reaction to go to comple- tion. The I 3 – remaining can then be titrated with thiosulfate, S 2 O 3 2– . I 3 – (aq)+2S 2 O 3 2– (aq) t S 4 O 6 2– (aq)+3I – (aq) This type of titration is called a back titration. Calcium ion plays an important role in many aqueous environmental systems. A useful direct analysis takes advantage of its reaction with the ligand ethylenedi- aminetetraacetic acid (EDTA), which we will represent as Y 4– . Ca 2+ (aq)+Y 4– (aq) t CaY 2– (aq) Unfortunately, it often happens that there is no suitable indicator for this direct titration. Reacting Ca 2+ with an excess of the Mg 2+ –EDTA complex Ca 2+ (aq) + MgY 2– (aq) t CaY 2– (aq)+Mg 2+ (aq) releases an equivalent amount of Mg 2+ . Titrating the released Mg 2+ with EDTA Mg 2+ (aq)+Y 4– (aq) t MgY 2– (aq) gives a suitable end point. The amount of Mg 2+ titrated provides an indirect mea- sure of the amount of Ca 2+ in the original sample. Since the analyte displaces a species that is then titrated, we call this a displacement titration. When a suitable reaction involving the analyte does not exist it may be possible to generate a species that is easily titrated. For example, the sulfur content of coal can be determined by using a combustion reaction to convert sulfur to sulfur dioxide. S(s)+O 2 (g) → SO 2 (g) Passing the SO 2 through an aqueous solution of hydrogen peroxide, H 2 O 2 , SO 2 (g)+H 2 O 2 (aq) → H 2 SO 4 (aq) produces sulfuric acid, which we can titrate with NaOH, H 2 SO 4 (aq) + 2OH – (aq) t SO 4 2– (aq)+2H 2 O(l) providing an indirect determination of sulfur. 9 A. 3 Titration Curves To find the end point we monitor some property of the titration reaction that has a well-defined value at the equivalence point. For example, the equivalence point for a titration of HCl with NaOH occurs at a pH of 7.0. We can find the end point, Chapter 9 Titrimetric Methods of Analysis 275 back titration A titration in which a reagent is added to a solution containing the analyte, and the excess reagent remaining after its reaction with the analyte is determined by a titration. displacement titration A titration in which the analyte displaces a species, usually from a complex, and the amount of the displaced species is determined by a titration. 1400-CH09 9/9/99 2:12 PM Page 275 Figure 9.1 Acid–base titration curve for 25.0 mL of 0.100 M HCl with 0.100 M NaOH. therefore, by monitoring the pH with a pH electrode or by adding an indicator that changes color at a pH of 7.0. Suppose that the only available indicator changes color at a pH of 6.8. Is this end point close enough to the equivalence point that the titration error may be safely ignored? To answer this question we need to know how the pH changes dur- ing the titration. A titration curve provides us with a visual picture of how a property, such as pH, changes as we add titrant (Figure 9.1). We can measure this titration curve ex- perimentally by suspending a pH electrode in the solution containing the analyte, monitoring the pH as titrant is added. As we will see later, we can also calculate the expected titration curve by considering the reactions responsible for the change in pH. However we arrive at the titration curve, we may use it to evaluate an indica- tor’s likely titration error. For example, the titration curve in Figure 9.1 shows us that an end point pH of 6.8 produces a small titration error. Stopping the titration at an end point pH of 11.6, on the other hand, gives an unacceptably large titration error. The titration curve in Figure 9.1 is not unique to an acid–base titration. Any titration curve that follows the change in concentration of a species in the titration reaction (plotted logarithmically) as a function of the volume of titrant has the same general sigmoidal shape. Several additional examples are shown in Figure 9.2. Concentration is not the only property that may be used to construct a titration curve. Other parameters, such as temperature or the absorbance of light, may be used if they show a significant change in value at the equivalence point. Many titra- tion reactions, for example, are exothermic. As the titrant and analyte react, the temperature of the system steadily increases. Once the titration is complete, further additions of titrant do not produce as exothermic a response, and the change in temperature levels off. A typical titration curve of temperature versus volume of titrant is shown in Figure 9.3. The titration curve contains two linear segments, the intersection of which marks the equivalence point. 