1. Solubility Equilibria
Will it all dissolve, and if not, how much?

2. All dissolving is an equilibrium.
If there is not much solid it will all dissolve.
As more solid is added the solution will become saturated.
Solid dissolved
The solid will precipitate as fast as it dissolves .
Equilibrium

3. General equation M+ stands for the cation (usually metal).
Nm- stands for the anion (a nonmetal).
MaNmb(s) aM+(aq) + bNm- (aq)
K = [M+]a[Nm-]b/[MaNmb]
But the concentration of a solid doesn’t change.
Ksp = [M+]a[Nm-]b
Called the solubility product for each compound.

4. Common misconceptions
Solubility is not the same as solubility product.
Solubility product is an equilibrium constant.
it doesn’t change except with temperature.
Solubility is an equilibrium position for how much can dissolve.
A common ion can change this.

5. Calculating Ksp
1.The solubility of iron(II) oxalate FeC2O4 is 65.9 mg/L . Determine Ksp
(mm of FeC2O4 is g/mol)
2.The solubility of Li2CO3 is 5.48 g/L Determine Ksp
(mm of Li2CO3 is g/mol)

6. FeC2O4(s) Fe2+(aq) + C2O4 2-(aq) Ksp = [Fe2+] [C2O4 2-]
65.9 mg/L = x 10-4 mol/L
At equilibrium:
[Fe2+] = 7.49 x 10-4 mol/L
[C2O4 2-]= 7.49 x 10-4 mol/L
Ksp = [7.49 x 10-4 mol/L] 2
= 5.6 x 10-7 M2

7. Ksp = [Li+]2 [CO3 2-] Li2CO3(s) 2Li+(aq) + CO3 2-(aq)
5.48 g/L= 7.28 x 10-2 mol/L
At equilibrium:
[Li+] = 1.46 x 10-1 mol/L
[CO3 2-]= 7.28 x 10-2 mol/L
Ksp = [1.46 x 10-1 mol/L ] 2[7.28 x 10-2 mol/L ]
= x 10-3 M3

8. Calculating Solubility
The solubility is determined by equilibrium.
Its an equilibrium problem.
Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M and g/L.
(mm of SrSO4 is g/mol)
Calculate the solubility of Ag2CrO4, with a Ksp of 9.0 x in M and g/L.
(mm of Ag2CrO4 is g/mol)

9. SrSO4(s) Ksp = [Sr2+] [SO4 2-] = 3.2 x 10-7 M2 Sr2+(aq) + SO4 2-(aq)
Let x(M) rep. the solubility or the amount
of Sr2+ present at equilibrium
At equilibrium:
[Sr2+] = x
[SO4 2-]= x
Ksp = [x] 2
x = 3.2 x 10-7 M2
= 5.7 x 10-4 mol/L = x 10-1 g/L

10. Ksp = [Ag+]2 [CrO4 2-] = 9.0 x 10-12 M3 Ag2CrO 4(s)
2Ag+(aq) CrO4 2-(aq)
Ksp = [Ag+]2 [CrO4 2-] = 9.0 x M3
Let x(M) rep. the solubility or the amount
of CrO42- present at equilibrium
At equilibrium:
[CrO42-] = x
[Ag+]= 2x
Ksp = [2x] 2 x
4x3 = 9.0 x 10-12
x = x 10-4 mol/L = x 10-4 mol/L = 4.74x10-2 g/L

11. Homework: p 486 #1-4 p.493 # 4-9

12. Common Ion Effect
If we try to dissolve a solid in a solution with either the cation or anion already present, less will dissolve.
Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M and g/L in a) water and b) solution of M Na2SO4.
(mm is g/mol)
Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M and g/L in a) water and b) solution of M Sr(NO3.)2

13. a) solved before….answer was = 5.7 x 10-4 mol/L = 1.05 x 10-1 g/L
Common ion causes a decrease in solubility as per Le Chatelier’s principle! Sulphate is the common ion
a) solved before….answer was
= 5.7 x 10-4 mol/L = x 10-1 g/L
b) Ksp = [Sr+2][SO4-2]
Ksp = [x][0.01 +x]
3.2 x 10-7 = [x][0.01 +x]
x= 3.2 x 10-5M (using quadratic or simplifying for x is small when added to 0.01)
Solubility is now 3.2 x 10-5M or 5.88 x 10-3 g/L which is much less than in a)

