Solubility Lesson 8 Review Notes. Adding a Crystal to a Saturated Solution Consider the saturated...

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Transcript of Solubility Lesson 8 Review Notes. Adding a Crystal to a Saturated Solution Consider the saturated...

  • SolubilityLesson 8Review Notes

  • Adding a Crystal to a Saturated SolutionConsider the saturated solution

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated-

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

    More solid AgCl is added to the saturated solution above.

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

    More solid AgCl is added to the saturated solution above.

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

    More solid AgCl is added to the saturated solution above.

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

    More solid AgCl is added to the saturated solution above.The solution is full already so the ion concentrations remain constant!

    Cl-

    Ag+

  • Adding a Crystal to a Saturated SolutionNote that the solution is saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

    More solid AgCl is added to the saturated solution above.The solution is full already so the ion concentrations remain constant!The new solid dissolves at a greater rate, however crystallizes at an equally greater rate so there is no net change in ion concentration!

    Cl-

    Ag+

  • Describe the change in each of the following when more solid AgCl is added to a saturated solution.1.[Ag+]

  • Describe the change in each of the following when more solid AgCl is added to a saturated solution.1.[Ag+]Constant2.[Cl-]

  • Describe the change in each of the following when more solid AgCl is added to a saturated solution.1.[Ag+]Constant2.[Cl-]Constant3.Rate of dissolving

  • Describe the change in each of the following when more solid AgCl is added to a saturated solution.1.[Ag+]Constant2.[Cl-]Constant3.Rate of dissolvingIncreases4.Rate of crystallizing

  • Describe the change in each of the following when more solid AgCl is added to a saturated solution.1.[Ag+]Constant2.[Cl-]Constant3.Rate of dissolvingIncreases4.Rate of crystallizing Increases

    Replay the last frame if you don't get this!

  • 5.The ions in hard water are Mg2+ and Ca2+

    Na2CO3 can be added to water to remove these ions

    6.Calculate the total ion concentration of 1.0 M AlCl3

    AlCl3Al3+ + 3Cl-

    1.0 M1.0 M3.0 M

    Total4.0 M

    7.What is the solubility of CaC2O4?

    Not on page 4On page 5Must be low! They all are!

  • 8.The solubility is 7.1 x 10-5 M. The compound is

    A.CaSO4

    B.CaCO3

    Ksp = s2=5.0 x 10-9B

    9.A solution of AgNO3 is added slowly to each of the following 0.10 M solutions. Which forms a precipitate first?

    A.NaCl

    B.NaIO3

  • 10.The solubility is 7.1 x 10-5 M. The compound is

    A.CaSO4

    B.CaCO3

    Ksp = s2=5.0 x 10-9B

    11.A solution of AgNO3 is added slowly to each of the following 0.10 M solutions. Which forms a precipitate first?

    A.NaClksp = 1.8 x 10-10

    B.NaIO3ksp = 3.2 x 10-8

  • 12.The solubility is 7.1 x 10-5 M. The compound is

    A.CaSO4

    B.CaCO3

    Ksp = s2=5.0 x 10-9B

    13.A solution of AgNO3 is added slowly to each of the following 0.10 M solutions. Which forms a precipitate first?

    A.NaClksp = 1.8 x 10-10smaller

    B.NaIO3ksp = 3.2 x 10-8

  • 14.Small amounts of AgNO3 are added to three solutions that have the same concentration. If only one solution does not form a precipitate, which one is it ?

    A.NaCl

    B.NaIO3

    C.NaBr

  • 14.Small amounts of AgNO3 are added to three solutions that have the same concentration. If only one solution does not form a precipitate, which one is it ?

    A.NaClksp = 1.8 x 10-10

    B.NaIO3ksp = 3.2 x 10-8

    C.NaBrksp = 5.4 x 10-8

  • 14.Small amounts of AgNO3 are added to three solutions that have the same concentration. If only one solution does not form a precipitate, which one is it ?

    A.NaClksp = 1.8 x 10-10

    B.NaIO3ksp = 3.2 x 10-8

    C.NaBrksp = 5.4 x 10-8

  • 14.Which solution has the greatest conductivity?

