Analytical Chem istry - DePauw University

518 Analytical Chemistry 2.00.0316 M HCl required 38.58 mL to reach the end point. Determinethe ppm CO 2 in the sample of air given that the density of CO 2 at thetemperature of the sample is 1.98 g/L.21. The purity of a synthetic preparation of methylethyl ketone, C 3 H 8 O, isdetermined by reacting it with hydroxylamine hydrochloride, liberatingHCl (see reaction in Table 9.8). In a typical analysis a 3.00-mL samplewas diluted to 50.00 mL and treated with an excess of hydroxylaminehydrochloride. The liberated HCl was titrated with 0.9989 M NaOH,requiring 32.68 mL to reach the end point. Report the percent purityof the sample given that the density of methylethyl ketone is 0.805 g/mL.Some of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction PotentialspH1412108640 10 20 30 40 50Volume of NaOH (mL)Figure 9.47 Titration curve for Problem9.24.22. Animal fats and vegetable oils are triesters formed from the reactionbetween glycerol (1,2,3-propanetriol) and three long-chain fatty acids.One of the methods used to characterize a fat or an oil is a determinationof its saponification number. When treated with boiling aqueousKOH, an ester saponifies into the parent alcohol and fatty acids (ascarboxylate ions). The saponification number is the number of milligramsof KOH required to saponify 1.000 gram of the fat or the oil.In a typical analysis a 2.085-g sample of butter is added to 25.00 mLof 0.5131 M KOH. After saponification is complete the excess KOH isback titrated with 10.26 mL of 0.5000 M HCl. What is the saponificationnumber for this sample of butter?23. A 250.0-mg sample of an organic weak acid is dissolved in an appropriatesolvent and titrated with 0.0556 M NaOH, requiring 32.58 mL toreach the end point. Determine the compound’s equivalent weight.24. Figure 9.47 shows a potentiometric titration curve for a 0.4300-g sampleof a purified amino acid that was dissolved in 50.00 mL of waterand titrated with 0.1036 M NaOH. Identify the amino acid from thepossibilities listed in the following table.amino acid formula weight (g/mol) K aalanine 89.1 1.36 × 10 –10glycine 75.1 1.67 × 10 –10methionine 149.2 8.9 × 10 –10taurine 125.2 1.8 × 10 –9asparagine 150 1.9 × 10 –9leucine 131.2 1.79 × 10 –10phenylalanine 166.2 4.9 × 10 –10valine 117.2 1.91 × 10 –10