1. Solubility Equilibria
Solubility Product Constant
Ksp
for saturated solutions at equilibrium

2. Comparing
values.

3. Solubility Product (Ksp) = [products]x/[reactants]y but.....
reactants are in solid form, so Ksp=[products]x
i.e. A2B3(s)  2A B2–
Ksp=[A3+]2 [B2–]3
Given: AgBr(s)  Ag+ + Br–
In a saturated the [Ag+] = [Br– ]= 5.7 x 10–7 M. Determine the Ksp value.

4. concentrations of each
Problem: A saturated solution of silver chromate was to found contain g/L of Ag2CrO4. Find Ksp
Eq. Expression:
Ag2CrO4 (s)  2Ag+ + CrO42–
Ksp = [Ag+]2[CrO42–]
So we must find the
concentrations of each
ion and then solve
for Ksp.

5. Problem: A saturated solution of silver chromate was to contain 0
Problem: A saturated solution of silver chromate was to contain g/L of Ag2CrO4. Find Ksp
Eq. Expression:
Ag2CrO4 (s)  2Ag+ + CrO42–
Ksp = [Ag+]2[CrO42–]
0.022 g Ag2CrO4 L
Ag+:
1.33 x 10–4
0.022g Ag2CrO4 L
CrO4–2:
6.63 x 10–5

6. Problems working from Ksp values.
Given: Ksp for MgF2 is 6.4 x 25 oC
Find: solubility in mol/L and in g/L
MgF2(s)  Mg F–
Ksp = [Mg2+][F–]2
N/A I. C. E.
N/A x x
N/A x x
Ksp= [x][2x]2 = 4x3
6.4 x 10–9 = 4x3
now for g/L:
7.3 x 10–2

7. The common ion effect “Le Chatelier”
overhead fig 17.16
What is the effect of adding NaF?
CaF2(s)  Ca F-

8. Solubility and pH
CaF2(s)  Ca F–
Add H+ (i.e. HCl)
2F– + H+  HF

9. Solubility and pH
Mg(OH)2(s)  Mg OH–
Adding NaOH?
Adding HCl?

10. The common ion effect “Le Chatelier”
Why is AgCl less soluble in sea water than in fresh water?
AgCl(s)  Ag+ + Cl–
Seawater contains
NaCl

11. Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M.
What is its solubility in seawater where the [Cl–] = 0.55 M?
(Ksp of AgCl = 1.8 x 10–10)
AgCl(s)  Ag Cl–
Ksp= [Ag+][Cl–] I. C. E.
N/A
N/A x x
N/A x x
Ksp= [x][ x]
try dropping this x
Ksp = 0.55x
1.8 x 10–10 = 0.55x
x = 3.3 x 10–10 = [Ag+]=[AgCl]
“AgCl is much less soluble in seawater”

12. more Common ion effect:
a. What is the solubility of CaF2 in M Ca(NO3)2?
Ksp(CaF2) = 3.9 x 10–11
CaF2(s)  Ca F–
Ksp=[Ca2+][F-]2
[Ca2+] [F–] I. C. E.
+x x
x x
Ksp= [ x][2x]2
 [0.010][2x]2 = 0.010(4x2)
3.9 x 10–11 = 0.010(4x2)
x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2
Now YOU determine the solubility of CaF2 in M NaF.

13. Answer:
3.9 x 10–7 M Ca2+
CaF2(s)  Ca F–
+x x
x x
Ksp = [x][ x]2
 x(0.010)2
3.9 x =x(0.010)2
x = 3.9 x 10-7
What does x tell us

14. Reaction Quotient (Q): will a ppt. occur?
Use Q (also called ion product) and compare to Ksp
Q < Ksp
reaction goes
Equilibrium
Q = Ksp
reaction goes
Q > Ksp

15. Problem: A solution is 1. 5 x 10–6 M in Ni2+
Problem: A solution is 1.5 x 10–6 M in Ni2+. Na2CO3 is added to make the solution 6.0 x 10–4 M in CO32–.
Ksp(NiCO3) = 6.6 x 10–9.
Will NiCO3 ppt?
We must compare Q to Ksp.
NiCO3  Ni CO32–
Ksp = [Ni2+][CO32–]
Q = [Ni2+][CO32–]
Q = [1.5 x 10–6][6.0 x 10–4] = 9.0 x 10–10
Q < Ksp
no ppt.