276 Modern Analytical Chemistry pH Volume NaOH (mL) 0.00 0.00 4.00 6.00 8.00 Equivalence point 10.00 12.00 14.00 10.00 20.00 30.00 50.00 2.00 40.00 titration curve A graph showing the progress of a titration as a function of the volume of titrant added. 1400-CH09 9/9/99 2:12 PM Page 276 Figure 9.2 Examples of titration curves for (a) a complexation titration, (b) a redox titration, and (c) a precipitation titration. 9 A. 4 The Buret The only essential piece of equipment for an acid–base titration is a means for deliv- ering the titrant to the solution containing the analyte. The most common method for delivering the titrant is a buret (Figure 9.4). A buret is a long, narrow tube with graduated markings, and a stopcock for dispensing the titrant. Using a buret with a small internal diameter provides a better defined meniscus, making it easier to read the buret’s volume precisely. Burets are available in a variety of sizes and tolerances Chapter 9 Titrimetric Methods of Analysis 277 pCd Volume of titrant (mL) 0.00 (a) 0.0 4.0 6.0 8.0 10.0 16.0 18.0 10.00 20.00 30.00 50.00 2.0 40.00 14.0 12.0 pCl Volume of titrant (mL) 0.00 (c) 0.0 2.0 4.0 8.0 10.00 20.00 30.00 50.00 40.00 6.0 Potential (V) Volume of titrant (mL) 0 (b) 0.000 0.400 0.600 0.800 1.000 1.600 1.800 20 40 60 100 0.200 80 1.400 1.200 Temperature (°C) Volume of titrant (mL) Equivalence point Figure 9.3 Example of a thermometric titration curve. buret Volumetric glassware used to deliver variable, but known volumes of solution. 1400-CH09 9/9/99 2:12 PM Page 277 Figure 9.4 Volumetric buret showing a portion of its graduated scale. 278 Modern Analytical Chemistry 0 1 2 Table 9 .1 Specifications for Volumetric Burets Volume Subdivision Tolerance (mL) Class a (mL) (mL) 5 A 0.01 ±0.01 B 0.01 ±0.02 10 A 0.02 ±0.02 B 0.02 ±0.04 25 A 0.1 ±0.03 B 0.1 ±0.06 50 A 0.1 ±0.05 B 0.1 ±0.10 100 A 0.2 ±0.10 B 0.2 ±0.20 a Specifications for class A and class B glassware are taken from American Society for Testing Materials (ASTM) E288, E542, and E694 standards. (Table 9.1), with the choice of buret determined by the demands of the analysis. The accuracy obtainable with a buret can be improved by calibrating it over several intermediate ranges of volumes using the same method described in Chapter 5 for calibrating pipets. In this manner, the volume of titrant delivered can be corrected for any variations in the buret’s internal diameter. Titrations may be automated using a pump to deliver the titrant at a constant flow rate, and a solenoid valve to control the flow (Figure 9.5). The volume of titrant delivered is determined by multiplying the flow rate by the elapsed time. Au- tomated titrations offer the additional advantage of using a microcomputer for data storage and analysis. 9 B Titrations Based on Acid–Base Reactions The earliest acid–base titrations involved the determination of the acidity or alka- linity of solutions, and the purity of carbonates and alkaline earth oxides. Before 1800, acid–base titrations were conducted using H 2 SO 4 , HCl, and HNO 3 as acidic titrants, and K 2 CO 3 and Na 2 CO 3 as basic titrants. End points were determined using visual indicators such as litmus, which is red in acidic solutions and blue in basic solutions, or by observing the cessation of CO 2 effervescence when neutraliz- ing CO 3 2– . The accuracy of an acid–base titration was limited by the usefulness of the indicator and by the lack of a strong base titrant for the analysis of weak acids. The utility of acid–base titrimetry improved when NaOH was first introduced as a strong base titrant in 1846. In addition, progress in synthesizing organic dyes led to the development of many new indicators. Phenolphthalein was first synthe- sized by Bayer in 1871 and used as a visual indicator for acid–base titrations in 1877. Other indicators, such as methyl orange, soon followed. Despite the increas- ing availability of indicators, the absence of a theory of acid–base reactivity made se- lecting a proper indicator difficult. Developments in equilibrium theory in the late nineteenth century led to sig- nificant improvements in the theoretical understanding of acid–base chemistry and, acid–base titration A titration in which the reaction between the analyte and titrant is an acid–base reaction. 