    A.1.0 M AgCl

    B.1.0 M CaCO3

    C.1.0 M AlCl3

    D.1.0 M CaCl2

  • 14.Which solution has the greatest conductivity?

    A.1.0 M AgCllow

    B.1.0 M CaCO3low

    C.1.0 M AlCl3high

    D.1.0 M CaCl2high

  • 14.Which solution has the greatest conductivity?

    A.1.0 M AgCllow

    B.1.0 M CaCO3low

    C.1.0 M AlCl3highAlCl3 Al3+ + 3Cl-D.1.0 M CaCl2highCaCl2 Al3+ + 2Cl-

  • 15.Which solution has the greatest conductivity?

    A.1.0 M AgCllow

    B.1.0 M CaCO3low

    C.1.0 M AlCl3highAlCl3 Al3+ + 3Cl-D.1.0 M CaCl2highCaCl2 Ca2+ + 2Cl-

    4 ions versus 3

  • 16.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

    AgCl(s) Ag+ + Cl-

    0.20 M

    Ksp=[Ag+][Cl-]

    1.8 x 10-10 = [0.20][Cl-]

    [Cl-] =9.0 x 10-10 M

    BaCl2(s) Ba2+ + 2Cl-9.0 x 10-10 M

  • 15.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

    AgCl(s) Ag+ + Cl- 0.20 M

    Ksp=[Ag+][Cl-]

    1.8 x 10-10 = [0.20][Cl-]

    [Cl-] =9.0 x 10-10 M

    BaCl2(s) Ba2+ + 2Cl-4.5 x 10-10 M9.0 x 10-10 M

  • 15.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution. .

    AgCl(s) Ag+ + Cl- 0.20 M

    Ksp=[Ag+][Cl-]

    1.8 x 10-10 = [0.20][Cl-]

    [Cl-] =9.0 x 10-10 M

    BaCl2(s) Ba2+ + 2Cl-4.5 x 10-10 M9.0 x 10-10 M0.50 L

  • 15.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

    AgCl(s) Ag+ + Cl- 0.20 M

    Ksp=[Ag+][Cl-]

    1.8 x 10-10 = [0.20][Cl-]

    [Cl-] =9.0 x 10-10 M

    BaCl2(s) Ba2+ + 2Cl-4.5 x 10-10 M9.0 x 10-10 M0.50 L x 4.5 x 10-10 mole1 L

  • 15.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

    AgCl(s) Ag+ + Cl- 0.20 M

    Ksp=[Ag+][Cl-]

    1.8 x 10-10 = [0.20][Cl-]

    [Cl-] =9.0 x 10-10 M

    BaCl2(s) Ba2+ + 2Cl-4.5 x 10-10 M9.0 x 10-10 M0.50 L x 4.5 x 10-10 molex 208.3 g 1 L 1 mole

  • 15.Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

    AgCl(s) Ag+ + Cl- 0.20 M

    Ksp=[Ag+][Cl-]

    1.8 x 10-10 = [0.20][Cl-]

    [Cl-] =9.0 x 10-10 M

    BaCl2(s) Ba2+ + 2Cl-4.5 x 10-10 M9.0 x 10-10 M0.50 L x 4.5 x 10-10 molex 208.3 g = 4.7 x 10-8 g1 L 1 mole

  • 16.In an experiment to determine the solubility of BaF2, 500 mL of the saturated solution was heated in an evaporating dish to remove the water. The evaporating dish and residue were heated two more times to ensure all of the water had been driven off.

    Volume of saturated solution500.0 mLMass of evaporating dish72.540 gMass of evaporating dish & BaF2 after first heating73.500 gMass of evaporating dish & BaF2 after second heating72.855 gMass of evaporating dish & BaF2 after third heating72.855 gUse this data to calculate the Ksp for BaF2.

  • [BaF2] = s = (72.855 - 72.540)gx1 mole175.3 g0.500 L

    s=0.0035938 M

    BaF2(s) Ba2+ + 2F- ss2sKsp=[Ba2+][F-]2Ksp=[s][2s]2Ksp=4s3Ksp=4(0.0035938)3

    Ksp=1.86 x 10-7

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