1. Solubility
Unit 3
Lesson 1

2. Unit Intro Our focus is on solutions of aqueous ions
As you know; acids, bases and salts form ionic solutions.
This unit is only concerned with salts.

3. Review
Electrolytes: substances that dissolve to give electrically conducting solutions that contain ions.
Ex: MgCl2(s) ® Mg+(aq) Cl-(aq)
NaNO3
CuSO4

4. Salts differ in their ability to dissolve in water
Salts differ in their ability to dissolve in water. Some salts dissolve well and produce a solution with a high ion concentration. These are strong electrolytes. When salt don’t dissolve as well, they produce solutions with lower ion concentrations and are weaker electrolytes.

5. Review
Non-electrolytes: a substance that dissolves to give non-conducting solutions containing only neutral molecules.
Ex: C2H2OH (ethanol)
C12H22O11 sucrose

6. Molecular vs Ionic solutions
metals and non-metals
compounds that contain polyatomic ions
Both are referred to as salts.

7. Molecular compounds They are covalent compounds
Non-metal and non-metal
For example.. all organic compounds.
(high carbon content. Carbon bonds covalently to other elements).

8. Why do salts dissolve so well in water
Why do salts dissolve so well in water? Because of the dipole in the water molecule. This allows for the water molecule to compete with the electrostatic attraction between the salt ions. This process is called dissociation.

9. Useful hint
When it comes to our chem12 course, you are only going to deal with ionic compounds that have one type of positive ion and one type of negative ion only.
- no KNaFCl ® K Na+ + F- + Cl-

10. Ionic Solutions Molecular/Covalent Solutions
Ionic Solutions Molecular/Covalent Solutions
NaCl(aq) C6H12O6(aq) metal or polyatomic ion
Ca(OH)2(aq) C12H22O11(aq) nonmetal or carbon
(NH4)3PO4(aq) CH3OH(aq) Molecular - covalently bonded
Ca(CH3COO)2(aq) O2(aq)
H2SO4(aq) N2H4(aq)
Conduct electricity Do not conduct electricity

11. Write equations to show the dissolving of the following substances in water
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)

12. Back to solubility In chem 11, you learnt that…
The solubility of a substance is the maximum amount of the substance which can dissolve in a given amount of solvent at a given temperature.
Units for this are….

13. Equilibrium Solubility
The solubility of a substance is the equilibrium concentration of the substance in solution at a given temperature.
when expressed in moles/L it’s called Molar Solubility.

14. Solubility at equilibirum
Solid MgCl2 dissolves and enters solution
dissolving reaction
Mg+2 + Cl-1 ions come together to form MgCl2
crystallization reaction

15. Solubility at equilibrium
When the rate of dissolving reaction equals the rate of crystallization reaction, we have equilibrium.
A solution at equilibrium is called a saturated solution.

16. Saturation exists when…
Equilibrium exists between the dissolved (ions) and the undissolved material (solid)
Some undissolved material is till present (crystal solids)

17. How to saturate a solution & determine solubility
To saturate a solution, add weighed portions of your solid to a volume of (solvent) water and stir until full.
A bit of excess solid will always be present at equilibrium saturation though.
In order to determine the solubility, you must completely fill or saturate the solution!

18. Determining The Solubility of MgCl2
Add measured portions of MgCl2 to 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
Mg2+
Cl-
saturated
33.0 g
equilibrium
MgCl2(s)
Rate of dissolving = Rate of crystallization

19. Calculate the solubility in units of g/L and mole/L
Molar Solubility =
Moles/L
33.0 g
x 1 mole
95.3g =
0.100 L
= 3.46 M

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

21. Un-Saturated, Saturated and Super-satured Review

22. Unsaturated Solutions
Not full – ( more solid can dissolve if you add it)
How does it look? Clear solution!
The rate of dissolving > the rate of crystallizing
Not at equilibrium

23. Full- ( adding more solid will not dissolve )
Saturated Solutions
Full- ( adding more solid will not dissolve )
How does it look? it always has crystals/solids in the solution.
The rate of dissolving = the rate of crystallizing
At equilibrium

24. Supersaturated video Supersaturated Solutions
Over full – ( adding more solid causes precipitation)
How does it look? Clear solution!
The rate of dissolving < the rate of crystallizing
Not at equilibrium
Supersaturated video

25. Predicting the solubility of salts
Page 332 in your textbook has the table
“ Solubility of common compounds in water”
Will be provided for you
Use it to predict solubility ( high or low )
Use it to predict if a precipitate will form.

26. Low Solubility means £ .1M
High Solubility means > .1 M
Na3PO4
High
CuCl2
High
Ca(NO3)2
High
CuCl
Low
K2CO3  
High
High
CuSO4
CaSO4
Low
Ag2SO4
Low
FeSO4
High
BaS
High

27. Precipitate Questions
Will a precipitate form when 0.2 M solutions of CaS and Na2SO4 are mixed?

28. Write the equation for 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

29. Write the equation for 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

30. Practice Time pg 74 #1-2, pg 76 #3-7, pg 77 #8-11 Pg 83 #21,22