1400-CH09 9/9/99 2:12 PM Page 278 Figure 9.5 Typical instrumentation for performing an automatic titration. Courtesy of Fisher Scientific. in turn, of acid–base titrimetry. Sørenson’s establishment of the pH scale in 1909 provided a rigorous means for comparing visual indicators. The determination of acid–base dissociation constants made the calculation of theoretical titration curves possible, as outlined by Bjerrum in 1914. For the first time a rational method ex- isted for selecting visual indicators, establishing acid–base titrimetry as a useful al- ternative to gravimetry. 9 B.1 Acid–Base Titration Curves In the overview to this chapter we noted that the experimentally determined end point should coincide with the titration’s equivalence point. For an acid–base titra- tion, the equivalence point is characterized by a pH level that is a function of the acid–base strengths and concentrations of the analyte and titrant. The pH at the end point, however, may or may not correspond to the pH at the equivalence point. To understand the relationship between end points and equivalence points we must know how the pH changes during a titration. In this section we will learn how to construct titration curves for several important types of acid–base titrations. Our Chapter 9 Titrimetric Methods of Analysis 279 1400-CH09 9/9/99 2:12 PM Page 279 approach will make use of the equilibrium calculations described in Chapter 6. We also will learn how to sketch a good approximation to any titration curve using only a limited number of simple calculations. Titrating Strong Acids and Strong Bases For our first titration curve let’s consider the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH. For the reaction of a strong base with a strong acid the only equilibrium reaction of importance is H 3 O + (aq)+OH – (aq) t 2H 2 O(l) 9.1 The first task in constructing the titration curve is to calculate the volume of NaOH needed to reach the equivalence point. At the equivalence point we know from re- action 9.1 that Moles HCl = moles NaOH or M a V a = M b V b where the subscript ‘a’ indicates the acid, HCl, and the subscript ‘b’ indicates the base, NaOH. The volume of NaOH needed to reach the equivalence point, there- fore, is Before the equivalence point, HCl is present in excess and the pH is determined by the concentration of excess HCl. Initially the solution is 0.100 M in HCl, which, since HCl is a strong acid, means that the pH is pH = –log[H 3 O + ] = –log[HCl] = –log(0.100) = 1.00 The equilibrium constant for reaction 9.1 is (K w ) –1 , or 1.00 × 10 14 . Since this is such a large value we can treat reaction 9.1 as though it goes to completion. After adding 10.0 mL of NaOH, therefore, the concentration of excess HCl is giving a pH of 1.30. At the equivalence point the moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissoci- ation of water. K w = 1.00 × 10 –14 =[H 3 O + ][OH – ]=[H 3 O + ] 2 [H 3 O + ] = 1.00 × 10 –7 M Thus, the pH at the equivalence point is 7.00. Finally, for volumes of NaOH greater than the equivalence point volume, the pH is determined by the concentration of excess OH – . For example, after adding 30.0 mL of titrant the concentration of OH – is [ ( . )( . ) ( )( . ) . HCl] moles excess HCl total volume M mL .200 M mL mL mL M aa bb ab == − + = − + = MV MV VV 0 100 50 0 0 10 0 50 0 10 0 0 050 VV MV M eq b aa b M mL M) mL== = = (. )( . ) (. . 