16. a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14
Problem: L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L of 1.0 x 10–3 M K2CrO4.
a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14
Pb(OAc)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KOAc(aq)
then:
PbCrO4(s)  Pb CrO42–
Ksp= [Pb2+][CrO42–]
[Pb2+]:
0.50 L
5.0 x 10–6
1 L
[CrO42-]:
0.50 L
5.0 x 10-4
1 L
Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9
compare to Ksp:
Q > Ksp
so a ppt. will occur

17. Try dropping the “+ x” term.
b. find the Eq. conc. of Pb2+ remaining in solution after the PbCrO4 precipitates.
Ksp(PbCrO4) = 1.8 x 10–14
Since [Pb2+] = 5.0 x 10-6 and [CrO42-] = 5.0 x 10-4 and there is a 1:1 stoichiometry, Pb2+ is the limiting reactant.
PbCrO4(s)  Pb CrO42–
I. (after ppt.) C. E.
x x 10-6 = 5.0 x 10-4
+x x
x x 10–4 + x
Ksp = [x][5.0 x 10–4 + x]
Try dropping the “+ x” term.
1.8 x 10–14 = x(5.0 x 10-4)
x = [Pb2+] = 3.6 x 10–11
This is the concentration of Pb2+ remaining in solution.

18. Complex ion formation:
AgCl(s)  Ag Cl–
Ksp= 1.8 x 10–10
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
Ag+(aq) + 2NH3(aq)  {Ag(NH3)2}+(aq)
formation or stability constant:
For Cu2+:
Cu NH3  [Cu(NH3)4]2+(aq)
K1 x K2 x K3 x K4 = Kf = 6.8 x 1012

19. Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
Solubility and complex ions:
Problem: How many moles of NH3 must be added to dissolve mol of AgCl in 1.0 L of H2O? (KspAgCl = 1.8 x 10–10 ; Kf[Ag(NH3)2]+ = 1.6 x 107)
AgCl(s)  Ag+ + Cl–
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
AgCl(s) + 2NH3  Ag(NH3)2+(aq) + Cl–
sum of RXNS:
= 2.9 x 10–3
Now use Koverall to solve the problem:
Koverall= 2.9 x 10–3 =
[NH3]eq = 0.93
but How much NH3 must we add?
[NH3]total= (2 x 0.050) = M
2 ammonia’s for each Ag+

20. [Cd2+] = 2.6 x 10–10 M when NiS starts to ppt.
Fractional Precipitation: “ppting” one ion at a time. Compounds must have different Ksp values (i.e. different solubilities)
Example: Ksp CdS = 3.9 x 10–29 and KspNiS = 3.0 x 10–21
? Which will ppt. first?
least soluble
Problem: A solution is M in both Cd2+ and Ni2+. Just before NiS begins to ppt., what conc. of Cd2+ will be left in solution?
Approach: Find conc. of S2– ion when Ni2+ just begins to ppt. since Cd2+ will already be ppting. Then use this S2– conc. to find Cd2+.
NiS(s)  Ni S2–
Ksp= x 10–21 = [0.020][S2–]
[S2–] = 1.5 x 10–19 M when Ni2+ just begins to ppt.
So:
CdS(s)  Cd S2–
Ksp= 3.9 x 10–29 = [Cd2+][1.5 x 10–19]
[Cd2+] = 2.6 x 10–10 M when NiS starts to ppt.