1 / 14

Solubility Unit III Lesson 1

Solubility Unit III Lesson 1. Solubility is the maximum amount of solute that will dissolve in a volume of water. Units:. g/L. mol/L. g/100 mL. The unit must have an amoun t on the top and volume on the bottom !.

archie
Download Presentation

Solubility Unit III Lesson 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Solubility Unit III Lesson 1

  2. Solubility is the maximum amount of solute that will dissolve in a volume of water. Units: g/L mol/L g/100 mL The unit must have an amount on the top and volume on the bottom! In order to determine the solubility you must completely fillor saturate the solution! To saturate a solution, add weighed portionsof the solid to a volume of waterand stir until full. A bit of excess solidis present when saturated.

  3. Determining The Solubility of MgCl2 Add measured portions of MgCl2 to 100.0 mL and stir to dissolve Amount MgCl2 Dissolved Rate of dissolving > Rate of crystallization 10.0 g MgCl2 MgCl2 MgCl2 MgCl2 MgCl2 unsaturated 10.0 g 10.0 g slow 100.0 mL 3.0 g very slow 0.0 g saturated 33.0 g equilibrium Mg2+ Cl- Rate of dissolving = Rate of crystallization MgCl2(s)

  4. Calculate the solubility in units of g/L and mole/L 33.0 g Solubility= = 330. g/L 0.100 L the molarity to saturate molar solubility = 33.0 g x 1 mole 95.3g s = 0.100 L = 3.46 M

  5. do not use the solid EquilibriumEquation MgCl2(s)⇌ Mg2+ + 2Cl- Expression: Keq = [Mg2+][Cl-]2 Ksp = [Mg2+][Cl-]2 The Ksp or solubility product is used for saturated solutions at equilibrium

  6. This unit is all about the Ksp and the solubility of saturated solutions.

  7. Unsaturated Solutions Not full -more solid dissolves The rate of dissolving > therateof crystallizing Not at equilibrium

  8. Saturated Solutions Full-more solid does not dissolve The rate of dissolving = therateof crystallizing At equilibrium

  9. Supersaturated Solutions Over full -more solid causes precipitation The rate of dissolving < therateof crystallizing Not at equilibrium

  10. What we just calculated was exact solubility The data book page 4 gives us relative solubility Low Solubility means  .1M High Solubility means > .1 M CuCl2 Na3PO4 High High Ca(NO3)2 CuCl Low High High K2CO3 High CuSO4 CaSO4 Low Ag2SO4 Low FeSO4 High BaS High

  11. Ionic Solutions Molecular Solutions NaCl(aq)C6H12O6(aq)metal Ca(OH)2(aq)C12H22O11(aq)nonmetal (NH4)3PO4(aq)CH3OH(aq) Ca(CH3COO)2(aq)O2(aq) H2SO4(aq)N2H4(aq) Conduct electricityDo not conduct electricity

  12. Write dissociation (ionic) or dissolving (molecular) equations for the first six compounds to show how each dissolves in water. The first five are solids and the sixth one is a liquid. NaCl(s)  Na+ + Cl- C6H12O6(s) C6H12O6(aq) Ca(OH)2(s) Ca2+ + 2OH- C12H22O11(s) C12H22O11(aq) (NH4)3PO4(s) 3NH4+ + PO43- CH3OH(l)  CH3OH(aq)

  13. Write the equationfor equilibrium present in a saturated solution of Al2(SO4)3(s) solution. Al2(SO4)3(s) ⇌ 2Al3+ + 3SO42- Equilibrium Expression Ksp = [Al3+]2[SO42-]3

  14. Write the equationfor equilibrium present in a saturated solution of Ca3(PO4)2(s) solution. Ca3(PO4)2(s) ⇌ 3Ca2+ + 2PO43- Equilibrium Expression Ksp = [Ca2+]3[PO43-]2 Supersaturated video

More Related