Ksp Problem Set - Widener University

More documents

Recommendations

Info

d.The mass of CaF 2 that will dissolve in 5000.0 L of 0.10 M potassium fluoride. Potassium fluoride is a salt, it dissociates completely according to KF --> K1+ + F1- − 1 C KF := 0.1⋅mole⋅liter C F_initial := C KF Next the equlibrium for the solubility of CaF 2 CaF2 Ca2+ + 2 F1- ⋅( ) 2 K sp_CaF2 = X C F_initial + 2X If you assume that X is small this rearanges to: 2 K sp_CaF2 = X⋅C F_initial X := K sp_CaF2 C F_initial 2 X 3.894 × 10 − 9 − 1 = mole⋅liter Which shows that the assumption was reasonable From this we can calculate the mass of CaF 2 that dissolves V := 5000⋅L mole CaF2 := V⋅ X mole CaF2 = 1.947 × 10 − 5 mol mass CaF2 := mole CaF2 ⋅MW CaF2 mass CaF2 = 1.52 × 10 − 3 gm ksp_a.mcd 3/25/2004 10 S.E. Van Bramer svanbram@science.widener.edu
4.Use the K sp values calculated in the previious problem set to determine the mass of precipitate formed and the concentration all ions remaining in soultion when: a.50.0 mL of 0.10 M barium acetate is mixed with 100.0 mL of 0.10 M sodium sulfate. Sodium sulfate is a salt, it dissociates completely according to Na 2 SO 4 --> 2Na 1+ + SO 4 2- − 1 C Na2SO4 := 0.1⋅mole⋅liter 50⋅mL C SO4_initial := C Na2SO4 ⋅ 100⋅mL − 1 C SO4_initial = 0.05 mole⋅liter Barium Acetate is a salt, it dissociates completely according to Ba(CH3COO)2 --> Ba2+ + 2 CH3COO- − 1 C BaCH3COO := 0.1⋅mole⋅liter 50⋅mL C Ba_initial := C BaCH3COO ⋅ 100⋅mL − 1 C Ba_initial = 0.05 mole⋅liter Since Barium Sulfate is insoluble, it will for a precipitate. In this direction the reaction has a large value for K, so we'll assume it goes to completion. Since the moles of barium and sulfate are equal there is no limiting reagent. V := 100⋅mL mole BaSO4 := mass BaSO4 := C Ba_initial ⋅V mole BaSO4 ⋅MW BaSO4 mole BaSO4 = 5 × 10 − 3 mol mass BaSO4 = 1.167 gm Next the equlibrium for the solubility of BaSO 4 BaSO4 Ba2+ + SO42- K sp_BaSO4 = X 2 ( ) 1 2 X := K sp_BaSO4 X 1.047 10 − 5 − 1 = × mole⋅liter From this we can give the concentration of all the ions in solution: C Ba := X C SO4 := X 50⋅mL C CH3COO := C BaCH3COO ⋅ 100⋅mL 50⋅mL C Na := C Na2SO4 ⋅ 100⋅mL C Ba 1.047 10 − 5 − 1 = × molliter C SO4 1.047 10 − 5 − 1 = × molliter − 1 C CH3COO = 0.05 molliter − 1 C Na = 0.05 molliter ksp_a.mcd 3/25/2004 11 S.E. Van Bramer svanbram@science.widener.edu
Page 1 and 2: Ksp Problem Set By S.E. Van Bramer
Page 3 and 4: c.4.807*10-4 g AgCl dissolves in 25
Page 5 and 6: 2.Use the K sp values calculated ab
Page 7 and 8: 3.Use the K sp values calculated ab
Page 9: c.The mass of AgCl that will dissol
Page 13 and 14: c.75.0 mL of 0.015 M silver nitrate
Page 15: Since Calcium Fluoride is insoluble

Ksp Problem Set - Widener University

Create successful ePaper yourself

Delete template?

Save as template?