Lecture05

Lecture05 - Example: Dissolve a 0.8864 g sample of solid...

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Unformatted text preview: Example: Dissolve a 0.8864 g sample of solid containing Na2CO3 + NaHCO3 and titrate with 0.1000 M HCl. Vphth = 10.25 mL; Vbcg = 37.38 mL At phenolphthalein E.P.: Na2CO3 + HCl -----> NaHCO3 + NaCl mmol Titrant = mmol Titrated x S.R. 1 mmol HCl mmol HCl = mmol Na CO x 1 mmol Na CO 2 3 2 3 0.1000 M x 10.25 mL = Wt Na2CO3 1 mmol HCl x 106.0 mg/mmol 1 mmol Na2CO3 Wt Na2CO3 = 108.6 mg 108.6 mg Wt% Na2CO3 = x 100% = 12.25% 886.4 mg At bromocresol green E.P.: 1) Reaction of NaHCO3 NaHCO3 + HCl -----> NaCl + CO2 + H2O mmol Titrant = mmol Titrated x S.R. mmol HCl = mmol NaHCO x 3 1 mmol HCl 1 mmol NaHCO 3 2) Reaction of Na2CO3 Na2CO3 + 2HCl -----> 2NaCl + CO2 + H2O mmol HCl = mmol Na CO x 2 3 2 mmol HCl 1 mmol Na CO 2 3 For mix of Na2CO3and NaHCO3: mmol HCl = mmol Na CO x 2 3 2 mmol HCl 1 mmol Na CO 2 3 3 + mmol NaHCO x 1 mmol HCl 1 mmol NaHCO 3 M HCl xV bcg = Wt Na CO 2 3 106.0 mg/mmol Wt NaHCO x 2 mmol HCl 1 mmol Na CO 2 3 1 mmol HCl + x 84.01 mg/mmol 1 mmol NaHCO 3 3 0.1000 M x 37.38 mL = 108.6 mg 2 mmol HCl x 106.0 mg/mmol 1 mmol Na CO 2 3 + Wt NaHCO 3 84.01 mg/mmol x 1 mmol HCl 1 mmol NaHCO 3 Wt NaHCO3 = 141.9 mg 141.9 mg Wt% NaHCO3 = x 100% 886.4 mg = 16.01% ! Back-Titration: e.g., determine Na2CO3 1) Add excess HCl to Na2CO3 sample: Na CO + 2HCl -----> 2NaCl + CO + H O 2 3 2 2 2) Back-Titrate excess HCl: HCl + NaOH -----> NaCl + H2O At equiv. point of back-titration: 1 NaOH mmol NaOH = mmol HCl x 1 HCl Where is Na2CO3? ! mmol HCl = mmol Na2CO3 x 2 HCl 1 Na 2CO3 1 HCl + mmol NaOH x 1 NaOH ! Back Titrations: React analyte with excess (known amount) of first standard solution. Back-titrate excess with second standard solution. At E.P., amount of first standard is chemically equivalent to total of analyte and back-titrant: Moles E = Moles A x S.R.A + Moles B x S.R.B E = First Standard, added in Excess = ME x VE A = Analyte (or something derived from it) = MA x VA or WtA m.m.A B = Back-Titrant = MB x VB Example (14-25): Antabuse is analyzed by burning it to generate SO2: C10H20N2S4 (Antabuse) + 21O2 -----> 10CO2 + 10H2O + 2NO2 + 4SO2 SO2 gas is absorbed in a solution of H2O2 to generate sulfuric acid: SO2(g) + H2O2 -----> H2SO4 Sulfuric acid is then titrated with standardized sodium hydroxide: H2SO4 + 2NaOH -----> Na2SO4 + 2H2O If a 0.4329 g sample requires 22.13 mL of 0.03736 M NaOH to reach the end point, what is the Wt% Antabuse in the drug? (1) Start with analytical rxn: mol NaOH = mol H2SO4 x (2) Work backwards: 2 NaOH 1 H2SO4 ! x 1 H2SO4 mol H2SO4 = mol SO2 1 SO2 4 SO2 mol SO2 = mol Antabuse x 1 Antabuse ! Substitute stoichiometric ratios: mol H2SO4 = ! 4 SO2 1 H2SO4 mol Antabuse x x 1 Antabuse 1 SO2 mol NaOH = mol Antabuse x ! 4 SO2 x 1 H2SO4 x 2 NaOH 1 Antabuse 1 H2SO4 1 SO2 ! ! ! ! Substitute numbers: 0.03736 M x 22.13 mL = Wt Antabuse 8 NaOH x 296.54 g/mol 1 Antabuse Wt Antabuse = 30.65 ! mg ! 30.65 mg Wt% Ant = x 100% = 7.079% 432.9 mg ! Sequence of Reactions: Analyte not titrated directly. Two steps to solve: (1) Start with Analytical Rxn (last rxn) (2) Work backwards to desired analyte For Example - Determination of A: aA + bB -----> cC rC + sD -----> tE **Analytical Reaction Fund. Rel. Titr. Analysis: sD mol D = mol C x rC From Previous Rxn: cC mol C = mol A x aA Substitute: cC sD mol D = mol A x x aA rC cxsD mol D = mol A x axrA Gravimetric Methods of Analysis Volatilization: Reaction evolves a gas. OR Weigh sample before and after rxn. Precipitation: Reaction forms a sparingly soluble salt. Filter, dry, and weigh. 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[Ba2+][IO3-]2 [Ba(IO3)2(s)] At equilibrium: K= But [Ba(IO3)2(s)] is constant, so: Ksp = K[Ba(IO3)2(s)] = [Ba2+][IO3-]2 Solubility Product: In general for salt MxAy: Ksp = [M]x[A]y Units = Molarityx+y (optional) NOTE: pKsp = -log(Ksp) ...
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