Titration. Lecture # 8. Titrations in Analytical Chemistry. Other Forms of Titration. End Point vs. Equivalence Point. Minimizing Titration Error

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1 Lecture # MnO 5 H C O H CO Mn H O (purple) (colorless) (colorless) (colorless) Volumetric Analysis 0 ml s in Analytical Chemistry 11 Gravimetric Analysis Other Forms of End Point vs. Equivalence Point e.g. Titer Plates EnzymeLinked Immunosorbent Assay (ELISA) Equivalence Point occurs when the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte End Point > Equivalence Point (titration error) End Point 1. lank 1. lank. ack lank (No ) 1

2 Direct Excess Titrant ack ack 1. lank. ack 3. Choose a property that is easily observable! 1. Observing Color Change of an Indicator. Precipitation/Dissolution 3. Detection of Color Change (Absorption of Light = Spectrophotometry). Detection of Change in Voltage or Current (Electrochemistry) Spectrophotometric Spectrophotometric Corrected = total volume x Observed Absobance initial volume Absorbance Absorbance End Point Absorbance Volume of Titrant Added Concentration KMnO 5 H O 3 H SO MnSO K SO 5 O H O 0.00 M KMnO H O H SO (5 ml) Calculations Standard Solution = 0.00 M KMnO.3 ml of KMnO added What is the molar concentration of H O? 0.00 mol/l x 0.03 ml = mol KMnO Moles of H O = 5 mol H O x (0.001 mol KMnO ) mol KMnO = mol H O Molar Concentration = mol/(0.05 L) = 0. M Purity of Titrant Primary Standard: pure enough to be weighed and used directly Primary Standard Technical Grade ACS Reagent Grade Ultra Pure Trace Analysis (1 00 ppm) Ultratrace Analysis (<1 ppm) Standardization prepare titrant with approximately the desired concentration and use it to titrate a primary standard

3 Standardization Example:We make an approximately 0.0 M solution of KMnO. We dissolve mg of a primary standard of sodium oxalate in 50.0 ml of water, and use this to standardize. A volume of ml is required to titrate the standard solution. What is the actual concentration of MnO? MnO 5 H C O H CO Mn H O (purple) (colorless) (colorless) (colorless) ( g/(13.00 g/mol) = mol C O ( mol MnO / 5 mol C O ) x mol C O = mol MnO mol / L = M MnO Johan Kjeldahl (191900) Kjeldahl Digestion Kjeldahl Digestion Organic C, H, N NH CO H O (Sample) H SO ( K SO ) 33 C NaOH NH NH 3 (g) H O Steam Distillation to Collect NH 3 Kjeldahl Digestion (continued) NH 3 HCl (excess) HCl NaOH NH Cl HCl Na Cl H O (titration) Example: Johan Kjeldahl digest 1 gram of grain by his Kjeldahl Method. He distills the resulting NH 3 (following addition of base) into ml of 0.00 M HCl. The unreacted HCl requires 3. ml of 0.00 M NaOH. What is the concentration of nitrogen in the gram of grain? (0.0 L)(0.0 mol/l HCl) = mol HCl (in receiver) (0.003 L)(0.0 mol/l NaOH) = mol NaOH = mol NH 3 = mol N mol x (1.007 g/mol) = g N/g grain Volhard Ag H NO 3 Cl AgCl (s) Ag Filter and Wash Ag K SCN Fe 3 (filtrate) AgSCN (s) FeSCN (end point) Ag (initial) Ag (filtrate) = Ag (precipitated with Cl ) Fajans efore Equivalance Point After Equivalance Point 3

4 Fajans Dichlorofluorescein pfunction: px = log [X] Ag I Ag I Ag I Ag I K sp = 1. x 1 K sp =.3 x 17 K sp = 1. x 1 K sp =.3 x ml of 0.00 M I titrated with M Ag V e = Volume added at end point ( ml)(0.00 mol/l) = V e ( mol L) V e = L = ml efore equivalence point: e.g..00 ml of Ag added. What is [Ag ]? Moles of I = ( L)(0.00 mol/l) (0.000 L)( mol/l) = mol mol = mol [I ] = ( mole)/( L L) = M [Ag ] = K sp /([I ] =.3 x 17 /(0.0571) = 1. 5 x 15 pag = log (1. 5 x 15 ) = 1. Ag I Ag I K sp = 1. x 1 K sp =.3 x 17 At the equivalence point: e.g ml of Ag added. What is [Ag ]? [Ag ] = [I ] = x [Ag ][I ] = (x)(x) = x = K sp =.3 x 17 x =.3 x 17 = 9.1 x 9 pag = log (9.1 x 9 ) =.0 Ag I Ag I K sp = 1. x 1 K sp =.3 x 17 After the equivalence point: e.g ml of Ag added. What is [Ag ]? V Ag = 5.00 ml ml =.00 ml Moles of Ag = ( L)( mol/l) = mol [Ag ] = ( mol)( L L L) = ( mol)/( L) = 1.30 x 3 M pag = log (1.30 x 3 ) =.9

5 Curve Curve A A (s) A A (s) efore Equivalence Point [ V ] = e V A [ V ] V i e V V A After Equivalence Point [A ] = V A V e [A ] V i V A [I ml.00 ml ] = (0.0 M) ml 5.00 ml 5.00 ml.00 ml [A ] = 5.00 ml ml (0.050 M) 5.00 ml 5.00 ml = (/5)(0.0 M)(5/35) = M = (/77)(0.050 M) = 1.30 x 3 M pa = log (K sp /[ ]) pa = log [A ] pag = log (.3 x 17 /0.0571) = 1. pag = log (1.30 x 3 ) =.9 A pag 1 1 Equivalance Point pa 1 1 C A A [ ] A > [ ] > [ ] C C A V Ag (ml) V A (ml) pa 1 1 D D A A C A D C D V A (ml) A AC AD 5