Lecture 9 anal chem.pdf

1 Analytical chemistry (BPH: 102) for the first year of Pharmacy Dr. Mustafa Benamer, Faculty of Pharmacy, Zawia Lecture 9 Acid-Base Titration Curves For acid-base titration, the equivalence point characterizes...

1 Analytical chemistry (BPH: 102) for the first year of Pharmacy Dr. Mustafa Benamer, Faculty of Pharmacy, Zawia Lecture 9 Acid-Base Titration Curves For acid-base titration, the equivalence point characterizes by a pH level that is a function of the acid-base strengths and concentrations of the analyte and titrant. However, the pH at the endpoint may or may not correspond to the pH at the equivalence point. To understand the relationship between endpoints and equivalence points we must know how the pH changes during a titration. In this lecture, we will learn how to construct titration curves for several important types of acid–base titrations. Our approach will make use of the equilibrium calculations described in the previous lectures. We also will learn how to sketch a good approximation to any titration curve using only a limited number of simple calculations. Titration curve A graph showing the progress of titration as a function of the volume of titrant added. It provides us with a visual picture of how a property, such as pH, changes as we add titrant. 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 H3O+(aq) + OH– (aq) 2H2O(l)  2 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 the above reaction that Moles HCl = moles NaOH or MaVa = MbVb The volume of NaOH needed to reach the equivalence point, therefore, is Before the equivalence point, HCl is present in excess and the pH is determined by the concentration of unreacted HCl. At the start of the titration, the solution is 0.100 M in HCl, which, because HCl is a strong acid, means that the pH is pH= -log[H3O+] = -log[HCl] = -log(0.100) = 1.00 After adding 10.0 mL of NaOH the concentration of excess HCl is and the pH increases to 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 dissociation of water. Kw = 1x10-14 = [H3O+][OH-] = [H3O+]2 [H3O+] = 1 x10-7 M Thus, the pH at the equivalence point is 7.00.  3 For volumes of NaOH greater than the equivalence point, the pH is determined by the excess OH–. For example, after adding 30.0 mL of titrant the OH– concentration is To find the concentration of H3O+ we use the Kw expression Table 9.1: Titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH The volume of NaOH (mL) pH The volume of NaOH (mL) pH 0.00 1.00 26.0 11.42 5.00 1.14 28.0 11.89 10.0 1.30 30.0 12.10 15.0 1.51 35.0 12.37 20.0 1.85 40.0 12.52 22.0 2.08 45.0 12.63 24.0 2.57 50.0 12.70 25.0 7.00  4 Figure 9.2: Titration curve for the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH. Acid-Base Applications Although many quantitative applications of acid-base titrations have been replaced by other analytical methods, a few important applications continue to be relevant. In this lecture, we review the general application of acid-base titrations to the analysis of inorganic and organic compounds, with an emphasis on applications in environmental and clinical analysis. Inorganic Analysis Acid–base titrimetry is a standard method for the quantitative analysis of many inorganic acids and bases. A standard solution of NaOH can be used to determine the concentration of inorganic acids (H3PO4, H3BO3, and H3AsO4) and inorganic bases, such as Na2CO3, NaHCO3, and the mixture of both can be analyzed using a standard solution of HCl.  5 Acid–base titrimetry continues to be listed as a standard method for the determination of alkalinity (OH–, HCO3–, and CO32–), acidity (HCl, HNO3, and H2SO4, and free CO2 in waters and wastewaters. Determination of carbonate in a mixture of carbonate and bicarbonate using two indicators and a standard HCl solution Na2CO3 + HCL → NaHCO3 + NaCl (pH=8.3) NaHCO3 + HCL → NaCl + CO2 + H2O (pH≈4) For carbonate: The volume of HCl required for ½ of carbonate (1/2 V1) = 9.5mL. The volume of HCl required for of carbonate = 2V1 = 2x 9.5mL= 19.0mL. For bicarbonate: The volume of HCl required for of bicarbonate present in a mixture = V2-2V1 = 23-19.0 = 4mL Determination of carbonate in a mixture of Na2CO3 and NaOH using two indicators and a standard HCl solution 1- When a known volume of the mixture is titrated with HCl in presence of ph. ph., the acid reacts with all the sodium hydroxide and with only half of the carbonate. V1 = all hydroxide + 1/2 the carbonate 2- When a known mixture volume is titrated with HCl in the presence of M.O., the acid reacts with all the hydroxide and carbonate.  6 V2 = all hydroxide + all carbonate Volume of HCl = 1/2 carbonate = V2 – V1 = V ml Volume of HCl = all carbonate = 2V ml Volume of HCl = NaOH = V2 - 2V ml Organic Analysis Acid–base titrimetry continues to have a small, but important role in the analysis of organic compounds in pharmaceutical, biochemical, agricultural, and environmental laboratories. Perhaps the most widely employed acid–base titration is the Kjeldahl analysis for organic nitrogen. Examples of analytes determined by a Kjeldahl analysis include caffeine and saccharin in pharmaceutical products, proteins in foods, and the analysis of nitrogen in fertilizers, sludges, and sediments. Any nitrogen present in a –3 oxidation state is quantitatively oxidized to NH4+. Determination of Protein in Bread Description of the Method. This quantitative method of analysis for proteins is based on a determination of the %w/w N in the sample. Since different cereal proteins have similar amounts of nitrogen, the experimentally determined %w/w N is multiplied by a factor of 5.7 to give the %w/w protein in the sample (on average there are 5.7 g of cereal protein for every gram of nitrogen). Cereals contain between 8-15% of different kinds of proteins such as albumins, globulins, prolamines, gliadins, glutelins, and glutenins. The conversion factors used for the Kjeldahl analysis are the commonly used conversion factor 6.25 and species-specific conversion factors 5.4 for the flours (16% of protein content), 5.6 for fish and shrimp, and 4.59 for seaweed. As described here, nitrogen is determined by the Kjeldahl method. The protein in a sample of bread is oxidized in hot concentrated H2SO4, converting the nitrogen to NH4+. After  7 making the solution alkaline, converting NH4+ to NH3, the ammonia is distilled into a flask containing a known amount of standard strong acid. Finally, back titration with a standard strong base titrant determines the excess strong acid. The benefits of the Kjeldahl method:  High level of precision and reproducibility  High productivity  Worldwide official method  Reliable and easy method  Time-saving  Affordable equipment cost  Moderate running costs