14. a) solved before….answer was = 5.7 x 10-4 mol/L = 1.05 x 10-1 g/L
Common ion causes a decrease in solubility as per Le Chatelier’s principle! Strontium is the common ion
a) solved before….answer was
= 5.7 x 10-4 mol/L = x 10-1 g/L
b) Ksp = [Sr+2][SO4-2]
Ksp = [ x][x]
3.2 x 10-7 = [ x][x]
x= 3.2 x 10-5M (using quadratic or simplifying for x is small when added to 0.01)
Solubility is now 3.2 x 10-5M or 5.88 x 10-3 g/L which is much less than in a)

15. pH and solubility Why are hydroxides more soluble in acid?
For other anions, if they come from a weak acid (CO32-, PO43-, CH3COO, etc) they are more soluble in a acidic solution than in water.
CaC2O Ca+2 + C2O4-2
H+ + C2O HC2O4-
Reduces C2O4-2 in acidic solution.

16. Precipitation Ion Product, Q =[M+]a[Nm-]b
If Q>Ksp a precipitate forms.
If Q<Ksp No precipitate.
If Q = Ksp equilibrium.
A solution of mL of 4.00 x 10-3M Ce(NO3)3 is added to mL of x 10-2M KIO3. Will Ce(IO3)3(S) (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?

17. A precipitate ONLY forms if the salt formed by double displacement is slightly soluble and the concentration of its ions reaches saturation point
In the solubility charts, the salt would be identified as insoluble or a precipitate
The precipitate forms if the concentration of its ions reaches the minimum Ksp value
Calculate the concentration of the ions of the possible precipitate after the two aqueous reactant solutions are mixed (and thus are diluted each of them in the process of mixing)

18. Calculate the trial ion product – the Q and compare to the Ksp
750.0 mL of 4.00 x 10-3M Ce(NO3)3 plus300.0 mL of 2.00 x 10-2M KIO3
Total is mL or 1.050L
c1V1=c2v2 for Ce+3
(0.004M)(750)= c2 (1050)
c2= M
c1V1=c2v2 for IO3-1
(0.02M)(300)=c2(1050)
c2= M
Q= [Ce+3][IO3-1]3=( )( )3
Q= 5.3 x …Q is greater than Ksp (1.9 x 10-10)…a precipitate forms

19. Ce(IO3)3(s) = Ce IO3-1 (other salt is potassium nitrate which is soluble no exception…it would appear as aqueous (aq) in bce)
Q= [Ce+3][IO3-1]3
Q=( M)( M)3
Q= 5.3 x M4
…Q is greater than Ksp (1.9 x 10-10)
…a precipitate forms

20. Relative solubilities
Visual Ksp comparison will only allow us to compare the solubility of solids that fall apart into the same number of ions.
The bigger the Ksp of those the more soluble.
If they fall apart into different number of pieces you have to do the math and solve for solubility to compare
Example: two binary salts have a Ksp of 4.1 x 10-5 and 7.6 x Which is less soluble and will precipitate first?

21. Selective Precipitations
Used to separate mixtures of metal ions in solutions.
Add anions that will only precipitate certain metals at a time.
Used to purify mixtures.
Often use H2S because in acidic solutions Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

22. Selective precipitationAP
Selective precipitation
Follow the steps first with insoluble chlorides (Ag, Pb, Ba)
Then sulfides in Acid.
Then sulfides in base.
Then insoluble carbonate (Ca, Ba, Mg)
Alkali metals and NH4+ remain in solution.

23. Complex ion EquilibriaAP
Complex ion Equilibria
A charged ion surrounded by ligands.
Ligands are Lewis bases using their lone pair to stabilize the charged metal ions.
Common ligands are NH3, H2O, Cl-,CN-
Coordination number is the number of attached ligands.
Cu(NH3)4+2 has a coordination # of 4

24. The addition of each ligand has its own equilibriumAP
The addition of each ligand has its own equilibrium
Usually the ligand is in large excess.
And the individual K’s will be large so we can treat them as if they go to equilibrium.
The complex ion will be the biggest ion in solution.

25. AP
Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)2 -3 in a solution made by mixing mL of AgNO3 with mL of 5.00 M Na2S2O3
Ag+ + S2O Ag(S2O3)- K1=7.4 x 108
Ag(S2O3)- + S2O Ag(S2O3)2 -3 K2=3.9 x 104

26. AP
500mL of a 0.20M solution of AgNO3 is mixed with 200mL of a 0.10M Sr(NO3)2.
Al2(CO3)3 solid is added slowly. Which will precipitate first? How many milligrams of solid will just begin precipitation?
Ksp Ag2CO3 = 8.46 x Ksp SrCO3 = 5.6 x 10-10