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VO-9

VO-9. Übungsbeispiele. Beispiel 1:. Welche Masse BaSO 4 (in Milligramm ) löst sich in 450 mL Wasser ? ( K sp = 1 . 10 -10 , M(BaSO 4 ) = 233 g/ mol ). K sp = [Pb 2+ ][SO 4 2- ] = 1 . 10 -10. x 2 = 1 . 10 -10  x = 1.10 -5

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VO-9

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  1. VO-9 Übungsbeispiele

  2. Beispiel 1: Welche Masse BaSO4 (in Milligramm) löstsich in 450 mL Wasser? (Ksp = 1.10-10, M(BaSO4) = 233 g/mol)

  3. Ksp = [Pb2+][SO42-] = 1.10-10 x2 = 1.10-10 x = 1.10-5 (1.10-5mol/L)(0.450 L)(233 g/mol) = 0.001 g = 1 mg

  4. Beispiel 2: BestimmenSie die Masse an gelöstem PbI2 in a) 500 mL Wasser b) 500 mL einer 0.1 M KI Lösung c) 500 mL einerLösung die 1.33 g Pb(NO3)2 enthält. M(PbI2) = 461 g/mol Ksp = [Pb2+][I-]2 = 1.4.10-8

  5. PbI2Pb2+ +2I- Ksp = [Pb2+][I-]2 = 1.4.10-8 ad. a: bei x = [Pb2+] ist [I-] = 2x 4x3 = 1.4.10-8 x = 1.5.10-3mol/L  (1.5.10-3 mol/L) (0.500 L) (461 g/mol) = 0.35 mg ad. b: [Pb2+](0.10)2 = 1.4.10-8  [Pb2+] = 1.4.10-6 M m(PbI2) = (1.4.10-6 mol/L)(0.500 L)(461 g/mol) = 3.2.10-4 g = 0.32 mg c) c(Pb(NO3)2) wenn die Konzentration der PbI2 yist, dannist [I-] = 2y (8.04.10-3)(2y)2 = 1.4.10-8 y = 6.6.10-4 M m(PbI2) = (6.6.10-4mol/L)(0.500 L)(461 g/mol) = 0.15 g = 150 mg 

  6. Beispiel 3: Welche Masse an K2Cr2O7wirdbenötigt, um auseinemÜberschuß an Oxalsäure (H2C2O4), 5.00 L CO2bei 75 °C and 1.07 atmDruckherzustellen? Das Cr2O72-wirdzu Cr3+ reduziert. M(K2Cr2O7) = 294.2 g/mol

  7. 8H+ + Cr2O72- + 3H2C2O4 6CO2 + 2Cr3+ + 7H2O n = pV/RT = (1.07 atm) (5.00 L)/(0.0821 L.atm/mol.K) (348 K)= 0.187 mol CO2 n(Cr2O72-) = 0.187 mol CO2/6 = 0.0312 Cr2O72- m(K2Cr2O7) = 0.0312 mol.294.2 g/mol = 9.18 g

  8. Beispiel 4: 10 g einerMischungaus Cu2S und CuSwirdmit 200 mL 0.75 M MnO4- in saurerLösungbehandelt und produziert SO2, Cu2+ und Mn2+. Das SO2wirdabgekocht und der Überschuss an KMnO4wirdmit 175.0 mL 1 M Fe2+ Lösungtitriert. M(CuS) = 95.606 g/mol M(Cu2S) = 159.15 g/mol SchreibenSiebittealleReaktionsgleichungen an und berechnenSieden Gehalt an CuS in der ursprünglichenMischung.

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