0 100 50 0 0 200 25 0 280 Modern Analytical Chemistry 1400-CH09 9/9/99 2:12 PM Page 280 To find the concentration of H 3 O + , we use the K w expression giving a pH of 12.10. Table 9.2 and Figure 9.1 show additional results for this titra- tion curve. Calculating the titration curve for the titration of a strong base with a strong acid is handled in the same manner, except that the strong base is in excess before the equivalence point and the strong acid is in excess after the equivalence point. Titrating a Weak Acid with a Strong Base For this example let’s consider the titra- tion of 50.0 mL of 0.100 M acetic acid, CH 3 COOH, with 0.100 M NaOH. Again, we start by calculating the volume of NaOH needed to reach the equivalence point; thus Moles CH 3 COOH = moles NaOH M a V a = M b V b VV MV M eq b aa b M mL M mL== = = (. )( . ) (. ) . 0 100 50 0 0 100 50 0 [] [] . . .HO OH 3 w + − − − == × =× K 100 10 0 0125 800 10 14 13 [] (. )( . ) (. )( . ) . OH moles excess NaOH total volume M mL M mL mL mL M bb aa ab − == − + = − + = MV MV VV 0 200 30 0 0 100 50 0 50 0 30 0 0 0125 Chapter 9 Titrimetric Methods of Analysis 281 Table 9 .2 Data for Titration of 50.00 mL of 0.100 M HCl with 0.0500 M NaOH Volume (mL) of Titrant pH 0.00 1.00 5.00 1.14 10.00 1.30 15.00 1.51 20.00 1.85 22.00 2.08 24.00 2.57 25.00 7.00 26.00 11.42 28.00 11.89 30.00 12.50 35.00 12.37 40.00 12.52 45.00 12.62 50.00 12.70 1400-CH09 9/9/99 2:12 PM Page 281 Before adding any NaOH the pH is that for a solution of 0.100 M acetic acid. Since acetic acid is a weak acid, we calculate the pH using the method outlined in Chapter 6. CH 3 COOH(aq)+H 2 O(l) t H 3 O + (aq)+CH 3 COO – (aq) x =[H 3 O + ] = 1.32 × 10 –3 At the beginning of the titration the pH is 2.88. Adding NaOH converts a portion of the acetic acid to its conjugate base. CH 3 COOH(aq)+OH – (aq) t H 2 O(l)+CH 3 COO – (aq) 9.2 Any solution containing comparable amounts of a weak acid, HA, and its conjugate weak base, A – , is a buffer. As we learned in Chapter 6, we can calculate the pH of a buffer using the Henderson–Hasselbalch equation. The equilibrium constant for reaction 9.2 is large (K = K a /K w = 1.75 × 10 9 ), so we can treat the reaction as one that goes to completion. Before the equivalence point, the concentration of unreacted acetic acid is and the concentration of acetate is For example, after adding 10.0 mL of NaOH the concentrations of CH 3 COOH and CH 3 COO – are giving a pH of A similar calculation shows that the pH after adding 20.0 mL of NaOH is 4.58. At the equivalence point, the moles of acetic acid initially present and the moles of NaOH added are identical. Since their reaction effectively proceeds to comple- tion, the predominate ion in solution is CH 3 COO – , which is a weak base. To calcu- late the pH we first determine the concentration of CH 3 COO – . The pH is then calculated as shown in Chapter 6 for a weak base. [] (. )( . ) .CH COO moles CH COOH total volume M mL mL mL M 3 3 − == + = 0 100 50 0 50 0 50 0 0 0500 pH =+ =476 0 0167 0 0667 416. log . . . [] (. )( . ) .CH COO M mL mL mL M 3 − = + = 0 100 10 0 50 0 10 0 0 0167 [ (. )( . ) (. )( . ) .CH COOH] M mL M mL mL mL M 3 = − + = 0 100 50 0 0 100 10 0 50 0 10 0 0 0667 [CH COO ] moles NaOH added total volume 3 bb ab − == + MV VV [CH COOH] moles unreacted CH COOH total volume 3 3aabb ab == − + MV MV VV pH p A HA] a =+ − K log [] [ K xx x a == − =× − − [][ ] [ ()() . . H O CH COO CH COOH] 3 + 3 3 0 100 175 10 5 282 Modern Analytical Chemistry 1400-CH09 9/9/99 2:12 PM Page 282 [...]... equivalence point is small *See Chapter 11 for more details about pH electrodes 1400-CH 09 9 /9/ 99 2:12 PM Page 291 Chapter 9 Titrimetric Methods of Analysis Table 9. 5 Data for the Titration of a Weak Acid with 0.100 M NaOH Normal Titration First Derivative 0.270 0.138 11.00 0.130 13.00 0.240 14.50 0.450 15.25 1.34 15.55 6.50 15.65 8 .90 15.75 7.80 15 .90 2 .90 16.50 0. 690 17.50 0. 290 19. 00 0.185 22.00 0.103 ∆2pH... methyl red and methylene blue neutral red and methylene blue blue-violet red-violet violet-blue green green green 3.2–3.4 5.2–5.6 6.8–7.3 HIn Figure 9. 11 Ladder diagram showing the range of pH levels over which a typical acid–base indicator changes color 1400-CH 09 9 /9/ 99 2:12 PM Page 290 290 Modern Analytical Chemistry 14.0 12.0 pH 10.0 Figure 9. 12 Titration curve for 50.00 mL of 0.100 M CH3COOH with 0.100... reach the equivalence point is 50 mL, so we draw a vertical line intersecting the x-axis at this volume (Figure 9. 7b) *Question 4 in the end-of -chapter problems asks you to consider why these pH limits correspond to approximately 10% and 90 % of the equivalence point volume 1400-CH 09 9 /9/ 99 2:12 PM Page 285 Chapter 9 Titrimetric Methods of Analysis Percent titrated 0 50 100 Percent titrated 150 200... initial point in the first derivative titration curve for the data in Table 9. 5 is ∆pH 4.52 − 2. 89 = = 0.815 ∆V 2.00 − 0.00 and is plotted at the average of the two volumes (1.00 mL) The remaining data for the first derivative titration curve are shown in Table 9. 5 and plotted in Figure 9. 13b 291 1400-CH 09 9 /9/ 99 2:12 PM Page 292 Modern Analytical Chemistry 14 12 10 8 6 4 2 0 10 ∆pH/∆V 8 0 10 5 15 20 Volume... linearized form of a titration curve 293 1400-CH 09 9 /9/ 99 2:12 PM Page 294 Modern Analytical Chemistry Temperature Table 9. 6 Titration branch Excess titration branch Volume (mL) Vtitr Temperature (a) Vtitr (b) Figure 9. 14 Figure 9. 15 Titration curves for 50.00 mL of 0.0100 M H3BO3 with 0.100 M NaOH determined by monitoring (a) pH, and (b) temperature 604 348 1 29 85.0 33.8 12.5 2.76 0.621 0.0805 0.0135... the Data in Table 9. 5 (a) 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.00 1.00 2.00 3.00 4.00 5.00 6.00 Volume of titrant Temperature 294 25.100 25.060 25.020 24 .98 0 –1 (b) 0 2 3 4 5 1 Volume of titrant 6 1400-CH 09 9 /9/ 99 2:12 PM Page 295 Chapter 9 Titrimetric Methods of Analysis for H3BO3 versus –55.6 kJ/mol for HCl), resulting in a favorable thermometric titration curve (Figure 9. 15b) 9B.3 Titrations in... – 5.3 = 8.7 The change in pH when the titration passes from 90 % to 110% completion is ∆pH = 8.7 – 5.3 = 3.4 295 1400-CH 09 9 /9/ 99 2:12 PM Page 296 296 Modern Analytical Chemistry 20.0 pH 15.0 (b) 10.0 (a) pH = pKs – pOH = 20.0 – 5.3 = 14.7 5.0 0.0 0.00 In this case the change in pH of 20.00 40.00 60.00 80.00 100.00 Volume of titrant Figure 9. 16 10–4 Titration curves for 50.00 mL of M HCl with 10–4 M... cannot be used to prepare carbonate-free solutions of NaOH Solutions of carbonate-free NaOH can be prepared from 50% w/v NaOH since Na2CO3 is very insoluble in concentrated NaOH When CO2 is absorbed, Na2CO3 299 1400-CH 09 9 /9/ 99 2:12 PM Page 300 300 Modern Analytical Chemistry precipitates and settles to the bottom of the container, allowing access to the carbonate-free NaOH Dilution must be done with... CO2(g) + 2OH–(aq) → CO32–(aq) + H2O(l) 9. 7 When CO2 is present, the volume of NaOH used in the titration is greater than that needed to neutralize the primary standard because some OH– reacts with the CO2 *The nominal concentrations are 12.1 M HCl, 11.7 M HClO4, and 18.0 M H2SO4 1400-CH 09 9 /9/ 99 2:12 PM Page 299 Chapter 9 Titrimetric Methods of Analysis Table 9. 7 Selected Primary Standards for the... feasible in a different solvent 9B.4 Representative Method Although each acid–base titrimetric method has its own unique considerations, the following description of the determination of protein in bread provides an instructive example of a typical procedure 1400-CH 09 9 /9/ 99 2:12 PM Page 297 Representative Methods Chapter 9 Titrimetric Methods of Analysis Method 9. 1 297 Determination of Protein in . acid–base titrations. Our Chapter 9 Titrimetric Methods of Analysis 2 79 1400-CH 09 9 /9/ 99 2:12 PM Page 2 79 approach will make use of the equilibrium calculations described in Chapter 6. We also will. 12.41 90 .00 12.46 95 .00 12. 49 100.00 12.52 Figure 9. 6 Titration curve for 50.0 mL of 0.100 M acetic acid (pK a = 4.76) with 0.100 M NaOH. 1400-CH 09 9 /9/ 99 2:12 PM Page 283 284 Modern Analytical Chemistry The. blue blue-violet green 3.2–3.4 methyl red and methylene blue red-violet green 5.2–5.6 neutral red and methylene blue violet-blue green 6.8–7.3 1400-CH 09 9 /9/ 99 2:12 PM Page 2 89 Figure 9. 12 Titration